= Solution
Path (b) changes <pressure> before <temperature>. Its <pressure> integral is unchanged, but the thermal expansion occurs at $P_2$:
$$
\boxed{W_b=\frac12V_1\kappa_T(P_2^2-P_1^2)-V_1\alpha P_2(T_2-T_1).}
$$
Along the straight path (c), $P=P_1+(P_2-P_1)(T-T_1)/(T_2-T_1)$. Thus $\int_{T_1}^{T_2}P\,dT=(P_1+P_2)(T_2-T_1)/2$, while $\int P\,dP$ depends only on the <pressure> endpoints. Consequently
$$
\boxed{W_c=\frac12V_1\kappa_T(P_2^2-P_1^2)
-\frac12V_1\alpha(P_1+P_2)(T_2-T_1)=\frac{W_a+W_b}{2}.}
$$
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.
\Image[/past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1-pressure-temperature-paths.png]
{title=The three pressure-temperature integration paths have the same endpoints but different work}
{description=Original schematic with symbolic endpoint coordinates. Path a heats first, path b compresses first, and path c interpolates linearly.}
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