A gas expands from an initial volume of 0.015 m³ to a final volume of 0.075 m³, while its pressure increases linearly with the volume (so that the process follows a straight-line path in a p-V diagram) from 80 kPa to 275 kPa. How much work is done by the system?

Question

Quizplus · Accepted Answer

The work done by the system during a process where pressure changes linearly with volume can be calculated as the area under the p-V curve, which in this case is a trapezoid. The formula for the work done in such a process is $W = \frac{1}{2}(P_{1} + P_{2})(V_{2} - V_{1})$, where $P_{1}$ and $P_{2}$ are the initial and final pressures, and $V_{1}$ and $V_{2}$ are the initial and final volumes, respectively. Substituting the given values: $W = \frac{1}{2}(80\,kPa + 275\,kPa)(0.075\,m^3 - 0.015\,m^3) = \frac{1}{2}(355\,kPa)(0.060\,m^3) = \frac{1}{2}(355 \times 10^3\,Pa)(0.060\,m^3) = \frac{1}{2}(21300\,Pa \cdot m^3) = 10650\,J = 10.65\,kJ$, which rounds to 10.6 kJ.

A Gas Expands from an Initial Volume of 0