Practice question

**Work calculation** isobaric W = P(V₂ - V₁), e.g., P=1 atm=1.013×10⁵ Pa ΔV=0.01 m³ W=1013 J. Isothermal W = n R T ln(V₂/V₁), for n=1 mol T=300 K V₂/V₁=2 W=1×8.314×300×ln2=1729 J, work done by gas during expansion, area under P-V curve. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.4 , R = 8.3 , T = 420 , V₂ = 21 , V₁ = 7 . W = 0.4 × 8.3 × 420 × ln((21)/(7)) = 1394.4 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 1394.4 × 1.0986 ≈ 1532 J . Using first law ΔU = Q - W,