This is a Physical Chemistry Question and comes from the Atkins Physical Chemistry book.

Suppose that $2.0~\mathrm{mol}$ $\ce{H2}$ at $2.0~\mathrm{atm}$ and $25~^\circ\mathrm{C}$ and $4.0~\mathrm{mol}$ $\ce{N2}$ at $3.0~\mathrm{atm}$ and $25~^\circ\mathrm{C}$ are mixed at constant volume. Calculate $\Delta_\text{mix}G$. What would be the value of $\Delta_\text{mix}G$ had the pressures been identical initially?

The answers are given but I don't understand how to get to $-9.7~\mathrm{kJ}$ for the first part of the question (when initial pressures are different. The gases are initially at different pressures but does that make a difference. Can I still use Raoult's Law?


1 Answer 1

  • Let’s start with the last question: Calculate $\Delta_\text{mix}G$ when the pressures are equal in both compartments before mixing. Most physical chemistry textbooks treat this case (including Atkins), and the expression of $\Delta_\text{mix}G$ is: $$\Delta_\text{mix}G= nRT(x_1\ln x_1 +x_2\ln x_2)$$ In this exercise, $1$ denotes hydrogen, $2$ denotes nitrogen $n=6\ \text{mol}$, $x_1=\frac{2}{6}$ and $x_2=\frac{4}{6}$. So, by substituting these values in the above expression, we find: $$\Delta_\text{mix}G= 6\ \text{mol}\times 8.314\ \frac{\text{J}}{\text{mol}\cdot\text{K}}\times 298\ \text{K}\times(\frac{2}{6}\ln\frac{2}{6} +\frac{4}{6}\ln\frac{4}{6})= -9462\ \text{J}$$ As $\Delta_\text{mix}G<0$ at constant temperature and pressure, this means that the mixing of two ideal gases is a spontaneous process .

    • Let’s now treat the first question with two different pressure in the two compartments. The initial Gibbs energy is $$G_\text{i}=n_1(\mu_1^0 +RT\ln p_1)+n_2(\mu_2^0 +RT\ln p_2)$$ Where $n_1=2\ \text{mol}$, $n_2=4\ \text{mol}$, $p_1=2\ \text{atm}$, $p_2=3\ \text{atm}$. $$G_\text{i}=2\ \text{mol}\times(\mu_1^0 +RT\ln2)+4\ \text{mol}\times(\mu_2^0 +RT\ln3)$$ When the partition is removed, and each gas occupies the volume $V_1+V_2$. The partial pressure of hydrogen is $$p^\prime_1=\frac{2\ \text{mol}\times RT}{V_1+V_2}$$ The partial pressure of nitrogen is $$p^\prime_2= \frac{4\ \text{mol}\times RT}{V_1+V_2} $$ On the other hand: $$V_1=\frac{2\ \text{mol}\times RT}{2\ \text{atm}}=1\frac{\text{mol}}{\text{atm}}\times RT$$ $$V_2=\frac{4\ \text{mol}\times RT}{3\ \text{atm}}=\frac{4}{3}\frac{\text{mol}}{\text{atm}}\times RT$$ i.e. $$V_1+V_2=\frac{7}{3}\frac{\text{mol}}{\text{atm}}\times RT$$ So $$p^\prime_1=\frac{2\ \text{mol}\times RT}{(7/3\ \text{mol/atm})RT}=\frac{6}{7}\ \text{atm}$$ $$p^\prime_2= \frac{4\ \text{mol}\times RT}{(7/3\ \text{mol/atm})RT}=\frac{12}{7}\ \text{atm} $$ The final Gibbs energy is $$G_\text{f}=n_1(\mu_1^0 +RT\ln p^\prime_1)+n_2(\mu_2^0 +RT\ln p^\prime_2)$$ $$G_\text{f}=2\ \text{mol}\times(\mu_1^0 +RT\ln\frac{6}{7})+4\ \text{mol}\times(\mu_2^0 +RT\ln\frac{12}{7})$$ The Gibbs energy of mixing is the difference between $G_\text{f}$ and $G_\text{i}$ $$\Delta_\text{mix}G= 2\ \text{mol}\times RT\ln\frac{6/7}{2} +4\ \text{mol}\times RT\ln\frac{12/7}{3}$$ $$\Delta_\text{mix}G= 2\ \text{mol}\times RT\ln\frac{6}{14} +4\ \text{mol}\times RT\ln\frac{12}{21}$$ $$\Delta_\text{mix}G= 2\ \text{mol}\times 8.314\frac{\text{J}}{\text{mol}\cdot\text{K}}\times 298\ \text{K}\times(\ln\frac{3}{7} +2\times\ln\frac{4}{7})= -9744\ \text{J}\approx{-9.7\ \text{kJ}}$$ In this case, the value of $\Delta_\text{mix}G$ is the sum of two contributions: the mixing itself and the changes of pressure of the two gases to their final total pressure.

I hope it’s clear now.

  • 1
    $\begingroup$ Why isn't zero? I mean at constant $T$ and $p$ it follows that $dG=-SdT + Vdp=0$. $\endgroup$
    – ado sar
    Commented Jun 17, 2020 at 12:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.