 Solution

ID: past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/section-b/b3/c/solution

Path (b) changes pressure before temperature. Its pressure integral is unchanged, but the thermal expansion occurs at

P_{2}

W_{b} = \frac{1}{2} V_{1} κ_{T} (P_{2}^{2} - P_{1}^{2}) - V_{1} α P_{2} (T_{2} - T_{1}) .

Along the straight path (c),

P = P_{1} + (P_{2} - P_{1}) (T - T_{1}) / (T_{2} - T_{1})

. Thus

\int_{T_{1}}^{T_{2}} P d T = (P_{1} + P_{2}) (T_{2} - T_{1}) /2

, while

\int P d P

depends only on the pressure endpoints. Consequently

W_{c} = \frac{1}{2} V_{1} κ_{T} (P_{2}^{2} - P_{1}^{2}) - \frac{1}{2} V_{1} α (P_{1} + P_{2}) (T_{2} - T_{1}) = \frac{W _{a} + W _{b}}{2} .

This pressure-temperature path-work identity is a consequence of the linearized volume law and the straight interpolation. It need not hold for strongly varying compressibility or expansion coefficient.

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