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(C) The work done in a cyclic process is equal to the area enclosed by the cycle on the $p-V$ diagram. The cycle is $ABCA$. $1$. Work done in process

Vedclass · Accepted Answer

$-300$

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(C) The work done in a cyclic process is equal to the area enclosed by the cycle on the $p-V$ diagram. The cycle is $ABCA$. $1$. Work done in process $AB$ (isobaric expansion): $W_{AB} = p_A(V_B - V_A) = 200 	imes (V_B - 2)$. Since $BC$ is isothermal,$p_B V_B = p_C V_C$. $200 	imes V_B = 500 	imes 2 \Rightarrow V_B = 5 \, m^3$. $W_{AB} = 200 	imes (5 - 2) = 600 \, kJ$. $2$. Work done in process $BC$ (isothermal compression): $W_{BC} = \int_{V_B}^{V_C} p \, dV = \int_{5}^{2} \frac{p_B V_B}{V} \, dV = 1000 \ln(2/5) = 1000 	imes (-0.916) \approx -916 \, kJ$. $3$. Work done in process $CA$ (isochoric compression): $W_{CA} = 0$ (since volume is constant). Total work done $W = W_{AB} + W_{BC} + W_{CA} = 600 - 916 + 0 = -316 \, kJ$. Rounding to the nearest option,the work done is approximately $-300 \, kJ$.

$A$ constant amount of an ideal gas undergoes the cyclic process $ABCA$ in the $p-V$ graph shown below. The path $BC$ is an isothermal process. The work done by the gas during one complete cycle,beginning and ending at $A$,is nearly .......... $kJ$.

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