Question

One mole of an ideal monatomic gas at pressure \(P\), volume \(V\) and temperature \(T\) is expanded isothermally to volume \(4V\). Thereafter, the gas is heated isochorically (at constant volume) till its pressure becomes \(P\). If \(R\) is the universal gas constant, the total heat transfer in the process, in units of \(RT\), is ................... (Round off to 2 decimal places)

Hint:

For multi-step processes, calculate heat for each path separately using the respective thermodynamic relations, and add them to get total heat transfer.

Answer 1

Correct Answer: 5.88 - 5.94

Step 1: For the isothermal expansion (A → B). 
\[ W_1 = nRT \ln{\frac{V_2}{V_1}} = RT \ln{4} \] \[ Q_1 = W_1 = RT \ln{4} \]

Step 2: For the isochoric heating (B → C). 
Since volume is constant, \(Q_2 = nC_V (T_2 - T_1)\). For an ideal gas: \[ \frac{P_2}{T_2} = \frac{P_1}{T_1} \] At the end of isothermal expansion, \(P_B = \frac{P}{4}\). After heating at constant volume, final pressure \(P_C = P\). \[ \Rightarrow \frac{T_2}{T_1} = \frac{P_C}{P_B} = \frac{P}{P/4} = 4 \] For a monatomic gas, \(C_V = \frac{3R}{2}\). \[ Q_2 = nC_V (T_2 - T_1) = \frac{3R}{2} (4T - T) = \frac{9RT}{2} \]

Step 3: Total heat transfer. 
\[ Q_{\text{total}} = Q_1 + Q_2 = RT \ln{4} + \frac{9RT}{2} \] \[ \frac{Q_{\text{total}}}{RT} = \ln{4} + 4.5 = 1.386 + 4.5 = 5.886 \]

Step 4: Conclusion. 
Hence, the total heat transfer in the process in units of \(RT\) is \(5.89\).