Question

Two-thirds mole of an ideal diatomic gas is taken around the cyclic process \(ABCA\) shown in the figure. What is the amount of heat rejected by the gas to the surrounding in the path \(CA\)?

• A. \(P_0V_0\)
• B. \(\dfrac{3}{2}P_0V_0\)
• C. \(\dfrac{5}{2}P_0V_0\)
• D. \(\dfrac{7}{2}P_0V_0\)

Hint:

For an isobaric process: \[ Q=nC_P\Delta T \] For a diatomic gas, \[ C_P=\frac{7R}{2}. \] Using this directly, \[ Q = n\frac{7R}{2} \left( -\frac{P_0V_0}{nR} \right) = -\frac{7}{2}P_0V_0. \] Hence the heat rejected is \[ \frac{7}{2}P_0V_0. \]

Answer 1

The Correct Option is C

Concept: For an ideal gas, \[ Q=\Delta U+W \] and for a diatomic gas, \[ C_V=\frac{5R}{2}. \] Along \(CA\), the process is isobaric at pressure \(P_0\).

Step 1: Determine the temperatures at \(A\) and \(C\). Using \[ PV=nRT \] At point \(A\), \[ P=P_0,\qquad V=V_0 \] \[ T_A=\frac{P_0V_0}{nR} \] At point \(C\), \[ P=P_0,\qquad V=2V_0 \] \[ T_C=\frac{2P_0V_0}{nR} \] Hence, \[ \Delta T=T_A-T_C = -\frac{P_0V_0}{nR} \]

Step 2: Calculate the change in internal energy. \[\begin{aligned} \Delta U &= nC_V\Delta T \\ &= n\left(\frac{5R}{2}\right) \left( -\frac{P_0V_0}{nR} \right) \\ &= -\frac{5}{2}P_0V_0 \end{aligned}\]

Step 3: Calculate the work done. During \(C\rightarrow A\), \[ W=P_0(V_A-V_C) \] \[ =P_0(V_0-2V_0) \] \[ =-P_0V_0 \]

Step 4: Find the heat exchanged. \[\begin{aligned} Q &= \Delta U+W \\ &= -\frac{5}{2}P_0V_0-P_0V_0 \\ &= -\frac{7}{2}P_0V_0 \end{aligned}\] The negative sign indicates heat is rejected. Therefore, \[ \text{Heat rejected} = \left|Q\right| = \frac{7}{2}P_0V_0. \] \[\begin{aligned} \boxed{\frac{7}{2}P_0V_0} \end{aligned}\] Hence, option \(\mathbf{(D)}\) is the physically correct answer.

Note: The marked option in the image appears to be (C), but applying the first law of thermodynamics gives \[ \boxed{\frac{7}{2}P_0V_0}. \]