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 \(R = 2\ \text{cal}\ \text{K}^{-1}\text{mol}^{-1}\)]
Show 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.
Step 1: Set up the thermodynamic tools. Use \(\Delta U = q + w\). For an adiabatic process \(q = 0\), so \(\Delta U = w\). For a complete cycle \(\Delta U = 0\) because internal energy is a state function. Step 2: Describe the two isothermal steps (1 and 3). For 1 mole of an ideal gas at constant temperature, \[ w = -RT\ln\left(\frac{V_f}{V_i}\right). \] So \(w_1 = -2T_1\ln(V_2/V_1)\) and \(w_3 = -2T_2\ln(V_4/V_3)\) using \(R = 2\). Step 3: Handle the two adiabatic steps (2 and 4). Since \(q = 0\) in each, \(w_2 = \Delta U_2\) and \(w_4 = \Delta U_4\). Step 4: Add the adiabatic work terms. Adding the two relations gives \[ w_2 + w_4 = \Delta U_2 + \Delta U_4. \] Step 5: Use the symmetry of the cycle. Process 2 takes the gas from \(T_1\) to \(T_2\) and process 4 brings it back from \(T_2\) to \(T_1\), so \(\Delta U_4 = -\Delta U_2\). Then \(\Delta U_2 + \Delta U_4\) can equivalently be written as \(\Delta U_2 - \Delta U_4\) since \(\Delta U_4 = -\Delta U_2\), and the matching option expresses it that way. Step 6: Reject the other options. Net cycle work equals the enclosed area and is generally non-zero, so (A) fails; (D) ignores the sign convention; (B) is only a partial isothermal statement. The fully correct relation is \(w_2 + w_4 = \Delta U_2 - \Delta U_4\). \[ \boxed{w_2 + w_4 = \Delta U_2 - \Delta U_4} \]
