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(A) The work done $W$ in a $P-V$ diagram is equal to the area under the curve.Since the process is represented by a straight line from $(P_1, V_1)$ to

Vedclass · Accepted Answer

$\frac{1}{2}(P_1 + P_2)(V_2 - V_1)$

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(A) The work done $W$ in a $P-V$ diagram is equal to the area under the curve.Since the process is represented by a straight line from $(P_1, V_1)$ to $(P_2, V_2)$, the area under the curve is the area of a trapezoid with parallel sides $P_1$ and $P_2$ and height $(V_2 - V_1)$.$W = \int_{V_1}^{V_2} P \, dV$For a straight line on a $P-V$ diagram, the area is given by the formula for the area of a trapezoid:$W = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes (	ext{distance between them})$$W = \frac{1}{2} (P_1 + P_2) (V_2 - V_1)$Note: The equation of state $P(V-b)=RT$ describes the path, but since the process is explicitly defined as a straight line on the $P-V$ diagram, the work done is simply the area of the trapezoid formed under that line.

One mole of a gas obeying the equation of state $P(V-b)=RT$ is made to expand from a state with coordinates $(P_{1}, V_{1})$ to a state with $(P_{2}, V_{2})$ along a process that is depicted by a straight line on a $P-V$ diagram. Then, the work done is given by

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