A definite mass of a monoatomic ideal gas at 1 bar and expands against vacuum from to . The change i — Thermodynamics and Thermochemistry Chemistry Question
Question
A definite mass of a monoatomic ideal gas at 1 bar and $27^\circ\text{C}$ expands against vacuum from $1.2 \text{ dm}^3$ to $2.4 \text{ dm}^3$. The change in free energy of the gas, $\Delta G$, is ($R = 0.08 \text{ bar-L/K-mol}, \ln 2 = 0.7$)
Answer: D
💡 Solution & Explanation
Free expansion is isothermal, meaning $\Delta H = 0$. Thus $\Delta G = -T\Delta S = -T(nR \ln(V_2/V_1)) = -P_1 V_1 \ln(V_2/V_1)$. Substituting initial values: $\Delta G = -(1 \text{ bar})(1.2 \text{ L}) \ln 2 = -1.2(0.7) = -0.84 \text{ bar-L}$. Since $1 \text{ bar-L} = 100 \text{ J}$, $\Delta G = -84 \text{ J}$.
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