Question

Verify that the area under the p-V curve has dimensions of work.

Answer 1

\[ \begin{array}{l} \text{Area under the } P-V \text{ curve is } \displaystyle\int_{V_i}^{V_f} P\,dV \text{, where } P \text{ is the} \\[6pt] \text{pressure and } V \text{ is the volume.} \\[8pt] [\text{pressure}] = \dfrac{[\text{force}]}{[\text{area}]} = \dfrac{[\mathrm{MLT^{-2}}]}{[\mathrm{L^2}]} \\[10pt] [\text{Volume}] = [\mathrm{M^0L^3T^0}] \\[10pt] \begin{aligned} \therefore\ [\text{Pressure} \times \text{Volume}] &= \frac{[\mathrm{MLT^{-2}}]}{[\mathrm{L^2}]} \cdot [\mathrm{L^3}] \\[4pt] &= [\mathrm{MLT^{-2}}]\ [\mathrm{L}] = [\text{force}]\ [\text{distance}] \\[4pt] &= [\text{work}] \end{aligned} \end{array} \]