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. Δ Q = m s Δ T . m = 0.2 , s = 127.7 , Δ T = 80 - 50 = 30 . Δ Q = 0.2 × 127.7 × 30 = 766.2 J ≈ 766 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T