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.

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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…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

An ideal gas undergoes isothermal expansion from volume 1 L to 2 L at 300 K a…
Calculate the work done on an ideal gas when it is compressed from 4 L to 1 L…
One mole of an ideal gas at standard temperature and pressure (STP) undergoes…
Calculate the change in internal energy of 2 moles of an ideal gas when its t…
In a cyclic process, 500 J of heat is absorbed by the gas and 200 J of work i…
A heat engine absorbs 1000 J of heat from a hot reservoir and rejects 600 J t…
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
An ideal gas undergoes isothermal expansion from volume 1…	✓ Solved
Calculate the work done on an ideal gas when it is compre…	✓ Solved
One mole of an ideal gas at standard temperature and pres…	✓ Solved
Calculate the change in internal energy of 2 moles of an …	✓ Solved
In a cyclic process, 500 J of heat is absorbed by the gas…	✓ Solved
A heat engine absorbs 1000 J of heat from a hot reservoir…	✓ Solved

Showing 6 of 8 questions

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 = 20...

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 a...

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

Showing 6 of 8 questions. Visit the full page for complete solutions.