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

Answer

Total work done by the gas from D to E to $\mathrm{F}=$ Area of $\triangle \mathrm{DEF}$ Area of $\triangle \mathrm{DEF}=(1 / 2) \mathrm{DE} \times \mathrm{EF}$ Where, $\mathrm{DF}=$ Change in pressure $=600 \mathrm{~N} / \mathrm{m}^2-300 \mathrm{~N} / \mathrm{m}^2=300 \mathrm{~N} / \mathrm{m}^2 \mathrm{FE}=$ Change in volume $=5.0 \mathrm{~m}^3-2.0 \mathrm{~m}^3=3.0 \mathrm{~m}^3$ Area of $\triangle \mathrm{DEF}=$ $(1 / 2) \times 300 \times 3=450$ J Therefore, the total work done by the gas from $D$ to $E$ to $F$ is 450 J.

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