Practice question

**Adiabatic process** no heat exchange Q=0, first law ΔU = -W, for ideal gas P V^γ = constant, T V^{γ-1}= constant, P^{1-γ} T^{γ}= constant, γ = C_p/C_v = (f+2)/f, monatomic γ=5/3, diatomic γ=7/5. Work done W = (P₁V₁ - P₂V₂)/(γ-1), temperature changes due to work. First Law: Δ Q = Δ U + Δ W . Δ Q = -300 J (heat released), Δ W = -150 J (work done on system, negative by convention). -300 = Δ U + (-150) . Δ U = -300 + 150 = -150 J . Using first law ΔU = Q - W, W = ∫ P dV,