Consider the reversible processes for 1.0 mol of an ideal gas as shown in the figure. Processes 2 and 4 are adiabatic. w_1, w_2, w_3 and w_4 represent work done (in calories) in the processes 1, 2, 3 and 4, respectively; U_2 and U_4 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
B. w_1+w_3=-2T_1 V_2 V_1 -2T_2 V_4 V_3
Using the IUPAC sign convention for thermodynamics, work done is given by w = - p dV. For the isothermal reversible process 1 (from volume V_1 to V_2 at constant temperature T_1): w_1 = -nRT_1 ( V_2 V_1 ) Given n = 1 mol and R = 2 cal K^ -1 mol^ -1 , we have: w_1 = -2T_1 ( V_2 V_1 ) For the isothermal reversible process 3 (from volume V_3 to V_4 at constant temperature T_2): w_3 = -nRT_2 ( V_4 V_3 ) = -2T_2 ( V_4 V_3 ) Adding the work done in these two processes: w_1 + w_3 = -2T_1 ( V_2 V_1 ) - 2T_2 ( V_4 V_3 ) Let us also verify the other options to be sure. For option (1): The net work done in a reversible cycle is the area enclosed by the p-V curve, which is non-zero. Thus, w_1 + w_2 + w_3 + w_4 0. For option (3): In adiabatic processes, q = 0, so from the first law of thermodynamics, U = w. Therefore, w_2 = U_2 and w_4 = U_4. This gives w_2 + w_4 = U_2 + U_4. Since internal energy is a state function, for the complete cycle U_ cycle = U_1 + U_2 + U_3 + U_4 = 0. For isothermal processes 1 and 3, U_1 = 0 and U_3 = 0. Thus, U_2 + U_4 = 0, which means w_2 + w_4 = 0. However, U_2 - U_4 = U_2 - (- U_2) = 2 U_2 0. Hence, w_2 + w_4 U_2 - U_4. For option (4): w_1 + w_2 = -2T_1 ( V_2 V_1 ) + U_2 2T_1 ( V_2 V_1 ). Answer: w_1+w_3=-2T_1 V_2 V_1 -2T_2 V_4 V_3
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