An ideal gas expands isothermally at 460 K from 3 L to 9 L with 0.5 moles . What is the work done by the gas? ( R = 8.3
**Isochoric work** zero because dV=0, so W=∫ P dV=0, internal energy change equals heat added, Q = n C_v ΔT, C_v molar specific heat at constant volume, for monatomic 3/2 R, for diatomic 5/2 R, temperature change directly from heat input. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.5 , R = 8.3 , T = 460 , V₂ = 9 , V₁ = 3 . W = 0.5 × 8.3 × 460 × ln((9)/(3)) = 1909 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 1909 × 1.0986 ≈ 2097 J . Using first law ΔU = Q - W,
Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature