Class 11 Physics

Kinetic Theory Of Gases

Molecular Concept Of Matter And Its Applications

WBBSE Class 11 Kinetic Theory of Gases Notes

For centuries, scientists have studied the structure of matter—how it is formed and which are its smallest entities, possessing properties identical to that of the matter itself.

• In the early nineteenth century, Dalton and Avogadro for the first time proposed the theory of molecular structure of matter.
• A molecule is the smallest entity and matter is composed of molecules having all the chemical properties of the matter itself.
• The physical quantities related to molecules are number of molecules in a material body, molecular velocities, intermolecular distances, intermolecular forces, etc.
• These are the internal microscopic properties of a body. Unfortunately, these properties cannot be measured directly through experiments. As a result, we have to start with certain basic assumptions (postulates) about the behavior of molecules.
• This gives us a picture of how molecules behave in a body. This is known as the molecular model. The next step is the application of Newton’s laws of motion on the molecules.

Now, the experimentally determined properties of a body as a whole, like pressure, temperature, and internal energy, are expected to be intimately related to the molecular model.

So Newton’s Jaws should give us expressions leading to these Bulk properties. This is essentially the object of the kinetic theory of matter.

• In short, the subject of study in which theoretical expressions of the bulk properties of a body are tamed from the application of Newton’s laws of motion on the internal molecular behaviour is called the kinetic theory of matter.
• Naturally, the values from the theoretical expressions of kinetic theory should match the experimental values. The formulas obtained from kinetic theory should be identical to the experimentally obtained thermodynamic formulas. This is actually the pre-condition for the success of kinetic theory.

This condition is beautifully obeyed in case of gases, leading to the very successful and advanced theory of the kinetic theory of gases. But this is not the case with liquids or solids. Partially successful molecular models exist for solids, but almost none so far has been developed for liquids.

Evidence of molecular motion: Molecules cannot be observed directly, but some natural phenomena clearly indicate the existence of molecular motion.

1. Diffusion: Let a gas jar filled with hydrogen gas be held upside down on another gas jar filled with carbon dioxide gas. Now the lids are removed. After an interval of time, it will be observed that the two gases will produce a homogeneous mixture in the two jars, ignoring gravitation.

• This phenomenon is called diffusion of the two gases. Clearly, molecular motion is evident in this phenomenon. Though carbon dioxide is heavier than hydrogen, CO₂ molecules move up and the hydrogen molecules move down to produce the mixture.
• The molecules of a gas randomly move at different velocities in all directions.
• So the molecules of the two gases mix and produce a homogeneous mixture. Diffusion takes place in liquids and solids also.
• If a few granules of copper sulphate are dropped at the bottom of a container filled with water, the blue colour gradually spreads throughout the whole volume of water.
• The density of copper sulphate solution is more than the density of water.
• But the solution moves up ignoring gravitation and after an interval of time, the whole mixture turns blue.
• It is an example of the motion of copper sulphate molecules diffusing into water. In a similar manner, solid phosphorus or boron can be diffused at high temperatures into solid silicon crystals to produce extrinsic semiconductors.
• In general, diffusion can be defined as the phenomenon by virtue of which movement of molecules occurs from a region of higher concentration to a region of lower concentration in a mixture till a homogeneity of concentration is established.

WBCHSE Class 11 Physics Notes

2. Vaporisation and vapour pressure: Liquid molecules are in motion inside the liquid. They move randomly at different velocities. Some molecules rise to the liquid surface with sufficient kinetic energy and overcome the attraction of other molecules inside the liquid.

• As a result, they may escape from the liquid. This phenomenon is called evaporation. Again, kinetic energies of molecules may be increased by applying heat. Hence, more molecules may escape from the liquid and vaporisation may occur.
• If a liquid is enclosed in a container, molecules leaving the liquid move randomly above the liquid surface. They collide with each other and hit the surface again and again. Some molecules may enter the liquid again. This leads to the vapour pressure corresponding to the vapour above the liquid surface.
• At a particular higher temperature, the kinetic energies of the molecules of the liquid become very high. In comparison, the potential energies due to intermolecular attractions become negligible. So the molecules are effectively free and all of them try to come out of the liquid at the same time. This phenomenon is called boiling of the liquid.

3. Expansion of gas: A gas spreads throughout the whole volume of its container. If the volume of the container increases, the gas spreads again to occupy the whole volume. This shows the property of random and unrestricted motion of the gas molecules.

4. Brownian motion: Very small, but still visible particles are often present as impurities in a liquid or in a gas. Observations through microscopes show that these particles move in a very random manner in all possible directions. This is known as Brownian motion.

This phenomenon can be explained by the concept of molecular motion. Molecules inside a liquid or a gas move randomly in all directions and collide time and again with small foreign particles (called Brownian particles). These collisions are directly responsible for the Brownian motion.

Understanding Kinetic Theory of Gases

Kinetic Theory Of Gases Brownian Motion

British scientist, Robert Brown, first observed this continuous and irregular motion of the particles with a powerful microscope. He put some pollen grains in water.

• These grains, being very light, remained suspended in water. Brown noticed the random, continuous, and to and fro motion of these grains. However he was not able to determine the mechanisms that caused this motion.
• Albert Einstein published a paper in 1905 that explained in precise detail how the motion that Brown had observed was a result of the pollen being moved by individual water molecules.
• Brownian motion of a particle is shown. Colloids in a colloidal solution, very small feathers suspended in air, etc., are examples of randomly moving Brownian particles.

Explanation of the origin of Brownian motion: Just after the discovery of Brownian motion, scientists assumed that the reason of the origin of Brownian motion was a chemical reaction, irregular change of temperature, surface tension of liquid, etc. However from various experiments, it was proved that these explanations were not correct.

• It became possible to explain Brownian motion with the help of kinetic theory. We know that liquid or gas molecules are moving randomly and colliding with floating particles at every instant and from all directions.
• The force exerted on a floating particle of big size in any direction, due to some colliding molecules, is cancelled by the equal and opposite force due to some other molecules.
• As a result, the resultant force acting on the floating particle becomes zero and it has no Brownian motion. But if the floating particle is very tiny in size, the resultant force on it does not become zero as the colliding particles do not exert force equally from all directions.
• Hence, the floating particle moves along the direction of the resultant force and Brownian motion is observed. As molecular thrusts are random, the magnitude of the resultant force is not equal always.
• Also, the resultant force does not act always in the same direction. So the floating particle moves in a random manner in all possible directions.

Characteristics of Brownian motion:

1. The motion is perpetual, spontaneous, random and continuous. Itoo Brownian particles, even at close proximity, do not have identical motion.
2. The motion does not depend on the motion of the container.
3. The velocities of the particles increase with rise in temperature.
4. Smaller particles have higher average velocities.
5. The velocities of the particles become higher in liquids with lower viscosity.
6. The motion depends only on the mass and the size of the particle and not on the material it is made of.

WBCHSE Class 11 Physics Notes

Kinetic Theory Of Gases – Basic Assumptions Of Kinetic Theory Of Gases

A gas is made of atoms and molecules. The three variables—volume, pressure, and temperature, are all consequences of the motion of the molecules.

• The kinetic theory of gases relates the motion of the molecules to the volume, pressure, and temperature of the gas. Actually, the kinetic theory of gases explains the macroscopic properties of the gases, though it is a microscopic mode.
• Rudolf Clausius and James Clark Maxwell developed the kinetic theory of gases and explained gas laws in terms of motion of the gas molecules.
• In order, to formulate the kinetic theory of gases some simplifying assumptions are made about the behavior of the molecules of the ideal gas. The assumptions are

1. A gas is composed of a large number of molecules. For a particular gas, the molecules are identical. But they are different for different gases.
2. Every gas molecule behaves as a point mass. So the sum of their volumes is negligible compared to the volume of the container. The intermolecular space in the container is an empty space.
3. The molecules are in continuous and random motion in all possible directions. The value of molecular velocities varies from zero to infinity.
4. During random motion, the molecules collide with each other and with the walls of the container. These collisions are perfectly elastic. This means that the velocity changes due to collision, but the net momentum and kinetic energy remain unchanged.
5. The average distance between the molecules is sufficiently large, so that the attractive or repulsive forces between them are negligible except during collision. As a result,
   • Molecular motion is unrestricted and the gas spreads throughout the inner volume of its container
   • The potential energy of a molecule is negligible, the total energy comes from its kinetic energy only
   • Between two collisions, a molecule moves with uniform velocity (according to Newton’s first law of motion). The straight line path between two successive collisions is called a free path.
6. Every collision is instantaneous the time of a collision is negligible compared to the time taken by a molecule to describe a free path.
7. The gas is homogeneous and isotropic. This means that the properties of the gas in any small portion are identical to those in any other equivalent portion anywhere inside the container.

A gas obeying the properties outlined in these assumptions is called an ideal gas or a perfect gas. Real gases show some deviations from these properties. Real gases available to us are good approximation of an ideal gas at low pressure and high temperature.

The kinetic theoretical definitions of mass and volume of a gas are obtained directly from the above assumptions:

1. The mass of a gas is defined as the sum of the masses of the constituent molecules.
2. The volume of a gas is defined as the inner volume of the gas container.

Mean free path: The straight line path described by a molecule between two collisions is called a free path.

• The gas molecules move randomly. As a result, the lengths of the free paths vary in an irregular manner. So, to get a concrete picture, the Idea of a mean-free path becomes essential.
• It is the average distance that a molecule can travel between two successive collisions.

Mean free path Definition: The mean value of the distance travelled by a molecule between two successive collisions Is called the mean free path (λ).

• In other words, if a molecule suffers N number of collisions with other molecules when it travels through a total distance d, then the mean free path is, \(\lambda=\frac{d}{N}\).
• The basic assumptions of kinetic theory describe every molecule as a point mass. But the value of mean free path becomes theoretically infinite if the molecules are treated as geometrical points.
• So some deviations from the assumptions, are necessary. Every molecule is assumed to be a very small hard sphere of non-zero radius. Then it is possible to get an effective theoretical value of the mean free path.

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An expression for the free path: Let σ = diameter of each molecule of the gas n = number of molecules per unit volume = number density of the molecules.

Let us consider a particular molecule of the gas moving with a velocity v at an instant of time. It will collide with all molecules whose centres come at a distance of cr or less along its line of motion. Now we choose a cylinder of radius cr and of length v. Clearly,

1. The axis of this cylinder is the path described by the chosen molecule in unit time, and
2. All other molecules, whose centers come within this cylinder, collide with the chosen molecule in that unit time.

The volume of this cylinder = cross-sectional area x length = πσ²v, and the number of molecules having centres in this volume = πσ²vn. So, the number of collisions per unit of time, N = πσ²vm.

The mean free path of the molecules is, therefore,

= \(\frac{\text { path described by a molecule }(d)}{\text { number of collisions }(N)}=\frac{\nu}{\pi \sigma^2 v n}\)

i.e., \(\lambda=\frac{1}{\pi \sigma^2 n}\)….(1)

This equation (1) shows that the mean free path (λ) of gas molecules is

1. Inversely proportional to the number of gas molecules (n) in a unit volume of the gas and
2. Inversely proportional to the square of molecular diameter (σ).

So, \(\lambda \propto \frac{1}{n} and \lambda \propto \frac{1}{\sigma^2}\),

i.e., \(\lambda \propto \frac{1}{\sigma^2 n} or, \lambda=\frac{k}{\sigma^2 n}\)

The constant k in equation (1) is \(\frac{1}{\pi}\). However, it has been estimated by different scientists by different other methods. The estimates give different results, but the value of k is always slightly greater or slightly less than 1. Then, the approximate expression for the mean free path is \(\lambda=\frac{1}{\sigma^2 n}\).

Kinetic Theory Of Gases – Mean Speed And Root Mean Square Speed of Gas Molecule

Assumptions of Kinetic Theory of Gases

Molecular velocity is a vector quantity. The number of gas molecules in any container is very large. So the velocity vectors are oriented randomly in all possible directions.

• As a result, the resultant velocity vector must be zero. Consequently, the mean velocity of the molecules is also zero. Clearly, this zero value is useless as it gives no information about the order of magnitude of the molecular velocities.
• Alternatively, we may take the magnitudes only of the molecular velocities to calculate the mean. Certainly, it is non-zero and a useful quantity. We also know the molecules move in straight lines between collisions.
• So the magnitude of molecular velocity is actually the molecular speed. The calculated mean velocity is essentially the mean speed of the molecules. However, mean speed is often loosely termed as mean velocity.

Let N be the number of molecules of a gas in a closed container and, at any instant, c₁, c₂, c₃…..c_N be the magnitudes of velocities of the N molecules, respectively.

So, mean velocity or mean speed of the molecules, \(\bar{c}=\frac{c_1+c_2+c_3+\cdots+c_N}{N}\)

Mean square velocity of the molecules is defined as the mean of the squares of velocities,

⇒ \(\overline{c^2}=\frac{c_1^2+c_2^2+c_3^2+\cdots+c_N^2}{N}\)

Root mean square speed or rms speed of the molecules, defined as the square root of mean square speed,

c = \(\sqrt{\overline{c^2}}=\sqrt{\frac{c_1^2+c_2^2+c_3^2+\cdots+c_N^2}{N}}\)

The mean velocity \(\bar{c}\) and the rms speed c of the gas molecules in a container are not equal. For example, let us take three molecules with velocities 40 m · s^-1, 80 m · s^-1, and 120 m · s^-1.

Then, \(\bar{c}=\frac{40+80+120}{3}=80 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

c = \(\sqrt{\frac{(40)^2+(80)^2+(120)^2}{3}}=86.4 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

In general, the rms speed is slightly greater than the mean velocity. In kinetic theory, the role of the rms speed is comparatively more important than that of the mean velocity.

Unit 9 Behavior Of Perfect Gas And Kinetic Theory Chapter 1 Kinetic Theory Of Gases

Deduction Of Difference Gas Laws From Kinetic Theory Of Gases

We have so far developed the kinetic theory of gases to a certain stage. Now it Is possible to see that the theoretical result of kinetic theory matches exactly with the thermodynamic gas laws obtained from experiments. This proves the success of the kinetic theory of gases.

1. Boyle’s law: According to the kinetic theory of gases, \(T \propto E \text { and } E=\frac{3}{2} p V . \text { So, } p V \propto T\). If T = constant for a certain amount of gas, then pV = constant. This is Hoyle’s law.

2. Charles’ law: According to the kinetic theory of gases, \(T \propto E \text { and } E=\frac{3}{2} p V \text {. So } p V \propto T\). If p = constant for a certain amount of gas, then V ∝ T, This is Charles’ law.

3. Pressure law: According to the kinetic theory of gases, \(T \propto E \text { and } E=\frac{3}{2} p V \text {. So } p V \propto T\). If V = constant for a certain amount of gas, then p ∝ T. This is Charles’ law of pressure.

4. Joule’s law: According to the kinetic theory of gases, E = 3/2 RT. So E is a function of temperature only; it does not depend on the volume or pressure of the gas. This is Joule’s law or Mayer’s hypothesis, as discussed In the chapter First and Second Law of Thermodynamics,

5. Avogadro’s law: Let equal volumes of two gases be taken at the same pressure and temperature. The pressure, temperature, and volume are p, T, and V, respectively.

Now for the first gas,

N₁ = total number of molecules in the container,

m₁ = mass of each molecule,

c₁ = rms speed of the molecules.

Then, \(n_1=\frac{N_1}{V}\) = number of molecules per unit volume. According to kinetic theory, the pressure of the gas,

p = \(\frac{1}{3} m_1 n_1 c_1^2=\frac{1}{3} m_1 \frac{N_1}{V} c_1^2\)

Pressure being the same, so we get similarly for the second gas,

p = \(\frac{1}{3} m_2 \frac{N_2}{V} c_2^2 .\)

So, \(\frac{1}{3} m_1 \frac{N_1}{V} c_1^2=\frac{1}{3} m_2 \frac{N_2}{V} c_2^2\)

or, \(m_1 N_1 c_1^2=m_2 N_2 c_2^2\)…..(1)

Again, the temperature of the two gases is the same. So the average kinetic energy of a molecule is equal for the two gases. This means that

⇒ \(\frac{1}{2} m_1 c_1^2=\frac{1}{2} m_2 c_2^2 \text { or, } m_1 c_1^2=m_2 c_2^2\)….(2)

From relations (1) and (2), we get N₁ = N₂. So, equal So, pV = RT (1 mol of an ideal gas) volumes of different gases, at the same pressure, and Equation (4) is known as the ideal gas equation temperature, contains an equal number of molecules. This is Avognclro’s law.

6. Dalton’s law of partial pressure: The pressure of a gas mixture on the walls of its container is equal to the sum of the partial pressures exerted by constituent gases separately, at same temperature as that of the mixture, provided that the gases do not react chemically with each other—this is Dalton’s law.

Let several gases be mixed in a closed container. Their densities are, ρ₁, ρ₂, ρ₃…… and the molecular rms speeds are c₁, c₂, c₃ ….. respectively. Every gas molecule moves with its own kinetic energy, which does not depend on the motion of the other molecules. So, the net pressure on the container will be the sum of the pressures exerted by all individual molecules. Then, the pressure of the gas mixture is

p = \(\frac{1}{3} \rho_1 c_1^2+\frac{1}{3} \rho_2 c_2^2+\frac{1}{3} \rho_3 c_3^2+\cdots\)….(3)

= \(p_1+p_2+p_3+\cdots\)…..(3)

Here, p₁, p₂, p₃,……. are the partial pressures of the first second, third, … gases on the walls of the container. So equation (3) expresses Dalton’s law of partial pressure.

7. Graham’s law of diffusion: The rate of diffusion of a gas in a mixture is inversely proportional to the square root of the density of the gas—this is Graham’s law.

Let the densities of two gases be ρ₁ and ρ₂ and the rms velocities of the molecules be c₁ and c₂, respectively. The gases are allowed to diffuse with each other at the same temperature and same pressure. The diffusion is due to the motion of molecules so the rate of diffusion is clearly proportional to the rms speed of the molecules.

Now, p = \(\frac{1}{3} \rho_1 c_1^2=\frac{1}{3} \rho_2 c_2^2\)

or, \(\rho_1 c_1^2=\rho_2 c_2^2 or, \frac{c_1}{c_2}=\sqrt{\frac{\rho_2}{\rho_1}}\).

As, rate of diffusion r  ∝ c, we have \(\frac{r_1}{r_2}=\frac{c_1}{c_2}, i.e., \quad \frac{r_1}{r_2}\)= \(\sqrt{\frac{\rho_2}{\rho_1}}\)

or, \(r \propto \frac{1}{\sqrt{\rho}}\).

This is Graham’s law.

8. Ideal gas equation: According to the kinetic theory of gases, \(T \propto E \text { and } E=\frac{3}{2} p V \text {. So, } p V \propto T \text {. }\). Then, pV = kT, where k is a constant. For 1 mol of an ideal gas, this constant k is called the universal gas constant, denoted by R.

So, pV= RT (1 mol of an ideal gas)…(4)

Equation (4) is known as the ideal gas equation.

Kinetic Energy and Temperature Relation in Gases

Unit 9 Behavior Of Perfect Gas And Kinetic Theory Chapter 1 Kinetic Theory Of Gases

Limitations Of Ideal Gas Laws

If a gas obeys Boyle’s law and Charles’ law accurately, then it is called an ideal gas. But in practice, real gases do not obey these ideal gas conditions in all circumstances.

Usually, a real gas behaves as an ideal gas at high temperatures and low pressures. But at low temperatures and high pressures, the behaviour deviates from that of an ideal gas.

The equation of state for 1 mol of an ideal gas is pV = RT. A real gas deviates from this equation chiefly due to the following two reasons:

1. The kinetic theory assumes that gas molecules are point masses and the volume of the molecules is negligible compared to the total volume of the gas. But every molecule, however small, has a finite volume. So their total volume is not always negligible.
   • In particular, at low temperatures and high pressures, the volume of a gas is comparatively small. In this case, the volume of the gas molecules becomes an important factor.
   • The effective volume for the motion of the molecules inside the gas container of volume V is reduced by an amount b (say). Then the equation of state for 1 mol of the gas becomes p(V-b) = RT.
2. The kinetic theory further assumes that the molecules do not attract one another. But in particular, at low temperatures and high pressures, the molecules come closer to one another.
   • As a result, the attractive forces are no longer negligible. Now, consider a molecule near the wall of the container. It experiences a resultant force towards the interior due to attractions by the other molecules. So it collides with the wall at a comparatively less velocity.

As a result, the pressure on the wall becomes less. Let p’ be the reduction of the pressure on the wall due to all these molecules, van der Waals established that \(p^{\prime}=\frac{a}{v^2}\) where a is a constant for a particular gas. Now, if p is the effective pressure on the wall, then the pressure would be \(p+p^{\prime} \text { or } p+\frac{a}{V^2}\) considering that the gas was ideal.

So the equation of state for 1 mol of a real gas becomes \(\left(p+\frac{a}{V^2}\right)(V-b)=R T\)….(1)

Equation (1) is called the van der Waals equation for a real gas. The constants a and b are known as van der Waals constants; their values depend on the nature of the gas. For n mol of a real gas, let V be its volume. Then molar volume = V/n.

Then, we get \(\left\{p+\frac{a}{(V / n)^2}\right\}\left\{\frac{V}{n}-b\right\}=R T\)

or, \(\left(p+\frac{a n^2}{V^2}\right)(V-n b)=n R T\)…(2)

This is the form of van der Waals equation for n mol of a real gas. It is to be noted that the van der Waals equation is only one of several equations of state of real gases, proposed by different scientists.

Unit 9 Behavior Of Perfect Gas And Kinetic Theory Chapter 1 Kinetic Theory Of Gases

Degree Of Freedom (DOF)

Degree Of Freedom Definition: The minimum number of independent coordinates necessary to specify the instantaneous position of a moving body is called the degree of freedom of the body.

Degree Of Freedom Example:

1. Let us consider the motion of a particle falling freely under gravity. We take the initial point as the origin and the downward line as the z-axis. Then the position of the particle at any instant is specified by the z-coordinate only.

So, the number of degrees of freedom of the particle is 1. Essentially, the degree of freedom of every one-dimensional motion is, for example, the motion of a car along a road, an ant moving on a stationary rope, etc.

2. Motions of projectiles under gravity, orbital motion of planets around the sun, circular motion, motion of an ant on the floor of a room, etc., are examples of two-dimensional motions, in each case, any one point on the plane, Is chosen as the origin and two perpendicular axes x and y are considered. Then the position of the object at any Instant Is specified by the x- and y- coordinates. So, the number of degrees of freedom in two-dimensional motion Is 2.

3. According to the kinetic theory, all gas molecules of an Ideal gas are point masses and they are In a completely random to-and-fro motion. At least three coordinates (say x, y, z) are necessary to specify the position of a molecule at any Instant.

So, the number of degrees of freedom of an Ideal gas molecule is 3. Essentially, every three-dimensional motion has 3 degrees of freedom, for example, Brownian motion, motion of a fly in a room, etc.

• In the above examples, the particle, the object, and the gas molecule all are considered to be as a point mass, which cannot undergo rotation, if the body is rigid and has a finite size, it can undergo rotation also, about any axis, So, a rigid body will have degrees of freedom both due to its translatory motion and rotatory motion.
• Like translatory motion, the rotatory can also be resolved into three mutually perpendicular components. Thus a rigid body has six degrees of freedom, 3 for translatory motion and 3 for rotatory motion.
• Now, let us consider a system of two particles or two-point masses. Each particle has three degrees of freedom, so the system has six degrees of freedom. If the two particles remain at a fixed distance from each other, then there is one definite relationship between them.
• These definite relationships are known as constraints. As a result, the number of independent coordinates required to describe the configuration of the system is reduced by one. Hence, the system has (6-1) = 5 degrees of freedom.

These 5 degrees of freedom may be interpreted in another way: degrees of freedom for the translatory motion of the centre of mass = 3 and degrees of freedom for the rotatory motion of the two particles around the center of mass = 2.

In a system consisting of N particles, if the particles possess k independent relations i.e., constraints among them, then the number of degrees of freedom of the system is given by, f = 3N – k.

Degrees of freedom of different types of gases:

1. Monatomic gas: The molecule of a monatomic gas (for example, neon, helium, argon, etc.) consists of a single atom (a point mass). At least three coordinates (say x, y, z) are necessary to specify the position of the molecule at any instant in the three-dimensional space. So, the number of degrees of freedom of a monatomic gas molecule, f = 3×1-0 = 3.

2. Diatomic gas: The molecule of a diatomic gas like hydrogen, oxygen, nitrogen etc. has two atoms in it. Two point atoms have a fixed distance between them (neglecting the vibration of the atoms in the molecules)

i. e., the number of constraints is 1. Here, N = 2 and k = 1

∴ f = 3×2-1 = 5

3. Triatomic gas: Triatomic gas molecules are of two types:

(1) In a linear molecule such as CO₂, CS₂ > HCN, etc. the three atoms are arranged in a straight line. The number of independent relations between them is only two.

∴ f = 3×3-2 – 7 i.e., such a molecule has seven degrees of freedom.

2. In a non-linear molecule like H₂O, SO₂, etc. the three atoms are located at the three vertices of a triangle. Hence, there are three fixed distances among the three atoms.

∴ f = 3×3-3 = 6

Therefore, a non-linear triatomic molecule has six degrees of freedom.

Degrees of freedom in different cases:

WBCHSE Class 11 Physics Notes

 Kinetic Theory Of Gases Conclusion

The subject of study in which theoretical expressions on the bulk properties of a body are obtained from the application of Newton’s laws of motion on the internal molecular behavior is called the kinetic theory of matter. This approach has been highly successful for gases.

• The random and perpetual motion of very small particles present as impurities in a liquid or gas is known as Brownian motion. This happens due to random collisions of the Brownian particles with the molecules in matter. This motion furnishes evidence of the molecular model.
• The kinetic theory of gases relates the motion of the molecules of the gas to the volume, pressure, and temperature of the gas. Actually, this theory explains the macroscopic properties of the gases with the help of microscopic parameters.

Three basic assumptions of the kinetic theory of gases are—

1. Molecules are point masses; the intermolecular space is much larger than the space occupied by the molecules.
2. The attraction between molecules is negligible. As a result, the molecular potential energy is zero, i.e., the energy is purely kinetic.
3. Molecular motion is continuous and random. A molecule may have any velocity between zero and infinity in any direction. A molecule may have any velocity between zero and infinity in any direction.

Gas Laws Explained Through Kinetic Theory

The straight-line path described by a molecule between two successive collisions is called a firepath. The mean value of the lengths of different free paths of different molecules of a gas is called the mean free path.

The pressure of a gas in a container depends on

1. The mass of a molecule,
2. The number of molecules in unit volume and
3. The average velocity of the molecules.

Temperature is a property of a gas, which is proportional to the total kinetic energy of the gas molecules.

Absolute zero temperature is the temperature at which the internal energy of the gas becomes zero, i.e., the molecular motion stops entirely.

The velocity which is possessed by the highest number of gas molecules in a container is called the most probable velocity.

Gases obeying Charles’ law and Boyle’s law perfectly are called ideal or perfect gases.

In general, real gases deviate from this ideal behavior, especially at

1. High pressures and
2. Low temperatures.

The minimum number of independent coordinates necessary to specify the instantaneous position of a moving particle is called the degree of freedom of the tire particle.

Principle of equipartition of energy: The average molecular kinetic energy of any substance is equally shared among the degrees of freedom; the average kinetic energy of a molecule per degree of freedom is 1/2 kT (T = absolute temperature and k = Boltzmann constant (1.38 x 10^-23 J · K^-1).

Kinetic Theory Of Gases Useful Relations For Solving Numerical Examples

Mean free path of a gas molecule (λ):

1. \(\lambda=\frac{d}{N}\) where d = total distance traveled by a molecule, N = total number of collisions suffered by that molecule through the distance d.

2. \(\lambda=\frac{1}{\pi \sigma^2 n}\) where, σ = diameter of each molecule of the gas, n = the number of molecules per volume i.e., the number density of the molecule.

Let c₁, c₂,…..c_N be the magnitudes of instantaneous velocities of N molecules in a gas. Then,

1. Mean velocity, \(\bar{c}=\frac{c_1+c_2+\cdots+c_N}{N}\)

2. Root mean square speed or rms speed, \(c=\sqrt{\frac{c_1^2+c_2^2+\cdots+c_N^2}{N}}\)

At absolute temperature T,

\(\begin{array}{|c|c|c|}
\hline \begin{array}{c}
\text { Average speed of } \\
\text { a gas molecule } \\
(c)
\end{array} & \begin{array}{c}
\text { rms speed of a } \\
\text { gas molecule } \\
(c)
\end{array} & \begin{array}{c}
\text { Most probable } \\
\text { speed of a gas } \\
\text { molecule }\left(c_m\right)
\end{array} \\
\hline \sqrt{\frac{8 k T}{\pi m}} & \sqrt{\frac{3 k T}{m}} & \sqrt{\frac{2 k T}{m}} \\
\hline
\end{array}\)

∴ \(\bar{c}: c: c_m=\frac{2}{\sqrt{\pi}}: \sqrt{\frac{3}{2}}: 1\)

Pressure of an ideal gas, p = \(\frac{1}{3} \rho c^2=\frac{1}{3} m n c^2\), where m- mass of a molecule, n = number of molecules in unit volume and ρ = mn = density.

The internal energy of a gas in a container, which is equal to the sum of the kinetic energies of the molecules, is

U = E = \(\frac{3}{2} p V ; \quad \text { So, } p=\frac{2}{3} \frac{E}{V}=\frac{2}{3} u \text {, }\)

where u = energy per unit volume.

• If M is the molecular weight of a gas, the rms speed of the molecules is related to temperature T of the gas by the relation, \(c=\sqrt{\frac{3 R T}{M}} \quad \text { or, } \quad c \propto \sqrt{T}\)
• The equation of state for 1 mol of an ideal gas is p V = R T.
• For real gases, volume and pressure corrections lead to the van der Waals equation of state:

⇒ \(\left(p+\frac{a}{V^2}\right)(V-b)=R T\)

• For n mol of real gas, van der Waals equation of state is \(\left(p+\frac{a n^2}{V^2}\right)(V-n b)=n R T\) [where, a, b are small positive constants ‘a’ is related to the average force of attraction between the molecules and ‘b’ Is related to the total volume of the molecules.
• The average kinetic energy of a gas molecule, e = \(\frac{3}{2} k T\)
• If some amount of gas contains N molecules, the total internal energy i.e., total kinetic energy, E = \(\frac{3}{2}Nk T\)
• For 1 mol of a gas, N = N_A = Avogadro’s number.
• Then, \(E=\frac{3}{2} N_A k T=\frac{3}{2} R T \text {, where } R=N_A k \text {. }\)
• Molar specific heat at constant volume, \(C_v=\frac{d Q}{d T}=\frac{d E}{d T}\)

If n₁ mol of a gas is mixed with n₂ mol of another gas (they do not react with each other), then,

1. Molar specific heat of the mixture at constant volume, \(C_\nu=\frac{n_1 C_{\nu_1}+n_2 C_{\nu_2}}{n_1+n_2}\) where, \(C_{\nu_1}\) = molar specific heat at constant pressure of 1st and 2nd gases respectively.

2. Molar specific heat of the gas mixture at constant pressure, \(C_p=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1+n_2}\) where, \(C_{p_1}\) and \(C_{p_2}\) are the molar specific heat at constant pressure of 1st and 2nd gases respectively.

3. The ratio of Cp and Cv for the mixture is,

⇒ \(\gamma=\frac{C_p}{C_v}=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1 C_{v_1}+n_2 C_{v_2}}\)

For 1 mol of a monatomic gas, \(C_\nu=\frac{d E}{d T}=\frac{3}{2} R ; C_p=C_\nu+R=\frac{5}{2} R ; \gamma=\frac{C_p}{C_\nu}=\frac{5}{3}\)

In general cases, if f is the number of degrees of freedom of an ideal gas molecule then \(C_\nu=\frac{f_2}{2} R ; C_p=\left(1+\frac{f}{2}\right) R ; \gamma=\frac{C_p}{C_\nu}=1+\frac{2}{f}\)

WBCHSE Class 11 Physics Notes

Kinetic Theory Of Gases Very Short Answer Type Questions

Short Answer Questions on Kinetic Theory

Question 1. What is the name of the smallest entity of matter that exhibits all the properties of that matter?
Answer: Molecule

Question 2. Which one of the following has the highest intermolecular force—solid, liquid, or gas?
Answer: Solid

Question 3. If a gas jar filled with a light gas like hydrogen is held side down on another gas jar filled with carbon dioxide, it is observed that the two gases produce a homogeneous mixture in the jars. What is the name of this process?
Answer: Diffusion

Question 4. How does the velocity of Brownian particles change due to the movement of the vessel?
Answer: No change occurs

Question 5. What is the direction of velocities of gas molecules according to the kinetic theory of gases?
Answer: All possible directions

Question 6. Gas molecules collide with each other and with the walls of the container. What is the type of these collisions? .
Answer: Perfectly elastic

Question 7. What do you call the straight line path described by a gas molecule between two successive collisions?
Answer: Free path

Question 8. Both vaporization and vapor pressure prove the ________ of molecules of a liquid.
Answer: Mobility

Question 9. Brownian motion supports the _______ of the matter.
Answer: Kinetic theory

Question 10. In Brownian motion in a medium, if the particles decrease in size, how does their velocity vary?
Answer: Increases

Question 11. Does the velocity of the particles increase or decrease, when the viscosity increases in Brownian motion in a medium?
Answer: Decreases

Question 12. Which gases obey the basic assumptions of the kinetic theory?
Answer: Ideal gases

Question 13. According to the kinetic theory of gases, every gas molecule behaves as a ________
Answer: Point mass

Question 14. According to the kinetic theory of gases, the velocity of gas molecules varies from ______ to ______
Answer: zero, infinity

Question 15. Which is greater rms speed or mean velocity?
Answer: rms speed

Question 16. Does the pressure of a gas increase or decrease when the velocity of the gas molecules increases?
Answer: Increases

Question 17. On which other factor does the pressure of a gas depend, besides the number of molecules per unit volume and the temperature of the gas?
Answer: Mass of the gas molecules

Question 18. Is the most probable velocity of gas molecules higher or lower than the mean velocity?
Answer: Lower

Question 19. Will the rms speed of oxygen and hydrogen gas molecules be the same at equal temperatures?
Answer: No, hydrogen gas will have a higher rms speed

Question 20. What is the ratio of the rms speeds of O₃ and O₂ at a certain temperature?
Answer: √2: √3

Question 21. By how many times will the pressure of a gas kept in a gas container of constant volume increase to double the rms speed of the gas molecules?
Answer: 4 times

Question 22. The velocities of the three gas molecules are 4cm · s^-1, 8 cm · s^-1, and 12 cm · s^-1, respectively. Calculate their rms speed.
Answer: 8.64 cm · s^-1

Question 23. What is the relation between rms speed and molecular mass of a gas?
Answer: rms speed is inversely proportional to the square root of the molecular mass of that gas

Question 24. Hydrogen and oxygen gases are kept in two vessels at the same temperature and pressure. What is the ratio of the rms speed of their molecules?
Answer: 4:1

Question 25. If a gas molecule of mass m and velocity u collides perpendicularly with a wall of a container, what will be the value of momentum of the molecule after the collision?
Answer: mu

Question 26. If n number of molecules, each having mass m and velocity u, perpendicularly hit the walls of a container in every second, what will be the value of the applied force?
Answer: 2mnu

Question 27. In the kinetic theory of gases, _____ is more important than mean velocity.
Answer: rms speed

Question 28. What is the name of the force that acts among the molecules of matter?
Answer: Intermolecular force

Question 29. Under which conditions do real gases behave as ideal gases?
Answer: At low pressure and high temperature

Question 30. Which property of a gas is proportional to the net internal energy of the gas molecules?
Answer: Temperature of the gas

Question 31. At which temperature does the kinetic energy of gas molecules become zero?
Answer: Absolute zero temperature

Question 32. To which gases is the van der Waals’ equation applicable?
Answer: Real gases

Question 33. If the temperature of a gas is increased at constant volume, how will the number of collisions of the molecules per unit time change?
Answer: Increase

Question 34. What is the dimension of an in van der Waals’ equation \(\left(p+\frac{a}{V^2}\right)(V-b)=R T?\)
Answer: ML^-3T^-2

Question 35. What is the dimension of b in van der Waals’ equation \(\left(p+\frac{a}{V^2}\right)(V-b)=R T?\)
Answer: L³

Question 36. What is the relation between the pressure p of a gas and its energy density?
Answer:

Relation between the pressure p of a gas and its energy density

\(\left[p=\frac{2}{3} u\right]\)

Question 37. According to the kinetic theory of gases, as there is no attractive force between the gas molecules, the entire energy of them is ______
Answer: Kinetic energy

Kinetic Theory Of Gases  Assertion Reason Type Question And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 Is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: The root mean square speeds of the molecules of different ideal gases at the same temperature are the same.

Statement 2: The average translational kinetic energy of molecules of a different ideal gas is same at the same temperature.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 2.

Statement 1: The rms speed of oxygen molecules (O₂) at an absolute temperature T is c. If the temperature is doubled and oxygen gas dissociates into atomic oxygen, the rms speed remains unchanged.

Statement 2: The rms speed of the molecules of a gas is directly proportional to √T/M.

Answer: 1. Statement 1 is true, and statement 2 Is true statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and volume.

Statement 2: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Applications of Kinetic Theory in Real Life

Question 4.

Statement 1: The mean free path of gas molecules, varies inversely with the density of the gas.

Statement 2: The mean free path of gas molecules is defined as the average distance traveled by a molecule between two successive collisions.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 5.

Statement 1: The following show \(\frac{pV}{T}\) versus p graph for a certain mass of O₂ gas at two temperatures T₁ and T₂. It follows from the graph that T₁ > T₂.

Statement 2: At higher temperatures, real gas behaves more like an ideal gas.

Answer: 1. Statement 1 is true, and statement 2 Is true statement 2 is a correct explanation for statement 1.

Question 6.

Statement 1: For an ideal gas, at a constant temperature, the product of the pressure and the volume is constant.

Statement 2: The mean square velocity of die molecules is inversely proportional to mass.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 7.

Statement 1: The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and volume.

Statement 2: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Kinetic Theory Of Gases Match Column 1 with Column 2

Question 1.

Answer: 1. C, 2. B, D 3. A, 4. A

Question 2.

Answer: 1. C, 2. B, 3. A

Question 3.

R = universal gas constant, f = number of degrees of freedom, T = temperature.

Answer: 1. C, 2. B, 3. A

Question 4. Match the following columns according to the graph.

Answer: 1. C, 2. A, 3. B

Kinetic Theory Of Gases Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. The pressure exerted by an ideal gas is p = \(\frac{1}{3} \frac{M}{V} c^2\), where the symbols have their usual meanings. Using standard gas equation, pV = RT, we find that \(c^2=\frac{3 R T}{M} \quad \text { or } \quad c^2 \propto T\) Average kinetic energy of translation of 1 mol of gas = \(\frac{1}{2} M c^2=\frac{3 R T}{2}\)

1. Average thermal energy of a helium atom at room temperature (27°C) is (given, Boltzmann constant k= 1.38 x 10^-23 J · K^-1)

1. 2.16 x 10²¹ J
2. 6.21 x 10²¹ J
3. 6.21 x 10^-21 J
4. 6.21 x 10^-23 J

Answer: 3. 6.21 x 10²¹ J

2. Average thermal energy of 1 mol of helium at this temperature is (given, gas constant for 1 mol = 8.31 J · mol^-1 · K^-1)

1. 3.74 x 10³ J
2. 3.74 x 10^-3 J
3. 3.47 x 10⁶ J
4. 3.47X10^-6 J

Answer: 1. 3.74 x 10³ J

3. At what temperature, when pressure remains unchanged, will the rms speed of hydrogen double its value at STP?

1. 819 K
2. 819 °C
3. 1000K
4. 1000°C

Answer: 2. 819 °C

4. At what temperature, when pressure remains unchanged, will the rms speed of a gas be half its value at 0°C?

1. 204.75 K
2. 204.75 °C
3. -204.75 K
4. -204.75 °C

Answer: 4. -204.75 °C

Question 2. If c₁, c₂, c₃ …. are random speeds of gas molecules at a certain moment then average velocity c_av = \(\frac{c_1+c_2+c_3+\cdots+c_n}{n}\) and root mean square speed of gas molecules,

⇒ \(c_{\mathrm{rms}}=\sqrt{\frac{c_1^2+c_2^2+c_3^2+\cdots+c_n^2}{n}}=c.\)

Further, c² ∝ T or, c ∝ √T.

At 0 K, c =0, i.e., molecular motion stops,

1. If three molecules have velocities 0.5km · s^-1, 1 km · s^-1 and 2 km · s^-1, the ratio of rms speed and average velocity is

1. 0.134
2. 1.34
3. 1.134
4. 13.4

Answer: 3. 1.134

2. The temperature of a certain mass of a gas is doubled. The rms speed of its molecules becomes n times, where n is

1. \(\sqrt{2}\)
2. 2
3. \(\frac{1}{\sqrt{2}}\)
4. \(\frac{1}{2}\)

Answer: 1. \(\sqrt{2}\)

3. KE per molecule of the gas in the above question becomes x times, where x is

1. \(\frac{1}{2}\)
2. \(\frac{1}{4}\)
3. 4
4. 2

Answer: 4. 2

4. KE per mole of hydrogen at 100°C (given R = 8.31 J · mol^-1 · K^-1) is

1. 4946 J
2. 4649 J
3. 4496 J
4. 4699J

Answer: 2. 4649 J

5. At what temperature, when pressure remains constant, will the rms speed of the gas molecules be increased by 10% of? rms speed at STP?

1. 57.3 K
2. 57.3 °C
3. 557.3 K
4. -57.3 °C

Answer: 2. 57.3 °C

Question 3. A cubical box of side lm contains helium gas (atomic weight = 4) at pressure 10 N • m^-2. During an observation time of 1s, an atom traveling with rms speed parallel to one of the edges of the cube was found to make 500 hits with a particular wall without any collision with other atoms.

1. Evaluate the temperature of the gas.

1. 125K
2. 160K
3. 181K
4. 185K

Answer: 2. 160K

2. Evaluate the average kinetic energy per atom.

1. 3.31 x 10^-21 J
2. 3.75 x 10⁶ J
3. 3.81 x 10^-15 J
4. 3.22 x 10³ J

Answer: 1. 3.31 x 10^-21 J

3. Evaluate the total mass of the helium gas in the box.

1. 9×10^-4 kg
2. 5x 10^-3 kg
3. 3 x 10^-4 kg
4. 7 x 10^-3 kg

Answer: 3. 7 x 10^-3 kg

Kinetic Theory Of Gases Integer Answer Type Questions

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. Two identical cylinders contain helium at 3.5 standard atmospheres and argon at 2.5 standard atmospheres respectively. If both these gases are filled in one of the cylinders, what would be the pressure of the mixture?
Answer: 6

Question 2. The rms speed of molecules of a gas at -73 °C and 1 standard atmosphere pressure is 100 m · s^-1. The temperature of the gas is increased to 527°C and pressure is doubled. The rms speed becomes k times. What is the value of k?
Answer: 2

Question 3. The density of a gas is 6 x 10^-2 kg· m^-3 and the root mean square speed of the gas molecules is 500 m · s^-1. The pressure exerted by the gas on the walls of the vessel is n x 10³ N · m^-2. Find the value of n.
Answer: 5

Question 4. A gas has molar heat capacity c = 37.55 J · mol^-1 · K^-1, in the process pT = constant. Find the number of degrees of freedom of the molecules of the gas.
Answer: 5

Question 5. A vessel has 6g of hydrogen at pressure p and temperature 500 K. A small hole is made in it so that hydrogen leaks out How much hydrogen (in g) leaks out if the final pressure is \(\frac{p}{2}\) and temperature falls to 300K?
Answer: 1

 

Kinetic Theory Of Gases – Pressure Of A Perfect Gas According To The Kinetic Theory

Kinetic Theory of Gases Simplified

Gas molecules, due to their random motion, collide continuously with the inner walls of its container. As a result, an outward force or thrust acts on the walls. The pressure of a gas is defined as the force exerted normally by the gas molecules on unit area of the walls of its container.

There are a large number of molecules in the container and they have random velocities in all possible directions. Statistically, on unit surface of any wall of the container, equal number of molecules collide with equal average velocity. As a result, the gas exerts equal pressure on the walls in all directions.

It is evident that the pressure of a gas is high, if

1. The gas molecules are heavy so that a large force acts on the walls,
2. The molecules move fast, so that a big impact is exerted on the walls and
3. The molecules are densely situated so that the number of collisions with the walls is high.

This means that the pressure of a gas on the walls of its container depends on

1. The mass of each molecule,
2. Average velocity of the molecules and
3. The number of molecules in a unit volume inside the container.

Derivation of the Expression for Pressure of an Ideal Gas: Let us take a cubical container.

It is filled with a gas. The inside surfaces of the container are such that the molecules suffer elastic collisions with them.

Let l be the length of each side of the container, m be the mass of each gas molecule and N be the total number of molecules in the container.ρ

Sp, A = \(l^2=\text { area of each of the six inner surfaces, }\)

V = \(l^3=\text { volume of the gas of the container, }\)

M = \(m N=\text { mass of the gas, }\)

ρ = \(\frac{M}{V}=\frac{m N}{V}=\frac{m N}{l^3}=\text { density of the gas, }\)

n = \(\frac{N}{V}=\frac{N}{\beta}\)= number of molecules in unit volume

= number density of the gas

= molecular concentration

Then, ρ = mn

Now, one corner O of the cubical container is taken as the origin, and the x, y, and z axes are chosen along the three mutually perpendicular sides at O.

Let c be the velocity of a gas molecule and u, v, w be the velocity components along the three axes, respectively.

So, c² = u² + v² + w²

Now, let us consider only the parallel surfaces R and S, which are perpendicular to the x-axis. As the collisions are elastic, the molecule hits the surface R at velocity u and rebounds from this surface In the opposite direction, i.e., the velocity of the molecule will he -u.

But v and w do not change In this case. So, for surfaces R and s, only the velocity component u or -u should be considered.

Thus, mu = momentum of the molecule before collision,

mu = momentum of the molecule after the collision,

So, a change in the momentum of the molecule

= final momentum- initial momentum

= – mu -(mu) = -2mu

Kinetic Theory of Gases and Temperature Effects

After a collision at R, the molecule moves towards S, Again the collision is S brings the molecule back to R, The effective distance traveled by the molecule between two collisions with the same surface = 2l.

The time taken between these two collisions, t = \(\frac{2 l}{u}\)

So, the rate of change in momentum of the molecule \(\frac{-2 m u}{\frac{2 m}{u}}=-\frac{m u^2}{1}\)

From Newton’s second law of motion, die force exerted by the surface R on the molecule = –\(\frac{m u^2}{l}\);

Again, horn Newton’s third law, the equal and opposite force exerted by the molecule on the die surface R = +\(\frac{m u^2}{l}\).

Then, pressure on surface R due to the molecule

= \(\frac{\text { force }}{\text { surface area }}=\frac{\frac{m u^2}{1}}{p^2}=\frac{m u^2}{p^3}=\frac{m u^2}{V}\)

Now, we consider all the N molecules and die components u₁, u₂…., u_N respectively of velocities along the x-axis.

So, the net pressure on surface R, due to all the N molecules in the container, Is \(p_x=\frac{m_1}{v}\left(u_1^2+u_2^2+\cdots+u_N^2\right)\)

Similarly, the pressure on surfaces perpendicular to die y and z axes, respectively, are \(p_y =\frac{m}{v}\left(v_1^2+v_2^2+\cdots+v_N^2\right)\)

and \(p_2 =\frac{m}{v}\left(w_1^2+w_2^2+\cdots+w_N^2\right)\)

As the gas exerts equal pressure in all directions, \(p_{\mathrm{x}}=p_y=p_z=p=\) pressure of the gas.

So, p= \(\frac{1}{3}\left(p_x+p_y+p_z\right)\)

= \(\frac{1}{3}\frac{m}{V}[\left(u_1^2+u_2^2+\cdots+u_N^2\right)+\left(v_1^2+v_2^2+\cdots+v_N^2\right)\)

…. + \(\left.\left(w_1^2+w_2^2+\cdots+w_N^2\right)\right]\)

= \(\frac{1}{3} \frac{m}{V}\left[\left(u_1^2+v_1^2+w_1^2\right)+\left(u_2^2+v_2^2+w_2^2\right)+\right.\)

…… + \(\left(u_N^2+v_N^2+w_N^2)\right]\)

= \(\frac{1}{3} \frac{m}{V}\left(c_1^2+c_2^2+\cdots+c_N^2\right)\)

= \(\frac{1}{3} \cdot \frac{m N}{V} \cdot \frac{\left.\left.c_1^2+c_2^2+\cdots+v_N^2+w_N^2\right)\right]}{N}\)

= \(\frac{1}{3} m \frac{N c^2}{V}\left[c^2=\frac{c_1^2+c_2^2+\cdots c_N^2}{N}(c=\text { ms speed })\right]\)

= \(\frac{1}{3} \frac{M_0}{V} c^2 \text { [total mass of the gas, } M_0=m N]\)

= \(\frac{1}{3} \rho c^2\left[\text { density of the gas, } \rho=\frac{M_0}{V}\right]\)…(1)

This expression shows that pressure depends on the total volume V, but not on the length l of the cubical container. This means that the expression is true for containers of all shapes.

This is the expression for the pressure of an ideal gas, according to kinetic theory.

From equation (1), \(c=\sqrt{\frac{3 p}{\rho}}\)….(2)

In this relation, pressure p and density ρ are bulk thermodynamic properties that are measured by experiments. So, from these measured values, we get an estimate of the rms speed c of the molecules.

Molecular number density: The number of molecules in unit volume N/V = n, this ‘n’ is called the molecular number density of gas.

Now, density, \(\rho=\frac{M_0}{V}=\frac{m N}{V}=m n\)

∴ From equation (1) we get, p = \(\frac{1}{3} m n c^2\)….(3)

This equation is mainly used in chemistry. It can be understood that the pressure of gas depends on three quantities

1. Mass of gas molecule (m),
2. Number of gas molecules in unit volume (n) and
3. rms Speed of the molecules.

Kinetic Theory Of Gases – Pressure Of A Perfect Gas According To The Kinetic Theory Numerical Examples

Kinetic Energy and Gas Pressure Relationship

Example 1. The velocity of 10 gas molecules in a container, are 2, 3, 3, 4, 4, 4, 5, 5,7, and 10 km • s^-1, respectively. Find out the mean velocity and rms speed.
Solution:

Given

The velocity of 10 gas molecules in a container, are 2, 3, 3, 4, 4, 4, 5, 5,7, and 10 km • s^-1, respectively.

Mean velocity, \(\bar{c}=\frac{2+3+3+4+4+4+5+5+7+10}{10}=\frac{47}{10}\)

= \(4.7 \mathrm{~km} \cdot \mathrm{s}^{-1} ;\)

rms speed \(\tilde{c}=\sqrt{\frac{2^2+3^2+3^2+4^2+4^2+4^2+5^2+5^2+7^2+10^2}{10}}\)

= \(\sqrt{\frac{269}{10}}=\sqrt{26.9}=5.1865 \mathrm{~km} \cdot \mathrm{s}^{-1} .\)

Example 2. Find out the rms speed of a gas of density 2 g · L^-1 at 76 cm Hg pressure. Given, density of mercury = 13.6 g · cm^-3 and 980 cm · s^-2
Solution:

Given, density of mercury = 13.6 g · cm^-3 and 980 cm · s^-2

p = \(76 \times 13.6 \times 980 \mathrm{dyn} \cdot \mathrm{cm}^{-2},\)

ρ = \(2 \mathrm{~g} \cdot \mathrm{L}^{-1}=\frac{2}{1000} \mathrm{~g} \cdot \mathrm{cm}^{-3}\)

So, rms speed,

c = \(\sqrt{\frac{3 p}{\rho}}\)

= \(\sqrt{\frac{3 \times 76 \times 13.6 \times 980 \times 1000}{2}}\)

= \(3.9 \times 10^4 \mathrm{~cm} \cdot \mathrm{s}^{-1}=0.39 \mathrm{~km} \cdot \mathrm{s}^{-1} .\)

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Example 3. Determine the rms speed of air molecules at STP. Given, the density of mercury = 13.6 g · cm^-3; the density of air = 0.00129 g · cm^-3
Solution:

Given, the density of mercury = 13.6 g · cm^-3; the density of air = 0.00129 g · cm^-3

p = \(76 \times 13.6 \times 980 \mathrm{dyn} \cdot \mathrm{cm}^{-2},\)

ρ = \(0.00129 \mathrm{~g} \cdot \mathrm{cm}^{-3}\)

∴ c = \(\sqrt{\frac{3 p}{\rho}}=\sqrt{\frac{3 \times 76 \times 13.6 \times 980}{0.00129}}\)

= \(4.85 \times 10^4 \mathrm{~cm} \cdot \mathrm{s}^{-1}=0.485 \mathrm{~km} \cdot \mathrm{s}^{-1} .\)

Common Misunderstandings about Gas Pressure

Example 4. The rms speed of hydrogen molecules at STP 1.85 km s^-1. What is the density of hydrogen gas?
Solution:

Given

The rms speed of hydrogen molecules at STP 1.85 km s^-1.

Here, c = 1.85 km · s^-1 = 1.85 x 10⁵ cm · s^-1

p = 76 x 13.0 x 980 dyn · cm^-2

From the equation p = \(\frac{1}{3} \rho c^2\) we get,

ρ = \(\frac{3 p}{c^2}=\frac{3 \times(76 \times 13.6 \times 980)}{\left(1.85 \times 10^5\right)^2}\)

= \(0.000089 \mathrm{~g} \cdot \mathrm{cm}^{-3} .\)

Example 5. Find out the rms speed of nitrogen gas molecules at 0°C. The density of nitrogen gas at STP = 1.25 g · L^-1 and the density of mercury = 13.6 g · cm^-3.
Solution:

Given

The density of nitrogen gas at STP = 1.25 g · L^-1 and the density of mercury = 13.6 g · cm^-3

p = 76 x 13.6 x 980 dyn cm^-2,

ρ = 1.25 g · L^-1 = 1.25 x 10^-3 g · cm^-3

rms speed, = \(\sqrt{\frac{3 p}{\rho}}\)

= \(\sqrt{\frac{3 \times(76 \times 13.6 \times 980}{1.25 \times 10^{-3}}}\)

= \(4.93 \times 10^4 \mathrm{~cm} \cdot \mathrm{s}^{-1} .\)

Visualizing Gas Molecule Motion and Pressure

Example 6. Find out the ratio of the rms speeds of hydrogen and nitrogen molecules at STP.
Solution:

According to Avogadro’s law, at STP the volume of 1 mol gas is 22.4L.

∴ Density of hydrogen \(\rho_{\mathrm{H}}=\frac{2}{22.4} \mathrm{~g} \cdot \mathrm{L}^{-1}\)

and density of nitrogen, \(\rho_{\mathrm{N}}=\frac{28}{22.4} \mathrm{~g} \cdot \mathrm{L}^{-1}\)

∴ The ratio of the rms speeds of hydrogen and nitrogen is,

⇒ \(\frac{c_{\mathrm{H}_2}}{c_{\mathrm{N}_2}}=\frac{\sqrt{\frac{3 p}{\rho_{\mathrm{H}}}}}{\sqrt{\frac{3 p}{\rho_{\mathrm{N}}}}}=\sqrt{\frac{\rho_{\mathrm{N}}}{\rho_{\mathrm{H}}}}=\sqrt{\frac{28}{2}}=\sqrt{\frac{14}{1}}=3.74: 1\)

Kinetic Theory Of Gases – Principle Of Equipartition Of Energy

WBBSE Class 11 Law of Equipartition of Energy Notes

The principle of equipartition of energy in kinetic theory came essentially from the concept of degrees of freedom. we have seen that the pressure of an ideal gas is p = \(\frac{1}{3}\left(p_x+p_y+p_z\right)\)

The number of degrees of freedom of an ideal gas molecule = 3. Clearly, it led to a factor of 1/3.

So, we get \(p=\frac{1}{3} \rho c^2 \text { or, } c^2=\frac{3 p}{\rho}\)(c = rms speed of the gas molecules).

Again, the total kinetic energy of the molecules in 1 mol of an ideal gas is \(\frac{3}{2} p V=\frac{3}{2} R T\). So, it may be said that the molecular kinetic energy of 1 mol of an ideal gas for each degree of freedom is \(\frac{1}{2}\) RT Scientist Ludwig Boltzmann analysed the motion of a single molecule in an ideal gas and theoretically established the principle of equipartition of energy

Statement of the principle of equipartition of energy: The average molecular kinetic energy of any substance is equally shared among the degrees of freedom; the average kinetic energy of a single molecule associated with each degree of freedom is \(\frac{1}{2} k T\)(T= absolute temperature, k = Boltzmann constant = 1.38 x 10^-23 J · K^-1)

The number of degrees of freedom of an ideal gas molecule = 3. So, from the equipartition principle, the average kinetic energy of a molecule = 3 x \(\frac{1}{2}kT\) = \(\frac{3}{2}\)kT. The molecule has no potential energy. So, the average total energy of a molecule is e = \(\frac{3}{2}\)kT

Again, the total energy of 1 mol of an ideal gas is E = \(\frac{3}{2}\)RT.

So, the number of molecules in 1 mol of a gas = \(\frac{E}{e}\) = \(\frac{R}{k}\).

Clearly, this is the Avogadro number N_A, i.e., \(N_A=\frac{R}{k} \text { or, } R=N_A k\)

This is the relation between the universal gas constant R, the Avogadro number N_A, and the Boltzmann constant k. Then the total energy of 1 mol of an ideal gas is \(E=\frac{3}{2} N_A k T\)

For any amount of ideal gas containing N molecules, \(E=\frac{3}{2} N_A k T\). This relation is widely used particularly in chemistry.

It is to be noted that the equipartition principle is applicable to all substances, not only to gases. For any substance, solid, liquid, or gas, the average molecular kinetic energy associated with each degree of freedom is \(\frac{1}{2}\)kT.

For ideal gases, the molecular potential energy is zero; so it is easy to calculate the total energy. But for real gases, liquids, or solids, the potential energy calculations are not easy. However, for these substances, the kinetic energy strictly follows the principle of equipartition.

Specific heat of a gas: The first law of thermodynamics is written as

dQ = dE+ dW [E is taken for internal energy]

At constant volume, dV = 0; so dW = pdV = 0.

Then, dQ = dE.

So, the heat absorbed or released at constant volume for a temperature change dT of 1 mol of gas is dQ = C_vdT.

Here, C_v = molar specific heat at constant volume.

Then, \(C_v=\frac{d Q}{d T}=\frac{d E}{d T}\)

In the case of monatomic gas: The kinetic theory assumes that the ideal gas molecules are monatomic. Actually, gases like helium, neon, and argon are monatomic. For 1 mol of such a gas, E = \(\frac{3}{2}\)RT

So, \(C_v=\frac{d E}{d T}=\frac{3}{2} R\)

The molar-specific heat at constant pressure is C_p.

∴ \(C_p-C_\nu=R \quad \text { or, } C_p=C_\nu+R=\frac{3}{2} R+R=\frac{5}{2} R\)

The ratio between the two specific heats is \(\gamma=\frac{C_p}{C_v}=\frac{5}{3}=1.67\)

This value tallies with the experimentally determined values of γ in the case of helium, neon, etc.

In Case Of diatomic gas: The molecules of gases like oxygen, nitrogen, hydrogen, etc., are diatomic. The number of degrees of freedom of a diatomic molecule = 5; so for 1 mol of such a gas, E = \(\frac{5}{2}\) RT.

Then, \(C_\nu=\frac{d E}{d T}=\frac{5}{2} R ; C_p=C_\nu+R=\frac{5}{2} R+R=\frac{7}{2} R\)

∴ \(\gamma=\frac{C_p}{C_v}=\frac{7}{5}=1.4\)

This value of γ is also supported by experiments.

Understanding Equipartition of Energy in Physics

In general cases: if f is the number of degrees of free-dom of an ideal gas molecule then, the energy of 1 mol of such a gas E = \(\frac{f}{2}\)RT

⇒ \(C_\nu=\frac{f_R}{2} ; \quad C_p=\frac{f_2}{2} R+R=\left(\frac{f+2}{2}\right) R\)

So, \(\gamma=\frac{C_p}{C_v}=\frac{f+2}{f}=1+\frac{2}{f}\).

Specific heats of helium and hydrogen gases: We know that, R ≈ 2 cal • mol^-1 • °C^-1 For helium gas,

⇒ \(C_v=\frac{3}{2} R=\frac{3}{2} \times 2=3 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

⇒ \(C_p=\frac{5}{2} R=\frac{5}{2} \times 2=5 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

The molecular weight of helium =4.

So the specific heats are, \(c_\nu=\frac{C_v}{4}=\frac{3}{4}=0.75 \mathrm{cal} \cdot \mathrm{g}^{-1 \cdot{ }^{\circ} \mathrm{C}^{-1}}\)

⇒ \(c_p=\frac{C_p}{4}=\frac{5}{4}=1.25 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

For hydrogen gas,

⇒ \(C_v=\frac{5}{2} R=\frac{5}{2} \times 2=5 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

⇒ \(C_p=\frac{7}{2} R=\frac{7}{2} \times 2=7 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

The molecular weight of hydrogen = 2.

So, the specific heats of hydrogen gas are,

⇒ \(c_\nu=\frac{C_\nu}{2}=\frac{5}{2}=2.5 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

and \(c_p=\frac{C_p}{2}=\frac{7}{2}=3.5 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

It may be noted that the value of c_p for helium gas and the values of both c_v and c_p for hydrogen gas are greater than the specific heat of water (1 cal · g^-1 · C^-1).

Degrees of Freedom in Equipartition Theorem

c_v, c_p, and γ of a gas mixture: Let n₁ mol of a gas (molar specific heat at constant volume and constant pressure be \(C_{v_1} \text { and } C_{p_1}\) respectively) is mixed with n₂ mol of a gas (molar specific heat at constant volume and constant pressure be \(C_{\nu_2} \text { and } C_{p_2}\) respectively) such that they do not react chemically.

Therefore, the thermal capacity of (n₁ + n₂) mol gas of the mixture at constant volume is \(n_1 C_{v_1}+n_2 C_{v_2}\)

Hence, at constant volume effective molar specific heat of the mixture is, \(C_v=\frac{n_1 C_{v_1}+n_2 C_{v_2}}{n_1+n_2}\)

Similarly, at constant pressure, the effective molar specific heat of the mixture is \(C_p=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1+n_2}\)

∴ Ratio of the two molar specific heats of the mixure, \(\gamma=\frac{C_p}{C_v}=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1 C_{v_1}+n_2 C_{v_2}}\)

Now, if f₁ and f₂ be the degress of freedom of the molecules of two gases respectively then,

⇒ \(C_{v_1}=\frac{f_1}{2} R ; C_{p_1}=C_{v_1}+R=\left(\frac{f_1}{2}+1\right) R\)

and \(C_{v_2}=\frac{f_2}{2} R ; C_{p_2}=C_{v_2}+R=\left(\frac{f_2}{2}+1\right) R\)

Putting these values in equation(1), we can find γ of the gas mixure.

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Kinetic Theory Of Gases – Principle Of Equipartition Of Energy Numerical Examples

Example 1. The rms speed of the molecules of an ideal gas at STP is 0.5 km · s^-1. Find the density of the gas. What will be the density at 21°C if the pressure remains the same? Given, atmospheric pressure = 10⁵ N · m^-2
Solution:

Given

The rms speed of the molecules of an ideal gas at STP is 0.5 km · s^-1.

Atmospheric pressure = 10⁵ N · m^-2

rms speed, \(c=\sqrt{\frac{3 p}{\rho}}; so \rho=\frac{3 p}{c^2}\).

Here, p = \(10^5 \mathrm{~N} \cdot \mathrm{m}^{-2}\),

c = \(0.5 \mathrm{~km} \cdot \mathrm{s}^{-1}=0.5 \times 1000 \mathrm{~m} \cdot \mathrm{s}^{-1}=500 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ \(\rho=\frac{3 \times 10^5}{(500)^2}=1.2 \mathrm{~kg} \cdot \mathrm{m}^{-3}\)

At constant pressure, \(c \propto \frac{1}{\sqrt{\rho}}; also c \propto \sqrt{T}. So, \rho \propto \frac{1}{T}\)

Then, \(\frac{\rho_0}{\rho_{21}}=\frac{T_{21}}{T_0}\)

or, \(\rho_{21}= \rho_0 \frac{T_0}{T_{21}}\)

(\(T_0=0^{\circ} \mathrm{C}=273 \mathrm{~K}, T_{21}=21^{\circ} \mathrm{C}=(21+273) \mathrm{K}=294 \mathrm{~K}\))

= \(1.2 \times \frac{273}{294}=1.11 \mathrm{~kg} \cdot \mathrm{m}^{-3}\).

Example 2. Find out the energy of 1 mol of a gas and its average molecular kinetic energy at 27°C. Given, R = 8.3×10⁷ erg mol.Kl and N_A = 6.02 x 10²³ mol^-1.
Solution:

Given, R = 8.3×10⁷ erg mol.Kl and N_A = 6.02 x 10²³ mol^-1.

The energy of 1 mol of the gas is

E = \(\frac{3}{2} R T=\frac{3}{2} \times\left(8.3 \times 10^7\right) \times 300\left[T=27^{\circ} \mathrm{C}=300 \mathrm{~K}\right]\)

= 3.735 x 10¹⁰ erg

Average molecular kinetic energy is

e = \(\frac{3}{2} k T=\frac{3}{2} \frac{R}{N_A} T=\frac{3}{2} \times \frac{\left(8.3 \times 10^7\right) \times 300}{6.02 \times 10^{23}}\)

= \(6.2 \times 10^{-14} \mathrm{erg} .\)

Example 3. The average kinetic energy of a molecule in a gas at STP is 5.6 x 10^-14 erg. Find out the number of molecules per volume of the gas. Given, density of mercury = 13.6 g · cm^-3
Solution:

Given

The average kinetic energy of a molecule in a gas at STP is 5.6 x 10^-14 erg.

Density of mercury = 13.6 g · cm^-3

Pressure of the gas, p

= \(\frac{2}{3}\) x energy of gas molecules per unit volume

= \(\frac{2}{3}\) x n x average kinetic energy of 1 molecule

where n = number of molecules per unit volume of the gas

∴ n = \(\frac{3 p}{2 \times \text { average kinetic energy of } 1 \text { molecule }}\)

= \(\frac{3 \times(76 \times 13.6 \times 980)}{2 \times\left(5.6 \times 10^{-14}\right)}=2.71 \times 10^{19} .\)

Applications of Law of Equipartition of Energy

Example 4. Find out the temperature at which the rms speed of nitrogen molecules will be equal to the escape velocity from the earth’s gravity. Given, the mass of a nitrogen atom = 23.24 x 10^-24 g; average radius of the earth = 6390 km; g = 980 cm · s^-2; Boltzmann constant = 1.37 x 10^-16 erg · °C^-1.
Solution:

Given, the mass of a nitrogen atom = 23.24 x 10^-24 g; average radius of the earth = 6390 km; g = 980 cm · s^-2; Boltzmann constant = 1.37 x 10^-16 erg · °C^-1.

rms speed of a molecule = \(\sqrt{\frac{3 R T}{M}} ;\)

escape velocity = \(\sqrt{2 g R_1}\), where Rl = radius of the earth.

According to the question, \(\sqrt{\frac{3 R T}{M}}=\sqrt{2 g R_1}\)

T = \(\frac{2}{3} \frac{g M R_1}{R}\)

Now, R = Nk and M = mN

where m = mass of a nitrogen molecule

= 2 x (23.24 X 10^-24) g ;

N = number of nitrogen molecules

So, T = \(\frac{2}{3} \cdot \frac{g m N R_1}{N k}=\frac{2}{3} \cdot \frac{g m R_1}{k}\)

= \(\frac{2}{3} \times \frac{980 \times\left(2 \times 23.24 \times 10^{-24}\right) \times\left(6390 \times 10^5\right)}{1.37 \times 10^{-16}}\)

= \(1.42 \times 10^5 \mathrm{~K} .\)

Example 5. Find out the temperature at which the average kinetic energy of a gas molecule will be equal to the energy gained by an electron on acceleration across a potential difference of 1 V. Given, Boltzmann constant = 1.38 × 10^-23 J · K^-1; charge of an electron = 1.6 x 10^-19 C.
Solution:

Given, Boltzmann constant = 1.38 × 10^-23 J · K^-1; charge of an electron = 1.6 x 10^-19 C

Energy gained by the electron

= 1 eV = (1.6 x 10~19 C) x 1 V = 1.6 x 1019 J.

Average kinetic energy of a gas molecule

= \(\frac{3}{2}\)kT = \(\frac{3}{2}\) x (1.38 x 1^-23) x T J

∴ 1.6 x 10^-19 = \(\frac{3}{2}\) × (1.38 x 10^-23) x T

or, \(T=\frac{2}{3} \times \frac{1.6 \times 10^{-19}}{1.38 \times 10^{-23}}=7729 \mathrm{~K}=7456^{\circ} \mathrm{C}\)

Example 6. Find out the molecular kinetic energy of 1 mol of oxygen gas at STP. Given, the molecular weight of oxygen = 32 density of oxygen at STP = 1.43 g · L^-1 density of mercury = 13.6 g · cm^-3.
Solution:

Given, the molecular weight of oxygen = 32 density of oxygen at STP = 1.43 g · L^-1 density of mercury = 13.6 g · cm^-3.

The volume of 1 mol or 32 g oxygen

= \(\frac{32}{1.43} \mathrm{~L}=\frac{32 \times 10^3}{1.43} \mathrm{~cm}^3 \text {; }\)

At STP, pressure p = 76 x 13.6 x 981 dyn · cm^-2; temperature, T = 0°C = 273 K.

∴ The molecular kinetic energy of 1 mol oxygen gas at STP (diatomic gas) is

E = \(\frac{5}{2} R T=\frac{5}{2} \frac{p V}{T} T=\frac{5}{2} p V\)

= \(\frac{5}{2} \times(76 \times 13.6 \times 981) \times \frac{32 \times 10^3}{1.43}\)

= \(5.67 \times 10^{10} \mathrm{erg} .\)

Example 7. At what temperature will the rms speed of a hydrogen molecule be equal to that of an oxygen molecule at 47°C?
Solution:

The molecular weights of oxygen and hydrogen, respectively, are M₁ = 32 and M₂ = 2.

rms speed, c = \(\sqrt{\frac{3 R T}{M}}\)

∴ \(\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2} \cdot \frac{M_2}{M_1}}\)

Here, \(c_1=c_2 and T_1=47^{\circ} \mathrm{C}=320 \mathrm{~K}\)

∴ \(\frac{T_1}{T_2} \cdot \frac{M_2}{M_1}=1\)

or, \(T_2 =T_1 \cdot \frac{M_2}{M_1}=320 \times \frac{2}{32}\)

= \(20 \mathrm{~K}=(20-273)^{\circ} \mathrm{C}=-253^{\circ} \mathrm{C} .\)

Equipartition Theorem and Specific Heat Capacity

Example 8. 0.76 g of a mixture of hydrogen and oxygen gases has a volume of 2 L, temperature of 300 K, and pressure of 10⁵ N · m^-2. Find out the individual masses of hydrogen and oxygen in the mixture.
Solution:

Given

0.76 g of a mixture of hydrogen and oxygen gases has a volume of 2 L, temperature of 300 K, and pressure of 10⁵ N · m^-2.

Let the number of moles of hydrogen and oxygen be n₁ and n₂ respectively.

Then, the pressure of the mixture

p = \(\frac{n_1 R T}{V}+\frac{n_2 R T}{V}=\left(n_1+n_2\right) \frac{R T}{V}\)

or, \(n_1+n_2=\frac{p V}{R T}=\frac{10^5 \times\left(2 \times 10^{-3}\right)}{8.3 \times 300}=0.08\)…(1)

Now, the mass of hydrogen gas = 2n₁ and the mass of oxygen gas = 32n₂.

So, 2n₁ + 32n₂ = 0.76

or, n + 16n₂ = 0.38….(2)

(2)-(1)

15n₂ = 0.3 or, n₂ = 0.02

Then from (1), n₁ = 0.08 – 0.02 = 0.06

∴ Mass of hydrogen = 2 x 0.06 = 0.12 g and mass of oxygen = 32 x 0.02 = 0.64g

Example 9. Find out the number of molecules in a gas of volume 20cm³ at a pressure of 76 cm of mercury, and at 27°C. Given, the average molecule kinetic energy at 27°C = 2 x 10^-14 erg.
Solution:

Given

Given, the average molecule kinetic energy at 27°C = 2 x 10^-14 erg.

The pressure of the gas,

p = \(\frac{1}{3} \rho c^2=\frac{1}{3} \frac{m N}{V} {c^2}\)

[m = mass of a molecule, N = number of molecules]

or, N = \(\frac{3 p V}{m c^2}=\frac{3 p V}{2 \times \frac{1}{2} m c^2}\)

[average molecular kinetic energy = \(\frac{1}{2}\) = 2 x 10^-14 erg]

= \(\frac{3 \times(76 \times 13.6 \times 980) \times 20}{2 \times\left(2 \times 10^{-14}\right)}\)

= 1.52 x 10²¹

Example 10. Find the temperature at which the average kinetic energy of a gas molecule will be equal to the energy of a photon in 600Å radiation. Given, the Boltzmann constant, k = 1.38 X 10^-23 J · K^-1; Planck’s constant, h = 6.625 x 10^-34 J · s.
Solution:

Given

Given, the Boltzmann constant, k = 1.38 X 10^-23 J · K^-1; Planck’s constant, h = 6.625 x 10^-34 J · s.

Let the required temperature = T.

Average molecular kinetic energy = \(\frac{3}{2}\)kT;

Energy of a photon = hv = \(\frac{h c}{\lambda}\)

[Here, λ = 6000Å = 6000 x 10^-10 m]

According to the question, \(\frac{3}{2} k T=\frac{h c}{\lambda}\)

∴ T = \(\frac{2}{3} \frac{h c}{\lambda k}=\frac{2}{3} \times \frac{\left(6.625 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{\left(6000 \times 10^{-10}\right) \times\left(1.38 \times 10^{-23}\right)}\)

= 1.6 x 10⁴ K.

Short Answer Questions on Equipartition of Energy

Example 11. Some amount of oxygen gas contained in a vessel has a density of 1.429 kg · m^-3 at STP. The temperature is increased until the pressure is doubled. Neglecting the change in volume of the vessel, find the rms speed of the oxygen molecules.
Solution:

Given

Some amount of oxygen gas contained in a vessel has a density of 1.429 kg · m^-3 at STP. The temperature is increased until the pressure is doubled. Neglecting the change in volume of the vessel,

Mass and volume of the gas remain the same; so the density also remains the same.

So, ρ =1.429 kg · m^-3 = 1.429 x 10^-3 g · cm^-3

In the first case, rms speed of oxygen molecules,

⇒ \(c_1=\sqrt{\frac{3 p_1}{\rho}}=\sqrt{\frac{3 \times(76 \times 13.6 \times 980)}{1.429 \times 10^{-3}}}=46114 \mathrm{~cm} \cdot \mathrm{s}^{-1} .\)

In the second case, p₂ = 2p₁

∴ \(\frac{c_2}{c_1}=\sqrt{\frac{p_2}{p_1}}=\sqrt{2}\)

or, \(c_2=\sqrt{2} c_1=46114 \times \sqrt{2}=65215 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

Example 12. Two ideal gases at absolute temperatures T₁ and T₂ are mixed with each other. If the molecular mass and the number of molecules are m₁, n₁, and m₂, n₂, respectively, find out the temperature of the mixture.
Solution:

Given

Two ideal gases at absolute temperatures T₁ and T₂ are mixed with each other. If the molecular mass and the number of molecules are m₁, n₁, and m₂, n₂, respectively,

Molecular kinetic energy of the first gas = \(n_1 \cdot \frac{3}{2} k T_1\) and that of the second gas = \(n_2 \cdot \frac{3}{2} k T_2\)

So the net energy = \(\frac{3}{2} k\left(n_1 T_1+n_2 T_2\right)\)

The number of molecules in the mixture = n₁ + n₂.

Let the temperature of the mixture be T.

The total molecular kinetic energy = \(\left(n_1+n_2\right) \cdot \frac{3}{2} k T\)

From the principle of energy conservation \(\left(n_1+n_2\right) \cdot \frac{3}{2} k T=\frac{3}{2} k\left(n_1 T_1+n_2 T_2\right)\)

or, T = \(\frac{n_1 T_1+n_2 T_2}{n_1+n_2}\) .

Example 13. 2 mol of a monatomic gas is mixed with 1 mol of a diatomic gas, Find out the value of γ of the mixture.
Solution:

Given

2 mol of a monatomic gas is mixed with 1 mol of a diatomic gas,

For the monatomic gas, \(\gamma_1=1+\frac{2}{f}=1+\frac{2}{3}=\frac{5}{3}\)

where f = number of degrees of freedom

For the diatomic gas, \(\gamma_2=1+\frac{2}{f}=1+\frac{2}{5}=\frac{7}{5}\)

[f = 3 for a monatomic gas and f = 5 for a diatomic gas]

So, for the mixture, \(\gamma=\frac{n_1 \gamma_1+n_2 \gamma_2}{n_1+n_2}=\frac{2 \times \frac{5}{3}+1 \times \frac{7}{5}}{2+1}=1.58\)

Example 14. The mean free path for the collision of nitrogen molecules at STP is 6.44 x 10^-6 cm. What is the mean time interval between collisions? Given, R = 8.31 x 10⁷ erg · mol^-1 · K^-1; molecular mass of nitrogen = 28.
Solution:

Given

The mean free path for the collision of nitrogen molecules at STP is 6.44 x 10^-6 cm.

R = 8.31 x 10⁷ erg · mol^-1 · K^-1; molecular mass of nitrogen = 28.

rms speed, c = \(\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 \times\left(8.31 \times 10^7\right) \times 273}{28}} \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

∴ Mean time interval between collisions = \(\frac{\text { mean free path }}{\text { rms velocity }}\)

= \(6.44 \times 10^{-6} \times \sqrt{\frac{28}{3 \times\left(8.31 \times 10^7\right) \times 273}}\)

= \(1.306 \times 10^{-10} \mathrm{~s}\)

Example 15. The mass of a hydrogen molecule is 3.32 x 10^-27 kg. 10²³ such molecules hit every second on a rigid wall of area 2 cm² at an angle of 45° with horizontal with a velocity of 10³ m · s^-1. If the molecules are reflected with the same velocity, then what is the pressure exerted on the wall?
Solution:

Given

The mass of a hydrogen molecule is 3.32 x 10^-27 kg. 10²³ such molecules hit every second on a rigid wall of area 2 cm² at an angle of 45° with horizontal with a velocity of 10³ m · s^-1. If the molecules are reflected with the same velocity,

Suppose the molecules are incident along PO and reflected along OQ.

Here, ∠PON = ∠NOQ = 45°.

Change of momentum normal to the wall for each hydrogen molecule = 2mvcos45°.

∴ Net change of momentum in a second

= exerted force = n· 2mvcos45°

(where, n = number of molecules)

∴ Pressure exerted on the wall = \(\frac{\text { force }}{\text { area }}=\frac{n \cdot 2 m \nu \cos 45^{\circ}}{\text { area }}\)

= \(\frac{10^{23} \times 2 \times\left(3.32 \times 10^{-27}\right) \times 10^3 \times \frac{1}{\sqrt{2}}}{2 \times 10^{-4}}\)

= \(2.35 \times 10^3 \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Real-Life Examples Illustrating Equipartition of Energy

Example 16. 22 g of CO₂ gas at 27°C is mixed with 16g of O₂ gas at 37° C. What will be the temperature of the mixture?
Solution:

Given

22 g of CO₂ gas at 27°C is mixed with 16g of O₂ gas at 37° C.

22 g CO₂= \(\frac{22}{44}\) or  \(\frac{1}{2}\) mol of CO₂

16 g O₂ =  \(\frac{16}{32}\) or  \(\frac{1}{2}\) mol O₂

Temperature of the mixture, T = \(\frac{n_1 T_1+n_2 T_2}{n_1+n_2}=\frac{\frac{1}{2}(27+273)+\frac{1}{2}(37+273)}{\frac{1}{2}+\frac{1}{2}}\)

= \(305 \mathrm{~K}=32^{\circ} \mathrm{C}\)

Example 17. A mixture of 8g oxygen, 14g nitrogen, and 22g carbon dioxide is contained in a vessel of volume 4L. What will be the pressure of the gas mixture if the temperature of the mixture is 27°C? [Given R = 8.315 J · mol^-1 K^-1]
Solution:

Given

A mixture of 8g oxygen, 14g nitrogen, and 22g carbon dioxide is contained in a vessel of volume 4L.

We know, pV = nRT [n =number of moles]

or, \(p=\frac{n R T}{V}=\frac{g}{M} \cdot \frac{R T}{V}\)

[g = mass of the gas, M = atomic weight of the gas]

Pressure exerted by \(\mathrm{O}_2 \text { is } p_{\mathrm{O}_2}=\frac{8}{32} \cdot \frac{R T}{V}=\frac{1}{4} \frac{R T}{V}\)

Pressure exerted by N is \(p_{\mathrm{N}_2}=\frac{14}{28} \cdot \frac{R T}{V}=\frac{1}{2} \frac{R T}{V}\)

and pressure exerted by \(\mathrm{CO}_2 \text { is } p_{\mathrm{CO}_2}=\frac{22}{44} \cdot \frac{R T}{V}=\frac{1}{2} \frac{R T}{V}\)

According to Dalton’s law, the pressure of the gas mixture,

p = \(p_{\mathrm{O}_2}+p_{\mathrm{N}_2}+p_{\mathrm{CO}_2}=\frac{R T}{V}\left(\frac{1}{4}+\frac{1}{2}+\frac{1}{2}\right)=\frac{5}{4} \frac{R T}{V} \)

= \(\frac{5}{4} \times \frac{8.315 \times 300}{4 \times 10^{-3}}=7.795 \times 10^5 \mathrm{~N} / \mathrm{m}^2\)

Example 18.  1 mol of He at 57°C is mixed with 1 mol of Ar at 27°C. Find the temperature of the gas mixture.
Solution:

Given

1 mol of He at 57°C is mixed with 1 mol of Ar at 27°C.

Average kinetic energy of 1 mol gas

= \(\frac{3}{2} R T=\frac{3}{2} k N_A T\)

Average kinetic energy of 1 mol of He =

= \(\frac{3}{2} k N_A(273+57)=\frac{3}{2} k N_A \times 330\)

Average kinetic energy of 1 mol of Ar

= \(\frac{3}{2} k N_A(273+27)=\frac{3}{2} k N_A \times 330\)

After mixing the two gases, the number of atoms in the mixture = 2 N_A.

Let the temperature of the mixture is T.

Average kinetic energy of the mixture = \(\frac{3}{2} \times 2 N_A k T\)

∴ From the principle of conservation of energy,

⇒ \(\frac{3}{2} \times 2 N_A k T=\frac{3}{2} k N_A \times 330+\frac{3}{2} k N_A \times 300\)

or, T = \(\frac{1}{2}\)(330 + 300) = 315K = (315 – 273) = 42°C.

Example 19. Find the minimum radius of the planet of density 5.5 x 10³ kg/m³ and temperature 427°C which can hold O₂ in its atmosphere. [Given G = 6.67 x 10^-11 N · m^-2 · kg^-2 and R = 8.3 J · mol^-1 · K^-1
Solution:

Given G = 6.67 x 10^-11 N · m^-2 · kg^-2 and R = 8.3 J · mol^-1 · K^-1

The escape velocity of any object at the surface of the planet of radius r and mass M₁ is

v = \(\sqrt{\frac{2 G M_1}{r}}=\sqrt{\frac{2 G}{r} \cdot \frac{4}{3} \pi r^3 \rho}]\)

[where ρ is the material density of the planet]

= \(\sqrt{\frac{8}{3} G \pi r^2 \rho}\)

rms speed of a gas of molecular weight M at an absorb temperature T is \(c=\sqrt{\frac{3 R T}{M}}\)

Since the planet holds O₂, hence v_min = c

∴ \(\sqrt{\frac{8}{3} G \pi r_{\min } \rho}=\sqrt{\frac{3 R T}{M}}\)

(\(r_{\min }\)= minimum radius of the planet)

or, \(\frac{8}{3} G \pi r_{\min }^2 \rho=\frac{3 R T}{M}\)

or, \(r_{\min }^2=\frac{9 R T}{8 G \pi \rho M}\)

or, \(r_{\min }=\sqrt{\frac{9 R T}{8 G \pi \rho M}}\)

= \(\sqrt{\frac{9 \times 8.3 \times(427+273) \times 7}{8 \times 6.67 \times 10^{-11} \times 22 \times 5.5 \times 10^3 \times 32 \times 10^{-3}}}\)

= \(421 \times 10^3 \mathrm{~m}=421 \mathrm{~km} .\)

Example 20. 1 mol O₂ at temperature 27°C at STP (1.01 x 10⁵ N/m²) IJS kept in a vessel. Find the number of collisions the molecules experience (in SI) per second per unit area with the wall of the vessel. [Given Boltzmann constant k = 138 x 10¹³ J/K]
Solution:

Given

1 mol O₂ at temperature 27°C at STP (1.01 x 10⁵ N/m²) IJS kept in a vessel.

Number of molecules in 1 mol of 02 = 6.023 x 1023

∴ Mass of a molecule, m = \(\frac{32}{6.023 \times 10^{23} \times 1000}\)

= \(5.316 \times 10^{-26} \mathrm{~kg}\)

Momentum of the molecule, P = mv = \(m \sqrt{\frac{3 k T}{m}}=\sqrt{3 k T m}\)

Change in momentum due to each collision,

ΔP = \(2 P=2 \sqrt{3 k T m}\)

= \(2 \sqrt{3 \times 1.38 \times 10^{-23} \times 300 \times 5.316 \times 10^{-26}}\)

= \(5.139 \times 10^{-23} \mathrm{~kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-1}\)

If the molecules experience ‘n’ number of collisions per second per square meter of the wall, then the pressure of the gas will be

p = nΔP or, 1.01 x 10⁵ = n x 5.139 x 10^-23

or, n = 1.965 x 10²⁷.

Example 21. 0.014kg N₂ gas at 27°C is kept in a dosed vessel, How much heat is required to double the rms of the N₂ molecules?
Solution:

Given

0.014kg N₂ gas at 27°C is kept in a dosed vessel,

Heat received, ΔQ = nC_vΔT

[C_v = molar specific heat at constant volume

Here, \(n=\frac{0.014 \times 1000}{28}=\frac{1}{2}\)

For diatomic molecule C_v = \(\frac{5}{2}\)R

and rms speed of N₂ molecule, c ∝ √T

∴ To double the velocity c, the temperature should be 4 T.

∴ \(\Delta Q=\frac{1}{2} \times \frac{5}{2} R \times(4 T-T)\)

= \(\frac{5}{4} \times 2 \times(273+27) \times 3\left[because R=2 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\right]\)

= 2250 cal

Example 22. If 2 mol of a gas at constant pressure, requires 70 cal heat to increase its temperature from 30°C to 35°C, then find its degrees of freedom.
Solution:

Given

If 2 mol of a gas at constant pressure, requires 70 cal heat to increase its temperature from 30°C to 35°C,

Work done at constant pressure,

ΔW = pΔV = nRΔT =2x2x5

[ΔT = (308-303) = 5K] = 20 cal

Given that, heat received, ΔQ = nC_pΔT = 70 cal.

∴ \(C_p=\frac{70}{2 \times 5}=7 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

∴ \(C_v=C_p-R=7-2=5 \mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

∴ \(\gamma=\frac{C_p}{C_v}=\frac{7}{5}\)

Again, \(\gamma=1+\frac{2}{f}\)

∴ 1 + \(\frac{2}{f}=\frac{7}{5}\)

or, \(\frac{2}{f}=\frac{7}{5}-1=\frac{7-5}{5}=\frac{2}{5}\)

∴ f = 5

Degree of freedom of the gas molecule is 5.

 

Kinetic Theory Of Gases

Interpretation Of Temperature From Kinetic Theory

WBBSE Class 11 Kinetic Interpretation of Temperature Overview

Total energy (E) of gas molecules: According to the kinetic theory, the potential energy of gas molecules = 0; the entire energy of the gas comes from the kinetic energy of the molecules.

∴ E = \(\frac{1}{2} m c_1^2+\frac{1}{2} m c_2^2+\cdots+\frac{1}{2} m c_N^2\)

[m = mass of each molecule; c₁, c₂,….. c_N = velocity of the N molecules in the container]

= \(\frac{1}{2} m\left(c_1^2+c_2^2+\cdots+c_N^2\right)=\frac{1}{2} m N \frac{c_1^2+c_2^2+\cdots+c_N^2}{N}\)

As mN = M = mass of the gas,

and c = \(\sqrt{\frac{c_1^2+c_2^2+\cdots+c_N^2}{N}}\)

= rms speed of the molecules,

∴ E = \(\frac{1}{2} M c^2=\frac{1}{2} M \cdot \frac{3 p}{\rho}\left[\text { As } p=\frac{1}{3} \rho c^2, \text { we have } c^2=\frac{3 p}{\rho}\right]\)

Again, as \(\frac{M}{\rho}\) = V = volume of the gas, we get

E = \(\frac{3}{2} p V\)…(1)

This is the expression for the total energy of a gas.

Relation between pressure and energy density: Energy density of a gas (u) = energy per unit volume = \(\frac{E}{V}\)

From relation (1), \(\frac{E}{V}=\frac{3}{2} p\)…(1)

or, \(u=\frac{3}{2} p \quad \text { or, } \quad p=\frac{2}{3} u\)…(2)

So, the pressure of a gas is \(\frac{2}{3}\)rd of its energy density. This relation (2) is a fundamental one, valid for all ideal gases.

Effect of heat absorbed by a gas: when a gas absorbs heat from its surroundings, two effects occur simultaneously.

1. The temperature of the gas increases.
2. Heat is converted to some other form of energy inside the gas. It is nothing but the kinetic energy of the molecules. This means that the kinetic energy of the molecules increases.

These two effects suggest that the temperature of a gas and the kinetic energy of its molecules are intimately related with each other. The definition of temperature in kinetic theory comes from this concept.

Definition of Kinetic Temperature for Class 11

Idea of temperature: Temperature (T) is a property of a gas, which is proportional to the kinetic energy of the gas molecules.

So, T ∝ E or, T = aE where a is a constant.

The constant a may have any value. Different values of a will give different temperature scales. Usually, the value of a is so chosen that the temperature scale is an exact match with the experimental Kelvin scale of temperature.

Let us take 1 mol of an ideal gas. Then E is the kinetic energy of the molecules present in 1 mol of the gas. Then we choose \(a=\frac{2}{3 R}\), where

R = universal gas constant = 8.31 J · mol · K^-1

Then, for 1 mol of an ideal gas, \(T=\frac{2}{3 R} E \quad \text { or, } \quad E=\frac{3}{2} R T\)….(3)

With this choice of a, the quantity T becomes exactly the same as the Kelvin temperature.

• We know that the potential energy of an ideal gas molecule is zero. So the total internal energy is essentially the kinetic energy of the molecules. But for real gases and also for liquids and solids, the molecular potential energy is not zero.
• Then the total internal energy becomes the sum of the molecular kinetic and potential energies. Here, it must be noted that the temperature is taken to be proportional to the kinetic energy only of the molecules.
• In this way, the concept of temperature is extended to all gases, liquids, and solids. This means that temperature is a property of all substances and is proportional to the kinetic energy of the molecules in the substance.

The proportionality constant is taken in such a way that the values of temperature exactly match with the values of the Kelvin scale. For this, the required absolute constant is the Boltzmann constant.

The idea of absolute zero of temperature: By definition, T ∝ E; so E = 0 when T = 0. This defines the absolute zero of temperature. It is the temperature at which the internal energy of the gas becomes zero, i.e., the molecular motions stop entirely.

Variation of the rms speed of gas molecules: Let M = molecular weight of a gas. Then, mass of 1 mol of the gas = M g.

The density of the gas is ρ = \(\frac{M}{V}\); so M = ρV.

Now, p = \(\frac{1}{3} \rho c^2 \text { or, } c^2=\frac{3 p}{\rho}=\frac{3 p V}{\rho V}=\frac{3 R T}{M}\)

∴ c = \(\sqrt{\frac{3 R T}{M}}\)….(3)

As R = constant and for a particular gas, M = constant, we get c ∝ √T. So, the rms speed of gas molecules is proportional to the square root of the temperature of the gas.

Average Kinetic Energy and Temperature Relationship

Most probable velocity of gas molecules: The kinetic theory assumes that a gas molecule may have a velocity between zero and infinity. But in reality, the number of molecules with very low or very high velocities is extremely small The majority of molecules have intermediate velocities.

• Maxwell analysed the phenomenon with his velocity distribution curve. This curve has a peak at P and the point P corresponds to a velocity c_m.
• Among all velocities, the velocity c_m is possessed by the highest number of gas molecules. This c_m is known as the most probable velocity.

The most probable velocity of gas molecules Definition: The velocity which is possessed by the highest number of gas molecules in a container is called the most probable velocity.

Shows that, in general, c_m is less than both the mean velocity \(\bar{c}\) and the rms speed c of gas molecules.

Actually, cm< \(\bar{c}\) <c.

Theoretically, we get an absolute temperature of T,

⇒ \(c_m=\sqrt{\frac{2 R T}{M}}, \bar{c}=\sqrt{\frac{8 R T}{\pi M}}, c=\sqrt{\frac{3 R T}{M}}\)

So, \(c_m: \bar{c}: c=\sqrt{2}: \sqrt{\frac{8}{\pi}}: \sqrt{3}=1: \frac{2}{\sqrt{\pi}}: \sqrt{\frac{3}{2}} .\)

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Interpretation Of Temperature From Kinetic Theory Numerical Examples

Examples of Kinetic Interpretation of Temperature

Example 1. At what temperature will the rms speed of molecules of nitrogen gas be twice of that at 0°C?
Solution:

T₁ = 0°C = 273 K; rms speed at 0°C = c₁

At temperature T₂, rms speed = c₂ = 2 c₁.

As \(c=\sqrt{\frac{3 R T}{M}}\),

∴ \(c \propto \sqrt{T}\), we have \(\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2}}\)

or, \(T_2 =T_1\left(\frac{c_2}{c_1}\right)^2=273 \times\left(\frac{2}{1}\right)^2=1092 \mathrm{~K}\)

= \((1092-273)^{\circ} \mathrm{C}=819^{\circ} \mathrm{C} .\)

Example 2. The temperature of a gas rises from 27°C to  327°C. Show that the rms speed of the gas molecules would be √2 times its initial value at die final higher temperature.
Solution:

Given

The temperature of a gas rises from 27°C to  327°C.

T₁ = 27°C = (27 + 273) K = 300 K;

T₂ = 327°C = (327 + 273) K = 600 K.

As \(c \propto \sqrt{T}\), we have \(\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2}}\)

or, \(c_2=c_1 \sqrt{\frac{T_2}{T_1}}=c_1 \sqrt{\frac{600}{300}}=\sqrt{2} c_1\)

Mathematical Formulas for Kinetic Energy and Temperature

Example 3. The rms speed of oxygen gas molecules at STP is 4.5 x 10⁴ cm · s^-1. Find out the same for carbon dioxide gas molecules at STP. Given, the molecular weights of oxygen and carbon dioxide are 32 and 44, respectively.
Solution:

Given

The rms speed of oxygen gas molecules at STP is 4.5 x 10⁴ cm · s^-1.

We know, c = \(\sqrt{\frac{3 R T}{M}}\)

As the temperature is the same for both the gases, \(c \propto \frac{1}{\sqrt{M}}\)

So, \(\frac{c_{\mathrm{O}_2}}{c_{\mathrm{CO}_2}}=\sqrt{\frac{M_2}{M_1}}\)

[M₁ = molecular weight of O₂, M₂ = molecular weight of CO₂]

or, \(c_{\mathrm{CO}_2}=c_{\mathrm{O}_2} \sqrt{\frac{M_1}{M_2}}=4.5 \times 10^4 \times \sqrt{\frac{32}{44}}\)

= \(3.84 \times 10^4 \mathrm{~cm} \cdot \mathrm{s}^{-1} .\)

Example 4. Find out the kinetic energy of 2g of nitrogen gas at 27°C, Given, R = 8.3 x 10⁷ erg mol^-1 K^-1
Solution:

The Kinetic energy of the molecules of 1 mol gas = 3/2 RT

Here, T = 27°C = (27 + 273) K = 300 K ; mass of 1 mol nitrogen gas = 28 g.

So, the kinetic energy of the molecules in 2 g of nitrogen gas

= \(\frac{2}{28} \times \frac{3}{2} R T=\frac{3}{28} R T=\frac{3 \times\left(8.3 \times 10^7\right) \times 300}{28}\)

= \(2.668 \times 10^9 \mathrm{erg}=2.668 \times 10^2 \mathrm{~J}=266.8 \mathrm{~J} .\)

Example 5. At what temperature the average kinetic energy of the molecules of a perfect gas be doubled than that at 20°C?
Solution:

Here, T₁ = 20 °C = 293 K.

As the average kinetic energy of gas molecules is proportional to the absolute temperature of the gas, the required temperature,

T₂ = 2 x 293 = 586 K = (586 – 273) °C = 313 °C

Kinetic Interpretation of Temperature in Gases

Question 6. Find out the temperature at which the molecular rms speed of a gas would be 1/3rd its value at 100°C.
Solution:

Let the required temperature be T₂K, molecular rms speed at this temperature be c_2, and that at 100°C be c₁.

According to the question, \(c_2=\frac{1}{3} c_1\)

Here, \(T_1=100^{\circ} \mathrm{C}=(100+273) \mathrm{K}=373 \mathrm{~K}\)

As \(c \propto \sqrt{T}\), we have, \(\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2}}\)

or, \(T_2=T_1\left(\frac{c_2}{c_1}\right)^2 =373 \times\left(\frac{1}{3}\right)^2=\frac{373}{9}=41.44 \mathrm{~K}\)

= \((41.44-273)^{\circ} \mathrm{C}=-231.56^{\circ} \mathrm{C}\).

Kinetic Theory Of Gases Question And Answers

WBBSE Class 11 Kinetic Theory of Gases Q&A

Question 1. Will the rms speeds of molecules of different gases at the same temperature be the same?
Answer:

If c is the rms speed of 1 mol of a gas, then c = \(\sqrt{\frac{3 R T}{M}} \)

The value of gas constant R is the same for 1 mol of different gases. But different gases have different molecular weights M so at the same absolute temperature T, we have \(c \propto \frac{1}{\sqrt{M}}\)

This means that the rms speed will not be the same for molecules of different gases; gases with higher molecular weight walls have less molecular rms speed.

Question 2. 1 cm³ of hydrogen gas and 1 cm³ of oxygen gas are both at STP. Which one contains more number of molecules?
Answer:

Given

1 cm³ of hydrogen gas and 1 cm³ of oxygen gas are both at STP.

Avogadro’s law states that equal volumes of all gases at the same temperature and pressure contain an equal number of molecules. So, 1cm³ of hydrogen gas and 1cm³ of oxygen gas contain an equal number of molecules at STP.

Question 3. How does the kinetic theory explain the increase of temperature of a gas when heat is supplied from outside?
Answer:

Energy supplied in the form of heat is converted into the kinetic energy of the gas molecules. So the total kinetic energy of the molecules increases.

The kinetic theory states that the temperature of a gas is proportional to the kinetic energy of the gas molecules i.e.,  E ∝ T. Thus the tem¬perature of the gas increases.

Question 4. A porous container is filled with a gas mixture. Which gas would leak faster from the container when it is placed in a vacuum?
Answer:

If the temperature is constant, rms speed of gas molecules is inversely proportional to the molecular weight of the gas, i.e., \(c \propto \frac{1}{\sqrt{M}}\)

So, the rms speed of the lighter gas molecules is higher than that of heavier gas molecules. As a result, the lighter gas would leak faster through the pores of the container.

Short Answer Questions on Kinetic Theory of Gases

Question 5. An equal number of molecules of an ideal monatomic and an ideal diatomic gas are at the same temperature. Which gas will be more heated if an equal amount of heat is supplied from outside?
Answer:

Given

An equal number of molecules of an ideal monatomic and an ideal diatomic gas are at the same temperature.

Let, N = Number of molecules in each of the gases, kept at a temperature T. Degrees of freedom of an ideal monatomic molecule = 3.

So, the average kinetic energy of the monatomic gas, \(E_1=\frac{3}{2} N k T\) then its heat capacity at constant volume, \(C_{v_1}=\frac{3}{2} N k\)

[k = Boltzmann constant]

Similarly, for the diatomic gas, \(C_{v_2}=\frac{5}{2} N k\), as the number of degrees of freedom of a diatomic molecule = 5.

Now, an amount of energy E, in the form of heat, is supplied to each of the gases. If T₁ and T₂ be the increases in tem¬perature, respectively, then

E = \(C_{\nu_1} T_1=C_{\nu_2} T_2\),

or, \(\frac{T_1}{T_2}=\frac{C_{v_2}}{C_{v_1}}=\frac{\frac{5}{2} N k}{\frac{3}{2} N k}=\frac{5}{3}\)

So, T₁ >T₂, Le., this monatomic gas will be more heated.

Question 6. The motion of gas molecules ceases at the temperature of absolute zero. Explain.
Answer:

The average kinetic energy of a gas molecule at an absolute temperature T is 3/2 kT, where k is the Boltzmann constant At T = 0, this kinetic energy becomes zero. So, molecular motion ceases at the temperature of absolute zero.

Question 7. The velocity of a gas molecule is comparable to that of a rifle bullet. Yet a gas molecule spends a much longer time than a bullet does to travel equal distances. Explain.
Answer:

Given

The velocity of a gas molecule is comparable to that of a rifle bullet. Yet a gas molecule spends a much longer time than a bullet does to travel equal distances.

A gas molecule suffers multiple collisions with the other molecules in the gas. As a result, it cannot move straight but travels any finite distance along a random zigzag path. On the other hand, a rifle bullet is much heavier than gas molecules.

Collisions with very light air molecules cannot alter the straight path of the bullet. So the bullet travels a finite distance in a much shorter time.

Key Concepts in Kinetic Theory of Gases: Q&A Format

Question 8. How would the rms speed of an ideal gas change if

1. Temperature increases,
2. Density increases at constant pressure,
3. Density increases at constant temperature.

Answer:

Let, the pressure, density, temperature, and molecular weight of an ideal gas are p, ρ, T, and M respectively.

∴ rms speed, c = \(\sqrt{\frac{3 p}{\rho}}=\sqrt{\frac{3 R T}{M}} .\)

1. In the above equation, as c ∝ √T, the rms speed increases with the increase in temperature of the gas.
2. As \(c \propto \frac{1}{\sqrt{\rho}}\) at constant pressure, the rms speed decreases with the increase in density of the gas.
3. At constant temperature, V \(\frac{1}{p}\)

Again, as V \(\frac{1}{p}\)

∴ p ∝ ρ

Hence, \(\frac{P}{\rho}=\mathrm{constant}\)

∴ ρ = \(\sqrt{\frac{3 P}{\rho}}=\mathrm{constant}\)

∴ In this case, the rms speed does not change with an increase in the density of the gas.

Question 9. Why does a real gas obey Boyle’s and Charles’ laws at

1. High temperature and
2. Low pressure?

Answer:

The intermolecular force of attraction is not negligible for real gases. So, a real gas molecule has some potential energy in addition to its kinetic energy.

However, this potential energy can still be neglected in two extreme cases:

1. High temperature: In this case, the molecular kinetic energy is so high that the potential energy is negligibly small.
2. Low pressure: In this case, the distance between molecules is so high that the force of attraction among them is very small so the molecular potential energy may be neglected.

In these two cases, a real gas molecule essentially has a kinetic energy only. So it behaves as an ideal gas and obeys Boyle’s and Charles’ laws.

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Question 10. A gas mixture contains 1 mol each of two different gases. Would the average molecular kinetic energy of the two gases be equal? Would the rms speeds be equal?
Answer:

In the mixture, both gases are at the same temperature T. The average molecular kinetic energy is \(\frac{3}{2}\) kT.

So, this is equal for both gases.

But the rms speed is \(\sqrt{\frac{3 A T}{M}}\). As the molecular weight M is different for the two gases, the rms speed of the molecules is different.

Applications of Kinetic Theory: Questions and Answers

Question 11. State the conditions in which a real gas behaves as an ideal gas.
Answer:

Condition 1: Each molecule is effectively a point, i.e., molecular volume is negligible.

Condition 2: Intermolecular attraction is negligible.

Question 12. For a fixed mass of a gas at a constant temperature, the pressure falls when the volume increases, and vice versa. Explain according to the kinetic theory.
Answer:

When volume increases at a constant temperature, the intermolecular distance increases. So, there are less number of molecules in unit volume. As a result, the number of collisions of the molecules per second with the unit area of wall of the container decreases.

Thus, the change of momentum of the molecules i.e., force exerted by the molecules decreases. This is why the pressure of the gas falls. Conversely, the pressure rises due to an increased number of collisions per second when the volume of a gas decreases at a constant temperature.

Question 13. For a fixed mass of a gas at constant volume, pressure rises when temperature increases, and vice versa. Explain the kinetic theory.
Answer:

When the temperature increases at constant volume, the gas molecules move in the container with greater velocities. So the molecules collide with the wall with greater momenta. As a result, they exert greater force on the wall and the pressure of the gas rises.

Conversely, due to the opposite behavior of gas molecules, pressure falls with a decrease in temperature at constant volume.

Question 14. Find out the molecular kinetic energy of 1 mol of an ideal gas. Is it equal for all gases?
Solution:

Let the molecular weight, volume, pressure and temperature of an ideal gas be M, V, p, and T respectively. If the density of the gas is ρ and the rms speed of the molecules is c, then according to the kinetic theory of gases,

p = \(\frac{1}{3} \rho c^2=\frac{1}{3} \frac{M}{V} c^2\)

or, \(\frac{1}{3} M c^2=p V=R T\) (R= universal gas constant)

So, the molecular kinetic energy of 1 mol of the gas

= \(\frac{1}{2} M c^2=\frac{3}{2} \times \frac{1}{3} M c^2=\frac{3}{2} R T\)

In general, real gases do not obey the ideal gas conditions So, the value of molecular kinetic energy differs from the ideal gas value \(\frac{3}{2}\)RT.

Moreover, the value becomes different for different gases, However, when gases obey the Ideal gas conditions, the value of the molecular kinetic energy becomes the same for all gases at the same temperature T.

Question 15. In a closed container, the gas molecules have a highly random motion. Yet, the pressure throughout the container Is uniform at a constant temperature, Explain.
Answer:

The number of gas molecules in a container is extremely large, For example, 1 cm³ of a gas contains nearly 10²³ molecules.

So, the individual behavior of the molecules are no longer important rather, the gross statistical behavior dominates, On every unit area anywhere on the wall, the number of collisions of molecules per second, the velocity of impact, etc., are all equal on the average.

The value of the mean velocity does not change if the temperature remains constant, So the pressure remains uniform, as long as the temperature of the gas remains the same.

Question 16. Why does a piece of wood floating on water have no Brownian motion?
Answer:

The piece of wood is very large compared to the dimension of water molecules. At every instant, a very large number of moving water molecules collide with this piece. The force exerted in any direction due to some colliding molecules is canceled by the equal and opposite force due to some other molecules. As a result, the resultant force on the piece of wood becomes zero and it has no Brownian motion.

Question 17. Light gases like hydrogen and arid helium are very rare in the Earth’s atmosphere. Why?
Answer:

Light gases like hydrogen and arid helium are very rare in the Earth’s atmosphere.

The escape velocity from the earth’s surface is 11.2 km · s^-1, approximately. At the upper atmosphere, it is still lower. The atmospheric temperature was very high at the time of the formation of the Earth.

• At that temperature, the rms speed of hydrogen gas molecules was 5 km · s^-1 or higher. As this is an average velocity, a large number of molecules were moving with velocities higher than the escape velocity.
• As a result, those molecules left the earth’s field of gravity forever. This incident occurred over a long period of time. So light gases like hydrogen and helium are rare in the earth’s atmosphere.

Question 18. If n is the number of degrees of freedom of the molecules of an Ideal gas, show that the ratio \(\frac{G_p}{C_p}\) is \(1+\frac{2}{n}\).
Answer:

Average kinetic energy of a gas molecule

= \(n \cdot \frac{1}{2} k T\) (k = Boltzmann constant)

As the molecules have no potential energy, the total Internal energy of 1 mol of a gas Is

E = \(N_0 n \cdot \frac{1}{2} k T=\frac{n}{2} N_0 k T=\frac{n}{2} R T\)

(\(N_0 k=R, \text { where } N_0=\text { Avogadro number }\))

∴ \(C_v=\frac{d E}{d T}=\frac{n}{2} R \text { and } C_p=C_v+R=\left(\frac{n}{2}+1\right) R,\) [for ideal gas]

Then, \(\frac{C_p}{C_p}=\frac{\frac{n}{2}+1}{\frac{n}{2}}=1+\frac{2}{n}\).

Mathematical Problems in Kinetic Theory: Q&A

Question 19. 1 mol of an ideal monatomic gas \(\left(\gamma=\frac{5}{3}\right)\) is mixed with 1 mol of an ideal diatomic gas \(\left(\gamma=\frac{7}{3}\right)\). Find the value of γ for the mixture.
Answer:

For the monatomic gas, \(C_v=\frac{3}{2} R\) and  \(C_p=\frac{5}{2} R\)

For the diatomic gas, \(C_v=\frac{5}{2} R \text { and } C_p=\frac{7}{2} R \text {. }\).

So, for 2 mol of the mixture,

⇒ \(C_v=1 \times \frac{3}{2} R+1 \times \frac{5}{2} R=4 R ;\)

⇒ \(C_p=1 \times \frac{5}{2} R+1 \times \frac{7}{2} R=6 R\)

or, \(\gamma=\frac{C_p}{C_v}=\frac{6 R}{4 R}=\frac{3}{2}=1.5 .\)

Question 20. The ratio between the specific heats of an ideal gas is γ. Show that the number of degrees of freedom of the gas molecules is n = \(\frac{2}{γ-1}\).
Answer:

Given

The ratio between the specific heats of an ideal gas is γ.

⇒ \(C_\nu=\frac{n}{2} R ; C_p=C_\nu+R=\frac{n}{2} R+R=\left(\frac{n}{2}+1\right) R\)

∴ \(\gamma=\frac{C_p}{C_v}=\frac{\frac{n}{2}+1}{\frac{n}{2}}=1+\frac{2}{n} ;\)

or, \(\frac{2}{n}=\gamma-1 \quad$ or, \quad n=\frac{2}{\gamma-1}\)

Question 21 If the absolute temperature of a perfect gas rises to j four times Its initial value, estimate the changes of

1. Molecular rms speed and
2. Total kinetic energy.

Answer:

1. Molecular rms speed, \(c \propto \sqrt{T}\)

∴ \(T_2=4 T_1\)

So, \(\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2}}\)

or, \(c_2=c_1 \sqrt{\frac{T_2}{T_1}}=c_1 \sqrt{\frac{4 T_1}{T_1}}=2 c_1\)

2. Total kinetic energy, \(E \propto T\).

So, \(\frac{E_1}{E_2}=\frac{T_1}{T_2} \quad or, \quad E_2=E_1 \frac{T_2}{T_1}=4 E_1\)

Question 22. Some gas cylinders are kept on a running vehicle. What will be the change in temperature of the gas molecules inside the cylinders?
Answer:

The motion of the cylinders is an external motion. It does not alter the internal motion of the molecules. So the molecular kinetic energy does not change. As a result, the temperature of the gas remains the same.

Question 23. Find the dimension of the constant a in the van der Waals equation \(\left(p+\frac{a}{V^2}\right)(V-b)=R T\).
Answer:

The expression \(p+\frac{a}{V^2}\) shows that

⇒ \(\left[\frac{a}{V^2}\right]=[p]\)

So, \(\quad[a]=\left[p V^2\right]=\mathrm{ML}^{-1} \mathrm{~T}^{-2} \cdot\left(\mathrm{L}^3\right)^2=\mathrm{ML}^5 \mathrm{~T}^{-2}\).

WBBSE Class 11 Sample Questions on Gas Behavior

Question 24. Find the dimension of the constant b in the van der Waals equation \(\left(p+\frac{a}{V^2}\right)(V-b)=R T\).
Answer:

The expression V- b shows that [b] = [V].

So, [b] = L³.

Question 25. We have a sample of a gas characterized by p, V, T and another sample by 2p, \(\frac{V}{4}\),2T. What is the ratio of the number of molecules in the two samples?
Answer:

For the first sample, pV = n₁RT (n₁ = number of moles)

For the second sample, \(2 p \cdot \frac{V}{4}=n_2 R \cdot 2 T \quad\left(n_2=\text { number of moles }\right)\)

or, \(p V=4 n_2 R T\)

∴ \(n_1 R T=4 n_2 R T \quad \text { or, } \frac{n_1}{n_2}=4 .\)

So, the ratio of the number of molecules in the two samples Is also 4:1.

Question 26. Find out the ratio between the absolute temperatures of two samples of hydrogen and oxygen gases, if their molecular rms speeds are equal.
Answer:

rms speed, c = \(\sqrt{\frac{3 R T}{M}}\)

For hydrogen, molecular weight, M₁ = 2, and for oxygen, M₂ = 32.

∴ \(\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2} \cdot \frac{M_2}{M_1}} \text { or, } \frac{T_1}{T_2}=\frac{M_1}{M_2} \cdot\left(\frac{c_1}{c_2}\right)^2=\frac{2}{32} \cdot\left(\frac{1}{1}\right)^2=\frac{1}{16} \text {. }\)

Question 27. At equilibrium, the volume, pressure, and temperature of a gas are V, p, and T, respectively. If the gas is divided into two parts by a partition, what will be the value of these quantities in each part?
Answer:

Given

At equilibrium, the volume, pressure, and temperature of a gas are V, p, and T, respectively. If the gas is divided into two parts by a partition,

The rms speed of gas molecules does not depend on the volume of the container. So, speed does not change when the volume of the gas is halved by using a partition.

As a result, the temperature remains the same. As the density of gas does not undergo any change and as p = \(\frac{1}{3} \rho c^2\), pressure will also remain the same. Thus only the volume becomes half, but pressure and temperature remain the same.

So, the values of volume, pressure, and temperature will be \([\frac{V}{2}\), p and T, respectively.

Interactive Q&A on Kinetic Theory for Students

Question 28. In a gas-filled container, a molecule of speed 200 m/s collides at an angle of 30° with the horizontal face of this container and rebounds with the same speed. Is the collision elastic or inelastic? In this momentum conserved in this collision?
Answer:

Given

In a gas-filled container, a molecule of speed 200 m/s collides at an angle of 30° with the horizontal face of this container and rebounds with the same speed.

As the molecule rebounds with the same speed, the collision is elastic. The momentum is always conserved in a collision, irrespective of whether the collision is elastic or inelastic.

Question 29. While considering the motion of gas molecules in a container, why do we use rms speed instead of the average speed of molecules?
Answer:

Since a large number of gas molecules is present in a container, therefore for the velocity of any molecule, there exists another molecule with an equal and opposite velocity.

As velocity is a vector quantity, the resultant velocity of all the molecules becomes zero. Hence average velocity also vanishes. So we could not derive any conclusion about the velocity of the gas molecules from it.

On the other side, if only the magnitudes of velocities (scalar) are considered to find the average, the average speed does not vanish.

But in the kinetic theory of gases, we find that the pressure, temperature, and molecular kinetic energy of a gas are proportional to the rms speed and not with the molecular velocity. Hence, rms speed of a molecule is preferable to the average speed in kinetic theory.

Unit 9 Behavior Of Perfect Gas And Kinetic Theory Chapter 1 Kinetic Theory Of Gases Multiple Choice Questions And Answers

WBBSE Class 11 Kinetic Theory of Gases MCQs

Question 1. If the volume of a body is V₁ and the total volume of the molecules of the body is V₂, then

1. V₁ = V₂
2. V₁ < V₂
3. V₁ >V₂
4. V₁ < V₂ or V₁ > V₂ for different bodies

Answer: 3. V₁ >V₂

Question 2. The molecules of all solids

1. Are relatively closer than those of liquids or gases
2. Are relatively farther than those of liquids or gases
3. Move faster than those of liquids or gases
4. Are stationary as they cannot move inside the solid

Answer: 1. Are relatively closer than those of liquids or gases

Question 3. Which of the following statements is inconsistent with the characteristics of Brownian motion?

1. The velocity of a particle increases as its size decreases
2. The velocity of the particles increases as the temperature increases
3. The velocity of the particles increases as the viscosity of the medium decreases
4. The velocity of the particles increases when the container is shaken

Answer: 4. The velocity of the particles increases when the container is shaken

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 4. A piece of wood floating on water does not have any Brownian motion, because

1. A part of the wooden piece is above the water
2. The resultant of the applied forces by the water molecules is zero
3. An adhesive force acts between the molecules of wood and water
4. The viscosity of water is comparatively less

Answer: 2. The resultant of the applied forces by the water molecules is zero

Question 5. The velocities of two particles moving towards the east are 4 m · s^-1 and 6 m · s^-1, respectively. The velocities of three other particles moving towards the west are 2 m · s^-1, 3 m · s^-1, and 5 m · s^-1, respectively. The root mean square speed of these 5 particles is

1. 0
2. 4m · s^-1
3. 1.667 m · s^-1
4. 4.242 m · s^-1

Answer: 4. 4.242 m · s^-1

Conceptual MCQs on Kinetic Theory for Class 11

Question 6. The pressure and density of hydrogen gas, kept in a vessel, are 1.013 x 10⁶ dyn · cm^-2 and 0.089 g · L^-1, respectively. The rms speed of the gas molecules will be

1. 18.5m · s^-1
2. 185m · s^-1
3. 1.85 km · s^-1
4. 18.5 km · s^-1

Answer: 3. 1.85 km · s^-1

Question 7. If the mean velocity, rms speed, and maximum probable velocity of gas are c, c, and cm, respectively, then

1. \(c_m<\bar{c}<c\)
2. \(\bar{c}<c<c_m\)
3. \(c_m>\bar{c}>c\)
4. None of these

Answer: 1. \(c_m<\bar{c}<c\)

Question 8. There is a mixture of hydrogen and oxygen gases in a vessel. The root mean square speed of the oxygen molecules is

1. 4 times that of hydrogen molecules
2. 16 times that of hydrogen molecules
3. 1/4 times that of hydrogen molecules
4. 1/16 times that of hydrogen molecules

Answer: 3. 1/4 times that of hydrogen molecules

Question 9. A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds is

1. 0.32
2. 0.45
3. 2.24
4. 3.16

Answer: 4. 3.16

Practice Questions on Ideal Gas Behavior

Question 10. At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be 1930 m/s. The gas is

1. H₂
2. F₂
3. O₂
4. Cl₂

Answer: 1. H₂

Question 11. If the volume of a container is V, the pressure on the walls of the container by a gas is p and the internal energy of the gas is U, then

1. U = p V
2. U = \(\frac{1}{3} p V\)
3. U = \(\frac{2}{3} p V\)
4. U = \(\frac{3}{2} p V\)

Answer: 4. U = \(\frac{3}{2} p V\)

Question 12. A certain amount of gas is at 27°C. The rms speed of the gas molecules becomes doubled at

1. 327°C
2. 600°C
3. 927°C
4. 1200°C

Answer: 3. 1200°C

Question 13. If a gas of a particular mass is expanded at a constant temperature, the variable which undergoes a change is

1. Pressure of the gas
2. Internal energy of the gas
3. Rms speed of the gas molecules
4. Kinetic energy of the gas molecules

Answer: 1. Pressure of the gas

Key MCQs on Gas Laws and Kinetic Theory

Question 14. At equilibrium conditions, the volume, pressure, and temperature of a gas kept in a closed container are V, p, and T, respectively. If the container is divided into two equal parts by a partition, the value of these quantities for each part will be

1. \(\frac{V}{2}, \frac{p}{2}, \frac{T}{2}\)
2. \(\frac{V}{2}, \frac{p}{2}, T\)
3. \(\frac{V}{2}, p, \frac{T}{2}\)
4. \(\frac{V}{2}, p, T\)

Answer: 4. \(\frac{V}{2}, p, T\)

Question 15. According to the kinetic theory of gases, there are no intermolecular attractions, so these molecules do not have

1. Linear momentum
2. Kinetic energy
3. Potential energy
4. Mechanical energy

Answer: 3. Linear momentum

Question 16. If k is Boltzmann constant and T is temperature, the average kinetic energy of each molecule of a gas will be

1. \(\frac{2}{3} k T\)
2. \(\sqrt{\frac{2}{3}} k T\)
3. \(\frac{3}{2} k T\)
4. \(\sqrt{\frac{3}{2}} k T\)

Answer: 3. \(\frac{3}{2} k T\)

Question 17. The rms speed of oxygen molecules at 47°C will be equal to the rms speed of hydrogen molecules at

1. 80K
2. -83K
3. 3K
4. 20K

Answer: 4. 20K

WBBSE Class 11 Revision MCQs for Kinetic Theory

Question 18. The pressure, volume, and temperature in two samples of a gas are p, V, T, and 2p,V/4, 2 T, respectively The ratio of the number of molecules in the two samples is

1. 2:1
2. 4:1
3. 8:1
4. 16:1

Answer: 2. 4:1

Question 19. The rms speed of gas molecules at 0°C will be reduced to half at

1. 0°C
2. -273°C
3. 32°C
4. -204°C

Answer: 4. -204°C

Question 20. A container of 5 L contains 10²⁶ number of molecules of a gas. If the mass and rms speed of each molecule are 2.4 x 10^-25 g and 3.5 x 10⁴ cm · s^-1, respectively the pressure of the gas will approximately be

1. 2 x 10⁶ dyn · cm^-2
2. 10⁶ dyn · cm^-2
3. 3 x 10⁶ dyn · cm^-2
4. 5 x 10⁶ dyn · cm^-2

Answer: 1. 2 x 10⁶ dyn · cm^-2

Sample Questions on Molecular Speed and Temperature

Question 21. Air is filled in two heat-insulated vessels 1 and 2 having pressure, volume, and temperature p₁, V₁, T₁ and p₂, V₂, T₂ respectively. If the intermediate valve between the two vessels is opened, the temperature of the air at equilibrium will be

1. \(T_1+T_2\)
2. \(\frac{T_1 T_2\left(p_1 V_1+p_2 V_2\right)}{p_1 V_1 T_2+p_2 V_2 T_1}\)
3. \(\frac{T_1+T_2}{2}\)
4. \(\frac{T_1 T_2\left(p_1 V_1+p_2 V_2\right)}{p_1 V_1 T_1+p_2 V_2 T_2}\)

Answer: 3. \(\frac{T_1+T_2}{2}\)

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Question 22. A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300 K. The ratio of the average rotational kinetic energy per O₂ molecule to per N₂ molecule is

1. 1:1
2. 1:2
3. 2:1
4. Depends on the moment of inertia of the two molecules

Answer: 1. 1:1

Question 23. 70 cal of heat is required to raise the temperature of 20 moles of an ideal diatomic gas at constant pressure from 30°C. The amount of heat required (in cal) to raise the temperature of the same gas through the same range (30°C to 35°C) at constant volume is

1. 30
2. 50
3. 70
4. 90

Answer: 2. 50

Question 24. Three closed vessels A, B, and C at the same temperature T contain gases that obey the Maxwellian distribution of velocities. Vessel A contains only O₂, B only N₂, and C a mixture of equal quantities of O₂ and N₂. If the average velocity of the O₂ molecule in vessel A is v₂ and that of the N₂ molecule in vessel B is v₂, the average velocity of the O₂ molecule is vessel C is

1. \(\frac{\left(v_1+v_2\right)}{2}\)
2. \(v_1\)
3. \(\left(v_1 \nu_2\right)^{\frac{1}{2}}\)
4. \(\sqrt{\frac{3 k T}{M}}\)

Answer: 2. \(v_1\)

WBBSE Class 11 Practice Tests on Kinetic Theory

In this type of question, more than one option are correct

Question 25. From the following statements concerning ideal gas at any given temperature T, select the correct one(s).

1. The coefficient of volume expansion at constant pressure is the same for all ideal gases
2. The average translational kinetic energy per molecule of oxygen gas is 3kT, k being the Boltzmann constant
3. The mean free path of molecules increases with a decrease in pressure
4. In a gaseous mixture, the average translational kinetic energy of the molecules of each component

Answer:

1. The coefficient of volume expansion at constant pressure is the same for all ideal gases

3. The mean free path of molecules increases with a decrease in pressure

Question 26. Let \(\bar{v}, v_{\mathrm{rms}} \text { and } v_p\) respectively, denote the mean velocity, root mean square speed, and most probable velocity of the molecules in an feed monatomic gas at absolute temperature T. The mass of a molecule is m. Then

1. No molecule can have a speed greater than \(\sqrt{2} v_{\mathrm{rms}}\)
2. No molecule can have speed less than \(\frac{v_p}{\sqrt{2}}\)
3. \(v_p<\bar{v}<v_{\mathrm{rms}}\)
4. The average kinetic energy of a molecule is \(\frac{3}{4} m v_p^2\)

Answer:

3. \(v_p<\bar{v}<v_{\mathrm{rms}}\)

4. The average kinetic energy of a molecule is \(\frac{3}{4} m v_p^2\)

Question 27. For a jar containing H₂ gases which of the following statements are correct?

1. Both the gas molecules have same average energy
2. Both the gas molecules have same average translational kinetic energy
3. Hydrogen molecules have greater average energy than helium molecules
4. Both the molecules have same average velocity

Answer:

3. Hydrogen molecules have greater average energy than helium molecules

4. Both the molecules have same average velocity

Question 28. The root mean square speed of the perfect gas molecules will be doubled if

1. Pressure is doubled at constant volume
2. Pressure is made 4 times at constant volume
3. Volume is made 4 times at constant pressure
4. Volume is increased by 41.4% at constant pressure

Answer:

2. Pressure is made 4 times at constant volume

3. Volume is made 4 times at constant pressure

Interactive MCQs on Gas Properties and Kinetic Theory

Question 29. According to the kinetic theory of gases, which of the following statements are true?

1. Real gas behaves as ideal gas at high temperatures and low pressure
2. The liquid state of an ideal gas is impossible
3. At any temperature and pressure, ideal gas obeys Boyle’s law and Charles’ law
4. The molecules of a real gas do not exert any force on one another

Answer:

1. Real gas behaves as ideal gas at high temperatures and low pressure
2. The liquid state of an ideal gas is impossible
3. At any temperature and pressure, ideal gas obeys Boyle’s law and Charles’ law

Energy

A body can do work by the transfer of some amount of energy.

Energy Definition: The total amount of work that a body can do is the same as its total amount of energy. Hence, energy and work are equivalent physical quantities, having the same units and dimensions. Both are scalar quantities.

Energy is manifested in nature in different forms:

1. Mechanical energy
2. Heat energy
3. Light energy
4. Sound energy
5. Magnetic energy
6. Electrical energy
7. Chemical energy
8. Atomic energy

In all-natural events, energy changes from one form to another. In fact, every natural phenomenon can be interpreted as a sequence of energy transformations.

Read and Learn More: Class 11 Physics Notes

Work-Energy Theorem Explained

Energy Examples:

1. In an electric bulb, electrical energy is converted into heat energy at first, and then into light energy.
2. In an electric fan, electrical energy changes into mechanical energy.
3. In a telephone, sound energy is converted into electrical energy (during speaking), and electrical energy into sound energy (during listening).
4. In hydroelectric plants, water stored in a high reservoir flows downwards and its potential energy changes into kinetic energy. This kinetic energy rotates the turbines which generate electrical energy.
5. Water from sea, river and other water bodies evaporate and forms clouds. This water eventually returns to the earth in the form of rain. The potential energy of the clouds converts into the kinetic energy of the raindrops.
6. On burning coal, chemical energy is converted into heat and light energy.

We observe such numerous incidents of energy transformations in daily life. But neither energy is created nor destroyed in such incidents. Energy can only get transferred from one body to another or can change its form. As soon as a body loses energy, another body gains it in equal amounts. Hence the total energy in the universe remains the same. This is the law of conservation of energy.

Law Of Conservation Of Energy: Energy cannot be created or destroyed. It can only be converted from one form to another.

• It means that the total energy in the universe remains constant.
• After the introduction of Einstein’s theory of relativity, the law of conservation of energy was corrected and the law of conservation of mass energy was established.
• This means that mass and energy in the universe are not conserved individually.
• Only mechanical energy is discussed in detail in the following sections.

Work And Energy – Conservation Force

Conservation Force Definition: A system in which the total mechanical energy i.e., the sum of its kinetic and potential energy [remains conserved, is called a conservative system. The forces acting in such a system are called conservative forces.

Conservation Force Example: Gravitational force, restoring force of spring or elastic force, electrostatic force, force between the magnetic poles, etc., are conservative forces.

Work is done to lift a body against gravity and is stored as potential energy. While returning to its initial position, the body uses this stored energy and does work that is exactly equal to the initial work done while lifting. Hence energy is conserved.

Work Done In A Dosed Path Under A Conservative Force: For a conservative system, the work done depends only on the initial and final positions. It does not depend on the path taken in reaching the final position from the initial position.

• This means that the sum of the kinetic and potential energies remains constant throughout.
• It also implies that if a body goes around a complete loop so that its final and initial positions are the same, then the total work done is zero. For any conservative force, the work done is reversible.
• Hence if work done against a force can be restored, the force is called conservative. Also if work done by a force in a closed path equals to zero, the force is called conservative.

Thus in a conservative system, it is possible to restore the initial work done.

Suppose a body gets displaced by dx along the x-axis under the action of a conservative force F (which may vary with position) also acting along the x-axis. Therefore the work done by the conservative force dW = F(x)dx.

Since change in potential energy is negative of work done, then change in potential energy for this displacement dx is given by dU = -F(x)

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In a three-dimensional reference frame, this equation is dU = \(-\vec{F} \cdot d \vec{r}\)….(1)

WBBSE Class 11 Work and Energy Revision Notes

Work And Energy – Conservation Of Mechanical Energy Of A Particle In A Conservative System

Suppose a conservative force \(\vec{F}\), acting on a particle of mass m moves, it from point A to point B.

If the velocity of the particle at any point of its motion in the force field is v, then \(\vec{F} \cdot d \vec{r}=m \frac{d \vec{v}}{d t} \cdot \vec{v} d t=m \vec{v} \cdot d \vec{v}=m d\left(\frac{1}{2} v^2\right)\)

∴ Work done, to take the particle from A to B, by force \(\vec{F}\),

W = \(\int_A^B \vec{F} \cdot d \vec{r}=\left[\frac{1}{2} m v^2\right]_A^B=\frac{1}{2} m v_B^2-\frac{1}{2} m v_A^2\)…(1)

where v_A and v_B are the velocities of the particle at A and B respectively.

Equation (1), we know \(\vec{F} \cdot d \vec{r}=-d U\)

∴ \(\int_A^B \vec{F} \cdot d \vec{r}=-\int_A^B d U=U_A-U_B\)

= difference in potential energy between the points A and B…(2)

Comparing (1) and (2), \(U_A-U_B=\frac{1}{2} m v_B^2-\frac{1}{2} m v_A^2\)…(2)

Equation (3) establishes that in a conservative field when the kinetic energy of a particle increases, its potential energy decreases.

∴ \(\frac{1}{2} m v_A^2+U_A=\frac{1}{2} m v_B^2+U_B\)…(4)

If K_A and K_B are the kinetic energies of the particle at A and B respectively, then

K_A+U_B = K_B+ U_B or, K + U = constant (E).

Hence in a conservative field i.e., in a conservative system (where no dissipation of energy occurs due to forces like friction, etc.), at each point sum of potential energy and kinetic energy remains constant.

In other words, total mechanical energy (E) of a particle in a conservative system remains constant. This is the statement of the principle of conservation of mechanical energy.

Total mechanical energy In a free fall under gravity remains constant: Let a body of mass m be at rest at point A. DE, the earth’s surface is taken as the reference plane and the height of point A from DE is h. As the body is at rest it has potential energy only and no kinetic energy.

Potential energy at A = mgh

Kinetic energy at A = 0

Total mechanical energy at A = mgh + 0 = mgh

Now the body is released and it starts falling freely. When the body reaches a point B, it acquires a velocity v.

At B, the body has both potential energy (due to its height) and kinetic energy (due to its motion).

Let AB = x

∴ Potential energy at B = mg(h- x)

Kinetic energy at B = \(\frac{1}{2} m v^2=\frac{1}{2} m \cdot 2 g x\)

= mgx (because \(v^2=u^2+2\) as and u=0)

∴ Total mechanical energy at B = mg(h-x) + mgx = mgh

= total energy at A.

Also at C, when the body is about to touch the reference plane, its potential energy becomes zero. If the velocity acquired at that point is V, then kinetic energy at C

= \(\frac{1}{2} m V^2=\frac{1}{2} m \cdot 2 g h\left[because V^2=2 g h\right]\)

= mgh

Total mechanical energy at C = mgh + 0 = mgh

∴ Total energy at A = total energy at B = total energy at C.

Therefore, the total mechanical energy (= potential energy + kinetic energy) of a freely falling body, under the action of i gravity, remains conserved at all positions.

As the body touches the ground, both its potential energy and kinetic energy become zero; the entire mechanical energy transforms into heat, sound, and other forms of energy.

Key Concepts in Work and Energy Notes

Total Mechanical Energy Of A Body, Falling Under Gravity Along A Frictionless Inclined Surface, Remains Constant: Let a body of mass m be at rest at a point A on a frictionless plane of inclination θ. Earth’s surface CD is taken as the reference plane.

Let the height of point A from the reference plane, GA = h. Being at rest, the body has no kinetic energy at A.

The potential energy at A = mgh Kinetic energy at A = 0

∴ Total mechanical energy at A = mgh + 0 = mgh

On releasing the body, it falls along the incline and reaches a point B such that, AB = x. Let the velocity of the body at B be v.

Acceleration of the body along the incline, a = gsinθ.

∴ \(v^2=0+2 g \sin \theta \cdot x=2 g x \sin \theta \quad \text { [as } v^2=u^2+2 a s \text { ] }\)

∴ Kinetic energy at \(B=\frac{1}{2} m v^2=\frac{1}{2} \cdot m \cdot 2 g x \sin \theta\)

= mgx sinθ

The perpendicular height of B from CD plane = FB = GE = GA-EA = h-x sinθ

(because \(\sin \theta=\frac{E A}{A B} \quad \text { or, }E A=A B \sin \theta=x \sin \theta\))

∴ The potential energy at B = mg(h-x sinθ)

∴ Total mechanical energy at B

= mg(h- xsinθ) + mgx sinθ

= mgh = total energy at A

At point C, i.e., where the body just touches the plane CD, it has no potential energy. If the velocity of the body at that moment is V, then

V² = 0 + 2g sinθ · AC = 2gsinθ · AC

and hence, kinetic energy at C = \(\frac{1}{2}\) m · 2g sinθ · AC = mgh

[sinθ = \(\frac{G A }{A C}\) = \(\frac{h}{A C}\) or, AC sinθ = h]

∴ Total energy at C = 0 + mgh = mgh

Thus, total mechanical energy at A = total mechanical energy at B = total mechanical energy at C.

Hence, the total mechanical energy of a body, moving along a frictionless inclined plane under gravity, is conserved.

Total Mechanical Energy Of A Hydrogen Gas-Filled Balloon Rising Upwards Remains Constant: Let us assume that air and earth form a system. If the earth’s surface is taken as the plane of reference, when the balloon is at rest on the ground, both its potential and kinetic energy = 0, i.e., total mechanical energy = 0.

Now the balloon is released. Upthrust, T due to air on the balloon is more than the weight, mg of the balloon filled with gas and so the balloon rises up. Let the mass of the gas-filled balloon = m and the mean resultant upward force on it = F – T – mg.

∴ Acceleration of the balloon = \(\frac{F}{m}\). If the velocity of the balloon is v at a height h from the earth surface, then

v2² = 2\(\frac{F}{m}\)h [according to the formula v² = u² + 2as]

Hence, kinetic energy at h = \(\frac{1}{2}\)= \(\frac{1}{2}\)m 2 \(\frac{F}{m}\)h = Fh

We know that when a stone falls freely under gravity, the work done is negative. Similarly, when the balloon rises by itself (without the help of any external agent) under the effect of the mean upward resultant force F, the work done is also negative. Here, the upthrust is not applied by an external agent.

So, at height h the change in potential energy = -Fh. As the initial potential energy of the balloon on the ground was zero, the total potential energy at a height h = -Fh.

∴ The total mechanical energy of the balloon at height h = potential energy + kinetic energy = – Fh + Fh = 0 = total energy of the balloon on the ground before its release.

Thus, the total mechanical energy remains conserved for a hydrogen gas-filled balloon when it is rising up. It is to be noted that, with the increase in height h, the kinetic energy increases but the potential energy decreases equally.

Actually, the expression mgh for potential energy should be modified for a balloon as (mg- T)h, where T is the upward thrust exerted on the balloon by the air surrounding it. For a hydrogen-filled balloon, the upward thrust is greater than its weight, i.e., T> mg. So the potential energy (mg-T)h is negative; this negative value goes on increasing with an increase in height h.

Another similar event is observed when a piece of wood is held at the bottom of a bucket full of water. When it is released, it floats up by itself. It can be shown that the total mechanical energy remains conserved. As the piece of wood rises upwards, its kinetic energy gradually increases and the potential energy decreases equally.

Work And Energy – Conservation Of Mechanical Energy Of A Particle In A Conservative System Numerical Examples

Common Questions on Work and Energy

Example 1. After falling from a height of 200 m, water flows horizontally with a certain velocity. Ignoring any energy dissipation, find the velocity of flow.
Solution:

Potential energy changes into kinetic energy during the free fall of water.

Let mass of water = m, height = h, final velocity = v

Here, at a height of 200 m

potential energy (P.E.)_i = mgh and kinetic energy (K.E.)_i = 0

when water falls and flows horizontally, potential energy, (P.E.)_f = 0 and kinetic energy (K.E.)_f = \(\frac{1}{2}\)mv²

∴ From the conservation of mechanical energy we,

∴ mgh = \(\frac{1}{2}\)mv

or, \(\nu=\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 200}\)

= \(62.61 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

= 62.61 m · s^-1

Example 2. The mass of the bob of a simple pendulum is 10 g and the effective length is 13 cm. The bob is pulled 5 cm away from the vertical and then released. What will be the kinetic energy of the bob when it passes through the lowest point?
Solution:

We observe that, the energy of the bob at point B = potential energy = mgh = 10 x 980 x AC.

Here OA = OB = 13 cm and CB = 5 cm.

∴ OC = \(\sqrt{13^2-5^2}\)

= \(\sqrt{144}=12 \mathrm{~cm}\)

∴ AC = OA-OC = 13 – 12 = 1 cm.

∴ Potential energy of the bob at B = 10 x 980 x 1 = 9800 erg

∴ The kinetic energy of the bob at the lowest point A = potential energy of the bob at B = 9800 erg.

Example 3. After a collision with an ideal spring, a body of mass 8 g, moving with a constant velocity of 10 cm· s^-1 comes to rest Force constant of the spring is 200 dyn · cm^-1. If the total kinetic energy of the body is spent in compressing the spring, find the compression.
Solution:

Here, the kinetic energy of the body transforms into the potential energy of the spring.

∴ \(\frac{1}{2} m v^2=\frac{1}{2} k x^2\), where x = compression of the spring

∴ x = \(\sqrt{\frac{m v^2}{k}}=\sqrt{\frac{8 \times 10^2}{200}}=2 \mathrm{~cm}\) .

Example 4. The effective length of a pendulum is 50 cm and the mass of the bob is 4 g. The bob is drawn to one side until the string is horizontal and is then released. When the string makes an angle 60° with the vertical, what is the velocity and the kinetic energy of the bob?
Solution:

OB is the horizontal position of the string.

At B, the total energy of the bob

= potential energy = mgh = (4 x 980 x 50) erg

At C, the total energy of the bob = kinetic energy + potential energy

= kinetic energy + 4 x 980 x AD

(cos 60° = \(\frac{OD}{OC}\) or, OD = \(\frac{50}{2}\) = 25 cm

∴ AD = OA – OD – 50-25 = 25 cm)

According to the law of conservation of energy, the energy of the bob at C = energy of the bob at B

or, K.E. of the bob at C + P.E. of the bob at C = K.E. of the bob at B + P.E. of the bob at B

or, \(\frac{1}{2} m v^2+4 \times 980 \times 25=0+4 \times 980 \times 50\) (v = velocity of bob at C)

or, \(\frac{1}{2} m v^2=4 \times 980 \times 25=98000\)

∴ \(v^2=\frac{98000 \times 2}{4} \quad \text { or, } v=\sqrt{49000}=222.36 \mathrm{~cm} \cdot \mathrm{s}^{-1}\).

Example 5. The mass of the bob of a pendulum is 100 g and the length of the string is 1 m. The bob is initially held in such a way that the string is horizontal. The bob is then released. Find the kinetic energy of the bob when the string makes an angle of

1. 0° and
2. 30° with the vertical.

Solution:

When the string is horizontal, the height of the bob above its lowest position = 1 m = 100 cm.

The energy of the bob at point P = potential energy of the bob

= mgh = 100 x 980 x 100 = 98 x 10⁵ erg.

1. At 0° angle with the vertical, the string holds the bob at its lowermost position. Hence, energy at B = kinetic energy of the bob = initial potential energy = 98 x 10 erg.

2. When the string makes a 30° angle with the vertical, the height of the bob from its lowermost position

= BD = BA – DA = 100- AC cos30°

= \(100\left(1-\frac{\sqrt{3}}{2}\right)=13.4 \mathrm{~cm}\)

Potential energy of the bob at C = 100 x 980 x 13.4 erg;

kinetic energy at this position = initial potential energy- potential energy at C

= 98 x 10⁵ -(100 x 980 x 13.4) = 8486800 erg.

Example 6. A body of mass 1 kg falls to the ground from the roof of a building 20 m high. Find its

1. Initial potential energy,
2. Velocity when it reaches the ground,
3. Maximum kinetic energy and
4. Kinetic and potential energies at a position 2 m above the earth’s surface.

Solution:

1. Initial potential energy of the body = mgh = 1 x 9.8 x 20 = 196 J.

2. Suppose the body touches the ground with velocity v. Potential energy at roof level = kinetic energy just before touching the ground.

∴ 196 = \(\frac{1}{2}\) x 1 x v²

∴ v² = 392 and v = 19.8 m · s^-1

3. Maximum kinetic energy = initial potential energy = 196 J

4. Potential energy at a height of 2 m = 1×9.8×2 =19.6J

∴  Kinetic energy at that height of 2 m = decrease in initial potential energy = 196 – 19.6 = 176.4 J.

Example 7. A pump lifts 200 L of water per minute through a height of 5 m and ejects it through an orifice 2 cm in diameter. Find the velocity of efflux of water and the power of the pump.
Solution:

Volume of water lifted by the pump in is = \(\frac{200}{60}=\frac{10}{3} \mathrm{~L}=\frac{10^4}{3} \mathrm{~cm}^3 \text {. }\)

Mass of this volume of water = \(\frac{10^4}{3} g=\frac{10}{3} \mathrm{~kg}\)

∴ Increase in potential energy in 1 second

= \(\frac{10}{3}\) x 9.8×5 J · s^-1 = 163.33 J

∴ Power of the pump to raise water up to the height of 5m = 163.33 W.

If the velocity of efflux is v, then \(\pi r^2 \times v=\frac{10^4}{3}\)

or, \(\nu=\frac{10^4 \times 7}{3 \times 22 \times(1)^2}=1061 \mathrm{~cm} \cdot \mathrm{s}^{-1}=10.61 \mathrm{~m} \cdot \mathrm{s}^{-1} \text {. }\)

The kinetic energy of the amount of water thrown out per second

= \(\frac{1}{2}\) x \(\frac{10}{3}\) x(10.61)² J =187.62 J

∴ Power of efflux = 187.62 W

∴ Total power of the pump = 163.33 + 187.62 = 350.95 W.

Example 8. A body of mass 10 kg is raised to a height of 10 m with an upward force of 196 N. Find the work done by the upward force and the work done against gravitation. Show that the total energy in this case is equal to the work done by the upward force. [g = 9.8 m · s^-2]
Solution:

Work done by the upward force, W = force x displacement

= 196 N x 10 m = 1960 N · m = 1960 J.

Upward acceleration of the body in the absence of gravity,

a’ = \(\frac{\text { upward force }}{\text { mass }}=\frac{196}{10}=19.6 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

∴ Effective upward acceleration, a = a’ -g = 19.6-9.8 = 9.8 m · s^-2

If the body starts from rest, and attains a velocity v at the height of 10 m, then from v² = u² + 2 as, v² = 2 x 9.8 x 10 m² · s^-2

∴  Kinetic energy of the body at this height,

K = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) x 10 x 2 x 9.8 x 10 = 980 J

Work done against gravitational force,

W’ = force due to gravity x displacement = mass x acceleration due to gravity x displacement = 10 x 9.8 x 10 = 980 J

This work done against gravitational pull gets stored as potential energy V of the body. Hence, V = 980 J.

∴ Total mechanical energy of the body at a height of 10 m = V + K= 980 + 980 = 1960 J = W

So, a part of the work done by the upward force changes into kinetic energy of the body, and the other part transforms itself into the stored potential energy. Thus, the total energy and work done by the upward force are equal.

Example 9. A body of mass 10 kg moving with a speed of 2.0 m · s^-1 on a frictionless table strikes a mounted spring and comes to rest. If the force constant of the spring be 4 x 10⁵ N · m^-1, then what will be the compression on the spring?
Solution:

The kinetic energy of the body is E = \(\frac{1}{2}\)mv², where m and v be the mass and the speed of the body respectively.

On striking the spring, the kinetic energy of the spring due to compression is completely converted into the potential energy of the spring.

If the spring is compressed through a distance x then its potential energy is U = \(\frac{1}{2}\)kx²

∴ \(\frac{1}{2}\) mv2 = \(\frac{1}{2}\)kx2

or, x = \(v \sqrt{\frac{m}{k}}=2 \times \sqrt{\frac{10}{4 \times 10^5}}=10^{-2} \mathrm{~m}=1 \mathrm{~cm}\)

Example 10. Shows two blocks of masses m₁ = 3 kg and m₂ = 5 kg, both moving towards the right on a frictionless surface with speeds u₁ = 10 m · s^-1 and u₁ = 4 m · s^-1 respectively. To the back side of m₂ an ideal spring of force constant 1000 N · m^-1 is attached. Calculate the maximum compression of the spring when the blocks collide.

Solution:

m₁ = 3kg, m₂ = 5kg, u₁ = 10 m · s^-1, u₂ = 4 m ·  s^-1

Force constant, k = 1000 N · m^-1

Let v be the speed of the combination.

Using the law of conservation of linear momentum, m₁u₁ + m₂u₂ =(m₁ + m₂)v

∴ v = \(\frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{3 \times 10+5 \times 4}{3+5}=6.25 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Also, let x be the maximum compression of the spring when the blocks collide.

∴ From conservation of mechanical energy, we get \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2}\left(m_1+m_2\right) v^2+\frac{1}{2} k x^2\)

or, \(500 x^2=380-312.5=67.5\)

∴ x = \(\sqrt{\frac{67.5}{500}} \approx 0.367 \mathrm{~m}\)

Work And Energy – Non-Conservative Force

Non-Conservative Force Definition: In the presence of resistive forces in a system, mechanical energy does not remain conserved and gets dissipated. Such a system is called a non-conservative system and the resistive force is called a non-conservative force or dissipative force.

Non-Conservative Force Example: Frictional force is a non-conservative force.

• Frictional force resists the motion of an object. Thus on slid¬ing a body over a rough surface, work done against friction does not get stored in the body as potential energy.
• Frictional force always acts opposite to the direction of motion. To move a body from its initial position to a final position, some amount of work is done to overcome friction.
• If the body is brought back to its initial position along the same path, again some work is done to overcome friction. Thus, each time the body moves, some energy is lost. It can therefore be stated that, it is not possible to restore the initial work done in a non-conservative system.

Work Done In A Closed Path Under A Non-Conservative Force: The total work done to move a body under a non-conservative force along a closed path once completely, is positive or negative but never zero. For example, to slide a body over a rough surface from one point to another, work has to be done against friction.

To return the body to its initial position by sliding it over the same surface following any path, work has to be done against friction again. So, the total amount of work done is not zero in a closed path. Hence, a force is called non-conservative when work done against it cannot be restored. Alternatively, when the work done by a force in a closed path is not zero, the force is called nonconservative.

Work And Energy – Mass Energy Equivalence

Definitions of Work, Energy, and Power

The principle of mass-energy equivalence can be obtained from Einstein’s famous theory of relativity.

Mass is a form of energy. Mass can be converted into energy and vice versa. If m amount of mass of a substance is completely converted to energy, then the amount of energy liberated is, E = mc² (c = velocity of light in vacuum = a constant)

According to this equation, the equivalent energy of mass m is E, and the equivalent mass of energy E is m

Mass Energy Equivalence Example:

1. In CGS system c = 3 x 10¹⁰ cm · s^-1

∴ Equivalent energy of 1 g mass

= 1 x (3 x 10¹⁰)²= 9 x 10²⁰ erg = 9 x 10¹³ J

Again in \(\mathrm{SI}\), \(c=3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Hence equivalent energy of 1 kg mass = \(1 \times\left(3 \times 10^8\right)^2=9 \times 10^{16} \mathrm{~J} .\)

2. Mass of an electron = \(9.1 \times 10^{-28} \mathrm{~g}\)

Equivalent energy of the mass of 1 electron = \(9.1 \times 10^{-28} \times\left(3 \times 10^{10}\right)^2 \mathrm{erg}\)

= \(\frac{9.1 \times 10^{-28} \times 9 \times 10^{20}}{1.6 \times 10^{-12}} \mathrm{eV}\)

= \(0.511 \times 10^6 \mathrm{eV}=0.511 \mathrm{MeV} .\)

The increase or decrease in energy of an electron while crossing a potential difference of 1V is called 1 electronvolt (eV). 1 eV = 1.6 x 10^-12 erg.

Rest Mass: Einstein’s theory of relativity also informs us that the mass of a substance is not a constant quantity but depends on the velocity of the substance. Especially if the velocity of an object becomes comparable to that of light, then its mass increases significantly.

That is why to measure the true mass of an object as an intrinsic property, the object should be at rest with respect to the observer. The mass of the object thus measured is called its rest mass and the equivalent energy is called rest energy. For example, the rest mass of an electron =9.1 x 10^-28 g, and the rest energy is 0.511 MeV.

Unit Of Mass And Energy: As mass and energy are equivalent to each other, their units are also the same. Sometimes mass is given in units of energy, and energy is given in units of mass. For example, ‘energy of 1g’ denotes 9 x 10²⁰ erg amount of energy; or mass of ‘9 x 10¹⁶ J’ represents a mass of 1 kg. From this equivalence, it can be stated that the rest mass of an electron is 0.511 MeV.

Law Of Conservation Of Mass-Energy: During the mutual transformation of mass and energy, the law of conservation of mass or the law of conservation of energy cannot be applied separately. It becomes the law of conservation of mass energy.

The total amount of mass energy in nature Is constant, it can never be created or destroyed. It can only change from one form to another.

Mass can be transformed into energy only within an atom. The energy thus obtained from mass is the source of atomic energy. When gamma rays with an energy of a few MeV or more enter the electric field of a heavy nucleus, it can be transformed into an electron and a positron (an electron-like particle but of positive charge). This is an example of the transformation of energy to mass.

Two-Dimensional Collisions Numerical Examples

Example 1. Two particles of masses m₁ and m₂, moving with velocities u₁ and u₁, respectively, and making an angle θ between them, collide with each other. After the collision, the 1st particle travels in the initial direction of motion of the 2nd, and vice-versa. Find the velocities of the two particles after collision. Under what condition, would this collision be elastic?
Solution:

Suppose v₁, v₂ are the velocities of the two particles, respectively, after collision. The particles before and after collision move as shown. It also shows the chosen directions of the x and the y-axis.

For momentum conservation along the x-axis, we get, \(m_1 u_1 \cos \frac{\theta}{2}+m_2 u_2 \cos \frac{\theta}{2}=m_1 v_1 \cos \frac{\theta}{2}+m_2 v_2 \cos \frac{\theta}{2},\)

or, \(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2\)….(1)

Similarly, along the y-axis, we get, \(-m_1 u_1 \sin \frac{\theta}{2}+m_2 u_2 \sin \frac{\theta}{2}=m_1 v_1 \sin \frac{\theta}{2}-m_2 v_2 \sin \frac{\theta}{2}\)

or, \(-m_1 u_1+m_2 u_2=m_1 v_1-m_2 v_2\)……(2)

Adding equations (1) and (2), \(2 m_2 u_2=2 m_1 v_1\)

or, \(v_1=\frac{m_2}{m_1} u_2\)…(3)

Subtracting equation (2) from (1), \(2 m_1 u_1=2 m_2 v_2\)

or, \(v_2=\frac{m_1}{m_2} u_1\)…..(4)

The kinetic energy before collision is, \(K_1=\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2\)

and that after collision is \(K_2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2=\frac{1}{2} m_1\left(\frac{m_2}{m_1} u_2\right)^2+\frac{1}{2} m_2\left(\frac{m_1}{m_2} u_1\right)^2\)

= \(\frac{1}{2} \frac{m_2}{m_1} m_2 u_2^2+\frac{1}{2} \frac{m_1}{m_2} m_1 u_1^2\)

Here, K₁ ≠K₂; so the collision is inelastic, in general. As a m special case, it would be an elastic collision if K₁ = K₂. It is possible only when m₁ = m₂, i.e., the two particles are of equal masses.

Real-Life Examples of Work Done by Forces

Example 2. A bomb explodes and splits up Into three fragments. Two fragments, each of mass 200 g, move away from each other making an angle of 120°, at a speed of 100 m · s^-1. Find the direction and velocity of the third fragment whose mass is 500 g. Also, find out the energy released in an explosion.
Solution:

The velocity of fragments A and B along OA and OB. A and B have equal mass and speed. From the law of conservation of linear momentum, the third piece must move along OD, in the direction opposite to the resultant of OA and OB.

If the velocity of the third piece is v, then taking the components along the line CD in the CGS system, 500 v = 200 x 10⁴ cos60° + 200 x 10⁴ cos60°

or, v = \(\frac{200 \times 10^4}{500}=4 \times 10^3 \mathrm{~cm} \cdot \mathrm{s}^{-1}=40 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Hence, the velocity of the third fragment is 40 m · s^-1. It moves so as to make an angle 120° with each of OA and OB. The energy released due to the explosion is the kinetic energy of the three fragments.

∴ Energy released = \(\frac{1}{2} \times 200 \times\left(10^4\right)^2\)

+ \(\frac{1}{2} \times 200 \times\left(10^4\right)^2+\frac{1}{2} \times 500 \times\left(4 \times 10^3\right)^2\)

= \(2400 \times 10^7 \mathrm{ergs}=2400 \mathrm{~J} .\)

Step-by-Step Solutions to Work and Energy Problems

Example 3. A spaceship while flying in space, splits up into three equal parts, due to an explosion. One fragment keeps moving in the same direction; the other two fly off at 60° to the original direction, on either side. If the energy released due to the explosion is twice the kinetic energy of the spaceship, find the kinetic energy of each of the fragments.
Solution:

Let the mass of the spaceship be 3m and the initial speed be u.

Hence, mass of each fragment = m.

Let their velocities be v₁, v_2, and v₃ after the explosion.

Applying the conservation of momentum law along the direction perpendicular to the original direction of motion,

0 = mv₂ sin60°- mv₃ sin60°

or, mv₂ sin60° – mv₃ sin60°

or, v₂ = v₃ = v (say)

Applying the law of conservation of linear momentum along the original direction of motion,

3 mu = mv₁ + 2mv cos60°

or, 3u = v₁ + v…(1)

The kinetic energy of the spaceship before explosion E= 1/2 x 3mu²

Energy released during the explosion, \(E_r=\frac{1}{2} m\left(v_1^2+2 v^2\right)-\frac{1}{2} \cdot 3 m u^2\)

According to the problem, \(E_r=2 E \quad \text { or, } \frac{1}{2} m\left(v_1^2+2 v^2\right)-\frac{1}{2} \cdot 3 m u^2=2 \times \frac{1}{2} \cdot 3 m u^2\)

or, \(v_1^2+2 v^2-3 u^2=6 u^2\)

or, \(v_1^2+2 v^2=9 u^2\)….(2)

Solving (1) and (2), v₁ =u and v = 2 u

Hence, the kinetic energy of the first fragment after the explosion

= \(\frac{1}{2} m v_1^2=\frac{1}{2} m u^2=\frac{1}{2} \cdot \frac{2}{3} \cdot E=\frac{1}{3} E\)

the kinetic energy of each of the other two fragments

= \(\frac{1}{2} m v^2=\frac{1}{2} m(2 u)^2=2 m u^2=2 \cdot \frac{2}{3} E=\frac{4}{3} E\)

 

Work And Energy Conclusion

If a body undergoes a displacement due to a force acting on it, work is said to be done, and the product of force and displacement is the measure of the work done.

Work Done Against A Force: Consider a particle on which some forces are acting. When an external agent causes a displacement opposite to the direction of the resultant of these forces, work is said to be done against the force.

Work Done By A Force: Consider a particle on which some forces are acting. The resultant of these forces can cause a displacement. This is said to be work done by a force.

• If a force acts at right angles to the direction of displacement of a body, no work is done by the force. This force is called a no-work force.
• The rate of doing work with respect to time is called power. That is work done in unit time is called power.
• Energy is the capacity to do work.
• The ability of a body to do work due to its speed, special position, or special configuration, or all of these, is called its mechanical energy.
• Mechanical energy is of two types— kinetic energy and potential energy.
• The ability of a body to do work due to its speed is called its kinetic energy.
• The ability of a body to do work due to its special position or configuration is called its potential energy.
• The ability of a body to do work, acquired due to its rise against gravity, is called its gravitational potential energy.
• Gravitational potential energy depends on the chosen plane of reference. This energy may also have a negative value.
• The ability of a body to do work, gained due to its special shape, is called elastic potential energy.

Law Of Conservation Of Energy: Energy can neither be created nor destroyed.

Mass-Energy Equivalence: In this universe, the total sum of mass and energy is a constant. This is the law of conservation of mass energy. Mass and energy are equivalent; one can be converted into the other.

• If the total momentum and the total kinetic energy of a system are conserved, the collision is termed as an elastic collision.
• If the total momentum is conserved, but the total kinetic energy is not, it is an inelastic collision.

Coefficient Of Restitution: The coefficient of restitution is defined as the ratio of the velocity with which the two bodies separate after a collision to their velocity of approach before the collision.

1. For elastic collision, e = 1.
2. For perfectly inelastic collision, e = 0.
3. For partially elastic collision, 0 < e < 1.

A system in which the total mechanical energy remains conserved is called a conservative system. Forces acting in such a system are called conservative forces.

In a system where resistive forces are present, the mechanical energy is not conserved. Such systems are called non-conservative systems. Forces of resistance are called dissipative forces.

Work And Energy Useful Relations For Solving Numerical Problems

When a force \(\vec{F}\) acting on a body is associated with a displacement \(\vec{s}\), work done, W = \(\vec{F}\) · \(\vec{s}\) = Fs cosθ, where θ is the angle between \(\vec{F}\) and \(\vec{s}\).

For a variable force \(\vec{F}\), if the displacement of a particle is from A to B, the total work done is

W = \(\int_A^B \vec{F} \cdot d \vec{s}\)

= \(\int_A^B F \cos \theta d s\)

Power (P) = \(\frac{\text { work }(W)}{\text { time }(t)}\) = force (F) x velocity of the body(v)

Kinetic energy = \(\frac{1}{2}\) mv²

For a moving object of mass m, and kinetic energy E, momentum p = √2mE

Gravitational potential energy = mgh

Within the elastic limit, the elastic potential energy of a spring (stretched or compressed by x) = \(\frac{1}{2}\)kx², where k is the spring constant.

Efficiency of a machine

= \(\frac{\text { work output from the machine }}{\text { energy input to the machine }} \times 100 \%\)

In a conservative field, total energy (E) = kinetic energy (K) + potential energy (V) = constant.

The energy equivalent of a mass m is, E = mc², where c is the velocity of light in a vacuum.

Law of conservation of momentum during a linear collision between two bodies: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

In case of elastic collision

1. \(u_1-u_2=v_2-v_1\) and
2. \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\)

Coefficient of restitution, e = \(\frac{v_2-v_1}{u_1-u_2}\)

In case of collision of a falling body with a fixed horizontal plane, total distance traveled before coming to rest, d = \(h \frac{1+e^2}{1-e^2}\)

Work And Energy Very Short Answer Type Questions

Question 1. A force is acting on a body in motion, but is not doing any work. Give an example of such a force.
Answer: Centripetal force on a body, in a uniform circular motion

Question 2. Is work a vector or a scalar quantity?
Answer: Scalar quantity

Question 3. What is the amount of work done by a force when the body moves in a circular path?
Answer: Zero

Question 4.In a tug-of-war game, which of the teams does effective work?
Answer: The stronger team

Question 5. A person is carrying a bucket of water and is in a lift moving up with uniform velocity. Is the person doing any work on the bucket of water? Will the energy of the bucket and water remain constant?
Answer: No

Question 6. A motor drives a belt at a constant velocity of v m · s^-1. If m kg of sand falls on the belt per second, what is the rate of work done by the force exerted by the belt on the sand?
Answer: 1/2mv³W

Question 7. A boy tried to lift a bucketful of water but failed. What is the work done by him?
Answer: Zero

Question 8. What is the work done by the tension in the string during the oscillation of a simple pendulum?
Answer: Zero

Question 9. A box was lifted vertically through a height of 6 m in 3 s. If the box had been lifted in a zig-zag way in 5 s, the work done would have been the same. Is the statement true or false?
Answer: True

Question 10. 1 kg · m = ________ J.
Answer: 9.8

Question 11. A force \(\vec{F}=(5 \hat{i}+3 \hat{j}+2 \hat{k}) \mathrm{N}\) acts on a particle, and the particle moves from the origin to a point \(\vec{r}=(2 \hat{i}-\hat{j}) \mathrm{m}\). What will be the work done on the particle?
Answer: 7

Question 12. How many joules are in 1 MeV?
Answer: 1.6 x 10^-13J

Question 13. What is the unit of energy?
Answer: Joule

Question 14. Does the kinetic energy of a ball, thrown inside a moving train, depend on the speed of the train?
Answer: No

Question 15. Which type of energy is lost in doing work against friction?
Answer: Mechanical energy

Question 16. A small car and a lorry are moving with the same kinetic energy. Brakes are applied to produce the same force against the motion. Which one will cover a greater distance before stopping?
Answer: Both cover the same distance

Question 17. When a body falls on the ground from a height, it becomes slightly warm—why?
Answer: Kinetic energy changes to heat energy

Question 18. Is the resistance due to air a conservative force?
Answer: No

Question 19. What happens to internal energy, when the temperature of the body increases?
Answer: Increases

Question 20. What type of energy is stored in the spring of a watch?
Answer: Potential energy

Question 21. The kinetic energies of a heavy and a light object are the same. Momentum of which object will be higher?
Answer: Heavy

Question 22. Momenta of a light and a heavy body are the same. Which body has greater kinetic energy.?
Answer: Lighter

Question 23. If E is the kinetic energy of a body of mass m, what will be its momentum?
Answer: \(\sqrt{2 m E}\)

Question 24. An object breaks up into two masses m₁ and m₂ due to explosion. The two fragments move in opposite directions. What will be the relation between the kinetic energy and the masses?
Answer: Inversely

Question 25. What is the loss of KE of a freely falling body of mass m, during the t th second?
Answer: [1/2 mg²(2t – 1)]

Question 26. The increase in momentum of a body is 100%. What will be the increase in its kinetic energy?
Answer: 300

Question 27. The increase in kinetic energy of a body is 69%. What will be the increase in its momentum?
Answer: 30

Question 28. Which physical quantity in conserved during both the elastic and inelastic collisions?
Answer: Momentum

Question 29. Two objects coalesce after a collision with each other. What is the coefficient of restitution?
Answer: 0

Question 30. The coefficient of restitution between a ball and a horizontal floor is e = 1/2. If the ball falls from a height of 10 m, after its impact with the floor, the ball bounces up to a height of ______.
Answer: 2.5m

Work And Energy Assertion Reason Type Questions And Answers

Comparative Analysis of Kinetic and Potential Energy

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: The absolute PE of a system as measured by two different persons at the same time can be different.

Statement 2: The value of the absolute PE of a system depends upon the reference value chosen.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 2.

Statement 1: Work done by normal contact force can be non-zero.

Statement 2: Normal contact force is always perpendicular to the displacement of the object. (Here displacement is measured with respect to a frame of reference attached to the two surfaces in contact.)

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: In a circular motion, work done by the centripetal force is not always zero.

Statement 2: If the speed of the particle increases or decreases in a circular motion, net force, acting on the particle is not directed toward the center.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 4.

Statement 1: When a body moves uniformly in a circle its momentum goes on changing but its kinetic energy remains constant.

Statement 2: \(\vec{p}=m \vec{v}, \mathrm{KE}=\frac{1}{2} m v^2\). In circular motion \(\vec{v}\) changes, v² does not change.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 5.

Statement 1: If a particle of mass m is connected to a light rod and whirled in a vertical circle of radius R, then to complete the circle, the minimum velocity of the particle at the lowest point is \(\sqrt{5 g R}\).

Statement 2: Mechanical energy is conserved and for the minimum velocity at the lowest point, the velocity at the highest point will be zero.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 6.

Statement 1: Work done by constant force is equal to the magnitude of force multiplied by displacement.

Statement 2: Work done is a scalar quantity. It may be positive, negative, or zero.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 7.

Statement 1: If work done by a conservative force is negative then potential energy associated with that force should increase.

Statement 2: This is from the reaction Δu = -W. Here Au is change in potential energy and W is work done by conservative force.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Work And Energy Match Column A With Column B.

Question 1. The displacement time graph of a body is shown.

Answer: 1. A, 2. B, C, D, 3. B, C, D, 4. C

Question 2. A force F = kx (where k is a positive constant) is acting on a particle. In Column A displacements (x) are given and in Column B work done by the force is given.

Answer: 1. B, 2. A, 3. C

Question 3.

Answer: 1. B, 2. D, 3. C

Question 4. The system is released from rest. Friction is absent and string is massless. In time t = 0.3 s (take g = 10 m · s^-2)

Answer: 1. C, 2. A, 3. D, 4. B

Question 5.

Answer: 1. A, B, C, 2. B, C, 3. A, C, 4. A, B, C

Question 6. A particle is suspended from a string of length R. It is given a velocity u = \(3 \sqrt{g R}\) at the bottom.

Answer: 1. C, 2. B, 3. A, 4. E

Work And Energy Question And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A body of mass 2 kg starts from rest and moves with uniform acceleration. It acquires a velocity 20 m · s^-1 in 4 s.

1. Power exerted on the body at 2 s is

1. 50 W
2. 100 W
3. 150 W
4. 200 W

Answer: 2. 100 W

2. Average power transferred to the body in the first 2 s is

1. 50 W
2. 100W
3. 150 W
4. 200 W

Answer: 1. 50 W

Question 2. A ball of mass m is dropped from a height H above a level floor as shown. After striking the ground it bounces back and reaches up to the height h.

1. During the collision, the part of the KE which appears in other forms (other than KE or PE) is (this part of the energy is termed as lost energy as it cannot be utilized properly)

1. mgH
2. mgh
3. mgH – mgh
4. Zero

Answer: 3. mgH – mgh

2. The speed of the ball just after the collision is

1. \(\sqrt{2 g H}\)
2. \(\sqrt{2} g h\)
3. \(\sqrt{2 g(H-h)}\)
4. None of these

Answer: 2. \(\sqrt{2} g h\)

3. If the lost energy in the collision is half of the value computed in Question (1), and H = \(\frac{3 h}{2}\), then the height attained by the ball after the collision is

1. \(\frac{7 h}{4}\)
2. \(\frac{3 h}{4}\)
3. \(\frac{3 h}{2}\)
4. \(\frac{9 h}{5}\)

Answer: 1. \(\frac{7 h}{4}\)

Question 3. A block of 2.5 kg is pulled 2.20 m along a frictionless horizontal table by a constant force of 16 N directed at 45° above the horizontal.

1. Work done by the applied force is

1. 25 J
2. 27 J
3. 24.9 J
4. 22.5 J

Answer: 3. 24.9 J

2. Work done by the normal force exerted by the table is

1. 24.9 J
2. Zero
3. 27 J
4. 27.5 J

Answer: 2. Zero

3. Work done by the force of gravity is

1. 24.9 J
2. 27 J
3. Zero
4. 27.5 J

Answer: 3. Zero

 

Dissipation Of Energy

Dissipation of Energy Notes for Class 11 WBCHSE

Energy

According to the law of conservation of energy, energy cannot be destroyed. But during the transformation of energy, some energy may change into such a form that has no practical utility and cannot be recovered in any usable form. This is called the dissipation of energy.

Dissipation Of Energy Example: Energy has to be supplied to a machine to make it work. But the work output (i.e., energy) is generally less than the energy supplied. This is because a part of the supplied energy is used to overcome friction and other resistive forces and this part transforms into heat or sound energy that cannot be used for practical purposes and is lost forever.

This is the dissipation of energy. It does not mean the destruction of energy. It simply denotes the transformation of energy into unusable form, also called unavailable energy. Various methods are used to reduce this dissipation of energy. But this could not be minimized to zero yet.

Efficiency Of A Machine: The ratio between the work output of a machine and the energy supplied to it, is called its efficiency. In real life, efficiency is less than 1, and it is often expressed in percentage by multiplying the ratio by 100.

Therefore, the efficiency of a machine = \(\frac{\text { work output of the machine }}{\text { energy supplied (input) }} \times 100 \%\)

For example, 90% efficiency of a machine means that, if 100 units of energy is supplied to the machine, work done by it will be 90 units.

Work Done Against Friction: The total mechanical energy of a body, falling under gravity along a frictionless inclined plane remains conserved. But, a frictionless surface is an ideal one, and cannot be obtained in practice. A frictional force always acts against the motion, and some work has to be done by the body against this force. As a result, some energy is dissipated.

Suppose a body of mass m begins to move from point A under gravity along a rough inclined plane towards C.

The height of point A above the reference plane CD is h. Hence, the potential energy of the body at A = mgh. The body is at rest so its kinetic energy is zero there. Thus, the total mechanical energy of the body at A = mgh + 0 = mgh.

Under the action of the component mg sin# of the weight, the body starts moving down along the incline. Then a frictional force, f = μR = μmg cosθ acts upwards on the body along the inclined plane, where μ = coefficient of friction.

Understanding Energy Dissipation in Physics

Hence, the resultant downward force along the plane,

F = mg sinθ- μmg cosθ = mg(sinθ – μ cosθ)

the acceleration along the plane, a = \(\frac{F}{m}\) = g(sinθ-μcosθ)

If the velocity of the falling body at B is v, where AB = x, then v² = 2ax = 2xg(sinθ- μcosθ)

Kinetic energy at B = \(\frac{1}{2}\) mv²

= \(\frac{1}{2}\) m · 2xg(sinθ – μcosθ)

= mgx(sinθ-μcosθ)

The potential energy at B = mg · DE = mg(DA-EA) =mg(h- xsinθ)

Hence, total mechanical energy at B

= mg(h- xsinθ) + mgx( sinθ- μcosθ)

= mgh-μmg cosθ · x = mgh – fx…(1)

Equation (1) shows that the mechanical energy at B is less than that at A by fx, which is the work done against the frictional force to cover a distance x along the plane. This amount fx of energy is transformed into unavailable form, in order to overcome the frictional force against motion.

So, the total energy dissipated during the sliding of the body along an inclined plane of length l (=AC) = fl = μmgl cosθ.

The above discussions show that, in the presence of dissipative forces like friction, mechanical energy does not remain conserved for a system. We see that,

Total mechanical energy at A = total mechanical energy at B +fx

The work fx, done against friction, actually transforms into heat energy at the surface of contact of the body with the plane. This heat can never be recovered in any usable form. However, taking this heat into consideration, we see that the total energy is certainly conserved.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
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Unit 4 Work Energy Power Chapter 1 Work And Energy

Dissipation Of Energy Numerical Examples

Examples of Energy Dissipation in Daily Life

Example 1. A block weighing 250 N Is pulled over a horizontal plane at a constant velocity up to a distance of 10 m. The coefficient of kinetic friction Is 0.2 and the force is applied by a string, attached to the block, Inclined at 60° with the vertical. Find the work done against friction.
Solution:

Given

A block weighing 250 N Is pulled over a horizontal plane at a constant velocity up to a distance of 10 m. The coefficient of kinetic friction Is 0.2 and the force is applied by a string, attached to the block, Inclined at 60° with the vertical.

Let the force applied on the block be F

Horizontal component of the applied force along the plane

= \(F \sin 60^{\circ}=\frac{\sqrt{3}}{2} F\) and its vertical component

= \(F \cos 60^{\circ}=\frac{F}{2} .\)

Since there is no vertical acceleration of the block, net force acting vertically is zero.

i.e., R + Fcos60° = W (where R is the normal reaction on the block)

∴ R = W – \(\frac{F}{2}\)

As the body is moving with a uniform velocity, the horizontal component of applied force = frictional force

or, \(\frac{\sqrt{3}}{2} F=\mu R=\mu\left(W-F \cos 60^{\circ}\right)=0.2\left(250-\frac{F}{2}\right)\) = 50-0.1 F

∴ F = \(\frac{50}{0.866+0.1}=\frac{50}{0.966} \mathrm{~N}\)

Hence, work done by the applied force

= \(F \sin 60^{\circ} \times 10=\frac{50}{0.966} \times \frac{\sqrt{3}}{2} \times 10=448.25 \mathrm{~J} .\)

Example 2. A particle Is sliding down along an inclined plane. The frictional force is 0.2 times the normal reaction, and the inclination of the plane is 60J. What Is the acceleration of the particle? If the mass of the particle is 1 g, find the change in the sum of potential and kinetic energies of the particle as it slides down the plane by 1 m.
Solution:

Given

A particle Is sliding down along an inclined plane. The frictional force is 0.2 times the normal reaction, and the inclination of the plane is 60J.

Let the acceleration of the particle along the inclined plane = a and the downward force on the particle along the plane = mg sinθ – f

∴ ma = mgsinθ – μR = mg sinθ – μmg cosθ

or, a = g (sinθ – nμcosθ) =9.8 (sin60° – 0.2cos60°)

= \(9.8\left(\frac{\sqrt{3}}{2}-0.2 \times \frac{1}{2}\right)=7.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

Change in mechanical energy

= work done against friction = μmg cosθ · s

= 0.2 x 0.001 x 9.8 x \(\frac{1}{2}\) x 1 = 0.00098 J.

Change in mechanical energy = 0.00098 J.

Practice Questions on Energy Dissipation for Class 11

Example 3. A box of mass 12 kg is pushed up by a distance of 10 m on the application of a 100 N force along a plane of inclination 30°. If the coefficient of friction is \(\frac{1}{\sqrt{3}}\), find the work done against friction, [g = 10 m · s^-2]
Solution:

Given

A box of mass 12 kg is pushed up by a distance of 10 m on the application of a 100 N force along a plane of inclination 30°. If the coefficient of friction is \(\frac{1}{\sqrt{3}}\),

Frictional force, f = μR = μmg cos30°

Hence work done against friction, W = fs

= μmg cos30° · s

= \(\frac{1}{\sqrt{3}} \times 12 \times 10 \times \frac{\sqrt{3}}{2} \times 10=600 \mathrm{~J}\)

Example 4. A car of mass 1000 kg moves up at 40 km · h^-1 along an inclined plane of slope \(\frac{1}{50}\). The coefficient of rolling friction between the road and the wheels of the car is 0.3. Find the power of the car engine.
Solution:

Given

A car of mass 1000 kg moves up at 40 km · h^-1 along an inclined plane of slope \(\frac{1}{50}\). The coefficient of rolling friction between the road and the wheels of the car is 0.3.

The angle of inclination = θ.

∴ Slope = tanθ = \(\frac{1}{50}\) ≈ sinθ [θ is small]

The resultant downward force on the car along the incline

= mg sinθ + f

= mg sinθ + μR

= mg sinθ + μmg cosθ = mg(sinθ + μcosθ)

= \(1000 \times 9.8\left(\frac{1}{50}+0.3 \times 1\right)\) (as is very small cosθ ≅ 1)

= 9800 x 0.32 = 3136 N

Velocity of the car =40 km · h^-1 = 11.11 m · s^-1

∴ Power of the car =3136 x 11.11 N · m · s^-1 =34840.96 W = 34.84 kW.

Example 5. An engine, working at a constant rate, is pulling a train of mass 500 tonne along a plane of inclination sin^-1\(\frac{1}{100}\).frictional force per metric tonne is 49 N and the train is moving with a velocity of 10 m · s^-1, what is the power of the engine in kilowatt? [1 tonne (metric ton) = 1000 kg]
Solution:

Given

An engine, working at a constant rate, is pulling a train of mass 500 tonne along a plane of inclination sin^-1\(\frac{1}{100}\).frictional force per metric tonne is 49 N and the train is moving with a velocity of 10 m · s^-1,

Downward effective force on the train along the inclined plane = mg sinθ + frictional force (f)

= 500 x 1000 x 9.8 x \(\frac{1}{100}\) + 500 x 49 = 7500 x 9.8 N

Velocity of the train = 10 m · s^-1

Since the car is moving upward with a constant velocity, the force applied by the car’s engine, F must exactly balance the net downward force.

∴ F = 3136 N

Hence, power of the engine = 7500 x 9.8 x 10 = 735000 J · s^-1 = 735000 W = 735 kW.

Example 6. A loaded lorry of total mass 5000 kg can come down from the top of a slope (1:40) effortlessly at 18 km · h^-1. What should be the horsepower of its engine so that it can go up with the same speed, from the base to the top? Resistance due to friction may be taken to be the same in both cases.
Solution:

Given

A loaded lorry of total mass 5000 kg can come down from the top of a slope (1:40) effortlessly at 18 km · h^-1.

Velocity of the lorry = \(18 \mathrm{~km} \cdot \mathrm{h}^{-1}=\frac{18 \times 1000}{60 \times 60} \mathrm{~m} \cdot \mathrm{s}^{-1}=5 \mathrm{~m} \cdot \mathrm{s}^{-1}\) \(\sin \theta=\frac{1}{40}\)

As the speed of the lorry remains constant for the downward journey, the friction just balances the downward active force mg sinθ.

∴ Frictional force, f = mg sinθ.

Effective force on the lorry for its upward journey = mg sinθ + f = mg sinθ + mg sinθ = 2mg sinθ

∴ Power of the engine = effective force on the lorry x velocity of the lorry

= 2mgsinθ x 5 = 2 x 5000 x 9.8 x \(\frac{1}{40}\) x 5

= 12250 W = \(\frac{12250}{746}\)hp = 16.42 hp.

Problems On Energy Dissipation

Example 7. A car of mass 500 kg is moving up along an inclined surface of slope \(\frac{1}{25}\) at a constant speed of 72 km · h^-1. If the coefficient of friction between the road and the car wheel is 0.1, find the power of the car engine (g = 9.8 m · s^-2).
Solution:

Given

A car of mass 500 kg is moving up along an inclined surface of slope \(\frac{1}{25}\) at a constant speed of 72 km · h^-1. If the coefficient of friction between the road and the car wheel is 0.1,

Frictional force, f = μmg cosθ, v = 72 km · h^-1 = 20 m · s^-1, m = 500 kg, g = 9.8m · s^-2, μ = 0.1 and tanθ = \(\frac{1}{25}\)

∴ sinθ  = \(\frac{1}{\sqrt{626}}\) and

cosθ = \(\frac{25}{\sqrt{626}}\)

F = force opposing the motion of the car = f + mg sinθ = mg(μcosθ + sinθ)

∴ Power of the car engine, P = Fv= mg(μcosθ + sinθ)v

= \(500 \times 9.8 \times\left(\frac{0.1 \times 25}{\sqrt{626}}+\frac{1}{\sqrt{626}}\right) \times 20\)

= \(13709 \mathrm{~J} \cdot \mathrm{s}^{-1}=13709 \mathrm{~W} .\)

Example 8. A 1.5 m long chain of mass 0.8 kg is kept on a horizontal table and a part of its length hangs from the edge of the table. When the length of the hanging part is one-third the total length of the chain, it starts sliding off the table. What will be the work done by friction when the whole length of the chain slides off the table?
Solution:

Given

A 1.5 m long chain of mass 0.8 kg is kept on a horizontal table and a part of its length hangs from the edge of the table. When the length of the hanging part is one-third the total length of the chain, it starts sliding off the table.

When \(\frac{1}{3}\) of the chain is hanging, it starts sliding.

In this condition, frictional force = weight of the hanging part of the chain

or, μ x \(\frac{2}{3}\)mlg= \(\frac{1}{3}\)mlg [μ = coefficient of friction, m = mass per unit length of the chain, and l = length of the chain]

∴ μ = \(\frac{1}{2}\) = 0.5

When the whole length of the chain slides off the table, the effective frictional force on the chain = 0.

∴ Effective average frictional force on the chain = \(\frac{\frac{2}{3} \mu \mathrm{mlg}+0}{2}=\frac{1}{3} \mu \mathrm{mlg} .\)

The chain moves through a distance of \(\frac{2}{3}\). l against the effective friction. Work done against friction is, therefore,

W = \(\frac{1}{3} \mu m g l \times \frac{2}{3} l=\frac{2}{9} \mu m g l^2\)

= \(\frac{2}{9} \times 0.5 \times \frac{0.8}{1.5} \times 9.8 \times(1.5)^2\)

= \(1.3 \mathrm{~J} .\)

Example 9. A body of mass 10 kg is pushed up 50 cm from the ground, along a plane inclined at 45° to the horizontal. if the coefficient of friction is 0.2, then calculate the work done.
Solution:

Given

A body of mass 10 kg is pushed up 50 cm from the ground, along a plane inclined at 45° to the horizontal. if the coefficient of friction is 0.2

Here, h = 50 cm = 0.5 m, m = 10 kg, g = 9.8 m · s^-2, θ = 45°, μ = 0.2

Let the friction acting on the body be f. Then, f = μmg cosθ

The force against which the body is pushed up is F = f + mg sinθ

= mg(μ cosθ + sinθ)

The body is pushed up by a distance \(\frac{h}{\sin \theta}\) along the inclined plane.

Therefore, the work done is, W = \(\frac{F h}{\sin \theta}=\frac{m g h}{\sin \theta}(\mu \cos \theta+\sin \theta)\)

= \(\frac{10 \times 9.8 \times 0.5}{\sin 45^{\circ}} \times\left(0.2 \times \cos 45^{\circ}+\sin 45^{\circ}\right)\)

= \(49 \times \sqrt{2} \times\left(0.2 \times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=58.8 \mathrm{~J} .\)

Example 10. By application of a force F, a body of mass m Is raised to the top of a hill. F is tangential along the whole path. If the height of the hill Is h, the length of the base of the hill is l, and the coefficient of friction is μ, then find the work done.
Solution:

Given

By application of a force F, a body of mass m Is raised to the top of a hill. F is tangential along the whole path. If the height of the hill Is h, the length of the base of the hill is l, and the coefficient of friction is μ,

The total path from the bottom to the top of the hill can be considered as an assembly of a large number of inclined planes of varying angles of inclination. Consider an incline of length Δs.

Its vertical height is Δh and the angle of inclination is α.

Total work done in lifting the body along this inclined surface = work done against gravity + work done against friction.

∴ \(\Delta W=m g \Delta h+\mu m g \cos \alpha \cdot \Delta s\)

= \(m g \Delta h+\mu m g \cos \alpha \frac{\Delta l}{\cos \alpha}=m g(\Delta h+\mu \Delta l)\)

∴ Total work done to lift the body up to the top of the hill,

W = \(\sum \Delta W=m g\left(\sum \Delta h+\mu \sum \Delta l\right)\)

= \(m g(h+\mu l)\)

Work

Work Definition: Work is said to be done when an object is displaced from its initial position under the action of a force. Work is measured as the scalar product of the applied force and the displacement of the object.

Let \(\vec{F}=\overrightarrow{O B}\) = force acting on a particle and \(\vec{s}=\overrightarrow{O A}\) corresponding displacement of the particle.

Then, by definition, the work done is W = \(\vec{F}\) \(\vec{s}\)…..(1)

If θ = ∠BOA = angle between the force and the displacement vectors, then from the property of the scalar product, we have W = Fscosθ……(2)

From the figure we get, \(F \cos \theta=O B \cdot \frac{O C}{O B}=O C\) = component of the force \(\vec{F}\) along the direction of the displacement \(\vec{s}\).

Read and Learn More: Class 11 Physics Notes

Hence, from equation (2), it may be written that, work done = displacement x component of force along the direction of displacement

If the components of \(\vec{F}\) and \(\vec{s}\) are known, equation (1) can be expressed as,

W = \(\left(F_x \hat{i}+F_y \hat{j}+F_z \hat{k}\right) \cdot\left(s_x \hat{i}+s_y \hat{j}+s_z \hat{k}\right)\)

= \(F_x s_x+F_y s_y+F_z s_z\)

It is very important to note that, force and displacement are both vector quantities, but their product, work, is a scalar quantity.

Work Discussions:

1. Force Without Displacement: If an object is not at all displaced under the action of a non-zero force \(\vec{F}\), then work done is considered to be zero,

i.e., if s = 0 but F≠ 0, then

W = Fscosθ = Fcosθ x 0 = 0

2. Displacement Without Force: If an object is displaced in the absence of any force acting on it, i.e., if s≠0 when F = 0, we find from equation (2), W = 0. For example, an object moving with uniform speed in a straight line maintains its state of motion in the absence of any external force on it (Newton’s 1st law).

But its displacement is not zero, because it is actually moving and as a result changing its position. In this case, there is no work done, only because F = 0.

3. No-Work Force: If the force and the displacement vectors are perpendicular to each other, then θ = 90° or cosθ = 0. So, from the equation (2), we have W = 0. In such cases, the force acting on an object is called a no-work force.

For example, an object in a uniform circular motion is displaced along the tangent of the circle, whereas the active force, called the centripetal force, acts in the radial direction. As the tangential and the radial directions are perpendicular to each other, the centripetal force is a no-work force.

WBBSE Class 11 Work Energy Power Study Notes

Sign Convention Of Work: Work done (W) is conventionally defined as W = \(\vec{F}\) – \(\vec{s}\) = Fscosθ….(3)

The different values of cosθ give rise to different cases:

1. If θ lies between 0 and 90°, cosθ is positive. So, from equation (3), the work done is positive.
2. If θ lies between 90° and 180°, cosθ is negative. So, from equation (3), the work done is negative.

It should be mentioned that the opposite convention, W = –\(\vec{F}\)– \(\vec{s}\), may also be used. No error will occur if this alternative relation is consistently followed throughout.

In many problems, we need to deal only with the absolute value of work done. Then, the positive or negative sign of W would not be important.

Sign Convention Of Work Discussion: When an object is displaced in a direction opposite to the force acting on it, it is said that the work is done against the force. In this situation, there must exist an external agent that is responsible for the displacement of this object against the force.

Lifting of an object upwards against the downward force of gravity acting on it, the motion of an object on a rough surface against the force of friction, etc., are examples of this type of work. In these cases, usually, some person or some machine acts as the external agent.

Let us take the example of lifting an object against gravity. This can be explained in two ways:

1. The external agent (which lifts the object) does positive work on the object by applying a force \(\vec{F}_2\) or
2. Gravity \(\vec{F}_1\) does negative work on the object.

Similarly, when an object falls under the action of gravity and an external agent opposes the motion, it can also be explained in two ways:

1. The external agent does negative work on the object or
2. Gravity does positive work on the object.

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Work Done On A System Of Bodies: The application of a force on a system, comprising a number of bodies, may produce different displacements for different bodies within the system, In such cases, to calculate the work done, the displacement of the point of application of the force is considered.

Hence, in this case, work done = force on the system of bodies x displacement of the point of application

Work Done By A System Of Forces: Since work is a scalar quantity work done due to a number of forces acting on a body is the algebraic sum of the work done due to each of the forces. The resultant of the forces produces the same amount of work.

Let a system comprising three forces act on a body, and the work done due to this system be W. Work done due to the individual forces are, w₁, w_2, and w₃. The resultant of the system of forces is \(\vec{F}\) and the displacement of the body is \(\vec{s}\). Hence W = w₁ + w₂+ w₃ = \(\vec{F}\) · \(\vec{s}\)

Work Done By A Varying Or A Variable Force: So far we have considered the work done by a force that is constant both in magnitude and direction. But often the force doing work is variable. A varying force means, a force whose magnitude or direction or both changes continuously.

• For example, when a rocket is fired upwards from the Earth, the force required to keep the rocket moving away from the Earth is not constant, it continuously decreases.
• Similarly, when a spring, whose end is fixed to a rigid support, is stretched by a force through a small distance x then the restoring force F developed in the spring increases with x.
• Let us consider a variable force F to be acting on a particle. Under the influence of this force, the particle is displaced along the x-axis from the initial position x_i to the final position x_j.
• The x-component of the force is denoted by F_x. In this case, the work done cannot be simply stated by W = F_x(x_f– x_i). This equation is applicable when the applied force is a constant.
• Here the force F varies with x and hence both F and F_x are functions of x. To calculate the work done in this case, the interval x_i to x_f is divided into a large number of small displacements Δx, so that F_x can be assumed to be constant over this displacement Δx.
• Hence for a very small displacement Ax, work done can be considered to be ΔW = F_xΔx.
• This can also be considered to be the area of the shaded region.

Therefore total work done when the particle is displaced from the initial position x_i to the final position x_f is given by the relation

W = \(\sum_{x_i}^{x_f} F_x \Delta x\)…(1)

If \(\Delta x \rightarrow 0\), then \(\lim _{\Delta x \rightarrow 0} \sum_{x_i}^{x_f} F_x \Delta x=\int_{x_i}^{x_f} F_x d x\)….(2)

From (1) and (2) we obtain, \(W=\int_{x_i}^{x_f} F_x d x\)

In this case, the total work done can be considered to be the area enclosed by the curve between x_i and x_f on the x-axis.

Key Concepts in Work Energy and Power Notes

Work Done On A Particle Moving Along A Curved Path: Let us consider a particle being moved along the curve AB as shown by a force \(\vec{F}\) which continuously changes in magnitude as well as in direction.

According to the diagram, if \(\vec{ds}\) is a small displacement of the particle under the action of the force, then the total work done in displacing the A particle from A to B is

W = \(\int_A^B \vec{F} \cdot d \vec{s}=\int_A^B F \cos \theta d s\)

It is to be noted that in this integration, neither F nor θ is a constant.

Graphical Representation Of Work: The relationship between forces acting on a body and its displacement can be represented in a force-displacement graph.

For A Constant Force: If the displacement of a body is s₀ under the action of a constant force F₀, we get a straight line graph AB. AB is parallel to the x-axis as F₀ is constant.

Obviously, the area under the graph, ABPO, is equal to F₀S₀ which is the magnitude of the work done.

For A Varying Force: If s₀ is the displacement of a body under a varying force, the force-displacement graph is not a straight line, but a curved line like AC or AD. In these cases also, it can be proved that the area under the curve (AC or AD) gives the work done.

Hence, work done corresponding to graph AC = area ACPO; work done corresponding to graph AD = area ADPO.

Thus, in general, work done = area under the force-displacement graph.

Absolute Units And Dimension Of Work: Work done = force x displacement of the body. Hence, a unit of work depends on units of force and displacement.

A unit work is said to be done when a unit force acting on a body produces a unit displacement along the direction of the force.

Dimension: Dimension of work = dimension of force x dimension of displacement = MLT^-2 x L = M²T^-2

Unit 4 Work Energy Power Chapter 1 Work And Energy

Work Numerical Examples

Comparative Analysis of Kinetic and Potential Energy

Example 1. To displace a body by 50 m, 150 J of work is done. What is the force applied in the direction of the displacement?
Solution:

Work done, W = Fs, where W = 150 J, s = 50 m

∴ 150 = 50F or, F = 3 N .

Example 2. A body of mass 10 kg is raised by 5m. What is the work done?
Solution:

Force, F= mg = 10 x 9.8 = 98 N ; displacement, s = 5 m

∴ Work done, W= Fs = 98 x5J = 490 J

Example 3. A cycle with the rider has a total mass of 80 kg; it rolls down 60 m on a plane of inclination 30°. What is the total work done by gravity on the cycle?
Solution:

Force acting vertically downwards, F = 80 kg-wt = 80 x 9.8 N

Displacement, s = 60 m

The angle between force and displacement, θ = 60°

∴ W=Fscosθ

= (80 x 9.8) x 60 x cos60°

= 80 x 9.8 x 60 x 0.5 = 23520 J

Example 4. A man of mass 100 kg climbs up a ladder of length 10 m. The ladder makes an angle 60° with the horizontal. Find the work done by the man against gravity in climbing up the ladder, [g = 9.8 m · s^-2]
Solution:

Work done to climb up the ladder by 10 m is equivalent to the work done to climb up a vertical height h.

Here, \(\sin 60^{\circ}=\frac{h}{10}\)

or, \(h=\frac{10 \sqrt{3}}{2}\)

∴ Work done, \(W=m g \times h\)

= \(100 \times 9.8 \times 5 \sqrt{3}=8487.04 \mathrm{~J}\).

Example 5. A body is constrained to move along the z-axis is subject to a constant force F = \((-\hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{N}\). Calculate the work done by this force in moving the body a distance of 4m along the z-axis.
Solution:

Displacement, \(\vec{s}=4 \hat{k} \mathrm{~m}\)

Force, \(\vec{F}=(-\hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{N}\)

∴ Work done, W = \(\vec{F} \cdot \vec{s}=(-\hat{i}+2 \hat{j}+3 \hat{k}) \cdot 4 \hat{k}\)

= \(-4 \hat{i} \cdot \hat{k}+8 \hat{j} \cdot \hat{k}+12 \hat{k} \cdot \hat{k}\)

= \(0+0+12 \cdot 1=12 \mathrm{~J} .\)

Common Questions on Work Energy Power

Example 6. A force F-acting on an object varies with distance x as shown here. The 2 force is in N and x in m. What is the amount of work done by the force in moving the object from x = 0 to x = 6m?

Solution:

Work done, W = area of the rectangle OABC + area of the triangle BCD,

= 3 x 3 + 1/2 x 3 x 3 = 13.5J

Example 7. The relationship between the force F and the position x of a body is as shown.  What will be the amount of work done in displacing the body from x = 1 m to x = 5m?

Solution:

Work done,

W = 10 x 1 5 x 1 + (-5 x 1) + 1/2 x 1 x 10

= 10 + 5 – 5 + 5

= 15J

Example 8. A position-dependent force F = (7 – 2x + 3x²) N acts on a body of mass 2 kg and displaces it from x = 0 to x – 5 m. Determine the amount of work done in i joule.
Solution:

Work done,

W = \(\int F d x=\int_0^5\left(7-2 x+3 x^2\right) d x\)

= \(\left[7 x-x^2+x^3\right]_0^5\)

= \(7 \times 5-5^2+5^3=35-25+125=135 \mathrm{~J}\)

Work And Energy

Power

Definitions Related to Power, and Energy

Power Definition: The time rate at which work is done by a force is called power due to that force or simply work done per unit time is called power.

If W is the amount of work done by a force in the time interval t, then power, \(\bar{P}=\frac{W}{t}=\frac{\vec{F} \cdot \vec{s}}{t}=\vec{F} \cdot \vec{v}\)…(1)

In general, work may not be done at a uniform rate throughout the time t. So, equation (1) actually represents the average power.

On the other hand, if work is done at a variable rate i.e., work W( t) done by a force is a function of time (t), then power at any given instant or instantaneous power is more useful. Let dW be the work done in an infinitesimal time interval dt, then the instantaneous power is P = \(\frac{dW}{dt}\)…(2)

If in that interval dt an infinitesimal displacement \(d \vec{s}\) takes place, then dW = \(\vec{F} \cdot d \vec{s}\)

or, P \(=\frac{d W}{d t}=\vec{F} \cdot \frac{d \vec{s}}{d t}=\vec{F} \cdot \vec{v}=F v \cos \theta\); where, \(\vec{v}=\frac{d \vec{s}}{d t}\) = instantaneous velocity and θ = the angle between the direction of force applied and displacement.

Power Example: Suppose a robot does a work of 1000 erg in 5 s and another robot takes 10 s to do the same work. Hence, the power of the first robot = 1000/5 = 200 erg · s^-1 and that of the second robot = 1000/10 = 100 erg · s^-1.

Though both of them do the same amount of work, the power delivered by the first robot is greater.

Absolute Units And Dimension Of Power: We know, power = \(\frac{work}{time}\). So the unit of power depends on the units of work and time.

The power to do unit amount of work in a unit time is the unit of power.

Abtokito Units:

Abtokito Units Relations:

1W = 1 J · s^-1 = 10⁷ erg ·  s^-1

1 kW = 10³ W, 1 MW = 10⁶ W

The practical unit of power in FPS system is horsepower (hp). When a mass of 550 lb is raised by 1 ft in 1 s against gravity, the power is 1 hp. The power of heat engines or electric motors are usually measured and expressed in horsepower.

So, 1 hp = 550 ft · lb · s^-1

Relation Between Horsepower And Watt: 1 hp =550 ft · lb · s^-1

= (500 x 30.48) cm x (453.6 x 981) dyn · s^-1

= 746 x 107 erg · s^-1 = 746 J · s^-1 = 746 W

∴ 1 kW = \(\frac{1000}{746}\) = 1.34 hp.

WBBSE Class 11 Work Power Energy Study Notes

Dimension:

Dimension of power = \(\frac{\text { dimension of work }}{\text { dimension of time }}\)

= \(\frac{M L^2 T^{-2}}{T}\)

= \(\mathrm{ML}^2 T^{-3}\)

Concept Of Power: Appliances used in our daily lives like electric bulbs, heaters, and motors are selected on the basis of their respective powers, and not of the total work that such devices can do. The power of a bulb is important as bulbs with higher wattage are more bright. The temperature of water can be raised faster with a high-power heater. The power ratings of appliances play a very important role in our practical lives.

Unit 4 Work Energy Power Chapter 1 Work And Energy

Power Numerical Examples

Example 1. A man of mass 50 kg climbs up 20 steps of a staircase in 5 s. Each step is 30 cm high. Find the power applied by the man.
Solution:

Height of 20 steps =20 x 30 = 600 cm = 6 m

∴ Work done = 50 x 9.8 x 6 = 2940 J [as m = 50 kg, g = 9.8 m · s^-2]

∴ Power = \(\frac{2940}{5}\) J · s^-1 = 588 W

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Example 2. Find the power applied by a man of mass 70 kg, carrying a load of 45 kg, moving up at 6.4 km per hour along a plane of inclination \(\frac{1}{10}\).
Solution:

Velocity of the man, \(v=6.4 \mathrm{~km} \cdot \mathrm{h}^{-1}=\frac{6.4 \times 5}{18} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Total weight of the man and the load = (70 + 45)kg = 115 x 9.8 N

While moving up, the man has to work against the force mg sinθ

Hence, the power applied by the man, P = mg sinθ x v

= \(115 \times 9.8 \times \frac{1}{10} \times \frac{6.4 \times 5}{18}\)

[Here, tanθ = 1/10 ≈ sinθ (as θ is small)]

= 200.36 W

Common Questions on Work Power Energy

Example 3. Water is lifted to a vertical height of 90 m using a 7.46 kW engine. If the efficiency of the engine is 80%, find the amount of water lifted in a minute.
Solution:

Let the mass of water lifted per minute be m.

Effective power of the engine = 7.46 x 10³ x 0.8 W

Now, work done per second = the effective power

∴ \(\frac{F s}{t}=\frac{m g s}{t}=7.46 \times 10^3 \times 0.8\)

or, m = \(7.46 \times 10^3 \times 0.8 \times \frac{t}{g s}\)

= \(\frac{7.46 \times 10^3 \times 0.8 \times 60}{9.8 \times 90}\)

(because \(t=1 \mathrm{~min}=60 \mathrm{~s}, g=9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}, s=90 \mathrm{~m}\)) \(\approx 406 \mathrm{~kg}\)