WBCHSE Class 11 Physics For First And Second Law Of Thermodynamics Long Answer Type Questions

First And Second Law Of Thermodynamics Long Answer Type Questions

Question 1. Water falls from the top to the bottom of a waterfall. Why does the temperature at the bottom become slightly higher?
Answer:

Water at the top has a potential energy due to its height. During free fall, the potential energy is converted into kinetic energy. On impact with the ground the kinetic energy of water is converted mainly into heat energy. The heat evolved increases the temperature of the water slightly.

Question 2. A bullet becomes hot on hitting a target. Why?
Answer:

A bullet is obstructed when it hits a target. As a result of the impact, the kinetic energy of the bullet is converted mainly into heat. This heat evolved increases the temperature of the bullet, as well as that of the target.

Read and Learn More Class 11 Physics Long Answer Questions

Question 3. Two balls of the same mass, one of iron and the other of copper, are dropped from the same height. Which one would become hotter?
Answer:

As the mass and the height of both balls are the same, the initial potential energy (= mgh) is the same for both the balls. So the kinetic energy on impact with the ground will also be the same. Then, the same amount of heat will be produced.

We know that, increase in temperature = \(\frac{\text { heat produced }}{\text { mass } \times \text { specific heat }}\)

However the specific heat of copper is lower than that of iron. So the copper ball will be hotter due to its higher rise in temperature.

Question 4. Show that the external work done by a gas in an isothermal expansion is equal to the heat supplied to the gas.
Answer:

For an ideal gas, the internal energy depends only on its temperature. For a real gas, the change in internal energy at constant temperature is negligible. So for an isothermal expansion (T = constant), U_f = U_i. Then, from the first law of thermodynamics,

U_f– U_i = Q-W or, 0 = Q-W or, W= Q

This means that the work done by the gas is equal to the heat supplied to it.

Question 5. ‘An isothermal process is essentially a very slow process’. Explain.
Answer:

The volume and the pressure of a system change in an isothermal process, but the temperature remains constant.

• For example, during the expansion of a gas, as the internal energy of the gas is converted into work, the temperature of the gas tends to fall.
• But if the expansion is very slow, the gas gets sufficient time to take heat from the surroundings. This heat does the work and the internal energy is not used up. Hence, the temperature remains constant.
• Similarly, during compression, work is done on the gas. Hence, heat is generated in the gas. This heat increases the temperature of the gas.
• But in a very slow compression, the gas gets sufficient time to lose this heat to the surroundings. As a result, the temperature remains constant.
• So, an isothermal process is essentially a very slow process.

Question 6. What is the source of the energy that does the work in an adiabatic expansion of a gas?
Answer:

From the first law of thermodynamics, U_f – U_i = Q-W, i.e., change of internal energy = heat gained – work done.

For an adiabatic expansion, Q= 0.

So, U_f -U_i = -W or, W = U_i – U_f

i.e., work done = decrease in internal energy. This means that internal energy is converted into external work in an adiabatic expansion and the temperature of the gas decreases.

Question 7. A gas is compressed to half its volume in two different ways:

1. Very rapidly and
2. Very slowly. In which process will the work done on the gas be higher?

Answer:

Let the initial state of the gas be represented by point A on a pV diagram.

1. When the gas is compressed very rapidly, it is an adiabatic process. This compression from volume V to volume \(\frac{V}{2}\) is represented by the curve AB. So the work done on the gas – area under the curve AB = area ABDFA.

2. When the gas is compressed very’ slowly, it is an isothermal process. This compression is represented by the curve AC. So the work done = area ACDEA.

The adiabatic curve is always steeper than the isothermal curve at a point. So line AB is above the line AC in the diagram. Then clearly, area ABDEA > area ACDEA. This means that a higher amount of work will be done on the gas in the rapid process.

Question 8. A gas is compressed to half its volume

1. Adiabatically,
2. Isothermally. In which process will the final temperature be higher?

Answer:

The temperature does not change in an isothermal process.

1. In adiabatic compression, work is done on the gas. This work increases its internal energy. As a result, the temperature of the gas also increases.
2. So, the final temperature would be higher for adiabatic compression.

Question 9. A gas expands from state 1 to another state 2 of higher pressure in two different processes as shown. Which process would require a greater supply of heat from the surroundings?

Answer:

Work done in each constant volume part of the two processes is zero.

For the first process, the work done in the constant pressure part = p₂ (V₂ -V₁).

From the first law of thermodynamics, the heat supplied from the surroundings during the first process, \(Q_1=\left(U_2-U_1\right)+W=\left(U_2-U_1\right)+p_2\left(V_2-V_1\right)\)

Similarly, the heat supplied from the surroundings in the second process, \(Q_2=\left(U_2-U_1\right)+p_1\left(V_2-V_1\right)\)

Now, (U₂ – U₁) is the same for the two processes as the initial and file final states are the same. As the final state is at a higher pressure, we have \(p_2>p_1 \text {. So, } Q_1>Q_2 \text {. }\).

So the first process would take a greater amount of heat from the surroundings.

Example 10. A screen divides a container of volume 1 m³ into two parts. One part is filled with an ideal gas at 300 K and the other part is empty. The system is isolated from the surroundings. Will there be any change in the temperature of the gas if the screen is suddenly removed?
Answer:

The system is isolated; so heat exchange with the surroundings, Q = 0.

There is no change in the volume of the container; so work done, W = 0

From the first law of thermodynamics,

U_f – U_i = Q- W= 0 or, U_f = U_i

This means that the internal energy of the gas does not change. For an ideal gas, the internal energy depends only on its temperature. So the temperature of the gas remains unchanged at 300 K.

Question 11. For an adiabatic process of an ideal monatomic gas, the pressure and the temperature are related as p ∝ T^c. Find out the value of C.
Answer:

For an adiabatic process of an ideal gas, \(p^{1-\gamma} T^\gamma= constant =k\) (say).

or, \(p^{1-\gamma}=k T^{-\gamma} or, \quad p=\left(k T^{-\gamma}\right)^{\frac{1}{1-\gamma}}=k^{\frac{1}{1-\gamma}} T^{\frac{\gamma}{\gamma-1}}\)

∴ \(p \propto T^{\frac{\gamma}{\gamma-1}}\)

As \(p \propto T^C, C=\frac{\gamma}{\gamma-1}\).

For a monatomic gas, \(\gamma=\frac{5}{3}\).

So, C = \(\frac{\frac{5}{3}}{\frac{5}{3}-1}=\frac{5}{2}\).

Question 12. Is the solar system in thermal equilibrium?
Answer:

No, the solar system is not in thermal equilibrium. The sun continuously radiates heat energy in all directions. Every planet, satellite, or object in the solar system absorbs this radiated heat. If the solar system is in thermal equilibrium, no heat flow will occur. This will cause the thermal death of the solar system.

Question 13. The pressure and the temperature of an ideal gas in an adiabatic process are related as p ∝ T³. What is the value of the ratio  C_P/C_v of the gas?
Answer:

For 1 mol of an ideal gas,

pV= RT or, T = \(\frac{pV}{R}\)

Now, \(p \propto T^3\)

or, p = \(kT^3=k\left(\frac{p V}{R}\right)^3\)[k= constant]

∴ p = \(k \frac{p^3 V^3}{R^3} or, p^2 V^3=\frac{R^3}{k}\)

or, \(p V^{3 / 2}=\left(\frac{R^3}{k}\right)^{1 / 2}\)

or, \(p V^{3 / 2}\)= constant

For an adiabatic process, \(p V^\gamma=\) constant

So, \(\gamma=\frac{C p}{C}=\frac{3}{2}\).

Question 14. At constant pressure, the temperature coefficient of volume expansion of an ideal gas is \(\delta=\frac{1}{V} \frac{d V}{d T}.\). What will be the nature of the graph relating δ with the temperature T?
Answer:

For 1 mol of an ideal gas,

pV = \(R T \text { or, } V=\frac{R T}{p}\).

So at constant pressure, \(\frac{d V}{d T} =\frac{R}{p}\)

∴ \(\delta =\frac{1}{V} \frac{d V}{d T}=\frac{1}{V} \frac{R}{p}=\frac{R}{p V}\)

= \(\frac{R}{R T}=\frac{1}{T}\)

or, δ T= constant

This equation is represented by a rectangular hyperbola on the δ-T graph.

Question 15. A rapid compression heats a gas, but a rapid expansion cools it—why?
Answer:

A rapid thermal process can be regarded as an adiabatic process because the gas does not get sufficient time to exchange heat with the surroundings.

So, Q = 0.

From the first law of thermodynamics, U_f – U_i = Q-W = 0 -W = -W.

∴ W= U_i – U_f

1. In adiabatic compression, W is negative. So \(U_i-U_f<0, \quad \text { or, } \quad U_i<U_f\). This means that the internal energy of the gas increases. The internal energy of a gas depends only on its temperature. So the temperature increases and the gas is heated.

2. Conversely, in an adiabatic expansion, W is positive. So, U_f < U_i i.e., the internal energy decreases. As a result, temperature decreases; so the gas is cooled.

Question 16. Why does a bicycle pump become hot when it pumps air in a bicycle tube?
Answer:

The pumping operation is very rapid. So, it is essentially an adiabatic compression. Then, Q = 0, and W is negative.

From the first law of thermodynamics, \(U_f-U_i=Q-W=0-W \text { or, } W=U_i-U_f\)

As W is negative, \(U_i-U_f<0 or, U_i<U_f\).

So, the internal energy of air increases.

As a result, the temperature also increases and the pump becomes hot.

Question 17. Adiabatic and isothermal processes of an ideal gas are represented on a pV diagram by two curves, A and B. Can we say that curve A represents the isothermal process?

Answer:

The statement is correct. We know that at every point on a pV diagram, the adiabatic curve is steeper than the isothermal curve. Curve B is steeper than curve A. So B represents the adiabatic process and A represents the isothermal process.

Question 18. In the V-T diagram, two different points show pressures p₁ and p₂. Which pressure would be higher p₁ or p₂?

Answer:

The V- T graph is a straight line.

Let the equation of the straight line be V = aT+ b.

For 1 mol of an ideal gas, pV = RT.

So, \(p=\frac{R T}{V}=\frac{R T}{a T+b}=\frac{R}{a+\frac{b}{T}}\)

According to this relation, p increases when T increases.

T₂ > T₁.

So, p₂ > p₁.

Question 19. The volume of a gas is doubled

1. Very rapidly, and
2. Very slowly. In which one of the processes the work done by the gas is higher?

Answer:

The rapid process is adiabatic and the slow process is isothermal. As the adiabatic curve is steeper than the isothermal curve, AB and AC represent the isothermal and adiabatic processes, respectively.

• Now, work done is given by the area under the curve corresponding to a process.
• As area ABDE > area ACDE, the work done in the isothermal process is greater than that in the adiabatic process.

Question 20. How will the temperature of an ideal gas change in an adiabatic expansion?
Answer:

In an adiabatic process, Q = 0

For the expansion of a gas, W is positive.

∴ Q = (U_f – U_i)+ W

or, 0 = (U_f – U_i) + W

or, U_f – U_i = -W

So, U_f< U_i, i.e., internal energy decreases. As the internal energy of a gas depends only on its temperature, the temperature decreases.