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

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

Short Answer Questions on First and Second Laws of Thermodynamics for Class 11

Question 1. An ideal gas undergoes a cyclic process (V, P) → (V, P) → (V, 4P) → (V, P) along straight lines in the P- V plane. Calculate the work done in the process.
Answer:

Given

An ideal gas undergoes a cyclic process (V, P) → (V, P) → (V, 4P) → (V, P) along straight lines in the P- V plane.

Area of the cycle = 1/2 x base x height

= 1/2(3V -V)(4P-P) = 3PV

Ad the cycle is anticlockwise, the work done = -3PV

Question 2. The temperature of a fixed mass of ideal gas is changed from T₁K to T₂K (T₂ > T₁). Calculate the work done if the change is brought about at

1. Constant pressure,
2. Constant volume.

Hence find from the first law of thermodynamics, in which case heat absorbed will be greater.

Answer:

1st law of thermodynamics: Q = (U₂-U₁)+ W

Work done at constant volume, W₁ = 0, Q₁ = C_v(T₂-T₁)

Change in internal energy = U₂ – U₁ = C_v(T₂ -T₁)

At constant pressure, Q₂ = C_p(T₂ – T₁)

For the same change in temperature, there will be some change in internal energy.

So the work done in this case, \(W_2=Q_2-\left(U_2-U_1\right)=C_p\left(T_2-T_1\right)-C_v\left(T_2-T_1\right)\)

= \(\left(C_p-C_v\right)\left(T_2-T_1\right)\)

As, \(C_p>C_v, W_2>0\). So, in this case \(Q_2=W_2+C_v\left(T_2-T_1\right)\)

∴ \(Q_2>Q_1\)

So in case of constant pressure heat absorbed will be greater.

Question 3. 1 mol of an ideal gas is compressed to half of its initial volume

1. Isothermallv and
2. Adiabatically.

Draw the p-V diagram in each case and hence state with reason, in which case the work done is less.

Answer:

The slope of the adiabatic curve is more than the isothermal curve.

Work done in the isothermal process = -area of ABCD and

the work done in the adiabatic process =- area of ABCU.

Comparing the two areas the work done in isothermal compression will be less.

Question 4. The second law of thermodynamics

1. Gives the definition of temperature.
2. Determines the direction of flow of heat during heat exchange between bodies.
3. Is another form of die principle of conservation of heat and other forms of energy.
4. Helps in computing the efficiency of the Camot’s engine.

Answer:

The options 1, 2, and 4 are correct.

Understanding First Law of Thermodynamics: Short Answers

Question 5. A certain amount of an ideal gas (γ= 1.4) performs 80 I of work while undergoing isobaric expansion. Find the amount of heat absorbed by the gas in the process and the change in its internal energy.
Answer:

Given

A certain amount of an ideal gas (γ= 1.4) performs 80 I of work while undergoing isobaric expansion.

For nmol of a gas,

pV = nRT or, pdV = nRdT [p constant]

or, \(d W=n R d T \quad or, d T=\frac{d W}{n R}\)….(1)

Also, \(\frac{C_p}{C_v}=\gamma\) or, \(\frac{C_p-C_v}{C_v}=\frac{\gamma-1}{1}\)

or, \(C_v=\frac{n R}{\gamma-1} \quad\left[because C_p-C_v=n R\right]\)

We know, change in internal energy, \(d U=C_v d T\)

From (1), (2) and (3)

dU = \(\frac{n R}{\gamma-1} \times \frac{d W}{n R}=\frac{80}{1-1.4}=200 \mathrm{~J}\)

Question 6. The efficiency of a Carnot engine is 50%. The temperature of the heat sink is 27 C.  Find the temperature of the heat source.
Answer:

Given

The efficiency of a Carnot engine is 50%. The temperature of the heat sink is 27 C.

The temperature of the heat sink

T₂ = 27 + 273 = 300K

Efficiency, \(\eta=50 \%=\frac{50}{100}=\frac{1}{2}\)

If the temperature of the heat source is T, \(\eta=1-\frac{T_2}{T_1}\)

i.e., \(T_1=\frac{T_2}{1-\eta}=\frac{300}{1-\frac{1}{2}}\)

= 600K = (600 – 273) °C =327 °C

Second Law of Thermodynamics: Short Answer Format

Question 7. During adiabatic expansion of 2 mol of a gas, the internal energy of the gas is found to decrease by 2J. The work done by the gas during the process is

1. 1J
2. -1J
3. 2J
4. -2J

Answer:

Given

During adiabatic expansion of 2 mol of a gas, the internal energy of the gas is found to decrease by 2J.

Work done in adiabatic process = decrease in internal energy

= 2J

The option 3 is correct.

Question 8. The slope of an isothermal curve is always

1. The same as that of an adiabatic curve
2. Greater than that of an adiabatic curve
3. Less than that of an adiabatic curve
4. None of these

Answer: 3. Less than that of an adiabatic curve

Then option 3 is correct.

Question 9. When 110 J of heat is supplied to a gaseous system, the internal energy of the system increases by 40 J. The amount of external work done (in f) is

1. 150
2. 70
3. 110
4. 40

Answer:

Given

When 110 J of heat is supplied to a gaseous system, the internal energy of the system increases by 40 J.

According to the first law of thermodynamics,

dQ = dU+ dW

or, dW= dQ-dU = 110-40 = 70

The option 2 is correct.

Examples of Short Answer Questions on Thermodynamic Laws

Question 10. One mole of an ideal monatomic gas is heated at a constant pressure from 0 °C to 100 °C. Then the change in the internal energy of the gas is (given R = 8.32 J · mol · k^-1)

1. 0.83 x 10³ J
2. 4.6 x 10³ J
3. 2.08 x 10³ J
4. 1.25 x 10³ J

Answer:

Given

One mole of an ideal monatomic gas is heated at a constant pressure from 0 °C to 100 °C.

Change in internal energy, \(\Delta U =n C_v\left(T_f-T_i\right)=1 \times \frac{3}{2} R \times 100\)

= \(1 \times \frac{3}{2} \times 8.32 \times 100=1.25 \times 10^3 \mathrm{~J}\)

The option 4 is correct.

Question 11. One mole of a van der Waals gas obeying the equation \(\left(p+\frac{a}{V^2}\right)(V-b)=R T\), undergoes the quasistatic cyclic process which is shown in the p- V diagram. The net heat absorbed by the gas in this process is

1. \(\frac{1}{2}\left(p_1-p_2\right)\left(V_1-V_2\right)\)
2. \(\frac{1}{2}\left(p_1+p_2\right)\left(V_1-V_2\right)\)
3. \(\frac{1}{2}\left(p_1+\frac{a}{V_1^2}-p_2-\frac{a}{V_2^2}\right)\left(V_1-V_2\right)\)
4. \(\frac{1}{2}\left(p_1+\frac{a}{V_1^2}+p_2+\frac{a}{V_2^2}\right)\left(V_1-V_2\right)\)

Answer:

Given

One mole of a van der Waals gas obeying the equation \(\left(p+\frac{a}{V^2}\right)(V-b)=R T\), undergoes the quasistatic cyclic process which is shown in the p- V diagram.

In the case of a cyclic process, the total heat absorbed by the gas

= work done in the total cycle

= area of the cycle

= \(\frac{1}{2}\left(p_1-p_2\right)\left(V_1-V_2\right)\)

The option 1 is correct.

Calculating Work and Heat Transfer: Short Answers

Question 12. A heating element of resistance r is fitted inside an adiabatic cylinder which carries a frictionless piston of mass m and cross-section A as shown in the diagram. The cylinder contains one mole of an ideal diatomic gas. The current flows through the element such that the temperature rises with time t as \(\Delta T=\alpha t+\frac{1}{2}\)βt² (α and β are constant), while the pressure remains constant. The atmospheric pressure above the piston is p₀, Then

1. The rate of increase in internal energy is \(\frac{5}{2} r(\alpha+\beta t)\)
2. The current flowing in the element is \(\sqrt{\frac{5}{2 r} r(\alpha+\beta t)}\)
3. The piston moves upwards with constant acceleration
4. The piston moves upwards at a constant speed

Answer:

Given

A heating element of resistance r is fitted inside an adiabatic cylinder which carries a frictionless piston of mass m and cross-section A as shown in the diagram. The cylinder contains one mole of an ideal diatomic gas. The current flows through the element such that the temperature rises with time t as \(\Delta T=\alpha t+\frac{1}{2}\)βt² (α and β are constant), while the pressure remains constant.

Increase in internal energy, \(\Delta U =n C_v \Delta T=1 \times \frac{5}{2} R \times\left(\alpha t+\frac{1}{2} \beta t^2\right)\)

= \(\frac{5}{2} R\left(\alpha t+\frac{1}{2} \beta t^2\right)\)

Rate of increase in internal energy, \(\frac{d(\Delta U)}{d t}=\frac{5}{2} R(\alpha+\beta t)\)

According to Charles’ law, \(\Delta V \propto \Delta T\) [when pressure constant]

∴ \(\Delta V=C\left(\alpha t+\frac{1}{2} \beta t^2\right)\) [C=constant]

So, displacement of the piston,

∴ \(\Delta x=\frac{\Delta V}{A}=C^{\prime}\left(\alpha t+\frac{1}{2} \beta t^2\right) \quad\left[C^{\prime}=\frac{C}{A}=\text { constant }\right]\)

Hence, the displacement of the piston obeys the equation of motion of a particle in constant acceleration.

The options 1 and 3 are correct.

Question 13. The pressure p, volume V, and temperature T for a certain gas are related \(p=\frac{A T-B T^2}{V}\), where A and B are constants. The work done by the gas when the temperature changes from T₁ to T₂ while the pressure remains constant is given by

1. \(A\left(T_2-T_1\right)+B\left(T_2^2-T_1^2\right)\)
2. \(\frac{A\left(T_2-T_1\right)}{V_2-V_1}-\frac{B\left(T_2^2-T_1^2\right)}{V_2-V_1}\)
3. \(A\left(T_2-T_1\right)-B\left(T_2^2-T_1^2\right)\)
4. \(\frac{A\left(T_2-T_2^2\right)}{V_2-V_1}\)

Answer:

Given

The pressure p, volume V, and temperature T for a certain gas are related \(p=\frac{A T-B T^2}{V}\), where A and B are constants.

Work done, \(W \int p d V=\int_{V_1}^{V_2} d V=p\left(V_2-V_1\right)\)

According to the given relation, \(V_1 =\frac{A T_1-B T_1^2}{p}, V_2=\frac{A T_2-B T_2^2}{p}\)

∴ W = \(p\left(\frac{A T_2-B T_2^2}{p}-\frac{A T_1-B T_1^2}{p}\right)\)

= \(A\left(T_2-T_1\right)-B\left(T_2^2-T_1^2\right)\)

The option 3 is correct.

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Question 14. 2 mol of an ideal monatomic gas is carried from a state (P₀, V₀) to a state (2P₀, 2V₀) along a straight line path in a P-V diagram. The amount of heat absorbed by the gas in the process is given by

1. \(3 P_0 V_0\)
2. \(\frac{9}{2} P_0 V_0\)
3. \(6 P_0 V_0\)
4. \(\frac{3}{2} P_0 V_0\)

Answer:

Given

2 mol of an ideal monatomic gas is carried from a state (P₀, V₀) to a state (2P₀, 2V₀) along a straight line path in a P-V diagram.

According to the graph, if temperatures of state A and state B of the gas are T₁ and T₂, then \(P_0 V_0=n R T_1 \text { and } 2 P_0 \times 2 V_0=n R T_2\)

∴ \(T_2-T_1=\frac{4 P_0 V_0}{n R}-\frac{P_0 V_0}{n R}=\frac{3 P_0 V_0}{n R}\)

From state A to B at the intermediate state, the pressure is changed from P₀ to 2P₀ by changing the temperature from T₁ to T₂ keeping the volume constant.

Then increase in internal energy of the gas,

ΔU = the amount of heat absorbed by n mol gas at constant volume

= n x C_v x(T₂-T₁)

= \(n \times \frac{3 R}{2} \times \frac{3 P_0 V_0}{n R}\) [for monatomic gas, \(C_\nu=\frac{3 R}{2}\)

= \(\frac{9 P_0 V_0}{2}\)

Work done by the gas absorbing the remaining heat, ΔW= area of ABDC

= \(\left(2 V_0+V_0\right) \frac{2 P_0-P_0}{2}=\frac{3 P_0 V_0}{2}\)

∴ \(\Delta Q =\Delta U+\Delta W\)

= \(\frac{9}{2} P_0 V_0+\frac{3}{2} P_0 V_0=6 P_0 V_0\)

The option 3 is correct.

Question 15. One mole of a monatomic ideal gas undergoes a quasistatic process, which is depicted by a straight line joining points (V₀, T₀) and (2V₀, 3T₀) in a V-T diagram. What is the value of the heat capacity of the gas at the point (V₀, T₀)?

1. R
2. 3/2 R
3. 2R
4. 0

Answer:

Given

One mole of a monatomic ideal gas undergoes a quasistatic process, which is depicted by a straight line joining points (V₀, T₀) and (2V₀, 3T₀) in a V-T diagram.

dQ = dU + pdV

or, \(d Q=n C_v d T+p d V\)

or, \(d Q=C_v d T+p d V\) [given n=1mol] ….(1)

For monatomic gas, \(\gamma=\frac{5}{3} \quad \text { or, } \frac{C_p}{C_v}=\frac{5}{3} \quad \text { or, } C_p=\frac{5}{3} C_v\)

Now, \(C_p-C_v=R\)

or, \(\frac{5}{3} C_v-C_v=R or, C_v=\frac{3}{2} R\)

∴ From (1) we get,

dQ = \(\frac{3}{2} R \times\left(3 T_0-T_0\right)+\frac{R T_0}{V_0} \times\left(2 V_0-V_0\right)\)

or, \(d Q=4 R T_0\)

or, \(C d T=4 R T_0\)[C is heat capacity]

or, \(C \times 2 T_0=4 R T_0\) or, C=2 R

The option 3 is correct.

Question 16. For an ideal gas with initial pressure and volume P_i and V_i respectively, a reversible isothermal expansion happens, when its volume becomes V₀. Then it is compressed to its original volume V_i by a reversible adiabatic process. If the final pressure is P_f then which of the following statement is true?

1. \(P_f=P_i\)
2. \(P_f>P_i\)
3. \(P_f<P_i\)
4. \(\frac{P_f}{V_0}=\frac{P_i}{V_i}\)

Answer:

Given

For an ideal gas with initial pressure and volume P_i and V_i respectively, a reversible isothermal expansion happens, when its volume becomes V₀. Then it is compressed to its original volume V_i by a reversible adiabatic process.

We know, the slope of the adiabatic curve > the slope of the isothermal curve.

When the gas of volume V₀ is compressed adiabatically to its initial volume V_i, the slope of the curve will be greater than an isothermal curve.

Here final pressure P_f at initial volume V_i will be greater than P_i.

∴ P_f > P_i

The option 2 is correct.

Practice Short Answer Questions on Thermodynamics for Class 11

Question 17. Which of the following statement(s) is/are true? “Internal energy of an ideal gas ______”

1. Decreases in an isothermal process
2. Remains constant in an isothermal process
3. Increases in an isobaric process
4. Decreases in an isobaric expansion

Answer:

The internal energy of an ideal gas depends only on the temperature of the gas.

The option 2 is correct.

Question 18. One mole of a diatomic ideal gas undergoes a cyclic process ABC as shown. The process BC is adiabatic. The temperatures at A, B, and C are 400 K, 800K, and 600K, respectively, Choose the correct answer

1. The change in internal energy in the whole cyclic process is 250 R
2. The change in internal energy in the process CA is 700R
3. The change in internal energy in the process AB is -350R
4. The change in internal energy in the process BC is -500R

Answer:

Given

One mole of a diatomic ideal gas undergoes a cyclic process ABC as shown. The process BC is adiabatic. The temperatures at A, B, and C are 400 K, 800K, and 600K, respectively,

Change in internal energy in

1. Process CA is \(U_A-U_C =C_\nu(400-600)=-200 C_v\)

= \(-200 \times \frac{5}{2} R\)

=-500 R

2. Process AB is \(U_B-U_A =C_\nu(800-400)=400 \times \frac{5}{2} R\)

= 1000 R

3. process BC is \(U_C-U_B=C_v(600-800)=-200 \times \frac{5}{2} R\)

=-500 R

4. The cycle ABCA is \(U_A-U_A=0\).

We used the property of an ideal gas that U is a function of temperature only.

Then from the 1st law of the hemodynamics, Q= (U₂ -U₁)+ W

We get for a process at constant volume, \(C_v\left(T_2-T_1\right)=\left(U_2-U_1\right)+0\)

i.e., \(U_2-U_1=C_v\left(T_2-T_1\right)\)

For a diatomic ideal gas, \(C_\nu=\frac{5}{2} R\)

The option 4 is correct.

Question 19. A solid body of constant heat capability of 1 J/°C, Is being heated by keeping it In contact with reservoirs in two ways:

1. Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies the same amount of beat.
2. Sequentially keeping In contact with 11 reservoirs such that each reservoir supplies the same amount of beat.

In both cases, body is brought from the initial temperature of 100°C to the final temperature of 200°C. Entropy change of the body in the two cases respectively Is

1. In 2, 4ln 2
2. In 2, In 2
3. In 2, 2ln 2
4. 21n 2, Bln 2

Answer:

100°C = 373K, 200°C = 473K

Increase in temperature 100°C= 100K

Heat capacity, C = 1 J/K

1. For each heat reservoir,

increase in temperature = 100/2 = 50 K

So, entropy change,

⇒ \(\Delta S_1=C\left[\int_{373}^{423} \frac{d T}{T}+\int_{423}^{473} \frac{d T}{T}\right]\)

= \(\ln \frac{423}{373}+\ln \frac{473}{423}\)

= \(\ln \frac{473}{373}\)

2. For each heat reservoir,

Increase in temperature = 100/8 = 12.5 K

So, entropy change,

⇒ \(\Delta S_2=C\left[\int_{373}^{385.5} \frac{d T}{T}+\cdots+\int_{460.5}^{473} \frac{d T}{T}\right]\)

= \(\ln \frac{473}{373}\)

(Note: The values of ΔS₁, and ΔS₂ do not match any of the given answers. If we take the temperatures to be 100 K and 200 K instead of 100°C and 200°C, then ΔS₁ = ΔS₂ = In2)

The option 2 is correct.

Question 20. An ideal gas undergoes a quasistatic, reversible process in which its molar heat capacity C remains constant. During tills process the relation of pressure p and volume V is given by pVⁿ – constant, then n is given by (Here C_p and C_v are molar sped He heat at constant pressure and constant volume, respectively)

1. \(n-\frac{C_p}{C_v}\)
2. \(n=\frac{C \cdot C_p}{C \cdot C_\nu}\)
3. \(n=\frac{C_p C}{C-C_v}\)
4. \(n-\frac{C-C_y}{C C_p}\)

Answer:

Given

An ideal gas undergoes a quasistatic, reversible process in which its molar heat capacity C remains constant. During tills process the relation of pressure p and volume V is given by pVⁿ – constant,

Here, pVⁿ = k (constant)….(1)

For 1 mol of an ideal gas, pV = RT

By (1)+(2) we get, \(V^{n-1} T=\frac{k}{R}\)

∴ \(\left(\frac{d V}{d T}\right)=-\frac{V}{(n-1) T}=\frac{V}{(1-n) T}\)

According to the first law of thermodynamics,

dQ = \(C_\nu d T+p d V\)

∴ \(\frac{d Q}{d T} =C_\nu+p\left(\frac{d V}{d T}\right)=C_\nu+\frac{p V}{(1-n) T}=C_\nu+\frac{R}{1-n}\)

So, heat capacity, \(C=C_v+\frac{R}{1-n}\)

or, \(1-n=\frac{R}{C-C_v}\)

or, \(n=1-\frac{R}{C-C_v}=\frac{C-\left(C_\nu+R\right)}{C-C_v}=\frac{C-C_p}{C-C_v}\)

(because \(C_p-C_v=R\))

The option 2 is correct.

Question 21. n mol of an ideal gas undergoes a process A→B as shown. The maximum temperature of the gas during the process will be

1. \(\frac{9 p_0 V_0}{4 n R}\)
2. \(\frac{3 p_0 V_0}{2 n R}\)
3. \(\frac{9 p_0 V_0}{2 n R}\)
4. \(\frac{9 p_0 V_0}{n R}\)

Answer:

Equation of straight line AB,

⇒ \(p-p_0=\frac{2 p_0-p}{V_0-2 V_0}\left(V-2 V_0\right)=-\frac{p_0}{V_0}\left(V-2 V_0\right)\)

or, \(p-p_0=-\frac{p_0 V}{V_0}+2 p_0\)

or, p = \(-\frac{p_0 V}{V_0}+3 p_0\)

We know, for n mol of an ideal gas, p V=n R T

∴ \(\frac{n R T}{V}=-\frac{p_0 V}{V_0}+3 p_0\)

or, \(T=-\frac{1}{n R}\left(\frac{p_0 V^2}{V_0}-3 p_0 V\right)\)

At the maximum temperature of the gas \(\frac{d T}{d V}=0\)

∴ \(-\frac{1}{n R}\left(\frac{2 p_0 V}{V_0}-3 p_0\right)=0\)

or, \(\frac{2 p_0 V}{V_0}=3 p_0\)

or, \(V=\frac{3}{2} V_0\)

∴ Maximum temperature,

⇒ \(T_{\text {max }}=-\frac{1}{n R}\left(\frac{9}{4} p_0 V_0-\frac{9}{2} p_0 V_0\right)=\frac{9 p_0 V_0}{4 n R}\)

The option A is correct.

Question 21. C_p and C_v are specific heats at constant pressure and constant volume respectively. It is observed that C_p-C_v= a for hydrogen gas C_p– C_v= b for nitrogen gas The correct relation between a and b is

1. a = 1/14 b
2. a = b
3. a = 14b
4. a = 28b

Answer:

Given

C_p and C_v are specific heats at constant pressure and constant volume respectively. It is observed that C_p-C_v= a for hydrogen gas C_p– C_v= b for nitrogen ga

Molar specific heat of the gas at constant pressure, \(X_p=M C_p\)

Molar specific heat of the gas at constant volume, \(X_\nu=M C_\nu because X_p-X_\nu=R\)

∴ \(C_p-C_\nu=\frac{R}{M}\)

Then for hydrogen gas, a = R/2 [M_H = 2]

and for nitrogen, b = \(\frac{R}{28}\) (\(M_{\mathrm{N}}=28\))

∴ \(\frac{a}{b}=\frac{28}{2}\) or, a=14 b

The option 3 is correct.

Question 22. A monatomic gas at a pressure p having a volume V expands isothermally to a volume 2 V and then adiabatically to a volume 16 V. The final pressure of the gas is (take γ = 5/3)

1. 64p
2. 32p
3. p/64
4. 16p

Answer:

Given

A monatomic gas at a pressure p having a volume V expands isothermally to a volume 2 V and then adiabatically to a volume 16 V.

1st part: p₁V₁ = p₂V₂

∴ pV = p₂ x 2 V or, p₂ = \(\frac{p}{2}\)

2nd part: for adiabatic process \(p V^\gamma=\text { constant } \quad \text { or, } p_1 V_1^\gamma=p_2{ }^{\prime} V_2^\gamma\)

Here, \(p_1=\frac{p}{2}, V_1=2 V, V_2=16 V, \gamma=\frac{5}{3}\)

∴ \(p_2{ }^{\prime}=p_1\left(\frac{V_1}{V_2}\right)^\gamma=\frac{p}{2}\left(\frac{2 V}{16 V}\right)^{5 / 3}=\frac{p}{64}\)

The option 3 is correct.

Question 23. A thermodynamic system undergoes the cyclic process ABCDA as shown. The work done by the system in the cycle is

1. P₀V₀
2. 2p₀V₀
3. \(\frac{p_0 V_0}{2}\)
4. Zero

Answer:

Work done by the system in the cycle = area under the p- V curve and V-axis

= \(\frac{1}{2}\left(2 p_0-p_0\right)\left(2 V_0-V_0\right)+\left[-\left(\frac{1}{2}\right)\left(3 p_0-2 p_0\right)\left(2 V_0-V_0\right)\right]\)

= \(\frac{p_0 V_0}{2}-\frac{p_0 V_0}{2}=0\)

The option 4 is correct.

Question 24. On observing light from three different stars P, Q, and R, it was found that the intensity of the violet color is maximum in the spectrum of P, the intensity of the green color is maximum in the spectrum of R and the intensity of red color in maximum in the spectrum of Q. If T_P, T_Q, and T_R are the respective absolute temperatures of P, Q, and R, then it can be concluded from the above observations that

1. T_P> T_Q> T_R
2. T_P>T_R>T_Q
3. T_P < T_R< T_Q
4. T_P<T_Q<T_R

Answer:

Light of violet color has the shortest wavelength, consequently, it has the highest frequency.

So the temperature of star P will be maximum. Similarly, the light red color has the longest length, and hence star Q has the minimum temperature.

The option 2 is correct.

Question 25. Shows two paths that may be taken by a gas to go from a state A to a state C.

In process, AB, 400 J of heat is added to the system, and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process of AC will be

1. 380 J
2. 500 J
3. 460 J
4. 300 J

Answer:

Change in internal energy over the entire cycle = 0

∴ Q = W

In the case of cycle ABC, Q = W

or, Q_AB + Q_BC+ Q_AC = area of triangle ABC

or, 400 + 100 + Q_CA

= \(\frac{1}{2} \times\left\{(6-2) \times 10^4\right\} \times\left\{(4-2) \times 10^{-3}\right\}\)

= 40

or, \( Q_{C A}=40-400-100=-460 \mathrm{~J}\)

∴ \(Q_{A C}=+460 \mathrm{~J}\)

The option 3 is correct.

Applications of First and Second Laws of Thermodynamics: Short Answers

Question 26. A Carnot engine, having the efficiency of as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at a lower temperature is

1. 100 J
2. 99 J
3. 90 J
4. 1J

Answer:

Given

A Carnot engine, having the efficiency of as heat engine, is used as a refrigerator. If the work done on the system is 10 J

For the heat engine, absorbed energy from the heat reservoir of higher temperature = Q₁ and work obtained, W= 10 J.

Now, \(\eta=\frac{W}{Q_1}=\frac{10}{Q_1}\) given, \(\eta=\frac{1}{10}\) then \(\frac{10}{Q_1}=\frac{1}{10}\) or, \(Q_1=100 \mathrm{~J}\)

So, the energy released at a lower temperature, \(Q_2=Q_1-W=100-10=90 \mathrm{~J}\)

The reverse phenomenon happens in the case of the refrigerator.

So, when the work done on the system is 10 J, then the amount of energy absorbed from the reservoir at a lower temperature = 90 J

The option 3 is correct.

Question 27. One mole of an ideal diatomic gas undergoes n transition from A to H along a path AH as shown,

The change in internal energy of the gas during the transition is

1. 20 KJ
2. -20 KJ
3. 20 J
4. -12KJ

Answer:

For an ideal diatomic gas, \(C_v=\frac{5}{2} R\)

So, change in internal energy,

⇒ \(\Delta U=C_U\left(T_B-T_A\right)=\frac{5}{2} R\left(\frac{p_B V_B}{R}-\frac{p_A V_A}{R}\right)\)

= \(\frac{5}{2}\left(p_B V_B-p_A V_A\right)\)

= \(\frac{5}{2}\left\{\left(2 \times 10^3\right) \times 6-\left(5 \times 10^3\right) \times 4\right\}\)

= \(-\frac{5}{2} \times 8 \times 10^3=-20 \times 10^3 \mathrm{~J}=-20 \mathrm{KJ}\)

The option 2 is correct.

Question 28. A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then

1. Compressing the gas through an adiabatic process will require more work to be done
2. Compressing the gas isothermally or dramatically will require the same amount of work
3. Which of the cases (whether compression through isothermal or through an adiabatic process) requires more work will depend upon the atomicity of the gas
4. Compressing the gas isothermally will require more work to be done

Answer:

Given

A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half.

Here, the AB curve represents adiabatic compression and the AC curve represents for isothermal compression.

Work done for adiabatic compression = area ABDE

Work done for isothermal compression = area ACDE

As the adiabatic curve is always steeper than the isother¬mal curve, then area ABDE> area ACDE

This means that a higher amount of work will be done in an adiabatic process.

The option 1 is correct.

Question 29. A refrigerator works between 4°C and 30°C. It is required to remove 600 cal of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (take 1 cal = 4,2 J)

1. 23.05 W
2. 236.5 W
3. 2365 W
4. 2.365 W

Answer:

Given

A refrigerator works between 4°C and 30°C. It is required to remove 600 cal of heat every second in order to keep the temperature of the refrigerated space constant.

The coefficient of performance of the refrigerator,

e = \(\frac{T_2}{T_1-T_2}=\frac{277}{303-277}=\frac{277}{26}\)

Here, \(T_2=273+4=277 \mathrm{~K}, T_1=273+30=303 \mathrm{~K}\)

Also, \(e=\frac{Q_2}{W}\) (where, \(Q_2=\) heat absorbed by the refrigerator from colder body

W= work done by refrigerator

or, W = \(\frac{Q_2}{e}=600 \times 4.2 \times \frac{26}{277}=236.53 \mathrm{~J}\)

As work done by the refrigerator is 236.53 ) per second, then the power required is 236.5)3 W.

The option 2 is correct.

Question 30. The volume 1 mol of an ideal gas with the adiabatic exponent y is changed according to the relation V = \(\frac{b}{T}\) where b = constant. The amount of heat absorbed by the gas in the process if the temperature is increased by ΔT will be

1. \(\left(\frac{1-\gamma}{\gamma+1}\right)_{R \Delta T}\)
2. \(\frac{R}{\gamma-1} \Delta T\)
3. \(\left(\frac{2-\gamma}{\gamma-1}\right) R \Delta T\)
4. \(\frac{R \Delta T}{\gamma-1}\)

Answer:

Given

The volume 1 mol of an ideal gas with the adiabatic exponent y is changed according to the relation V = \(\frac{b}{T}\) where b = constant.

V = \(\frac{b}{T}\)

∴ dV = \(-\frac{b}{T^2} d T\)

Work done, W = \(\int P d V=-\int \frac{R T}{V} \cdot \frac{b}{T^2} d T\)

= \(-\int \frac{R T}{V} \cdot \frac{V T}{T^2} d T=-R \int d T=-R \Delta T\)

Change in internal energy,

⇒ \(\Delta U=\int C_\nu d T=\frac{R}{\gamma-1} \int d T=\frac{R}{\gamma-1} \Delta T\)

(As \(\frac{R}{\gamma-1}=\frac{C_p-C_v}{\frac{C_p}{C_\nu}-1}=\frac{C_p-C_v}{\frac{C_p-C_\nu}{C_\nu}}=C_v\))

∴ Heat absorbed, Q = \(\Delta U+W=\left(\frac{R}{\gamma-1}-R\right) \Delta T\)

= \(R \times \frac{1-\gamma+1}{\gamma-1} \Delta T=\left(\frac{2-\gamma}{\gamma-1}\right) R \Delta T\)

The option 3 is correct.

Question 31. One mole of a gas obeying the equation of state P(V- b) = RT is made to expand from a state with coordinates (P₁, V₁) to a state with (P₂, V₂) along a process that is depicted by a straight line on a P-V diagram. Then, the work done is given by

1. \(\frac{1}{2}\left(P_2-P_1\right)\left(V_2+V_1+2 b\right)\)
2. \(\frac{1}{2}\left(P_1+P_2\right)\left(V_2-V_1\right)\)
3. \(\frac{1}{2}\left(P_2-P_1\right)\left(V_2-V_1\right)\)
4. \(\frac{1}{2}\left(P_1+P_2\right)\left(V_2-V_1+2 b\right)\)

Answer:

Given

One mole of a gas obeying the equation of state P(V- b) = RT is made to expand from a state with coordinates (P₁, V₁) to a state with (P₂, V₂) along a process that is depicted by a straight line on a P-V diagram

Work done = area below straight line 1 → 2

= \(\frac{1}{2}\left(P_1+P_2\right)\left(V_2-V_1\right)\)

The option 2 is correct.

Question 32. The volume V of a monatomic gas varies with its temperature T, as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

1. 1/3
2. 2/3
3. 2/5
4. 2/7

Answer:

Given

The volume V of a monatomic gas varies with its temperature T, as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it,

Here the curve represents an isobaric process.

In an isobaric process, dQ = nC_pdT….(1)

Degrees of freedom for monatomic gas, f = 3

∴ \(\frac{C_p}{C_v}=\gamma=1+\frac{2}{f}=\frac{5}{3} \text { or, } C_v=\frac{3 C_p}{5}\)

Also, we know, C_p – C_v = R

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

We get from equation (1), \(d Q=n\left(\frac{5}{2} R\right) d T\)

Work is done by gas in this process. dW = pdV = nRdT

∴ \(\frac{d W}{d Q}=\frac{n R d T}{n\left(\frac{5}{2} R\right) d T}=\frac{2}{5}\)

The option 3 is correct.

Question 33. The efficiency of an ideal heat engine working between the freezing point and boiling point of water is

1. 6.25%
2. 20%
3. 26.8%
4. 12.5%

Answer:

Efficiency of an ideal heat engine,

⇒ \(\eta=1-\frac{T_2}{T_1} \quad\left[\begin{array}{c}
T_2=\text { temperature of the sink } \\
T_1=\text { temperature of the source }
\end{array}]\right.\)

or, \(\eta =1-\frac{273}{373}=\frac{100}{373}\)

∴ \(\eta =\frac{100}{373} \times 100 \%\)

= 26.8%

The option 3 is correct

Question 34. A sample of 0.1 g of water at 100°C and normal pressure (1.013 x 10⁵ N · m^-2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is

1. 42.2J
2. 208.7 J
3. 104.3 J
4. 84.5 J

Answer:

Given

A sample of 0.1 g of water at 100°C and normal pressure (1.013 x 10⁵ N · m^-2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc

Initial volume = 0.1 cm

Final volume = 167.1 cm³

ΔQ = ΔU+ΔW

or, ΔQ = ΔU+pΔV

or, 54×4.2 = AU+1.013 X 10⁵ x (167.1 – 0.1) X 10^-6

or, ΔU = 208.78J

The option 2 is correct

Question 35. Give anyone the difference between a refrigerator and a heat engine. and What type of thermodynamic process is associated with the sudden bursting of an inflated balloon?
Answer:

A refrigerator is a device that transfers heat from lower to higher temperatures, whereas a heat engine is used to convert work into heat.

• In a sudden burst, the air inside the balloon expands to a much higher volume in a very short interval of time. The balloon air does not get sufficient time to exchange heat with its surroundings.
• So the associated thermodynamic process is an adiabatic expansion.

Question 36. Define the isochoric process. What is the work done in such a process?
Answer:

Isochoric process

If a closed thermodynamic system does not change its volume when it exchanges some energy with its environment, the process is known as an isochoric process.

In this process, the volume V = constant; so, dV = 0.

Then, work done, W = ∫pdV = 0, i,e., no work is done in such a process.

Question 36. The molar specific heat capacity at the constant volume of a mono-atomic gas is (3/2)R, where R is the universal gas constant. Find the value of molar specific heat capacity of the gas at constant pressure.
Answer:

Given,

The molar specific heat capacity at the constant volume of a mono-atomic gas is (3/2)R, where R is the universal gas constant.

C_v = \(\frac{3}{2}R\)

From the relation C_p -C_v = R, we have,

∴ \(C_p=C_v+R=\frac{3}{2} R+R=\frac{5}{2} R\)

Question 37. How are these overcome in the second law of thermodynamics? Explain briefly.
Answer:

The 2nd law of thermodynamics overcomes these problems by the introduction of a new variable called entropy (S), which is a property of all thermodynamic systems.

• There are techniques to measure the change in entropy (ΔS) of a system and its surroundings during a process if the entropy increases (ΔS > 0), the 2nd law asserts that the process occurs in nature.
• In the reverse process, the entropy necessarily decreases (ΔS < 0), and that process is not allowed. Moreover, the amount of unavailable energy in a process is directly proportional to ΔS and can be calculated using a simple formula.

Question 38. A refrigerator is maintained (stabilizer kept inside) at 9°C. If room temperature is 36°C, calculate the coefficient of performance.
Answer:

Given

A refrigerator is maintained (stabilizer kept inside) at 9°C. If room temperature is 36°C,

Temperature inside the refrigerator, T₁ = 9°C = (273 + 9)K = 282 K

Room temperature, T₂ = 36°C = (273 + 36)K = 309 K

The coefficient of performance of a Carnot refrigerator is

p = \(\frac{T_1}{T_2-T_1}=\frac{282}{309-282}=\frac{282}{27}=10.44\)

This value corresponds to an ideal Carnot refrigerator. All refrigerators in practical use have a lower value of p.

Question 39. State the type of thermodynamic process when

1. Temperature is constant,
2. Volume is constant.

Answer:

1. Isothermal process
2. Isochoric process

Question 40. An ideal gas is compressed at a constant temperature; will its internal energy increase or decrease?
Answer:

As we know, internal energy depends on its temperature. Here temperature is constant, so (dT) = 0, this means that there is no change in internal energy.