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CBSE · Class 11 · Physics

Thermodynamics

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
800 J of heat is given to a gas, and the gas does 600 J of work on its surroundings. The change in its internal energy is:
  1. −200 J
  2. 1400 J
  3. 800 J
  4. 200 J
2.
A gas expands at a constant pressure of 10 × 105 N/m2 from 4 L by 18 L more. The work done by the gas is:
  1. 18,000,000 J
  2. 18 J
  3. 18,000 J
  4. 180,000 J
3.
A heat engine takes in 4100 J from the hot source and does 600 J of work in each cycle. Its efficiency is:
  1. 683.33%
  2. 0.15%
  3. 85.37%
  4. 14.63%
4.
A Carnot engine works between 870 K and 250 K. Its efficiency is:
  1. 71.26%
  2. 28.74%
  3. 248%
  4. 0.71%
5.
The zeroth law of thermodynamics leads to the idea of:
  1. temperature
  2. internal energy
  3. work
  4. entropy
6.
A Carnot refrigerator keeps its inside at 270 K in a room at 310 K. Its coefficient of performance is:
  1. 7.75
  2. 6.75
  3. 0.15
  4. 0.87
7.
A gas goes round a rectangular cycle on a p–V diagram between 3 × 105 and 6 × 105 N/m2 and between 1 L and 5 L. The net work done in one cycle is:
  1. 1,200 J
  2. 0 J
  3. 3,000 J
  4. 2,400 J
8.
Assertion (A): A gas goes through a complete cycle and does 1000 J of work. The net heat it absorbs in the cycle is 1000 J.
Reason (R): In a cyclic process no heat is exchanged, because the gas returns to its initial state.
  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.
Section B — Short answer8 × 2 = 16 marks
9.
A system absorbs 1200 J of heat and does 1200 J of work. Find the change in its internal energy.
10.
4 mol of an ideal gas expands isothermally at 400 K to 2 times its volume. Find the work done by the gas. (R = 8.31 J/mol K, ln 2 = 0.693) (2 decimal places where needed.)
11.
A Carnot engine takes in 5,000 J per cycle from a source at 650 K and rejects heat to a sink at 580 K. Find its efficiency, the work done per cycle and the heat rejected. (2 decimal places where needed.)
12.
A refrigerator with coefficient of performance 5 uses 1700 J of electrical work. How much heat does it remove from the inside, and how much does it give out to the room?
13.
5 mol of a monatomic ideal gas is heated at constant volume through 80 K. Find the heat supplied, the work done and the change in internal energy. (R = 8.31 J/mol K) (2 decimal places where needed.)
14.
5 mol of a diatomic ideal gas is heated at constant pressure through 45 K. Find the work done by the gas, the change in internal energy and the heat supplied. (R = 8.31 J/mol K) (2 decimal places where needed.)
15.
In an adiabatic expansion, a gas goes from 8 × 105 N/m2, 1 L to 3 × 105 N/m2, 9 L. Taking γ = 7/5, find the work done by the gas. Does its temperature rise or fall? (2 decimal places where needed.)
16.
A gas is taken round a rectangular cycle: pressures 2 × 105 and 5 × 105 N/m2, volumes 1 L and 6 L, going clockwise on a p–V diagram. Find the net work done, the net heat absorbed and the change in internal energy per cycle.
Section C — Word problems4 × 4 = 16 marks
17.
A Carnot engine works between 820 K and 280 K and takes in 15,000 J per cycle. (a) Find its efficiency, the work done and the heat rejected per cycle. (b) To what temperature must the sink be lowered for an efficiency of 60%? (2 decimal places where needed.)
18.
2 mol of a monatomic ideal gas expands at constant pressure from 340 K to 690 K. Find (a) the work done by the gas, (b) the change in internal energy, (c) the heat supplied, (d) the ratio of work done to heat supplied. (R = 8.31 J/mol K) (2 decimal places where needed.)
19.
A Carnot refrigerator keeps food at 280 K in a kitchen at 290 K. It removes 16,000 J of heat from the food. Find (a) its coefficient of performance, (b) the work needed, (c) the heat given out to the kitchen. (2 decimal places where needed.)
20.
A gas is taken round a triangular cycle on a p–V diagram (V across, p up) with corners at (3 L, 5 × 105 Pa), (6 L, 6 × 105 Pa) and (6 L, 5 × 105 Pa), in that order and back (clockwise). Find the net work done by the gas in one cycle, the net heat absorbed and the change in internal energy.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 11 · Physics

Answer key

Thermodynamics · AW-11P-THD-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 200 JΔU = Q − W
2(c) 18,000 JW = pΔV, with 1 L = 10−3 m3
3(d) 14.63%η = W/Q1
4(a) 71.26%η = 1 − T2/T1
5(a) temperature—
6(b) 6.75β = T2/(T1 − T2)
7(a) 1,200 Jthe area enclosed: Δp × ΔV
8(c) A is true, but R is false.A is true: in a cycle ΔU = 0, so Q = W = 1000 J. R is false.
90 JΔU = Q − W
109,214.13 JW = nRT ln(V2/V1) = 4 × 8.31 × 400 × 1 × 0.693
1110.77%; 538.46 J; 4,461.54 Jη = 1 − T2/T1; W = ηQ1; Q2 = Q1 − W
128,500 J removed; 10,200 J given outQ2 = βW; Q1 = Q2 + W
13Q = ΔU = 4,986 J; W = 0Cv = (3/2)R; isochoric, so no work
14W = 1,869.75 J; ΔU = 4,674.38 J; Q = 6,544.13 JW = nRΔT; ΔU = nCvΔT; Q = nCpΔT
15−4,750 J; its temperature risesW = (p1V1 − p2V2)/(γ − 1)
16W = 1,500 J; Q = 1,500 J; ΔU = 0area of the rectangle; in a cycle ΔU = 0
17(a) 65.85%; 9,878.05 J; 5,121.95 J. (b) 328 Kη = 1 − T2/T1
18(a) 5,817 J; (b) 8,725.5 J; (c) 14,542.5 J; (d) 2/5W = nRΔT; ΔU = nCvΔT; Q = nCpΔT
19(a) 28; (b) 571.43 J; (c) 16,571.43 Jβ = T2/(T1 − T2); W = Q2/β; Q1 = Q2 + W
20Net work 150 J (the area of the triangle; positive because the cycle is clockwise); net heat absorbed 150 J; ΔU = 0area = ½ × base × height