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(B) The process is given on a $p-T$ diagram. For an ideal gas, $p V = \mu R T$, which implies $p = \frac{\mu R}{V} T$. Since $p \propto T$ for a const

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$20$

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(B) The process is given on a $p-T$ diagram. For an ideal gas, $p V = \mu R T$, which implies $p = \frac{\mu R}{V} T$. Since $p \propto T$ for a constant volume, the lines passing through the origin in a $p-T$ diagram represent isochoric (constant volume) processes. In the given figure, $A B$ and $C D$ are straight lines passing through the origin, so these are isochoric processes. Work done in an isochoric process is zero. Thus, $W_{A B} = 0$ and $W_{C D} = 0$. For the isobaric processes $B C$ and $D A$, the work done is $W = p \Delta V = \mu R \Delta T$. For process $B C$ (at constant pressure $p_2$): $W_{B C} = \mu R (T_C - T_B) = 3 	imes R 	imes (2400 - 800) = 3 R 	imes 1600 = 4800 R$. For process $D A$ (at constant pressure $p_1$): $W_{D A} = \mu R (T_A - T_D) = 3 	imes R 	imes (400 - 1200) = 3 R 	imes (-800) = -2400 R$. The total work done in the cycle is $W = W_{A B} + W_{B C} + W_{C D} + W_{D A} = 0 + 4800 R + 0 - 2400 R = 2400 R$. Substituting $R = 8.314 \, J / mol K$: $W = 2400 	imes 8.314 = 19953.6 \, J \approx 20 \, kJ$.

$3$ moles of an ideal monoatomic gas perform an $A B C D A$ cyclic process as shown in the figure. The gas temperatures are $T_A=400 \, K$, $T_B=800 \, K$, $T_C=2400 \, K$, and $T_D=1200 \, K$. The work done by the gas is (approximately) $(R=8.314 \, J / mol K)$. (in $ \, kJ$)

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