A Closer Look at Heat and Work — Question 2

Question 2

A curved pressure law requires an integral rather than an endpoint average

During quasistatic expansion from 1.00 L to 3.00 L, the numerical pressure in kPa is $p(v)=100+20v^2$, where $v=V/(1\,\mathrm L)$. Use $W=\int p\,dV$ and $1\,\mathrm{kPa\,L}=1\,\mathrm J$.

1. Find the exact work and the error of the endpoint-trapezoid estimate.

2. Find the constant pressure that would perform the same work over the same volume interval.

Question 2 - Solution

A curved pressure law requires an integral rather than an endpoint average

(a) Method and calculation

$$W=\int_1^3(100+20v^2)dv\,\mathrm J=\left[100v+\frac{20v^3}{3}\right]_1^3\mathrm J=\frac{1120}{3}\,\mathrm J.$$$$W_{\mathrm{trap}}=\frac{120+280}{2}(3-1)=400\,\mathrm J.$$

(b) Interpretation and further calculation

$$W_{\mathrm{trap}}-W=26.6667\,\mathrm J,\quad\frac{W_{\mathrm{trap}}-W}{W}=\frac1{14}\approx7.143\%.$$$$p_{\mathrm{equivalent}}=\frac{W}{\Delta V}=186.667\,\mathrm{kPa}.$$The chord lies above this convex curve, explaining the positive trapezoid error.

Physical check

The equivalent pressure is a volume average, not necessarily the arithmetic mean of endpoint pressures.