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(A) The total work done by the gas in a cyclic process is equal to the area enclosed by the $P-V$ diagram.For the process $D \rightarrow E \rightarrow

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$450$

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(A) The total work done by the gas in a cyclic process is equal to the area enclosed by the $P-V$ diagram.For the process $D ightarrow E ightarrow F$,the work done is the area under the curve $DE$ minus the area under the curve $EF$ (since the volume decreases from $E$ to $F$).This is equivalent to the area of the triangle $DEF$.Area of $\Delta DEF = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Base $EF = V_E - V_F = 5.0 \, m^3 - 2.0 \, m^3 = 3.0 \, m^3$.Height $DF = P_D - P_F = 600 \, N/m^2 - 300 \, N/m^2 = 300 \, N/m^2$.Work done $= 	ext{Area} = \frac{1}{2} 	imes 3.0 \, m^3 	imes 300 \, N/m^2 = 450 \, J$.Therefore,the total work done by the gas from $D$ to $E$ to $F$ is $450 \, J$.

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

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