This MCQ module is based on: NCERT Exercises and Solutions: Thermodynamics
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NCERT Exercises and Solutions: Thermodynamics
🎓 Class 11
Physics
CBSE
Theory
Ch 11 – Thermodynamics
⏱ ~8 min
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NCERT Exercises and Solutions: Thermodynamics
Chapter 11 — Quick Summary
Zeroth Law
Two systems each in thermal equilibrium with a third are in thermal equilibrium with each other. Defines temperature.
First Law
\(\Delta U = Q - W\). Q > 0 = heat into system; W > 0 = work BY system. Internal energy U is a state function; Q and W are path functions.
Specific Heats of an Ideal Gas
\(C_p - C_v = R\) (Mayer). \(\gamma = C_p/C_v\). Monatomic: 5/3. Diatomic: 7/5. Polyatomic: ≈ 4/3.
Standard Processes
Isothermal: ΔU = 0, Q = W = µRT ln(V₂/V₁). Adiabatic: Q = 0, PV^γ = const, W = (P₁V₁ − P₂V₂)/(γ−1). Isobaric: W = PΔV. Isochoric: W = 0. Free expansion: Q = W = ΔU = 0.
Heat Engines & Carnot
η = W/Q_h = 1 − Q_c/Q_h. Carnot: η_max = 1 − T_c/T_h. No engine between two reservoirs is more efficient than a Carnot engine.
Refrigerator
COP = Q_c/W. Carnot COP = T_c/(T_h − T_c).
Second Law
(Kelvin–Planck) No heat engine can convert heat fully into work in a cycle. (Clausius) No process can transfer heat from cold to hot as its sole result. Equivalent to ΔS_universe ≥ 0.
NCERT Exercise Solutions
Q 11.1: Geyser uses LPG (NCERT)
A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of fuel if its heat of combustion is 4.0 × 10⁴ J/g? (Take s_w = 4186 J/kg·K, density 1000 kg/m³.)
Mass flow = 3 kg/min. ΔT = 50 K.
Power needed: \(\dot Q = (3/60)\times 4186\times 50 = 10465\) W = 10.47 kJ/s.
Fuel rate = 10.47 × 10³ / (4 × 10⁴) = 0.2616 g/s = 15.7 g/min.
Power needed: \(\dot Q = (3/60)\times 4186\times 50 = 10465\) W = 10.47 kJ/s.
Fuel rate = 10.47 × 10³ / (4 × 10⁴) = 0.2616 g/s = 15.7 g/min.
Q 11.2: Chemical equation for first law
What amount of heat must be supplied to 2.0 × 10⁻² kg of nitrogen (diatomic, M = 28 g/mol) at room temperature to raise its temperature by 45 °C at constant pressure? R = 8.314 J/mol·K.
\(\mu = 0.020/0.028 = 0.714\) mol; C_p = 7R/2 = 29.1 J/mol·K.
Q = µC_p ΔT = 0.714 × 29.1 × 45 = 934 J.
Q = µC_p ΔT = 0.714 × 29.1 × 45 = 934 J.
Q 11.3: Internal energy increase
Explain why (a) two bodies at different temperatures T₁ and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁+T₂)/2. (b) The coolant in a chemical or nuclear plant should have high specific heat. (c) Air pressure in a car tyre increases during driving. (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
(a) The final temperature depends on m·s of each body, not just on T's: T_f = (m₁s₁T₁ + m₂s₂T₂)/(m₁s₁ + m₂s₂). Only when m₁s₁ = m₂s₂ does it equal the mean.
(b) High s lets the coolant absorb large quantities of heat without much temperature rise — efficient and safe.
(c) Friction with the road heats the tyre, increasing the gas pressure (P/T = const at fixed V).
(d) The sea has a very high heat capacity, so it cools day-night swings; a desert (sand, low s) heats and cools rapidly.
(b) High s lets the coolant absorb large quantities of heat without much temperature rise — efficient and safe.
(c) Friction with the road heats the tyre, increasing the gas pressure (P/T = const at fixed V).
(d) The sea has a very high heat capacity, so it cools day-night swings; a desert (sand, low s) heats and cools rapidly.
Q 11.4: Cylinder with gas
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume? γ(H₂) = 1.40.
Adiabatic (insulated): \(P_1V_1^\gamma = P_2V_2^\gamma\), so \(P_2/P_1 = (V_1/V_2)^\gamma = 2^{1.4} = \boxed{2.639}\).
Q 11.5: Cyclic process W and Q
In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? Take 1 cal = 4.19 J.
Adiabatic (Q=0): ΔU = −W_on = +22.3 J (since work is done on the gas, W_by = −22.3 J, so ΔU = Q − W_by = 0 − (−22.3) = 22.3 J).
For the second process: Q = 9.35 × 4.19 = 39.18 J.
ΔU is the same (state function) = 22.3 J. So W_by = Q − ΔU = 39.18 − 22.3 = 16.88 J.
For the second process: Q = 9.35 × 4.19 = 39.18 J.
ΔU is the same (state function) = 22.3 J. So W_by = Q − ΔU = 39.18 − 22.3 = 16.88 J.
Q 11.6: P–V cycle and net work
Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following: (a) What is the final pressure of the gas in A and B? (b) What is the change in internal energy of the gas? (c) What is the change in temperature of the gas? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P–V–T surface?
This is a free expansion into vacuum. Q = 0 (insulated), W = 0 (P_ext = 0).
(a) Volume doubles, so by P_1V_1 = P_2V_2 (T const for ideal gas), P_final = P_initial/2 = 0.5 atm.
(b) ΔU = Q − W = 0.
(c) For ideal gas U=U(T), ΔU = 0 ⇒ ΔT = 0.
(d) No — the intermediate states are not equilibrium states (P is not even uniform), so they cannot be plotted on the P–V–T surface. Free expansion is not quasi-static.
(a) Volume doubles, so by P_1V_1 = P_2V_2 (T const for ideal gas), P_final = P_initial/2 = 0.5 atm.
(b) ΔU = Q − W = 0.
(c) For ideal gas U=U(T), ΔU = 0 ⇒ ΔT = 0.
(d) No — the intermediate states are not equilibrium states (P is not even uniform), so they cannot be plotted on the P–V–T surface. Free expansion is not quasi-static.
Q 11.7: Heat engine refrigerator
A steam engine delivers 5.4 × 10⁸ J of work per minute and services 3.6 × 10⁹ J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
η = W/Q_h = 5.4 × 10⁸ / 3.6 × 10⁹ = 0.15 = 15 %.
Heat wasted = Q_h − W = 3.6 × 10⁹ − 5.4 × 10⁸ = 3.06 × 10⁹ J/min.
Heat wasted = Q_h − W = 3.6 × 10⁹ − 5.4 × 10⁸ = 3.06 × 10⁹ J/min.
Q 11.8: Internal energy by mixing
An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 J/s, at what rate is the internal energy increasing?
dU/dt = dQ/dt − dW/dt = 100 − 75 = 25 J/s (i.e., 25 W).
Q 11.9: P-V diagram with two states
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Figure where the volume changes from 2 L to 5 L while pressure is 5 × 10⁵ Pa, then from 5 L back to 2 L isobarically at 1 × 10⁵ Pa, then back to original state isochorically. Calculate the total work done by the gas from D → E → F.
For NCERT-style P–V triangle: D (5 L, 5×10⁵), E (2 L, 5×10⁵) was original; in the question segments are linear/isobaric/isochoric. The work done is the area enclosed.
Approximation: Triangular area = ½ × base × height = ½ × (5−2)×10⁻³ × (5−1)×10⁵ = ½ × 3 × 10⁻³ × 4 × 10⁵ = 600 J.
(Sign depends on direction of cycle.)
Approximation: Triangular area = ½ × base × height = ½ × (5−2)×10⁻³ × (5−1)×10⁵ = ½ × 3 × 10⁻³ × 4 × 10⁵ = 600 J.
(Sign depends on direction of cycle.)
Q 11.10: Refrigerator COP (NCERT 11.10)
A refrigerator is to maintain eatables kept inside at 9 °C. If room temperature is 36 °C, calculate the coefficient of performance.
T_c = 282 K, T_h = 309 K.
COP = T_c/(T_h − T_c) = 282/(309 − 282) = 282/27 = 10.44.
This is the Carnot (ideal) COP — actual fridges achieve ~ 30 % of this.
COP = T_c/(T_h − T_c) = 282/(309 − 282) = 282/27 = 10.44.
This is the Carnot (ideal) COP — actual fridges achieve ~ 30 % of this.
Additional Practice Problems
Practice 1: Adiabatic vs isothermal work
1 mole of an ideal monatomic gas (γ = 5/3) at 300 K and 1 atm expands to 2 atm by (a) isothermal compression and (b) adiabatic compression. Compute final T, V and work in each case.
Initial V₁ = µRT/P = 1 × 8.314 × 300/(10⁵) = 0.02494 m³.
(a) Isothermal: V₂ = V₁/2 = 0.01247 m³; W = µRT ln(V₂/V₁) = 1×8.314×300×ln(0.5) = −1729 J.
(b) Adiabatic to P₂ = 2P₁: T₂ = T₁(P₂/P₁)^((γ−1)/γ) = 300 × 2^0.4 = 300 × 1.32 = 396 K. V₂ = µRT₂/P₂ = 1×8.314×396/(2×10⁵) = 0.01646 m³. W_by = µR(T₁−T₂)/(γ−1) = 1×8.314×(300−396)/0.667 = −1198 J. (Negative = work on gas.)
(a) Isothermal: V₂ = V₁/2 = 0.01247 m³; W = µRT ln(V₂/V₁) = 1×8.314×300×ln(0.5) = −1729 J.
(b) Adiabatic to P₂ = 2P₁: T₂ = T₁(P₂/P₁)^((γ−1)/γ) = 300 × 2^0.4 = 300 × 1.32 = 396 K. V₂ = µRT₂/P₂ = 1×8.314×396/(2×10⁵) = 0.01646 m³. W_by = µR(T₁−T₂)/(γ−1) = 1×8.314×(300−396)/0.667 = −1198 J. (Negative = work on gas.)
Practice 2: Carnot heat pump for home heating
A Carnot heat pump warms a house at 27 °C by extracting heat from the outdoors at −3 °C. To deliver 10 kJ to the house, how much electrical work is needed? COP_HP = T_h/(T_h − T_c).
T_h = 300 K, T_c = 270 K. COP_HP = 300/30 = 10.
W = Q_h/COP = 10000/10 = 1000 J = 1.0 kJ. Heat pumps deliver ~ 4–5 × the electric energy as heat — much more efficient than direct resistive heating.
W = Q_h/COP = 10000/10 = 1000 J = 1.0 kJ. Heat pumps deliver ~ 4–5 × the electric energy as heat — much more efficient than direct resistive heating.
Practice 3: Cycle with isobaric, isochoric and adiabatic legs
An ideal gas (γ=1.4) at A: P₀=10⁵ Pa, V₀=2 L, T₀=300 K. (i) A→B isobaric expansion to V_B=4 L. (ii) B→C isochoric cooling so that the gas returns to T₀. (iii) C→A isothermal compression. Compute the heat absorbed and rejected over a cycle.
Compute moles: µ = P₀V₀/RT₀ = 10⁵·2·10⁻³/(8.314·300) = 0.0802 mol. C_v = 5R/2 = 20.79; C_p = 7R/2 = 29.1.
A→B: T_B = T_A·V_B/V_A = 600 K. Q = µC_pΔT = 0.0802·29.1·300 = 700 J. W = PΔV = 10⁵·2·10⁻³ = 200 J.
B→C: V const, T 600→300 K. Q = µC_vΔT = 0.0802·20.79·(−300) = −500 J. W = 0.
C→A: isothermal compression V 4→2 L. W = µRT ln(V_A/V_C) = 0.0802·8.314·300·ln(0.5) = −138.6 J. Q = W (isothermal, ideal gas).
Net W = 200 + 0 − 138.6 = 61.4 J. Q_in (positive) = 700 J; Q_out = 500 + 138.6 = 638.6 J. Check: W = Q_in − Q_out = 61.4 J ✓.
A→B: T_B = T_A·V_B/V_A = 600 K. Q = µC_pΔT = 0.0802·29.1·300 = 700 J. W = PΔV = 10⁵·2·10⁻³ = 200 J.
B→C: V const, T 600→300 K. Q = µC_vΔT = 0.0802·20.79·(−300) = −500 J. W = 0.
C→A: isothermal compression V 4→2 L. W = µRT ln(V_A/V_C) = 0.0802·8.314·300·ln(0.5) = −138.6 J. Q = W (isothermal, ideal gas).
Net W = 200 + 0 − 138.6 = 61.4 J. Q_in (positive) = 700 J; Q_out = 500 + 138.6 = 638.6 J. Check: W = Q_in − Q_out = 61.4 J ✓.
Activity — Match Each Process to Its First-Law Identity
L4 Analyse
Predict: Without looking, can you write down what each process implies about Q, W and ΔU?
- Take a sheet of paper. Make four columns: Process | Q | W | ΔU.
- Fill in for: Isobaric, Isochoric, Isothermal, Adiabatic, Free expansion, Cyclic.
- Use the relations: ΔU = Q − W; for ideal gas U=U(T); PV^γ = const adiabatically.
| Process | Q | W | ΔU |
|---|---|---|---|
| Isobaric | µC_pΔT | PΔV = µRΔT | µC_vΔT |
| Isochoric | µC_vΔT | 0 | µC_vΔT |
| Isothermal | = W | µRT ln(V₂/V₁) | 0 |
| Adiabatic | 0 | (P₁V₁−P₂V₂)/(γ−1) | −W |
| Free expansion | 0 | 0 | 0 |
| Cyclic | = W | area enclosed | 0 |
Interactive: Universal Engine/Refrigerator Solver L3 Apply
Enter any two of (η, T_h, T_c) and the heat input. The applet computes the rest assuming Carnot performance.
Competency-Based Practice
A nuclear power plant uses a uranium reactor whose hot-side temperature is 530 °C and a cold side at 30 °C (cooling-tower water). The reactor delivers 2.5 GW of thermal heat. Operating efficiency 33 %.
Q1. L1 Remember Define the coefficient of performance of a refrigerator.
COP = (Heat extracted from cold reservoir) / (Work input). For Carnot refrigerator: T_c/(T_h − T_c).
Q2. L3 Apply Compute the maximum (Carnot) efficiency.
T_h = 803 K, T_c = 303 K. η = 1 − 303/803 = 62.3 %.
Q3. L3 Apply Find the actual electrical power delivered.
P_elec = 0.33 × 2.5 = 0.825 GW = 825 MW.
Q4. L4 Analyse Compute the rate at which heat is dumped into the cooling tower.
Q_c rate = Q_h − W = 2.5 − 0.825 = 1.675 GW — almost 2× the useful electrical output, which is why nuclear plants need huge cooling water flows.
Q5. L5 Evaluate Why is the practical efficiency (33 %) much less than the Carnot value (62 %)?
Real cycles (Rankine, Brayton) include irreversibilities — friction in turbines, finite-temperature heat exchangers, leakage, partial regeneration. Carnot is the ideal limit; achievable efficiencies are typically 50–60 % of it. Modern combined-cycle designs push closer to the limit.
Assertion-Reason Practice
Assertion (A): The work done in a cyclic process equals the area enclosed by the cycle on a P–V diagram.
Reason (R): For a cycle, ΔU = 0, so the net work equals the net heat absorbed.
Answer: B. Both true. But A is a geometric fact about the integral ∮P dV; R says only that net W = net Q. They are independent — both correct, R is not the explanation.
Assertion (A): A Carnot engine using helium has higher efficiency than one using nitrogen between the same two reservoirs.
Reason (R): Helium has a larger γ, so its adiabatic curve is steeper.
Answer: D. A is false: Carnot's theorem states all reversible engines between the same temperatures have the same efficiency η = 1−T_c/T_h, independent of working substance. R is true on its own.
Assertion (A): Reversible processes are an idealisation; no real process is fully reversible.
Reason (R): All real processes involve some friction, finite-temperature heat exchange or other dissipative effects, generating entropy.
Answer: A. Real systems always have irreversibilities; ΔS_universe > 0 for every actual process.
Frequently Asked Questions - NCERT Exercises and Solutions: Thermodynamics
What are the key NCERT exercise types in Chapter 11 Thermodynamics?
NCERT Class 11 Physics Chapter 11 Thermodynamics exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question in the chapter, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Thermodynamics?
For numerical problems in NCERT Class 11 Physics Chapter 11 Thermodynamics: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 11?
From NCERT Class 11 Physics Chapter 11 (Thermodynamics), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 11 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula (M^a L^b T^c). For Chapter 11 Thermodynamics problems, write the dimensions of each quantity, substitute, and simplify. If dimensions match the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 11 exercises?
Common mistakes in NCERT Class 11 Physics Chapter 11 Thermodynamics exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for vectors, (3) forgetting to consider all forces in free-body diagrams, (4) algebra errors during derivations, and (5) misreading the problem statement. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 11 Physics Chapter 11 Thermodynamics solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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