How to Use This Page
Attempt each NCERT question on your own before reading the solution. For numericals, follow the formula-substitution-calculation-answer format carefully. Pay special attention to sign conventions: ΔQ positive when heat is supplied, ΔW positive when work is done by the system, and ΔU positive when internal energy increases.
Key Concepts (Quick Recap)
- First law: ΔQ = ΔU + ΔW
- Isothermal process: ΔT = 0, ΔU = 0, ΔQ = ΔW
- Isothermal work: W = nRT ln(V₂/V₁) = 2.303 nRT log₁₀(V₂/V₁)
- Adiabatic process: ΔQ = 0, ΔW = −ΔU
- Adiabatic relations: PV^γ = const, TV^(γ−1) = const, W = (P₁V₁ − P₂V₂)/(γ − 1)
- Isobaric work: W = PΔV = nRΔT
- Isochoric process: ΔV = 0, ΔW = 0, ΔQ = ΔU
- Carnot efficiency: η = 1 − T₂/T₁
- COP of refrigerator: COP = T₂/(T₁ − T₂)
- Second law (Kelvin-Planck): No process can convert heat entirely into work with no other effect
- Second law (Clausius): Heat cannot flow from cold to hot without external work
Intext Questions
Page 309
Q1. A gas is compressed isothermally to half its volume. How much work is done if the initial volume is 2 L at 1 atm?
W = nRT ln(V₂/V₁) = P₁V₁ ln(V₂/V₁) [since P₁V₁ = nRT for ideal gas]
V₂/V₁ = 1/2
W = 1 × 1.013 × 10⁵ × 2 × 10⁻³ × ln(0.5)
= 202.6 × (−0.693) = −140.4 J
Q2. In an adiabatic process, 1000 J of work is done on a system. What is the change in internal energy?
ΔQ = 0 (adiabatic), ΔW = −1000 J (work done on gas, so ΔW = −1000)
ΔQ = ΔU + ΔW → 0 = ΔU + (−1000) → ΔU = +1000 J
Page 314
Q3. Can a Carnot engine have efficiency of 1? What would that require?
Page 318
Q4. A refrigerator transfers heat from inside at 4°C to the room at 30°C. What is its COP?
T₂ = 4 + 273 = 277 K, T₁ = 30 + 273 = 303 K
COP = T₂/(T₁ − T₂) = 277/(303 − 277) = 277/26 = 10.65
Chapter-End Exercise
Q1. A geyser heats water flowing at 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 the fuel if its heat of combustion is 4.0 × 10⁴ J/g?
Mass of water per minute = 3.0 kg (1 L water ≈ 1 kg)
Q per minute = mcΔT = 3.0 × 4186 × (77 − 27) = 3.0 × 4186 × 50 = 627900 J/min
Rate of fuel = Q / (heat of combustion) = 627900 / (4 × 10⁴) = 15.7 g/min
Q2. What amount of heat must be supplied to 2.0 × 10⁻² kg of nitrogen to raise its temperature by 45°C at constant pressure? Molecular mass of N₂ = 28, C_p = 29.1 J mol⁻¹ K⁻¹.
n = mass/M = 2.0 × 10⁻² / (28 × 10⁻³) = 0.714 mol
Q = nC_pΔT = 0.714 × 29.1 × 45 = 934.8 J
Q3. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. How much work is done in compressing it to one-third its original volume: (a) isothermally (b) adiabatically? γ_H₂ = 1.4.
Q4. In changing the state of a gas adiabatically from an equilibrium state A to another state B, 600 J of work is done on the system. If the gas is taken from A to B via a process in which the net heat absorbed is 70 J, how much work is done by the gas?
Path A→B (adiabatic): ΔQ = 0, ΔW = −600 J (work done on system)
ΔU = ΔQ − ΔW = 0 − (−600) = +600 J
ΔU depends only on the endpoints, so ΔU_A→B = 600 J for any path.
Other path: ΔQ = 70 J, ΔU = 600 J
ΔQ = ΔU + ΔW → 70 = 600 + ΔW → ΔW = −530 J
Q5. Two cylinders A and B of equal capacity are connected by a stopcock. A contains a gas at STP and B is evacuated. The stopcock is opened. (a) What is the final pressure? (b) What is the change in internal energy? (c) What is the change in temperature? (d) Does the gas do work?
Q6. 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. Calculate the efficiency.
η = W/Q₁ = 5.4 × 10⁸ / (3.6 × 10⁹) = 0.15 = 15%
Q7. An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 J/s, at what rate is the internal energy increasing?
dQ/dt = 100 W, dW/dt = 75 J/s
dU/dt = dQ/dt − dW/dt = 100 − 75 = 25 J/s = 25 W
Q8. A thermodynamic system is taken from an original state A to state B via two paths ACB and ADB. When path ACB is used, 60 J of heat flows into the system and 30 J of work is done. Along ADB, 70 J of heat flows into the system. How much work is done along ADB?
ΔU_A→B is the same for both paths.
Path ACB: ΔU = ΔQ − ΔW = 60 − 30 = 30 J
Path ADB: ΔQ = 70 J, ΔU = 30 J
ΔW = ΔQ − ΔU = 70 − 30 = 40 J
Q9. The efficiency of a Carnot engine is 40%. If the temperature of the sink is 300 K, find the temperature of the source.
η = 1 − T₂/T₁ = 0.40
T₂/T₁ = 0.60 → T₁ = T₂/0.60 = 300/0.60 = 500 K
Q10. A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of the source be increased to raise the efficiency to 60%?
Original: T₁ = 300/0.60 = 500 K (from Q9)
New: T₁' such that 1 − 300/T₁' = 0.60 → T₁' = 300/0.40 = 750 K
Increase = 750 − 500 = 250 K
Q11. A refrigerator is to maintain eatables at 9°C. Room temperature is 36°C. The COP is 10. Calculate the heat extracted per unit work done.
COP = Q₂/W = 10
For each 1 J of work, heat extracted = 10 J
Heat extracted per unit work = 10 J
Q12. State the second law of thermodynamics in its two forms.
Q13. What is a reversible process? What are the conditions for reversibility?
Q14. What is an irreversible process? Give two examples.
Q15. Show that for an isothermal process, the work done by an ideal gas is W = nRT ln(V₂/V₁).
For an ideal gas: PV = nRT
Work done by gas: W = ∫ P dV = ∫ (nRT/V) dV
Since T is constant (isothermal): W = nRT ∫ dV/V = nRT [ln V] from V₁ to V₂
W = nRT ln(V₂/V₁)
Q16. Derive the expression for work done in an adiabatic process.
ΔQ = 0 → ΔW = −ΔU
For an ideal gas: dU = nC_v dT
W = −∫ dU = −nC_v (T₂ − T₁) = nC_v (T₁ − T₂)
Using C_v = R/(γ−1): W = nR(T₁ − T₂)/(γ − 1) = (P₁V₁ − P₂V₂)/(γ − 1)
Q17. Explain the Carnot cycle with a P-V diagram.
Internal Links
- [Chapter 12 Guide — Thermodynamics](/school/cbse/class/11/physics/cbse-thermodynamics)
- [All Class 11 Physics Solutions](/school/cbse/class/11/physics/solutions)
- [Class 11 Physics Subject Guide](/school/cbse/class/11/physics)
Frequently Asked Questions
The first law states that the heat supplied to a system (ΔQ) equals the change in internal energy (ΔU) plus the work done by the system (ΔW): ΔQ = ΔU + ΔW. It is a statement of conservation of energy applied to thermal processes.