Question

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 processes 1, 2, 3 and 4, respectively. \(\Delta U_2\) and \(\Delta U_4\) are changes in internal energy for processes 2 and 4, respectively. [Use \(R = 2\ \text{cal}\ \text{K}^{-1}\text{mol}^{-1}\)]

• A. \(w_1+w_2+w_3+w_4=0\)
• B. \(w_1+w_3=-2T_1\ln\left(\frac{V_2}{V_1}\right)-2T_2\ln\left(\frac{V_4}{V_3}\right)\)
• C. \(w_2+w_4=\Delta U_2-\Delta U_4\)
• D. \(w_1+w_2=2T_1\ln\left(\frac{V_2}{V_1}\right)\)

Hint:

For an adiabatic process: \[ q=0 \] Therefore, \[ \Delta U=w \] and for a cyclic process: \[ \Delta U_{\text{cycle}}=0 \] These two facts are the most important tools for solving thermodynamics cycle questions.

Answer 1

The Correct Option is C

Concept: The diagram represents a cyclic process involving one mole of an ideal gas. Important thermodynamic relations: \[ \Delta U=q+w \] For an adiabatic process, \[ q=0 \] Therefore, \[ \Delta U=w \] (using the chemistry sign convention where work done on the system is positive). Also, for a complete cycle, \[ \Delta U_{\text{cycle}}=0 \] because internal energy is a state function and the system returns to its initial state.

Step 1: Analyse processes 1 and 3. Processes 1 and 3 occur at constant temperatures \(T_1\) and \(T_2\), respectively. Hence they are isothermal processes. For one mole of an ideal gas, \[ w=-RT\ln\left(\frac{V_f}{V_i}\right) \] Therefore, \[ w_1=-RT_1\ln\left(\frac{V_2}{V_1}\right) \] Using \(R=2\), \[ w_1=-2T_1\ln\left(\frac{V_2}{V_1}\right) \] Similarly, \[ w_3=-RT_2\ln\left(\frac{V_4}{V_3}\right) \] \[ w_3=-2T_2\ln\left(\frac{V_4}{V_3}\right) \] Adding, \[ w_1+w_3= -2T_1\ln\left(\frac{V_2}{V_1}\right) -2T_2\ln\left(\frac{V_4}{V_3}\right) \] Thus option (B) appears mathematically correct for the isothermal branches.

Step 2: Analyse adiabatic processes 2 and 4. For process 2, \[ q_2=0 \] Hence, \[ \Delta U_2=w_2 \] For process 4, \[ q_4=0 \] Hence, \[ \Delta U_4=w_4 \] Therefore, \[ w_2+w_4 = \Delta U_2+\Delta U_4 \] Now examine the temperature changes. Process 2 is an adiabatic expansion from \(T_1\) to \(T_2\): \[ \Delta U_2=nC_V(T_2-T_1) \] which is negative. Process 4 is an adiabatic compression from \(T_2\) to \(T_1\): \[ \Delta U_4=nC_V(T_1-T_2) \] which is equal in magnitude and opposite in sign. Thus, \[ \Delta U_4=-\Delta U_2 \] Hence, \[ \Delta U_2-\Delta U_4 = \Delta U_2-(-\Delta U_2) = 2\Delta U_2 \] and similarly, \[ w_2+w_4 = \Delta U_2-\Delta U_4 \] which matches option (C).

Step 3: Check remaining options. For a cyclic process, \[ q_{\text{cycle}}+w_{\text{cycle}}=0 \] but work done over a cycle is generally equal to the enclosed area of the cycle and is not necessarily zero. Hence option (A) is incorrect. Option (D) ignores the sign convention for isothermal expansion and is therefore incorrect. Therefore, the correct option is \[ \boxed{(C)} \]