One mole of an ideal gas undergoes a reversible process in which the entropy of the gas changes with — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Standard entropy differential is $dS = C_V \frac{dT}{T} + R \frac{dV}{V}$. Differentiating the given function $S(T)$ yields $dS = a dT + C_V \frac{dT}{T}$. Equating them leaves $a dT = R \frac{dV}{V}$. Integrating from initial state ($T_o, V_o$) to ($T, V$) results in $a(T - T_o) = R \ln\left(\frac{V}{V_o}\right) \implies T = T_o + \frac{R}{a} \ln\left(\frac{V}{V_o}\right)$.