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Answer:

We are given a PV graph.

Let's first find V₁, V₂, and V₃.

Path B to C:

The relation between P and V is given by:

[tex]\to \sf P = e^{-V}[/tex]

→ At point C:

[tex]\to \sf \dfrac{1}{e^2 } = e^{-V_3 }[/tex]

[tex]\to \sf V_3 = 2[/tex]

→ At point B:

[tex]\to \sf \dfrac{1}{e } = e^{-V_2 }[/tex]

[tex]\to \sf V_2 = 1[/tex]

Path A to B:

AB meets the origin O. So the slope of OA is equal to the slope of AB.

[tex]\to \sf \dfrac{1/e^2 }{V_1 } = \dfrac{1/e}{V_2 }[/tex]

[tex]\to \sf V_1 = 1/e[/tex]

Subsequently, the equation of AB is given by:

[tex]\to \sf P = \dfrac{1}{e} \cdot V[/tex]

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Now, work done is given by:

[tex]\to \sf W = \int \sf P \cdot dv[/tex]

Work done on path AB:

[tex]\to \sf W_1 = \int\limits^{V_2 }_{V_1 } \sf \dfrac{V}{e} \cdot dv[/tex]

[tex]\to \sf W_1 = \dfrac{1}{e} \left[ \dfrac{1}{2} \left( 1 - \dfrac{1}{e^2 } \right) \right][/tex]

Work done on path BC:

[tex]\to \sf W_2 = \int\limits^{V_3 }_{V_2 } \sf e^{-V} \cdot dv[/tex]

[tex]\to \sf W_2 = - ( e^{-2} - e^{1} )[/tex]

[tex]\to \sf W_2 = \dfrac{1}{e} - \dfrac{1}{e^2 }[/tex]

Adding the two work together:

[tex]\to \sf W_T = \dfrac{1}{e} - \dfrac{1}{e^2 } + \dfrac{1}{2e} - \dfrac{1}{2e^3 }[/tex]

[tex]\to \bf W_T = \dfrac{3}{2e} - \dfrac{1}{e^2 } - \dfrac{1}{2e^3 }[/tex]