15.2 The first law of thermodynamics and some simple processes  (Page 3/13)

$\text{PV}$ diagrams clearly illustrate that the work done depends on the path taken and not just the endpoints . This path dependence is seen in [link] (a), where more work is done in going from A to C by the path via point B than by the path via point D. The vertical paths, where volume is constant, are called isochoric processes. Since volume is constant, $\mathrm{\Delta }V=0$ , and no work is done in an isochoric process. Now, if the system follows the cyclical path ABCDA, as in [link] (b), then the total work done is the area inside the loop. The negative area below path CD subtracts, leaving only the area inside the rectangle. In fact, the work done in any cyclical process (one that returns to its starting point) is the area inside the loop it forms on a $\text{PV}$ diagram, as [link] (c) illustrates for a general cyclical process. Note that the loop must be traversed in the clockwise direction for work to be positive—that is, for there to be a net work output.

Total work done in a cyclical process equals the area inside the closed loop on a PV Diagram

Calculate the total work done in the cyclical process ABCDA shown in [link] (b) by the following two methods to verify that work equals the area inside the closed loop on the $\text{PV}$ diagram. (Take the data in the figure to be precise to three significant figures.) (a) Calculate the work done along each segment of the path and add these values to get the total work. (b) Calculate the area inside the rectangle ABCDA.

Strategy

To find the work along any path on a $\text{PV}$ diagram, you use the fact that work is pressure times change in volume, or $W=P\mathrm{\Delta }V$ . So in part (a), this value is calculated for each leg of the path around the closed loop.

Solution for (a)

The work along path AB is

Since the path BC is isochoric, $\mathrm{\Delta }{V}_{\text{BC}}=0$ , and so $W_{\text{BC}}=0$ . The work along path CD is negative, since $\mathrm{\Delta }{V}_{\text{CD}}$ is negative (the volume decreases). The work is

Again, since the path DA is isochoric, $\mathrm{\Delta }{V}_{\text{DA}}=0$ , and so $W_{\text{DA}}=0$ . Now the total work is

Solution for (b)

The area inside the rectangle is its height times its width, or