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(B) In the $PT$ diagram, the lines passing through the origin represent isochoric processes (constant volume) because $P \propto T$ implies $V = 	ext

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$20$

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(B) In the $PT$ diagram, the lines passing through the origin represent isochoric processes (constant volume) because $P \propto T$ implies $V = 	ext{constant}$. Thus, for processes $1-2$ and $3-4$, the work done is zero.The processes $2-3$ and $4-1$ are isobaric (constant pressure) because they are horizontal lines on the $PT$ diagram.The work done in an isobaric process is $W = P \Delta V = nR \Delta T$.Total work done $W_T = W_{2-3} + W_{3-4} + W_{4-1} + W_{1-2}$.Since $W_{1-2} = 0$ and $W_{3-4} = 0$, we have:$W_T = W_{2-3} + W_{4-1} = nR(T_3 - T_2) + nR(T_1 - T_4)$.Given $n = 3$, $R \approx 8.314 \, J/(mol \cdot K)$ (using $R = 8.314$ for calculation, but often $R \approx 8.3$ is used in such problems to get $20 	imes 10^3$ exactly).$W_T = 3R(T_3 - T_2 + T_1 - T_4)$.$W_T = 3R(2400 - 800 + 400 - 1200)$.$W_T = 3R(800) = 2400R$.Using $R = 8.314 \, J/mol \cdot K$, $W_T = 2400 	imes 8.314 \approx 19953.6 \, J \approx 20 \, kJ$.

Three moles of an ideal monoatomic gas perform a cycle as shown in the figure. The gas temperatures in different states are: $T_1 = 400\, K, T_2 = 800\, K, T_3 = 2400\, K$ and $T_4 = 1200\, K$. The work done by the gas during the cycle is .... $kJ$.

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