Question 14

For 1 mole of monatomic gas, $C_V = \frac{3}{2}R$ and $C_P = \frac{5}{2}R$.

1. Process $J \to K$ (Isobaric): $P = P_0$, $T$ goes from $T_0$ to $3T_0$.

   • Work $W_{JK} = P_0 \Delta V = R \Delta T = R(3T_0 - T_0) = 2RT_0$.
   • $\Delta U_{JK} = n C_V \Delta T = 1 \cdot \frac{3}{2}R(2T_0) = 3RT_0$. (Matches Q $\to$ 3)

2. Process $K \to L$ (Isothermal): $T = 3T_0$, $P$ goes from $P_0$ to $2P_0$.

   • $\Delta U_{KL} = 0$.
   • $W_{KL} = nRT \ln\left(\frac{P_i}{P_f}\right) = 1 \cdot R(3T_0) \ln\left(\frac{P_0}{2P_0}\right) = -3RT_0 \ln 2$.
   • Heat $Q_{KL} = \Delta U_{KL} + W_{KL} = -3RT_0 \ln 2$. (Matches R $\to$ 5)

3. Process $L \to M$ (Isobaric): $P = 2P_0$, $T$ goes from $3T_0$ to $T_0$.

   • Work $W_{LM} = 2P_0 \Delta V = R \Delta T = R(T_0 - 3T_0) = -2RT_0$.
   • $\Delta U_{LM} = \frac{3}{2}R(-2T_0) = -3RT_0$.

4. Process $M \to J$ (Isothermal): $T = T_0$, $P$ goes from $2P_0$ to $P_0$.

   • $\Delta U_{MJ} = 0$. (Matches S $\to$ 2)
   • $W_{MJ} = nRT \ln\left(\frac{P_i}{P_f}\right) = R T_0 \ln\left(\frac{2P_0}{P_0}\right) = RT_0 \ln 2$.

Calculation for P: Total Work $W = W_{JK} + W_{KL} + W_{LM} + W_{MJ}$ $W = 2RT_0 - 3RT_0 \ln 2 - 2RT_0 + RT_0 \ln 2 = -2RT_0 \ln 2$. (Matches P $\to$ 4)

Correct Combination: P-4, Q-3, R-5, S-2. Correct Option: B