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21-Mat-A1 Thermodynamics · May 2014

Question 6 of 7: Gibbs Energy of Mixing of Ideal Gases

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2014 — 10-Met-A1 Metallurgical Thermodynamics. Three-hour, closed-book exam using an approved (Casio or Sharp) calculator; candidates were told to state any interpretive assumptions. Any five of the seven questions constitute a complete paper — all seven are solved below for completeness. All questions are of equal value (20 marks each out of 100).

Reference texts: Gaskell, D. R., Introduction to the Thermodynamics of Materials (2nd–5th ed.) — the exam's own Ellingham diagram (Fig. 9-3) is reproduced from this text, which also supplies the standard-state enthalpy/entropy data used below; supporting 298 K entropies for the elements, CO, H₂, H₂O(g) and CO₂ from the NIST-JANAF Thermochemical Tables.

Question 6: Gibbs Energy of Mixing of Ideal Gases (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Ideal gases O₂ and N₂ at $T=298$ K, 1 atm; three mixing scenarios as stated.

Find. $\Delta G_{mix}$ for each case, where case (c) is the additional Gibbs energy change of adding one more mole of O₂ to an already-mixed 1:1 O₂/N₂ mixture (not a mixing-from-scratch calculation).

Approach. Use the ideal-solution mixing formula $\Delta G_{mix}=RT\sum_i n_i\ln x_i$ directly for (a) and (b). For (c), compute $\Delta G_{mix}$ of the mixture built entirely from pure components at the final composition (2 mol O₂ + 1 mol N₂) and subtract the $\Delta G_{mix}$ already "spent" forming the initial 1:1 mixture — the difference is the Gibbs energy of the actual process asked for.

  1. (a) Equimolar mixing. $x_{O_2}=x_{N_2}=0.5$: $$\Delta G_{mix}=RT\big[n_{O_2}\ln x_{O_2}+n_{N_2}\ln x_{N_2}\big]=(8.314)(298)(2)\ln(0.5)=\boxed{-3.43\ \text{kJ}}.$$
  2. (b) 1 mol O₂ with 2 mol N₂. $x_{O_2}=1/3$, $x_{N_2}=2/3$: $$\Delta G_{mix}=(8.314)(298)\big[(1)\ln(1/3)+(2)\ln(2/3)\big]=\boxed{-4.73\ \text{kJ}}.$$
  3. (c) Adding 1 mol O₂ to an existing 1:1 mixture. The final state is 2 mol O₂ + 1 mol N₂ ($x_{O_2}=2/3$, $x_{N_2}=1/3$) — by symmetry with step 2 this mixture-from-pure-components has the identical $\Delta G_{mix,\,final}=-4.73$ kJ. The process asked for is (final mixture from pure components) minus (initial 1:1 mixture already formed, from part a) minus (pure O₂ being added, which contributes 0 since it starts and stays unmixed until this step): $$\Delta G_{(c)}=\Delta G_{mix,\,final}-\Delta G_{mix,\,initial}=-4.73-(-3.43)=\boxed{-1.30\ \text{kJ}}.$$
CaseΔGmix (kJ)
(a) 1 mol O₂ + 1 mol N₂−3.43
(b) 1 mol O₂ + 2 mol N₂−4.73
(c) 1 mol O₂ added to an existing 1:1 mixture−1.30