One mole of a monatomic ideal gas undergoes t | vedclass

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(B) From the $P-T$ diagram:$J: (P_0, T_0) \Rightarrow V_J = \frac{RT_0}{P_0}$$K: (P_0, 3T_0) \Rightarrow V_K = \frac{3RT_0}{P_0} = 3V_J$$L: (2P_0, 3T_

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$P \rightarrow 4 ; Q \rightarrow 3 ; R \rightarrow 5 ; S \rightarrow 2$

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(B) From the $P-T$ diagram:$J: (P_0, T_0) \Rightarrow V_J = \frac{RT_0}{P_0}$$K: (P_0, 3T_0) \Rightarrow V_K = \frac{3RT_0}{P_0} = 3V_J$$L: (2P_0, 3T_0) \Rightarrow V_L = \frac{3RT_0}{2P_0} = 1.5V_J$$M: (2P_0, T_0) \Rightarrow V_M = \frac{RT_0}{2P_0} = 0.5V_J$$(P)$ Work done in the cycle: $W_{net} = \oint P dV$. The area enclosed in the $P-V$ diagram is $W_{net} = (2P_0 - P_0)(V_M - V_K) = P_0(0.5V_J - 3V_J) = -2.5 P_0 V_J = -2.5 RT_0$. However,calculating path-wise: $W_{JK} = P_0(3V_J - V_J) = 2RT_0$; $W_{KL} = nRT_0 \ln(V_L/V_K) = 3RT_0 \ln(0.5) = -3RT_0 \ln 2$; $W_{LM} = 2P_0(0.5V_J - 1.5V_J) = -2RT_0$; $W_{MJ} = nRT_0 \ln(V_J/V_M) = RT_0 \ln 2$. Summing these: $W_{net} = 2RT_0 - 3RT_0 \ln 2 - 2RT_0 + RT_0 \ln 2 = -2RT_0 \ln 2$. Thus,$P \rightarrow 4$.$(Q)$ $\Delta U_{JK} = n C_v \Delta T = 1 \cdot \frac{3}{2} R (3T_0 - T_0) = 3RT_0$. Thus,$Q \rightarrow 3$.$(R)$ Process $KL$ is isothermal $(T=3T_0)$. $\Delta U = 0$,so $Q_{KL} = W_{KL} = nRT \ln(V_L/V_K) = 3RT_0 \ln(1.5/3) = -3RT_0 \ln 2$. Thus,$R \rightarrow 5$.$(S)$ Process $MJ$ is isothermal $(T=T_0)$. $\Delta U = 0$. Thus,$S \rightarrow 2$.Correct option is $B$ $(P \rightarrow 4, Q \rightarrow 3, R \rightarrow 5, S \rightarrow 2)$.

$List-I$	$List-II$
$(P)$ Work done in the complete cyclic process	$(1)$ $R T_0 - 4 R T_0 \ln 2$
$(Q)$ Change in the internal energy of the gas in the process $JK$	$(2)$ $0$
$(R)$ Heat given to the gas in the process $KL$	$(3)$ $3 R T_0$
$(S)$ Change in the internal energy of the gas in the process $MJ$	$(4)$ $-2 R T_0 \ln 2$
	$(5)$ $-3 R T_0 \ln 2$

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