One mole of an ideal diatomic gas () undergoes a transformation from an initial state at and to a fi — Thermodynamics and Thermochemistry Chemistry Question
Question
One mole of an ideal diatomic gas ($C_v = 5 \text{ cal K}^{-1} \text{mol}^{-1}$) undergoes a transformation from an initial state at $25^\circ\text{C}$ and $1 \text{ L}$ to a final state at $100^\circ\text{C}$ and $10 \text{ L}$. What is the total entropy change for this process? ($R = 2 \text{ cal K}^{-1} \text{mol}^{-1}$)
💡 Solution & Explanation
The entropy change for an ideal gas undergoing simultaneous temperature and volume changes is given by $\Delta S = nC_v \ln(\frac{T_2}{T_1}) + nR \ln(\frac{V_2}{V_1})$. Converting given temperatures to absolute Kelvin: $T_1 = 298 \text{ K}$, $T_2 = 373 \text{ K}$. Volume expands from $1 \text{ L}$ to $10 \text{ L}$. Thus, $\Delta S = (1)(5) \ln(\frac{373}{298}) + (1)(2) \ln(\frac{10}{1}) = 5 \ln(\frac{373}{298}) + 2 \ln 10$.