For the reversible processes of 1 mol of an ideal gas shown (a cycle p₁V₁T₁→ p₂V₂T₁ [process 1, isothermal] → p₃V₃T₂ [process 2, adiabatic] → p₄V₄T₂ [process 3, isothermal] → back [process 4, adiabatic]), with w₁,w₂,w₃,w₄ the work done (in calories) and Δ U₂,Δ U₄ the internal-energy changes in processes 2 and 4. [Use R=2 cal K⁻¹mol⁻¹.] The correct option is:

Solution & Explanation

Work sign convention: for a reversible isothermal step of an ideal gas, w = −nRT·ln(V_f/V_i). Here n = 1 and R = 2 cal K⁻¹mol⁻¹. Process 1 is isothermal at T₁ from V₁ to V₂, so w₁ = −2T₁·ln(V₂/V₁). Process 3 is isothermal at T₂ from V₃ to V₄, so w₃ = −2T₂·ln(V₄/V₃). Adding the two isothermal works: w₁ + w₃ = −2T₁·ln(V₂/V₁) − 2T₂·ln(V₄/V₃), which is exactly option A. Note: processes 2 and 4 are adiabatic (q = 0), where w = ΔU and is governed by temperature change, not by these log terms; the asked combination involves only the isothermal steps. Answer: option A.