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(B) The work done in a $P-V$ diagram is equal to the area under the curve.For the process $D \rightarrow E$,the work done $W_{DE}$ is the area of the

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(B) The work done in a $P-V$ diagram is equal to the area under the curve.For the process $D ightarrow E$,the work done $W_{DE}$ is the area of the trapezoid under the line $DE$:$W_{DE} = 	ext{Area of trapezoid} = \frac{1}{2} 	imes (P_D + P_E) 	imes (V_E - V_D)$$W_{DE} = \frac{1}{2} 	imes (600 + 300) 	imes (5.0 - 2.0) = \frac{1}{2} 	imes 900 	imes 3 = 1350 \, J$For the process $E ightarrow F$,the pressure is constant at $300 \, N/m^2$ (isobaric process),and the volume decreases from $5.0 \, m^3$ to $2.0 \, m^3$:$W_{EF} = P 	imes \Delta V = 300 	imes (2.0 - 5.0) = 300 	imes (-3.0) = -900 \, J$The total work done $W_{DEF} = W_{DE} + W_{EF} = 1350 - 900 = 450 \, J$.

$A$ thermodynamic system is taken from an original state $D$ to an intermediate state $E$ by the linear process shown in the figure. Its volume is then reduced to the original volume from $E$ to $F$ by an isobaric process. The total work done by the gas from $D$ to $E$ to $F$ will be $......J$

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