WBCHSE Class 11 Physics Kinetic Theory Of Gases Question and Answers

Kinetic Theory Of Gases Question and Answers

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:

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.

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:

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:

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.

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.

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.

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:

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

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:

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

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

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

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:

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.

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:

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.