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: \[ Q_T = \omega_1 + \omega_2 + \omega_3 = 2 P_0 V_0 \ln 2 - 0.75 P_0 V_0 + 0 = P_0 V_0 (2 \ln 2 - 0.75) \]

Answer 2

To solve this problem, we follow the steps of the thermodynamic process described: isothermal expansion, isobaric compression, and isochoric heating.

1. Isothermal Expansion: The gas expands isothermally from volume \( V_0 \) to \( 4V_0 \) at a constant temperature. In an isothermal process, the work done \( W \) and the heat exchanged \( Q \) can be calculated using the formula: \(Q_1 = W = nRT \ln \frac{V_f}{V_i}\) Since \(P_0 V_0 = nRT\) (initial state), and the final volume is \( 4V_0 \): \(Q_1 = P_0 V_0 \ln 4 = P_0 V_0 \cdot 2 \ln 2\)
2. Isobaric Compression: The gas is compressed isobarically (constant pressure) back to its original volume \( V_0 \). In an isobaric process, work done on the gas is given by: \(W = P \Delta V = P_0(V_0 - 4V_0) = -3P_0 V_0\) Since it's compression, the work is negative, and heat exchanged in this process is: \(Q_2 = \Delta U + W = 0 + (-3P_0 V_0) = -3P_0 V_0\)
3. Isochoric Heating: The volume is constant, so work done \( W = 0 \). The heat added or removed during an isochoric process only changes the internal energy: The system needs to return to its initial temperature and state, meaning it will require additional heat to recover energy lost during the isobaric phase.

To find the total heat exchanged, \(Q_{\text{total}}\), sum the heat exchanged in each process:

\(Q_{\text{total}} = Q_1 + Q_2 + Q_3 = P_0 V_0 \cdot 2 \ln 2 - 3P_0 V_0 + 2.25P_0 V_0\)

Simplifying, we obtain:

\(Q_{\text{total}} = P_0 V_0 (2 \ln 2 - 0.75)\)

Thus, the correct option is: \( P_0 V_0 (2 \ln 2 - 0.75) \). This matches with the option given in the question, confirming the step-by-step calculations.