Gibbs Energy of Mixing Ideal Gases: 21-Mat-A1 Thermodynamics, May 2015, Q5

Question 5: Gibbs Energy of Mixing Ideal Gases (20 marks)

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Given. Ideal gases at $T=298$ K, total $P=1$ atm throughout. (a) 1 mol O₂ + 1 mol N₂. (b) 1 mol O₂ + 2 mol N₂. (c) 1 additional mole of pure O₂ added into an existing 1:1 mol O₂/N₂ mixture (from part a), forming a final mixture of 2 mol O₂ + 1 mol N₂.

Find. $\Delta G_{mix}$ for each of the three mixing processes.

Approach. For ideal gases, $\Delta G_{mix}=RT\sum_i n_i\ln x_i$ when pure components are combined from scratch. Part (c) is different in kind — it adds pure O₂ to a mixture that is already mixed — so its $\Delta G$ is the difference between the Gibbs energy of forming the final 3-mole mixture from scratch and that of the initial 2-mole mixture (the pure O₂ being added contributes no mixing term by itself).

(a) 1 mol O₂ + 1 mol N₂. $x_{O_2}=x_{N_2}=0.5$. $$\Delta G_{mix}=RT\big[1\ln(0.5)+1\ln(0.5)\big]=(8.314)(298)(2\ln0.5)=\boxed{-3434.6\ \text{J}}.$$
(b) 1 mol O₂ + 2 mol N₂. $x_{O_2}=1/3$, $x_{N_2}=2/3$. $$\Delta G_{mix}=RT\big[1\ln(1/3)+2\ln(2/3)\big]=(8.314)(298)(-1.9095)=\boxed{-4731.0\ \text{J}}.$$
(c) Adding 1 mol O₂ to the 1:1 mixture from (a). The final mixture is 2 mol O₂ + 1 mol N₂ ($x_{O_2}=2/3$, $x_{N_2}=1/3$). Its own from-scratch mixing energy uses the identical pair of mole-fraction terms as part (b) (just the O₂/N₂ labels swapped, 2 mol at $x=2/3$ and 1 mol at $x=1/3$), so numerically $$\Delta G_{mix}(\text{final, from scratch})=RT\big[2\ln(2/3)+1\ln(1/3)\big]=-4731.0\ \text{J (same value as part b).}$$ The process actually asked for is this final mixture minus the initial 1:1 mixture from part (a) — the incoming mole of O₂ is pure, so it carries zero mixing-Gibbs-energy of its own: $$\Delta G=\Delta G_{mix}(\text{final})-\Delta G_{mix}(\text{initial, part a})=-4731.0-(-3434.6)=\boxed{-1296.4\ \text{J}}.$$

Case	ΔG_mix
(a) 1 mol O₂ + 1 mol N₂	−3434.6 J
(b) 1 mol O₂ + 2 mol N₂	−4731.0 J
(c) +1 mol O₂ into the (a) mixture	−1296.4 J

Concept & Theory

Topic. Gibbs energy of mixing for ideal gases.

Key relations. $\Delta G_{mix}=RT\sum_i n_i\ln x_i$ (always $\le0$ since $x_i\le1$); $\Delta H_{mix}=0$ and $\Delta S_{mix}=-R\sum_in_i\ln x_i$ for ideal gases, so $\Delta G_{mix}=-T\Delta S_{mix}$ exactly.

Why this works. Ideal-gas molecules do not interact, so mixing changes only the configurational entropy (more accessible microstates for a mixture than for separated pure gases) and not the enthalpy — hence $\Delta G_{mix}$ is purely an entropic, always-favourable ($\Delta G<0$) effect at any composition and any $T>0$.

Common pitfall. Treating part (c) as if it were a fresh three-component mixing calculation with $\Delta G=RT[2\ln(2/3)+\ln(1/3)]$ outright — that is the Gibbs energy of forming the final mixture from pure components, not the Gibbs energy of the process described (adding pure O₂ to an already-mixed system), which requires subtracting off the mixing energy the system already had.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 8 (thermodynamics of solutions/ideal-gas mixing).

Question 5 of 7: Gibbs Energy of Mixing Ideal Gases

Question 5: Gibbs Energy of Mixing Ideal Gases (20 marks)