Question

A poly-atomic molecule (C\(_3\)R, \(C_v = 4R\), where \(R\) is gas constant) goes from phase space point A (\(P_A = 10^4 \, \text{Pa}, V_A = 4 \times 10^{-3} \, \text{m}^3\)) to point B (\(P_B = 5 \times 10^4 \, \text{Pa}, V_B = 6 \times 10^{-7} \, \text{m}^3\)) to point C (\(P_C = 10^4 \, \text{Pa}, V_C = 8 \times 10^{-3} \, \text{m}^3\)). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is:

• A. 500R(\( \ln 3 + \ln 4 \))
• B. 450R(\( \ln 3 \))
• C. 500R(\( \ln 2 \))
• D. 400R ln 2

Hint:

For adiabatic processes, no heat is exchanged, and the internal energy change equals the work done. In isothermal processes, the heat absorbed is related to the work done, which can be calculated using \( \Delta Q = nRT \ln \left( \frac{V_C}{V_B} \right) \).

Answer 1

The Correct Option is B

We use the first law of thermodynamics for calculating the net heat absorbed: \[ \Delta Q = \Delta Q_{\text{adiabatic}} + \Delta Q_{\text{isothermal}} \] For the adiabatic process (A to B), no heat is exchanged (\( \Delta Q_{\text{adiabatic}} = 0 \)). For the isothermal process (B to C), the heat absorbed is given by: \[ \Delta Q_{\text{isothermal}} = W_{\text{isothermal}} = nRT \ln \left( \frac{V_C}{V_B} \right) = 450R \ln \left( \frac{V_C}{V_B} \right) = 450R(\ln 3) \] Thus, the net heat absorbed is \( 450R \ln 3 \).