Two moles of an ideal monoatomic gas undergo a cyclic process as shown in figure. The temperatures in different states are given as 6 ~T ₁=3 ~T ₂=2 ~T ₄= T ₃=2400 ~K . The work done by the gas during the complete cycle is ( R= Universal gas constant )

Step-by-step solution

For an ideal monoatomic gas undergoing a cyclic process, the work done equals the area enclosed in the P – V diagram. From the T – P grid, using unit pressure P_u , the pressures are P₁ = P_u , P₂ = 2P_u , P₃ = 3P_u , P₄ = 2P_u . Given T₁ = 400 K, T₂ = 800 K, T₃ = 2400 K, T₄ = 1200 K, and using V₀ = nR/P_u , the volumes from the ideal gas law are V₁ = 400 V₀ , V₂ = 400 V₀ , V₃ = 800 V₀ , V₄ = 600 V₀ . The work in each process is calculated as the area under the curve in P – V space. For isochoric process 1–2, W₁₂ =