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 isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 3 , T₁ = 400 , T₂ = 200 . (3)/(400) = (P₂)/(200) ⇒ P₂ = (3 × 200)/(400) = 1.5 atm . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η =