Consider the reversible processes for 1.0 mol of an ideal gas as shown in the figure. Processes 2 and 4 are adiabatic. w₁ , w₂ , w₃ and w₄ represent work done (in calories) in the processes 1, 2, 3 and 4, respectively; U₂ and U₄ are changes in the internal energy for the processes 2 and 4, respectively. [use R =2 cal K ⁻¹ mol ⁻¹ ] The correct option is

Step-by-step solution

Using the IUPAC sign convention for thermodynamics, work done is given by w = - p dV . For the isothermal reversible process 1 (from volume V₁ to V₂ at constant temperature T₁ ): w₁ = -nRT₁ ( V₂ V₁ ) Given n = 1 mol and R = 2 cal K ⁻¹ mol ⁻¹ , we have: w₁ = -2T₁ ( V₂ V₁ ) For the isothermal reversible process 3 (from volume V₃ to V₄ at constant temperature T₂ ): w₃ = -nRT₂ ( V₄ V₃ ) = -2T₂ ( V₄ V₃ ) Adding the work done in these two processes: w₁ + w₃ = -2T₁ ( V₂ V₁ ) - 2T₂ ( V₄ V₃ ) Let us also verify the other op