Discussion
A fluid undergoes a process where its pressure (P), temperature (T) and volume (V) changes from \((P_{1}, T_{1}, V_{1})\) to \((P_{2}, T_{2}, V_{2})\). During the process, volume expansivity \((\beta)\) and isothermal compressibility \((\kappa_{T})\) remains constant. Given that \(\beta = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P}\) and \(\kappa_{T} = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T}\), the ratio \(\left(\frac{V_{2}}{V_{1}}\right)\) is
Answer: C
The ratio is \(\frac{\exp[\beta(T_{2}-T_{1})]}{\exp[\kappa_{T}(P_{2}-P_{1})]}\).
Write the total differential of V(T, P) using the two definitions:
\[ \begin{aligned} \frac{dV}{V} &= \beta\,dT - \kappa_{T}\,dP \\ \ln\frac{V_{2}}{V_{1}} &= \beta(T_{2}-T_{1}) - \kappa_{T}(P_{2}-P_{1}) \end{aligned} \]
with \(\beta\) and \(\kappa_{T}\) constant. Exponentiating gives the keyed expression.
The difference \(\beta\Delta T - \kappa_{T}\Delta P\) is ln(V2/V1), not the ratio itself, and the product and quotient forms do not come from integrating the differential.
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