Thermodynamics — Class 11 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Physics chapter "Thermodynamics" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Thermodynamics" — 8 important questions with detailed answers for CBSE board exam prep…
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NCERT Solutions for Class 11 Physics — Thermodynamics. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Thermodynamics — All 8 Questions Solved
Q1. An ideal gas undergoes isothermal expansion from volume 1 L to 2 L at 300 K against a constant external pressure of 2 atm. Calculate the work done by the gas. (1 L·atm = 101.3 J)
Step 1: For isothermal expansion against constant external pressure:
W = P_ext × ΔV
Step 2: Calculate change in volume:
ΔV = V_f - V_i = 2 - 1 = 1 L
Step 3: Calculate work:
W = 2 atm × 1 L = 2 L·atm
Step 4: Convert to Joules:
W = 2 × 101.3 = 202.6 J
Alternatively, using W = nRT ln(V_f/V_i):
For isothermal process: PV = constant
P₁V₁ = 2 atm × 1 L = 2 atm·L = 202.6 J
nRT = 202.6 J
W = 202.6 × ln(2) = 202.6 × 0.693 = 140.4 J
(Note: 2 atm is likely external pressure for isobaric, giving W = 202.6 J)
Final Answer: Work done = 202.6 J (isobaric) or 140.4 J (isothermal reversible)
Q2. Calculate the work done on an ideal gas when it is compressed from 4 L to 1 L at a constant pressure of 2 atm. (1 L·atm = 101.3 J)
Step 1: For compression at constant pressure (isobaric process):
W = P × ΔV = P × (V_f - V_i)
Step 2: Since volume decreases (compression), ΔV is negative:
ΔV = V_f - V_i = 1 - 4 = -3 L
Step 3: Calculate work done ON the gas:
W = 2 atm × (-3 L) = -6 L·atm
Step 4: Negative sign indicates work done ON the gas:
W_on = -(-6) = 6 L·atm = 6 × 101.3 = 607.8 J
Or: Work done by gas = -607.8 J
Work done on gas = +607.8 J
Final Answer: Work done on gas = 607.8 J or 6.08 kJ
Q3. One mole of an ideal gas at standard temperature and pressure (STP) undergoes adiabatic expansion to 10 times its original volume. If γ = 1.4, calculate the final temperature.
Step 1: For adiabatic process: TV^(γ-1) = constant
T₁V₁^(γ-1) = T₂V₂^(γ-1)
Step 2: Rearrange for T₂:
T₂ = T₁ × (V₁/V₂)^(γ-1)
Step 3: Substitute values:
T₁ = 273 K (at STP)
V₂/V₁ = 10, so V₁/V₂ = 0.1
γ = 1.4, so γ - 1 = 0.4
T₂ = 273 × (0.1)^0.4
Step 4: Calculate (0.1)^0.4:
(0.1)^0.4 = (10⁻¹)^0.4 = 10^(-0.4) = 1/10^0.4
10^0.4 = 10^(2/5) = (10²)^(1/5) = 100^0.2 ≈ 2.512
(0.1)^0.4 = 1/2.512 = 0.398
T₂ = 273 × 0.398 = 108.6 K
Final Answer: Final temperature = 108.6 K or -164.4°C
Q4. Calculate the change in internal energy of 2 moles of an ideal gas when its temperature increases from 300 K to 400 K. Given: C_V = (5/2)R, where R = 8.314 J/mol·K.
Step 1: Change in internal energy for ideal gas:
ΔU = n × C_V × ΔT
where n = number of moles, C_V = heat capacity at constant volume
Step 2: Calculate ΔT:
ΔT = T_f - T_i = 400 - 300 = 100 K
Step 3: Calculate C_V:
C_V = (5/2) × 8.314 = 2.5 × 8.314 = 20.785 J/mol·K
Step 4: Calculate ΔU:
ΔU = 2 × 20.785 × 100
ΔU = 4157 J = 4.157 kJ
Final Answer: Change in internal energy ΔU = 4157 J or 4.16 kJ
Q5. In a cyclic process, 500 J of heat is absorbed by the gas and 200 J of work is done by the gas. Calculate the change in internal energy.
Step 1: Apply first law of thermodynamics:
ΔU = Q - W
where Q = heat absorbed by gas, W = work done by gas, ΔU = change in internal energy
Step 2: For a cyclic process, internal energy returns to initial state:
ΔU = 0 (for complete cycle)
But for the given process portion:
ΔU = Q - W = 500 - 200 = 300 J
Step 3: Verification:
Heat absorbed = 500 J (positive)
Work done by gas = 200 J (positive)
Change in internal energy = 300 J
Note: If this is part of complete cycle, eventually ΔU_total = 0 and Q_total = W_total
Final Answer: Change in internal energy ΔU = 300 J
Q6. A heat engine absorbs 1000 J of heat from a hot reservoir and rejects 600 J to a cold reservoir in each cycle. Calculate (a) the work done by the engine, (b) the efficiency of the engine.
Step 1: Work done by engine:
W = Q_H - Q_C
where Q_H = heat absorbed from hot reservoir, Q_C = heat rejected to cold reservoir
W = 1000 - 600 = 400 J
Step 2: Efficiency of engine:
η = W/Q_H = (Q_H - Q_C)/Q_H = 1 - Q_C/Q_H
η = 400/1000 = 0.4 = 40%
Or: η = 1 - 600/1000 = 1 - 0.6 = 0.4 = 40%
Final Answer: (a) Work done = 400 J; (b) Efficiency = 40% or 0.4
Q7. An ideal gas undergoes an isobaric (constant pressure) process where 100 J of heat is supplied. The gas expands and does 40 J of work. Calculate the change in internal energy.
Step 1: Apply first law of thermodynamics:
ΔU = Q - W
where Q = heat supplied to gas (positive), W = work done by gas (positive)
Step 2: Substitute values:
Q = 100 J (heat supplied)
W = 40 J (work done by gas during expansion)
ΔU = 100 - 40 = 60 J
Step 3: Interpretation:
Of the 100 J heat supplied, 40 J goes into doing work and 60 J goes into increasing internal energy (raising temperature).
Final Answer: Change in internal energy ΔU = 60 J
Q8. A Carnot engine operates between two reservoirs at temperatures 400 K and 300 K. It absorbs 2000 J of heat per cycle. Calculate (a) the maximum possible efficiency, (b) the heat rejected per cycle, (c) the work done per cycle.
Step 1: Carnot efficiency:
η_max = 1 - T_C/T_H
where T_H = hot reservoir, T_C = cold reservoir (in Kelvin)
η_max = 1 - 300/400 = 1 - 0.75 = 0.25 = 25%
Step 2: Heat rejected per cycle:
η = W/Q_H = (Q_H - Q_C)/Q_H
0.25 = (2000 - Q_C)/2000
0.25 × 2000 = 2000 - Q_C
500 = 2000 - Q_C
Q_C = 1500 J
Or: Q_C/Q_H = T_C/T_H
Q_C = Q_H × T_C/T_H = 2000 × 300/400 = 1500 J
Step 3: Work done per cycle:
W = Q_H - Q_C = 2000 - 1500 = 500 J
Or: W = η × Q_H = 0.25 × 2000 = 500 J
Final Answer: (a) Maximum efficiency = 25%; (b) Heat rejected = 1500 J; (c) Work done = 500 J
Thermodynamics — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. The Zeroth Law of Thermodynamics defines:
- A. Internal energy
- B. Work
- C. Heat
- D. Temperature
Answer: D. Temperature — The Zeroth Law states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other, which leads to the concept of temperature.
MCQ 2. According to the First Law of Thermodynamics, for an isolated system, the change in internal energy is:
- A. Positive
- B. Negative
- C. Zero
- D. Depends on the process
Answer: C. Zero — For an isolated system, no heat can enter or leave (Q=0) and no work can be done by or on the system (W=0), thus the change in internal energy is zero.
MCQ 3. An ideal gas undergoes an isothermal expansion. Which of the following is true?
- A. Its internal energy increases.
- B. Its internal energy decreases.
- C. No heat is exchanged with the surroundings.
- D. The work done by the gas is equal to the heat absorbed by it.
Answer: D. The work done by the gas is equal to the heat absorbed by it. — In an isothermal process, the temperature remains constant, so the change in internal energy (ΔU) for an ideal gas is zero. According to the First Law (ΔU = Q - W), this implies Q = W.
MCQ 4. During an adiabatic expansion of an ideal gas, its temperature:
- A. Increases
- B. Decreases
- C. Remains constant
- D. May increase or decrease depending on the gas
Answer: B. Decreases — In an adiabatic expansion, the gas does work on the surroundings without heat exchange. This work is done at the expense of its internal energy, leading to a decrease in temperature.
MCQ 5. The work done by an ideal gas during an isochoric process is:
- A. PΔV
- B. -PΔV
- C. nRT ln(V2/V1)
- D. Zero
Answer: D. Zero — In an isochoric process, the volume of the gas remains constant, meaning there is no displacement and thus no work done by or on the gas.
MCQ 6. 5 moles of an ideal gas expand isobarically from an initial volume of 0.1 m³ to 0.3 m³ at a constant pressure of 2 x 10⁵ Pa. The work done by the gas is:
- A. 4 x 10⁵ J
- B. 2 x 10⁵ J
- C. 4 x 10⁴ J
- D. 2 x 10⁴ J
Answer: C. 4 x 10⁴ J — In an isobaric process, work done is given by W = PΔV. Substituting the given values yields the result.
MCQ 7. For an ideal gas, the relationship between its molar specific heat capacities at constant pressure (C_p) and constant volume (C_v) is given by:
- A. C_p / C_v = R
- B. C_p - C_v = R
- C. C_v - C_p = R
- D. C_p * C_v = R
Answer: B. C_p - C_v = R — This is Mayer's relation, a fundamental relationship for ideal gases relating the molar specific heats at constant pressure and constant volume with the universal gas constant R.
MCQ 8. The internal energy of an ideal gas depends solely on its:
- A. Pressure
- B. Volume
- C. Temperature
- D. Density
Answer: C. Temperature — For an ideal gas, the internal energy is purely kinetic and depends only on the average kinetic energy of its molecules, which is directly proportional to the absolute temperature.
MCQ 9. An ideal gas undergoes a cyclic process shown in a P-V diagram (clockwise loop). The net work done by the gas in one cycle is:
- A. Positive
- B. Negative
- C. Zero
- D. Depends on the specific path
Answer: A. Positive — In a P-V diagram, the area enclosed by a cyclic process represents the net work done. For a clockwise cycle, the expansion (work done by gas) is greater than the compression (work done on gas), resulting in positive net work.
MCQ 10. A Carnot engine operates between 327°C and 27°C. Its efficiency is:
- A. 25%
- B. 50%
- C. 75%
- D. 100%
Answer: B. 50% — The efficiency of a Carnot engine is given by η = 1 - (T_cold / T_hot). Temperatures must be in Kelvin.
MCQ 11. A refrigerator has a coefficient of performance (COP) of 5. If it rejects 300 J of heat to the surroundings per cycle, the work input per cycle is:
- A. 25 J
- B. 50 J
- C. 100 J
- D. 200 J
Answer: B. 50 J — For a refrigerator, COP = Q_L / W. Also, Q_H = Q_L + W. Given Q_H and COP, we can deduce Q_L = COP * W, and substitute it into the first equation to find W.
MCQ 12. The Second Law of Thermodynamics, according to the Kelvin-Planck statement, implies that:
- A. Heat can spontaneously flow from a colder to a hotter body.
- B. It is impossible to construct a heat engine that operates in a cycle and converts all the heat absorbed into work.
- C. The total entropy of an isolated system can decrease.
- D. The coefficient of performance of a refrigerator can be infinite.
Answer: B. It is impossible to construct a heat engine that operates in a cycle and converts all the heat absorbed into work. — The Kelvin-Planck statement of the Second Law of Thermodynamics states that it is impossible to design a heat engine that produces net work from a single heat reservoir, meaning not all absorbed heat can be converted into work.
MCQ 13. For a monoatomic ideal gas, the ratio of specific heats (γ = C_p / C_v) is:
- A. 1.67
- B. 1.40
- C. 1.33
- D. 1.00
Answer: A. 1.67 — For a monoatomic gas, the number of degrees of freedom (f) is 3 (translational). Using γ = 1 + 2/f, we get γ = 1 + 2/3 = 5/3 ≈ 1.67.
MCQ 14. An ideal monoatomic gas (γ = 5/3) is compressed adiabatically such that its volume becomes 1/8th of its initial volume. If the initial temperature was 27°C, the final temperature will be:
- A. 108 K
- B. 600 K
- C. 900 K
- D. 1200 K
Answer: D. 1200 K — For an adiabatic process, TV^(γ-1) = constant. Convert initial temperature to Kelvin, apply the formula using γ=5/3, and calculate the final temperature.
MCQ 15. Which of the following is a state function?
- A. Heat
- B. Work
- C. Internal energy
- D. Both A and B
Answer: C. Internal energy — A state function depends only on the initial and final states of the system, not on the path taken. Internal energy is a state function, while heat and work are path functions.
MCQ 16. A heat engine absorbs 600 J of heat from a source and performs 200 J of work. The heat rejected to the sink is:
- A. 800 J
- B. 600 J
- C. 400 J
- D. 200 J
Answer: C. 400 J — According to the First Law of Thermodynamics for a cyclic process in a heat engine, the heat absorbed from the source (Q_H) minus the heat rejected to the sink (Q_L) is equal to the work done (W). So, Q_H - Q_L = W.
MCQ 17. The work done on 1 mole of an ideal gas during an adiabatic compression from T₁ to T₂ is:
- A. R(T₂ - T₁) / (γ - 1)
- B. R(T₁ - T₂) / (γ - 1)
- C. C_v(T₂ - T₁)
- D. C_p(T₂ - T₁)
Answer: A. R(T₂ - T₁) / (γ - 1) — For an adiabatic process, Q=0, so ΔU = -W_by (work done by the gas). The work done ON the gas is W_on = -W_by = ΔU. For 1 mole of ideal gas, ΔU = C_v(T₂ - T₁), and C_v = R/(γ-1).
MCQ 18. A Carnot engine working between a source at 400 K and a sink at 300 K absorbs 1200 J of heat from the source. The amount of heat rejected to the sink is:
- A. 300 J
- B. 600 J
- C. 900 J
- D. 1200 J
Answer: C. 900 J — For a Carnot engine, Q_L/Q_H = T_L/T_H. Using this relation with the given values, we can find the heat rejected to the sink (Q_L).
MCQ 19. A mixture of 3 moles of oxygen (O₂, diatomic) and 1 mole of argon (Ar, monoatomic) is kept at temperature T. The total internal energy of the mixture is: (R is the universal gas constant)
- A. 4.5 RT
- B. 6 RT
- C. 7.5 RT
- D. 9 RT
Answer: D. 9 RT — The total internal energy of a gas mixture is the sum of the internal energies of its components. For an ideal gas, U = (f/2)nRT, where f is the degrees of freedom.
MCQ 20. An ideal gas goes from state A to state B by two different paths, Path 1 and Path 2, as shown in a P-V diagram. If Q₁ and W₁ are heat and work for Path 1, and Q₂ and W₂ for Path 2, then:
- A. Q₁ = Q₂
- B. W₁ = W₂
- C. Q₁ - W₁ = Q₂ - W₂
- D. Q₁ + W₁ = Q₂ + W₂
Answer: C. Q₁ - W₁ = Q₂ - W₂ — Internal energy is a state function, so the change in internal energy (ΔU) between A and B is the same for both paths. According to the First Law of Thermodynamics, ΔU = Q - W, so Q₁ - W₁ = Q₂ - W₂.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.