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. A quasi-static process is infinitely slow, maintaining the system in equilibrium with its surroundings at every stage, allowing well-defined P and T . Option A is correct; it’s a condition for reversibility but not sufficient alone. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T