Question

The molar excess Gibbs free energy ($g^E$) of a liquid mixture of $A$ and $B$ is given by \[ \frac{g^E}{RT} = x_A x_B [C_1 + C_2(x_A - x_B)] \] where $x_A$ and $x_B$ are the mole fraction of $A$ and $B$, respectively, the universal gas constant, $R = 8.314$ J K$^{-1}$ mol$^{-1}$, $T$ is the temperature in K, and $C_1, C_2$ are temperature-dependent parameters. At 300 K, $C_1 = 0.45$ and $C_2 = -0.018$. If $\gamma_A$ and $\gamma_B$ are the activity coefficients of $A$ and $B$, respectively, the value of \[ \int_{0}^{1} \ln\left(\frac{\gamma_A}{\gamma_B}\right)\,dx_A \] at 300 K and 1 bar is \(\underline{\hspace{2cm}}\) (rounded off to the nearest integer).

Hint:

For binary mixtures, expressing $g^E/RT$ as a function of $x_A$ lets you use $\ln(\gamma_A/\gamma_B)=d(g^E/RT)/dx_A$, which often turns integrals into simple end–point evaluations.

Answer 1

Correct Answer: 0

Let \[ \Phi(x_A) = \frac{g^E}{RT} = x_A x_B [C_1 + C_2(x_A - x_B)], x_B = 1 - x_A . \] For a binary mixture with molar excess Gibbs energy $\Phi(x_A)$, the activity coefficients satisfy
\[ \ln\gamma_A = \Phi + (1 - x_A)\Phi', \ln\gamma_B = \Phi - x_A\Phi', \] where $\Phi' = \dfrac{d\Phi}{dx_A}$.
Therefore,
\[ \ln\left(\frac{\gamma_A}{\gamma_B}\right) = \ln\gamma_A - \ln\gamma_B = \big[\Phi + (1 - x_A)\Phi'\big] - \big[\Phi - x_A\Phi'\big] = \Phi'. \] Hence, the required integral becomes
\[ \int_{0}^{1} \ln\left(\frac{\gamma_A}{\gamma_B}\right)\,dx_A = \int_{0}^{1} \Phi'\,dx_A = \Phi(1) - \Phi(0). \] At $x_A = 1$, $x_B = 0$, so $\Phi(1) = 1\cdot0[\cdots] = 0$.
At $x_A = 0$, $x_B = 1$, so $\Phi(0) = 0\cdot1[\cdots] = 0$.
Thus,
\[ \int_{0}^{1} \ln\left(\frac{\gamma_A}{\gamma_B}\right)\,dx_A = 0 - 0 = 0. \] Rounded to the nearest integer, the value is $0$.