The molar internal energy () of a non-ideal gas confined in a closed rigid vessel follows the relati — Thermodynamics and Thermochemistry Chemistry Question
Question
The molar internal energy ($U$) of a non-ideal gas confined in a closed rigid vessel follows the relation $U = a + bT + cT^2$, where $a = 20 \text{ J mol}^{-1}$, $b = 10 \text{ J K}^{-1}\text{mol}^{-1}$, and $c = 2 \times 10^{-2} \text{ J K}^{-2}\text{mol}^{-1}$. Calculate the entropy change ($\Delta S$) when one mole of this gas is heated from $200 \text{ K}$ to $400 \text{ K}$. (Assume $\ln 2 = 0.7$)
💡 Solution & Explanation
Constant volume heating dictates $dU = C_v dT$. By differentiating $U(T)$ with respect to $T$, we obtain $C_v = \frac{dU}{dT} = b + 2cT$. The fundamental entropy change is $\Delta S = \int \frac{C_v}{T} dT = \int_{T_1}^{T_2} (\frac{b}{T} + 2c) dT = b \ln(\frac{T_2}{T_1}) + 2c(T_2 - T_1)$. Substituting the specific values: $\Delta S = 10 \ln(\frac{400}{200}) + 2(0.02)(400 - 200) = 10 \ln 2 + 0.04(200) = 10(0.7) + 8 = 15 \text{ J K}^{-1}$.