Question

An ideal gas exists in a state with pressure \( P_0 \), volume \( V_0 \). It is isothermally expanded to 4 times of its initial volume \( (V_0) \), then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is :

• A. \( P_0 V_0 (2 \ln 2 - 0.75) \)
• B. \( P_0 V_0 (\ln 2 - 0.75) \)
• C. \( P_0 V_0 (\ln 2 - 0.25) \)
• D. \( P_0 V_0 (2 \ln 2 - 0.25) \)

Hint:

For a cyclic thermodynamic process, the net heat exchanged is equal to the net work done by the system. Calculate the work done in each step of the cycle: isothermal expansion \( W = nRT \ln \frac{V_f}{V_i} \), isobaric process \( W = P(V_f - V_i) \), and isochoric process \( W = 0 \). Sum the work done in each step to find the total heat exchanged.

Answer 1

The Correct Option is A

The process consists of three steps forming a cycle. The total heat exchanged in a cyclic process is equal to the total work done by the system, since the change in internal energy for a cyclic process is zero (\( \Delta U_{cyclic} = 0 \)). 
The total heat \( Q_T = \omega_1 + \omega_2 + \omega_3 \), where \( \omega_1, \omega_2, \omega_3 \) are the work done in the isothermal expansion, isobaric compression, and isochoric heating, respectively. 

Step 1: Isothermal expansion from \( (P_0, V_0) \) to \( (P_1, 4V_0) \).
For an isothermal process, \( PV = constant \), 
so \( P_0 V_0 = P_1 (4V_0) \Rightarrow P_1 = \frac{P_0}{4} \). 
Work done \( \omega_1 = \int_{V_0}^{4V_0} P dV = \int_{V_0}^{4V_0} \frac{P_0 V_0}{V} dV = P_0 V_0 [\ln V]_{V_0}^{4V_0} = P_0 V_0 (\ln(4V_0) - \ln V_0) = P_0 V_0 \ln \frac{4V_0}{V_0} = P_0 V_0 \ln 4 = P_0 V_0 (2 \ln 2) \). 

Step 2: Isobaric compression from \( (\frac{P_0}{4}, 4V_0) \) to \( (\frac{P_0}{4}, V_0) \).
Work done \( \omega_2 = \int_{4V_0}^{V_0} P dV = P_1 (V_0 - 4V_0) = \frac{P_0}{4} (-3V_0) = -\frac{3}{4} P_0 V_0 = -0.75 P_0 V_0 \). 

Step 3: Isochoric heating from \( (\frac{P_0}{4}, V_0) \) to \( (P_0, V_0) \).
For an isochoric process, the volume is constant (\( dV = 0 \)). Work done \( \omega_3 = \int_{V_0}^{V_0} P dV = 0 \). 
The total heat exchanged in the process is the sum of the work done in each step: \[ Q_T = \omega_1 + \omega_2 + \omega_3 = 2 P_0 V_0 \ln 2 - 0.75 P_0 V_0 + 0 = P_0 V_0 (2 \ln 2 - 0.75) \]