Assuming ideal gas behaviour, the for the isothermal mixing of 0.8 mole and 0.2 mole of is () — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$\Delta S_{\text{mix}} = -R(n_1 \ln x_1 + n_2 \ln x_2) = -2 [0.8 \ln(0.8) + 0.2 \ln(0.2)]$. We have $\ln(0.8) = \ln(8/10) = 3\ln 2 - \ln 10 = 2.1 - 2.3 = -0.2$. $\ln(0.2) = \ln(2/10) = \ln 2 - \ln 10 = 0.7 - 2.3 = -1.6$. $\Delta S_{\text{mix}} = -2[0.8(-0.2) + 0.2(-1.6)] = -2[-0.16 - 0.32] = +0.96 \text{ cal/K}$.