NEET Physics: Heat, Thermodynamics and Kinetic Theory · Chapter 2

Thermodynamics

What to remember

First law of thermodynamics: ΔU = Q − W. Q is positive when heat is supplied to the system; W is positive when the system does work on the surroundings. Internal energy U is a state function; heat Q and work W are not.
Four basic processes of an ideal gas: isothermal (ΔU = 0, Q = W), adiabatic (Q = 0, PV^γ = constant), isochoric (W = 0, Q = ΔU), isobaric (W = PΔV). Work done equals the area under the curve on the P-V diagram; for a cycle it equals the area enclosed.
Cp − Cv = R (Mayer's relation) and γ = Cp/Cv.
Efficiency of a heat engine η = W/Q₁ = 1 − Q₂/Q₁. A Carnot engine has η = 1 − T₂/T₁ and no engine working between the same two temperatures is more efficient. The coefficient of performance of a refrigerator is Q₂/W.

Thermal equilibrium and the zeroth law

Two bodies are in thermal equilibrium when no heat flows between them, that is, when they are at the same temperature. The zeroth law states: if body A and body B are each in thermal equilibrium with a third body C, then A and B are in thermal equilibrium with each other. This law gives meaning to temperature and allows a thermometer to be used. Thermodynamics studies systems in terms of large-scale (macroscopic) variables such as pressure, volume and temperature.

Heat, work and internal energy

Internal energy U of a system is the sum of the kinetic and potential energies of its molecules, measured in the frame in which the system is at rest as a whole. For an ideal gas the molecules have no mutual potential energy, so U depends only on the temperature.
Heat Q is energy that flows because of a temperature difference.
Work W is energy transferred by a force moving through a displacement. For a gas at constant pressure P, W = PΔV. In general W = ∫P dV.
Heat and work are modes of transfer of energy. They depend on the path followed. They are not properties of a state.
U is a state function: its change depends only on the initial and final states, not on the path.

First law of thermodynamics

The heat supplied to a system equals the increase in its internal energy plus the work done by the system:

Q = ΔU + W, or ΔU = Q − W

Sign convention used here:

Quantity	Positive when	Negative when
Q	heat is supplied to the system	heat is given out by the system
W	work is done by the system (expansion)	work is done on the system (compression)
ΔU	internal energy (temperature of an ideal gas) increases	it decreases

Specific heat capacities of gases

For n moles of an ideal gas:

At constant volume, Q = nCvΔT. No work is done, so Q = ΔU. Therefore ΔU = nCvΔT for any process of an ideal gas.
At constant pressure, Q = nCpΔT.
Mayer's relation: Cp − Cv = R, with R = 8.314 J mol⁻¹ K⁻¹. Cp is greater than Cv because at constant pressure part of the heat does work of expansion.
γ = Cp/Cv. Then Cv = R/(γ − 1) and Cp = γR/(γ − 1).

State variables and equation of state

The state of a gas is described by P, V, T (and U). These are state variables. Intensive variables (pressure, temperature) do not depend on the amount of gas. Extensive variables (volume, internal energy) are proportional to the amount of gas. The equation of state of an ideal gas is PV = nRT.

Thermodynamic processes

A quasi-static process is carried out so slowly that the system stays almost in equilibrium at every stage. Only such processes can be drawn as a curve on a P-V diagram.

Process	Condition	W	ΔU	Q
Isothermal	T constant	nRT ln(V₂/V₁)	0	equal to W
Adiabatic	Q = 0	nR(T₁ − T₂)/(γ − 1)	−W = nCv(T₂ − T₁)	0
Isochoric	V constant	0	nCvΔT	nCvΔT
Isobaric	P constant	PΔV = nRΔT	nCvΔT	nCpΔT
Cyclic	back to initial state	area of loop	0	equal to W

Isothermal. PV = constant. W = nRT ln(V₂/V₁) = nRT ln(P₁/P₂). It takes place slowly, in contact with a heat reservoir. For an ideal gas, ΔU = 0.

Adiabatic. No heat enters or leaves (an insulated container, or a very fast process). PV^γ = constant. Also TV^(γ−1) = constant. Work done W = (P₁V₁ − P₂V₂)/(γ − 1). In adiabatic expansion the gas does work at the expense of its internal energy, so it cools. In adiabatic compression it warms.

Isochoric. The volume is fixed, so no work is done. All the heat goes into internal energy.

Isobaric. Pressure is fixed. W = P(V₂ − V₁).

Cyclic. The system returns to its initial state. ΔU = 0 so net Q = net W.

P-V diagrams in words.

The work done by the gas is the area under the curve between the initial and final volumes. It is positive for expansion and negative for compression.
For a closed cycle, the net work is the area enclosed by the loop. If the loop is traced clockwise, the net work is done by the gas (a heat engine). If it is traced anticlockwise, the net work is done on the gas (a refrigerator).
An isochoric process is a vertical line, and an isobaric process is a horizontal line.
An isothermal curve is a falling curve with slope −P/V. An adiabatic curve through the same point is steeper, with slope −γP/V, because γ is greater than 1.

Heat engines

A heat engine takes heat Q₁ from a hot reservoir at T₁, converts part of it into work W, and rejects heat Q₂ to a cold reservoir at T₂. Over one cycle, W = Q₁ − Q₂. The efficiency is

η = W/Q₁ = 1 − Q₂/Q₁

The efficiency is always less than 1. A working substance, a hot source and a cold sink are all needed.

Refrigerators and heat pumps

A refrigerator works as a heat engine in reverse. It takes heat Q₂ from a cold reservoir at T₂, receives work W from outside, and rejects heat Q₁ = Q₂ + W to the hot reservoir at T₁. The coefficient of performance is

α = Q₂/W = Q₂/(Q₁ − Q₂)

A heat pump uses the same principle to deliver heat to a warm space. The refrigerator is meant to cool a cold space; the heat pump is meant to warm the hot space. For a refrigerator α can be greater than 1. Also α = (1 − η)/η.

Second law of thermodynamics

Kelvin-Planck statement: no process is possible whose sole result is the absorption of heat from a reservoir and its complete conversion into work. In other words, no heat engine has 100% efficiency.
Clausius statement: no process is possible whose sole result is the transfer of heat from a colder body to a hotter body. A refrigerator needs external work.
The two statements are equivalent.

Reversible and irreversible processes

A process is reversible if it can be made to retrace its path exactly, so that both the system and the surroundings return to their initial states. It must be quasi-static and without friction or other dissipative effects. A process that does not satisfy this is irreversible. Real processes are irreversible: for example, free expansion of a gas, mixing of two gases, friction, heat flow across a finite temperature difference, and the release of a compressed spring without recovering the energy.

Carnot engine

A Carnot engine is a reversible engine working between two temperatures T₁ (source) and T₂ (sink). Its cycle has four steps:

1. Isothermal expansion at T₁ (gas absorbs Q₁).
2. Adiabatic expansion (temperature falls from T₁ to T₂).
3. Isothermal compression at T₂ (gas rejects Q₂).
4. Adiabatic compression (temperature rises from T₂ to T₁).

Efficiency η = 1 − T₂/T₁ (temperatures in kelvin), and Q₂/Q₁ = T₂/T₁. It does not depend on the working substance. No engine working between the same two temperatures can be more efficient. A Carnot refrigerator has α = T₂/(T₁ − T₂). Efficiency of 100% would need T₂ = 0 K, which cannot be reached.

Worked examples

Example 1 (first law). A gas takes 500 J of heat and does 200 J of work. ΔU = Q − W = 500 − 200 = 300 J.

Example 2 (isothermal work). 1 mol of an ideal gas at 300 K expands isothermally to double its volume. Take R = 8.3 J mol⁻¹ K⁻¹ and ln 2 = 0.69. W = nRT ln(V₂/V₁) = 1 × 8.3 × 300 × 0.69 = 1718 J, about 1.7 kJ. The heat absorbed is the same, as ΔU = 0.

Example 3 (adiabatic). A monatomic ideal gas (γ = 5/3) is compressed adiabatically to one-eighth of its volume. T₂/T₁ = (V₁/V₂)^(γ−1) = 8^(2/3) = 4, so the absolute temperature becomes 4 times. P₂/P₁ = (V₁/V₂)^γ = 8^(5/3) = 32, so the pressure becomes 32 times.

Example 4 (Carnot engine). A Carnot engine works between 500 K and 300 K. η = 1 − 300/500 = 0.4 = 40%. If it absorbs 1000 J per cycle, the work is 400 J and the heat rejected is 600 J. A Carnot refrigerator between 270 K and 300 K has α = 270/(300 − 270) = 9.

Formula summary

Quantity	Formula
First law	ΔU = Q − W
Work at constant pressure	W = PΔV
Isothermal work	W = nRT ln(V₂/V₁)
Adiabatic relations	PV^γ = constant; TV^(γ−1) = constant
Adiabatic work	W = (P₁V₁ − P₂V₂)/(γ − 1)
Mayer's relation	Cp − Cv = R
Ratio of heat capacities	γ = Cp/Cv
Efficiency of engine	η = 1 − Q₂/Q₁
Carnot efficiency	η = 1 − T₂/T₁
Refrigerator COP	α = Q₂/W = T₂/(T₁ − T₂) for Carnot

NEET traps

1. The sign convention here is Q positive when supplied and W positive when done by the gas, so ΔU = Q − W. Do not mix it with ΔU = Q + W.
2. Heat and work depend on the path; internal energy does not.
3. For an ideal gas ΔU = nCvΔT in every process, not only at constant volume.
4. In an isothermal process ΔU = 0, but Q is not zero. In an adiabatic process Q = 0, but ΔU is not zero.
5. An adiabatic curve is steeper than an isothermal curve at the same point.
6. Use kelvin in the Carnot formula; do not use Celsius.
7. Net work in a cycle is the enclosed area, and it is not zero in general.
8. The Carnot efficiency does not depend on the working substance.
9. A refrigerator's COP can be greater than 1, but an engine's efficiency is always less than 1.
10. The Kelvin-Planck statement is about engines (heat to work); the Clausius statement is about refrigerators (heat from cold to hot).

One-liners

1. The zeroth law defines temperature.
2. ΔU = Q − W.
3. For an ideal gas U depends on temperature only.
4. Isothermal: ΔU = 0 and Q = W.
5. Adiabatic: Q = 0 and ΔU = −W.
6. Isochoric: W = 0.
7. Mayer's relation: Cp − Cv = R.
8. γ = Cp/Cv, and it is greater than 1.
9. Adiabatic law: PV^γ = constant.
10. Work in a cycle equals the area of the loop.
11. A clockwise loop on a P-V diagram means an engine.
12. η = 1 − Q₂/Q₁.
13. Carnot efficiency is 1 − T₂/T₁.
14. A refrigerator's COP is Q₂/W.
15. All real processes are irreversible.

Practice questions

The zeroth law of thermodynamics leads to the concept of
1. work
2. internal energy
3. temperature
4. pressure
Answer
C. temperature
If A and B are each in thermal equilibrium with C, they are in equilibrium with each other. This defines temperature.
A gas absorbs 300 J of heat and does 120 J of work on its surroundings. The change in its internal energy is
1. 180 J
2. 300 J
3. 420 J
4. −180 J
Answer
A. 180 J
ΔU = Q − W = 300 − 120 = 180 J.
For an ideal gas undergoing an isothermal process, the change in internal energy is
1. equal to the work done
2. zero
3. equal to the heat absorbed
4. negative always
Answer
B. zero
U of an ideal gas depends only on T. Constant T means ΔU = 0.
For a gas, Cp = 29 J mol⁻¹ K⁻¹. Take R = 8.3 J mol⁻¹ K⁻¹. The value of γ = Cp/Cv is about
1. 1.33
2. 1.67
3. 1.29
4. 1.40
Answer
D. 1.40
Cv = Cp − R = 20.7. γ = 29/20.7 = 1.40.
A gas at a constant pressure of 2 × 10⁵ Pa expands from 3 L to 5 L. The work done by the gas is
1. 1000 J
2. 400 J
3. 200 J
4. 4000 J
Answer
B. 400 J
W = PΔV = 2 × 10⁵ × 2 × 10⁻³ = 400 J.
One mole of an ideal gas at 300 K expands isothermally to four times its volume. Take R = 8.3 J mol⁻¹ K⁻¹ and ln 2 = 0.7. The work done is about
1. 3.5 kJ
2. 1.7 kJ
3. 7.0 kJ
4. 0.62 kJ
Answer
A. 3.5 kJ
W = RT ln 4 = 8.3 × 300 × 1.4 = 3486 J ≈ 3.5 kJ. (Exact ln 4 gives 3451 J.)
A monatomic ideal gas (γ = 5/3) is compressed adiabatically to one-eighth of its initial volume. The final absolute temperature is
1. 2 times the initial value
2. 32 times the initial value
3. 8 times the initial value
4. 4 times the initial value
Answer
D. 4 times the initial value
TV^(γ−1) = constant gives T₂/T₁ = 8^(2/3) = 4. (32 is the pressure ratio.)
For a closed cycle on a P-V diagram, the net work done by the gas equals
1. the area under the upper curve only
2. the area enclosed by the loop
3. zero always
4. the slope of the loop
Answer
B. the area enclosed by the loop
Net work in a cycle is the area enclosed by the loop.
A Carnot engine works between 600 K and 300 K. Its efficiency is
1. 100%
2. 200%
3. 50%
4. 25%
Answer
C. 50%
η = 1 − T₂/T₁ = 1 − 300/600 = 0.5.
A Carnot engine has its source at 127 °C and its sink at 27 °C. Its efficiency is
1. 25%
2. 78.7%
3. 75%
4. 21%
Answer
A. 25%
T₁ = 400 K, T₂ = 300 K. η = 1 − 300/400 = 0.25. Using Celsius would be wrong.
A refrigerator with a coefficient of performance of 5 removes 1000 J of heat from the cold space in each cycle. The work needed per cycle is
1. 800 J
2. 5000 J
3. 200 J
4. 1200 J
Answer
C. 200 J
α = Q₂/W, so W = 1000/5 = 200 J. The heat rejected is 1200 J.
When heat is supplied to a gas at constant volume, it is used entirely to
1. do work on the surroundings
2. keep the temperature constant
3. reduce the pressure
4. increase the internal energy
Answer
D. increase the internal energy
At constant volume W = 0, so Q = ΔU.
An ideal gas expands adiabatically and does 500 J of work. The change in its internal energy is
1. 0
2. −250 J
3. −500 J
4. +500 J
Answer
C. −500 J
Q = 0, so ΔU = −W = −500 J.
For an ideal gas undergoing a reversible adiabatic process, which relation is correct?
1. TV^(γ−1) = constant
2. T^γ V = constant
3. TV^γ = constant
4. PV = constant
Answer
A. TV^(γ−1) = constant
From PV^γ = constant and PV = nRT, TV^(γ−1) = constant.
A heat engine with an efficiency of 25% absorbs 800 J of heat per cycle. The heat rejected per cycle is
1. 400 J
2. 600 J
3. 200 J
4. 1000 J
Answer
B. 600 J
W = 0.25 × 800 = 200 J, so Q₂ = 800 − 200 = 600 J.
A diatomic gas (γ = 7/5) at 300 K is compressed adiabatically to 1/32 of its volume. The final temperature is
1. 600 K
2. 2400 K
3. 9600 K
4. 1200 K
Answer
D. 1200 K
T₂ = T₁(V₁/V₂)^(γ−1) = 300 × 32^(2/5) = 300 × 4 = 1200 K.
Given below are two statements: Statement I: The internal energy of an ideal gas depends only on its temperature. Statement II: In an isothermal process the internal energy of an ideal gas stays constant.
1. Both Statement I and Statement II are incorrect
2. Both Statement I and Statement II are correct
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are correct
Both are correct; the second follows from the first.
Given below are two statements: Statement I: Heat and work are state functions of a system. Statement II: Internal energy is a state function of a system.
1. Both Statement I and Statement II are correct
2. Statement I is incorrect but Statement II is correct
3. Both Statement I and Statement II are incorrect
4. Statement I is correct but Statement II is incorrect
Answer
B. Statement I is incorrect but Statement II is correct
Heat and work depend on the path. Only internal energy is a state function.
Given below are two statements: Statement I: In adiabatic compression the temperature of a gas rises. Statement II: No heat leaves the gas, so the work done on it raises its internal energy.
1. Both Statement I and Statement II are incorrect
2. Statement I is correct but Statement II is incorrect
3. Statement I is incorrect but Statement II is correct
4. Both Statement I and Statement II are correct
Answer
D. Both Statement I and Statement II are correct
Q = 0 and W is negative, so ΔU = −W is positive. Both are correct.
Given below are two statements: Statement I: For a gas, Cp is greater than Cv. Statement II: At constant pressure part of the supplied heat does work of expansion.
1. Both Statement I and Statement II are incorrect
2. Both Statement I and Statement II are correct
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are correct
Cp − Cv = R. The extra heat at constant pressure goes into expansion work.
Which of the following statements are correct? 1. The efficiency of every heat engine is less than 1. 2. The coefficient of performance of a refrigerator can be greater than 1. 3. The net work in a complete cycle is always zero.
1. 1 and 3 only
2. 2 and 3 only
3. 1 and 2 only
4. 1, 2 and 3
Answer
C. 1 and 2 only
Statements 1 and 2 are correct. In a cycle ΔU = 0, but net work equals net heat and is not zero in general.
Which of the following statements are correct? 1. At the same point on a P-V diagram the adiabatic curve is steeper than the isothermal curve. 2. The slope of an isothermal curve of an ideal gas is −P/V. 3. The slope of an adiabatic curve is −P/(γV).
1. 2 and 3 only
2. 1 and 3 only
3. 1, 2 and 3
4. 1 and 2 only
Answer
D. 1 and 2 only
The isothermal slope is −P/V and the adiabatic slope is −γP/V, which is steeper. Statement 3 is wrong.
Given below are two statements: Statement I: A Carnot engine with its sink at 0 °C can have an efficiency of 100%. Statement II: A Carnot engine can have an efficiency of 100% only if its sink is at 0 K.
1. Both Statement I and Statement II are correct
2. Statement I is incorrect but Statement II is correct
3. Both Statement I and Statement II are incorrect
4. Statement I is correct but Statement II is incorrect
Answer
B. Statement I is incorrect but Statement II is correct
η = 1 − T₂/T₁ with T₂ = 273 K is less than 1. Efficiency 1 requires T₂ = 0 K.
Given below are two statements: Statement I: A reversible process is quasi-static and has no dissipative effects such as friction. Statement II: All real processes are reversible.
1. Statement I is correct but Statement II is incorrect
2. Both Statement I and Statement II are correct
3. Both Statement I and Statement II are incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Statement I is correct but Statement II is incorrect
Real processes are irreversible. Statement I is correct.
Given below are two statements: Statement I: For an isothermal expansion of an ideal gas, W = nRT ln(V₂/V₁). Statement II: In an isothermal expansion of an ideal gas, the heat absorbed equals the work done.
1. Both Statement I and Statement II are incorrect
2. Statement I is correct but Statement II is incorrect
3. Both Statement I and Statement II are correct
4. Statement I is incorrect but Statement II is correct
Answer
C. Both Statement I and Statement II are correct
ΔU = 0, so Q = W = nRT ln(V₂/V₁). Both are correct.
Assertion (A): In an isothermal expansion of an ideal gas, the gas absorbs heat. Reason (R): The internal energy does not change, so the heat absorbed equals the work done.
1. Both A and R are true but R is not the correct explanation of A
2. A is true but R is false
3. Both A and R are true and R is the correct explanation of A
4. A is false but R is true
Answer
C. Both A and R are true and R is the correct explanation of A
ΔU = 0, so Q = W > 0 for expansion. R explains A.
Assertion (A): An ideal gas cools during adiabatic expansion. Reason (R): The gas does work at the expense of its internal energy.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Q = 0, so W = −ΔU. Work done by the gas lowers U, and so T. R explains A.
Assertion (A): The adiabatic curve is steeper than the isothermal curve on a P-V diagram. Reason (R): The slope of the adiabatic curve is γ times that of the isothermal curve through the same point, and γ > 1.
1. Both A and R are true but R is not the correct explanation of A
2. A is true but R is false
3. A is false but R is true
4. Both A and R are true and R is the correct explanation of A
Answer
D. Both A and R are true and R is the correct explanation of A
Slopes are −γP/V and −P/V. R explains A.
Assertion (A): The net work done by a gas in a cyclic process is always zero. Reason (R): Internal energy is a state function.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is false but R is true
4. A is true but R is false
Answer
C. A is false but R is true
A is false: net work equals the enclosed area. R is true but ΔU = 0 only gives Q = W.
Assertion (A): In a cyclic process the change in internal energy is zero. Reason (R): In a cyclic process the heat absorbed by the system is zero.
1. A is true but R is false
2. Both A and R are true and R is the correct explanation of A
3. Both A and R are true but R is not the correct explanation of A
4. A is false but R is true
Answer
A. A is true but R is false
A is true as U is a state function. R is false: Q = W, which is not zero in general.
Match List I with List II. List I: a. Isothermal b. Adiabatic c. Isochoric d. Isobaric List II: i. Work done is zero ii. Heat exchanged is zero iii. Change in internal energy of an ideal gas is zero iv. Pressure is constant
1. a-iii, b-ii, c-i, d-iv
2. a-iii, b-i, c-ii, d-iv
3. a-ii, b-iii, c-i, d-iv
4. a-iii, b-ii, c-iv, d-i
Answer
A. a-iii, b-ii, c-i, d-iv
Isothermal: ΔU = 0 (iii). Adiabatic: Q = 0 (ii). Isochoric: W = 0 (i). Isobaric: P constant (iv).
Match List I with List II. List I: a. Mayer's relation b. Carnot efficiency c. COP of a refrigerator d. Isothermal work List II: i. nRT ln(V₂/V₁) ii. Cp − Cv = R iii. 1 − T₂/T₁ iv. Q₂/W
1. a-ii, b-iv, c-iii, d-i
2. a-ii, b-iii, c-i, d-iv
3. a-iii, b-ii, c-iv, d-i
4. a-ii, b-iii, c-iv, d-i
Answer
D. a-ii, b-iii, c-iv, d-i
a: Cp − Cv = R (ii). b: 1 − T₂/T₁ (iii). c: Q₂/W (iv). d: nRT ln(V₂/V₁) (i).
Match List I with List II. List I: a. Zeroth law b. First law c. Kelvin-Planck statement d. Clausius statement List II: i. Conservation of energy ii. Basis of the concept of temperature iii. Heat from a single reservoir cannot be fully converted into work in a cycle iv. Heat cannot flow by itself from cold to hot
1. a-ii, b-iii, c-i, d-iv
2. a-ii, b-i, c-iii, d-iv
3. a-i, b-ii, c-iii, d-iv
4. a-ii, b-i, c-iv, d-iii
Answer
B. a-ii, b-i, c-iii, d-iv
Zeroth: temperature (ii). First: energy conservation (i). Kelvin-Planck: (iii). Clausius: (iv).
Match List I with List II. List I: a. Isochoric heat for n moles b. Isobaric heat for n moles c. Isothermal heat for n moles d. Adiabatic heat List II: i. nCpΔT ii. 0 iii. nCvΔT iv. nRT ln(V₂/V₁)
1. a-iii, b-i, c-iv, d-ii
2. a-i, b-iii, c-iv, d-ii
3. a-iii, b-iv, c-i, d-ii
4. a-iii, b-i, c-ii, d-iv
Answer
A. a-iii, b-i, c-iv, d-ii
Isochoric Q = nCvΔT (iii); isobaric Q = nCpΔT (i); isothermal Q = nRT ln(V₂/V₁) (iv); adiabatic Q = 0 (ii).
Match List I with List II. List I: a. Heat supplied to the system b. Heat given out by the system c. Work done by the system d. Work done on the system List II: i. W is negative ii. Q is negative iii. W is positive iv. Q is positive
1. a-iv, b-iii, c-ii, d-i
2. a-ii, b-iv, c-iii, d-i
3. a-iv, b-ii, c-i, d-iii
4. a-iv, b-ii, c-iii, d-i
Answer
D. a-iv, b-ii, c-iii, d-i
With ΔU = Q − W: heat supplied Q > 0 (iv); heat given out Q < 0 (ii); work by system W > 0 (iii); work on system W < 0 (i).
Match List I with List II. List I: a. Heat engine b. Refrigerator c. Heat pump d. Carnot cycle List II: i. Removes heat from a cold space to cool it ii. Absorbs heat from a hot source and converts part of it to work iii. Delivers heat to a warm space iv. Two isothermal and two adiabatic steps
1. a-i, b-ii, c-iii, d-iv
2. a-ii, b-iii, c-i, d-iv
3. a-ii, b-i, c-iii, d-iv
4. a-ii, b-i, c-iv, d-iii
Answer
C. a-ii, b-i, c-iii, d-iv
Engine: ii. Refrigerator: i. Heat pump: iii. Carnot cycle: iv.
A gas at a constant pressure of 1 × 10⁵ Pa absorbs 500 J of heat and expands from 2 × 10⁻³ m³ to 4 × 10⁻³ m³. The change in its internal energy is
1. 700 J
2. 300 J
3. 500 J
4. 200 J
Answer
B. 300 J
W = PΔV = 1 × 10⁵ × 2 × 10⁻³ = 200 J. ΔU = 500 − 200 = 300 J.
One mole of a gas has Cv = 12.5 J mol⁻¹ K⁻¹. Take R = 8.3 J mol⁻¹ K⁻¹. The heat needed to raise its temperature by 50 K at constant pressure is
1. 625 J
2. 1455 J
3. 415 J
4. 1040 J
Answer
D. 1040 J
Cp = Cv + R = 20.8. Q = 1 × 20.8 × 50 = 1040 J. (Check: ΔU = 625 J plus W = RΔT = 415 J.)
A monatomic ideal gas (γ = 5/3) expands adiabatically from a state with PV = 600 J to a state with PV = 300 J. The work done by the gas is
1. 450 J
2. 225 J
3. 900 J
4. 300 J
Answer
A. 450 J
W = (P₁V₁ − P₂V₂)/(γ − 1) = 300/(2/3) = 450 J. Check: nR ΔT = 300 J and W = nCvΔT with Cv = 3R/2 gives 450 J.
A heat engine of efficiency 30% does 600 J of work per cycle. The heat it rejects to the sink per cycle is
1. 2000 J
2. 1800 J
3. 1400 J
4. 600 J
Answer
C. 1400 J
Q₁ = W/η = 600/0.3 = 2000 J. Q₂ = 2000 − 600 = 1400 J.
A Carnot engine with its sink at 300 K has an efficiency of 40%. To raise the efficiency to 50% with the same sink, the source temperature must be raised by
1. 200 K
2. 50 K
3. 150 K
4. 100 K
Answer
D. 100 K
η = 0.4 gives T₁ = 300/0.6 = 500 K. η = 0.5 gives T₁ = 600 K. The rise is 100 K.
A Carnot refrigerator works between −3 °C and 27 °C. Its coefficient of performance is
1. 9
2. 10
3. 0.9
4. 1/9
Answer
A. 9
T₂ = 270 K, T₁ = 300 K. α = T₂/(T₁ − T₂) = 270/30 = 9.
An inventor claims an engine working between 400 K and 300 K with an efficiency of 30%. This claim is
1. possible, because the maximum efficiency is 75%
2. not possible, because the maximum efficiency is 25%
3. possible, because it is below 100%
4. not possible, because the maximum efficiency is 30% only for 700 K
Answer
B. not possible, because the maximum efficiency is 25%
A Carnot engine gives the maximum: 1 − 300/400 = 25%. No engine can exceed it.

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अध्याय पढ़ने के बाद अभ्यास करें।

1.Which of the following is NOT a state function or state variable in a thermodynamic system?
2.Which thermodynamic quantity helps determine the spontaneity of a chemical reaction?
3.What is the function of a diathermic wall in a thermodynamic system?
4.In which system of thermodynamics, there is no exchange of energy or matter between the system and the surroundings?
5.Which of the following statements is correct about the open system of thermodynamics?
6._______ is the SI unit used for measuring thermodynamic temperature.
7.In which year did French physicist Sadi Carnot create the Carnot engine, which uses thermodynamic principles to function as an ideal heat engine?

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