The change in internal energy of an ideal gas (γ = 1.4) when 2 moles are heated from 300 K to 400 K at constant volume (R = 8.314 J K⁻¹ mol⁻¹) is approximately:
A2079 J
B3328 J
C4158 J
D5829 J
Answer & Solution
Correct answer: C. 4158 J
1. γ = Cp/Cv = 1.4 and Cp − Cv = R, so Cv = R/(γ−1) = 8.314/0.4 = 20.79 J K⁻¹ mol⁻¹.
2. ΔU = n Cv ΔT = 2 × 20.79 × 100 = 4158 J.
3. At constant V, no work is done, so Q = ΔU.
4. The distractor 2079 J is for n = 1; 5829 ≈ n Cp ΔT (constant-P heat), not ΔU.
_Source: NCERT Class 11 Physics Ch 11, §11.6 (Cv ↔ γ relation)_
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