Solution: T_C = 27°C = 300 K. Initial: 0.5 = 1 − 300/T_H1 → T_H1 = 600 K. For 70%: 0.7 = 1 − 300/T_H2 → T_H2 = 300/0.3 = 1000 K. Answer: (B) 1000 K

⚡ Always convert Celsius to Kelvin first: T_C = 27 + 273 = 300 K

Solution: W = nRT ln(V₂/V₁) = 1 × RT × ln(2V/V) = RT ln(2). Answer: (A) RT ln(2)

⚡ Don't confuse with isobaric work (W = nRΔT). For isothermal: W = nRT ln(V₂/V₁).

Solution: U = 3PV/2 = 3nRT/2 (using PV = nRT). So U = f/2 nRT → f/2 = 3/2 → f = 3 (monoatomic gas). γ = (f+2)/f = 5/3. Answer: (A) 5/3

Solution: Isothermal → ΔT = 0 → ΔU = 0 (ideal gas). First Law: Q = ΔU + W = 0 + W = W. So Q = W. Note: Q ≠ 0 — heat is exchanged to maintain temperature. Answer: (B) ΔU = 0 and Q = W

⚠️ Option A is the most common wrong answer. Students confuse isothermal with adiabatic.

Statement: It is impossible to construct a heat engine that, operating in a complete cycle, converts all the absorbed heat into work.

Calculation: η = 1 − T_C/T_H = 1 − 200/500 = 1 − 0.4 = 0.6 = 60%

For CBSE 3-mark: Write statement (1 mark) + substitution (1 mark) + answer (1 mark)

Solution:

Step 1→3 (isochoric, V = V₀): W₁₃ = 0. T₁ = P₀V₀/nR, T₃ = 2P₀V₀/nR = 2T₁. ΔU₁₃ = nCv(T₃-T₁) = n(3R/2)(T₁) = 3P₀V₀/2. Q₁₃ = ΔU = 3P₀V₀/2

Step 3→2 (isobaric, P = 2P₀): V goes from 2V₀ to ... wait, state 2 is (P₀, 2V₀) but step is from (2P₀,2V₀) to (P₀,2V₀). That's isochoric (V=2V₀)! W₃₂ = 0. ΔU₃₂ = nCv(T₂−T₃) = (3R/2)n × (−T₁) = −3P₀V₀/2. Q₃₂ = −3P₀V₀/2.

Total: W = 0, ΔU_total = 0 (returns to T₁), Q_net = 0. ✓ Consistent.