Thermodynamics of Mixing Ideal Gases

Introduction

Two different ideal gases placed in separate compartments can spread through one another when the partition is removed. No chemical bond has to form, and at fixed temperature the ideal-gas enthalpy does not change. Yet mixing is favourable because there are more ways to distribute each species through the available space. Chemical potentials turn that intuition into precise Gibbs-energy and entropy formulas.

Core explanation

Take amounts n i of distinct ideal gases initially separated at the same T and pressure p. After mixing at the same T and total p, component i has partial pressure p i = x i p, where x i = n i/n total. Its chemical potential changes from μ i° + RT ln(p/p°) to μ i° + RT ln(x i p/p°), a difference RT ln x i. Thus Δ mixG = Σ i n i RT ln x i = n total RT Σ i x i ln x i. For a true mixture, every 0 < x i < 1, so ln x i < 0 and Δ mixG < 0.

For ideal gases, enthalpy depends only on temperature. Because the initial and final temperature are equal, Δ mixH = 0. Using G = H − TS at fixed T gives Δ mixS = −Δ mixG/T = −RΣ i n i ln x i, which is positive. The entropy increase expresses the larger number of accessible arrangements after each gas can occupy the whole volume. Ideal mixing at fixed T can therefore be spontaneous with no exothermic heat of mixing.

For two equal amounts, x A = x B = 1/2 and Δ mixS = 2nR ln 2 if each initial amount is n. If one species is present only in a tiny fraction, its per-mole chemical-potential change is large in magnitude, but its contribution to the total mixture entropy is weighted by its small amount. The limit x ln x → 0 as x → 0 prevents an infinite total contribution from a disappearing component.

The result is for distinct, ideal, nonreacting gases and a clearly defined comparison state. Removing a partition between two portions of the same gas at the same T and p does not create a new distinguishable mixture or the same entropy gain. Mixing nonideal gases can have nonzero enthalpy changes and activity or fugacity corrections. Likewise, if initial temperatures or pressures differ, additional thermal or mechanical changes accompany mixing and cannot be assigned solely to the formula above.

Step-by-step reasoning

Write the initial pressure of each separated gas and final partial pressure after mixing. Use μ i(final) − μ i(initial) = RT ln(p i/final divided by p i/initial). Multiply by n i and sum to obtain Δ mixG. At common initial and final p, simplify to nRTΣx i ln x i. Then use ideal-gas Δ mixH = 0 to find Δ mixS = −Δ mixG/T and check signs.

Visual explanation

Draw a box split into red A and blue B sides. Remove the partition and show both colours dispersed throughout. Beneath, plot μ A and μ B each falling by RT ln x i from their pure-gas values. Put Δ mixG below zero, Δ mixS above zero and Δ mixH at zero on a sign chart.

Real-world analogy

Two groups assigned separate halves of a large room have fewer possible seating arrangements than when everyone may choose seats throughout the room. Removing the divider increases arrangement possibilities. Molecules likewise gain accessible translational configurations, although their statistics involve indistinguishability and energy weighting rather than conscious seating choices.

Real-world example

Nitrogen and oxygen form an approximately ideal gas mixture at ordinary low pressures. Mixing separate samples at the same temperature and pressure gives a negative Gibbs-energy change and positive entropy of mixing, with negligible ideal-gas enthalpy of mixing. Their partial pressures, not the shared total pressure, determine each species' final chemical potential.

Why?

Before mixing, each species occupies a restricted volume. Afterwards, its mole fraction lowers its partial pressure at the same total p, and its μ falls by RT ln x i. Adding those weighted decreases gives the negative Gibbs-energy change. Because ideal-gas intermolecular interaction energy is absent in the model, the driver is entropic rather than an enthalpy release.

Common misconception

“No heat of mixing” does not mean “no thermodynamic drive to mix.” Δ mixH = 0 while Δ mixG < 0 because TΔ mixS is positive. Another error is to put ln x i directly in a total free energy without multiplying by each component amount. The formula also should not be applied as though two identical portions of the same gas were distinguishable species.

Worked example

Mix 1.00 mol A with 1.00 mol B at 298 K, initially separate at the same 1-bar pressure and finally at 1 bar total. Both mole fractions are 0.500. Δ mixG = RT[1.00 ln 0.5 + 1.00 ln 0.5] = −2RT ln 2 ≈ −3.44 kJ. Δ mixS = 2R ln 2 ≈ 11.5 J K⁻¹ and Δ mixH = 0. The signs agree with spontaneous ideal mixing at fixed T and p.

Quick check

1. What is the sign of Δ mixG for two distinct ideal gases initially separate at the same T and p? Answer: Negative. Each final mole fraction is below one, so every n iRT ln x i term is negative; the total Gibbs energy falls on mixing.

Exam focus

State the initial comparison state and use final partial pressures. At fixed common T and p for ideal gases, write Δ mixG = RTΣn i ln x i, Δ mixH = 0 and Δ mixS = −RΣn i ln x i. Show units and distinguish total n from component n i. Check that entropy is positive and Gibbs energy negative.

Advanced insight

The term −RΣn i ln x i is a macroscopic form of compositional information entropy for an ideal mixture. It depends on the number of distinguishable species categories and their fractions. Statistical counting with the proper indistinguishability factors removes an artificial mixing entropy when the compartments contain the same gas.

Summary

Mixing distinct ideal gases at the same T and p lowers each component's chemical potential by RT ln x i. Hence Δ mixG = RTΣn i ln x i < 0, Δ mixS = −RΣn i ln x i > 0 and Δ mixH = 0. The spontaneous drive is entropic under these ideal conditions, and formulas change when interactions or initial states differ.

Practice questions

1. For 2 mol A and 1 mol B, write the ideal-gas Δ mixG expression at temperature T without evaluating logarithms. Answer: x A = 2/3 and x B = 1/3, so Δ mixG = RT[2 ln(2/3) + 1 ln(1/3)] in joules if R is in SI units. 2. Why can an ideal-gas mixture form spontaneously even though Δ mixH = 0? Answer: Entropy increases, making Δ mixG = Δ mixH − TΔ mixS = −TΔ mixS negative at positive temperature. 3. Does removing a partition between equal portions of the same ideal gas at the same T and p give 2R ln 2 per pair of moles as a physical mixing entropy? Answer: No. The gases are identical and the macroscopic state is unchanged by removing the partition. Treating them as distinct species would spuriously overcount arrangements.