Practice question

**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. For isothermal: Δ U = 0 , Δ Q = Δ W . Δ Q = -960 J (heat released). Δ W = -960 J (work by gas negative), so work on gas = 960 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W =