Chemical Potential of an Ideal Gas

Introduction

Compressing an ideal gas changes its tendency to move, mix and react even when temperature stays constant. Chemical potential measures that tendency. For an ideal gas, its pressure dependence has a simple logarithmic form. The pressure must be compared with a reference pressure so the logarithm receives a dimensionless argument; that reference defines the standard chemical potential.

Core explanation

For one pure component, μ is its molar Gibbs energy. The Gibbs differential at fixed composition is dG = −S dT + V dp. For one mole at constant T, dμ = V m dp. The ideal-gas molar volume is V m = RT/p, so dμ = RT dp/p = RT dln p. Integrate from a chosen standard pressure p° to p at the same temperature: μ(T,p) − μ°(T) = RT ln(p/p°). Therefore μ(T,p) = μ°(T) + RT ln(p/p°).

The usual standard pressure in modern thermodynamic conventions is p° = 1 bar. The standard chemical potential μ° depends on temperature and on the substance, but it does not change when the actual gas pressure changes at fixed T. The ratio p/p° is dimensionless, as a logarithm requires. If p doubles, Δμ = RT ln 2, regardless of the starting pressure within the ideal-gas model. If p is below p°, the logarithmic term is negative; this does not make chemical potential or the gas itself unphysical.

For component i in an ideal-gas mixture, replace total pressure with partial pressure p i = y i p total: μ i = μ i° + RT ln(p i/p°). This explains why mixing ideal gases at fixed T and overall p lowers each component's μ when its mole fraction falls below one. A pure-gas pressure and a mixture total pressure are not interchangeable for component chemical potential. The relevant variable is the species' own partial pressure.

The formula is the foundation for gas reaction equilibria. For reaction stoichiometric coefficients ν i, Δ rG = Σν iμ i = Δ rG° + RT ln Q p, where Q p is constructed from dimensionless partial-pressure activities (p i/p°)^ν i. At equilibrium Δ rG = 0, giving Δ rG° = −RT ln K. Thus the standard-state convention cancels consistently when calculating physical equilibria.

Step-by-step reasoning

Start from (∂μ/∂p) T = V m. Insert V m = RT/p only if the gas is ideal. Integrate over pressure at fixed T, explicitly choosing p° as the lower limit. Use a dimensionless ratio inside ln. For a mixture, calculate p i = y i p total first. Check whether the question asks for absolute μ relative to μ° or merely a difference between two pressures, where μ° cancels.

Visual explanation

Plot μ against ln p. The ideal-gas relation is a straight line with slope RT at fixed T, crossing μ° at p = p°. A second plot against p itself is curved, rising rapidly at low p and more slowly at high p. Mark a mixture component's partial pressure to show why it sits below the pure-gas total-pressure point.

Real-world analogy

An item's value in a crowded market can change when its availability changes; a logarithmic scale responds to ratios rather than fixed increments. Chemical potential likewise changes by the same amount for any pressure doubling at a fixed temperature. The analogy is about ratio sensitivity; it does not imply molecules make economic choices.

Real-world example

In a gas mixture at 1 bar total pressure, a species with mole fraction 0.10 has p i = 0.10 bar. Relative to its pure 1-bar standard state, its ideal-gas chemical potential is lower by RT ln 0.10. This contribution to Gibbs energy helps explain why ideal gases spontaneously mix and why reactant or product partial pressures affect reaction direction.

Why?

The logarithm arises because the molar volume of an ideal gas is inversely proportional to pressure. Each additional pressure increase contributes V m dp to μ, but V m shrinks as p rises. Integrating dp/p gives ln p. The reference pressure is a mathematical and thermodynamic convention that makes the integral's constant explicit.

Common misconception

Writing ln p when p carries units is incomplete; use ln(p/p°). In a mixture, substituting total pressure for each component's partial pressure loses the entropy-of-mixing contribution. Also, μ° is not a universal zero or the chemical potential of every substance at 1 bar; it is species- and temperature-specific.

Worked example

At 298 K, an ideal gas is compressed from 1.00 bar to 4.00 bar. Δμ = RT ln(4.00/1.00). With R = 8.314 J mol⁻¹ K⁻¹, RT ≈ 2478 J mol⁻¹, so Δμ ≈ 2478 × 1.386 = 3435 J mol⁻¹, or 3.44 kJ mol⁻¹. The result is positive: at the higher pressure, transferring an infinitesimal amount of gas into that phase is less favourable than at the lower pressure, all else equal.

Quick check

1. At the same T, what is μ(T,p°) in the ideal-gas expression? Answer: Since p/p° = 1 and ln 1 = 0, μ(T,p°) = μ°(T). The standard chemical potential is defined at that reference state.

Exam focus

Keep temperature fixed while integrating dμ = V m dp. Use partial pressure for a component in a mixture and 1 bar only if that is the stated standard pressure. The numerical logarithm must be dimensionless. A chemical-potential difference for two pressures is RT ln(p₂/p₁), which avoids an unnecessary μ°.

Advanced insight

The ideal-gas standard state can be defined as a hypothetical ideal gas at p°, even if a real substance at that pressure is measurably nonideal. This choice makes μ° a consistent reference for real-gas fugacity formulations. The physical chemical potential remains independent of the arbitrary convention when activities and standard values are transformed together.

Summary

At fixed temperature, ideal-gas V m = RT/p leads to μ = μ° + RT ln(p/p°). Pressure ratios determine chemical-potential differences. For an ideal mixture, each species uses its partial pressure. This logarithmic term connects gas mixing and reaction equilibria to measurable composition and pressure.

Practice questions

1. At 300 K, how much does an ideal-gas μ change when pressure is halved? Answer: Δμ = RT ln(1/2) = −RT ln 2 ≈ −(8.314)(300)(0.693) = −1.73 kJ mol⁻¹. 2. A component has y i = 0.20 at total pressure 5.0 bar. What pressure goes in its ideal-gas μ formula? Answer: Its partial pressure is p i = y i p total = 1.0 bar. With p° = 1 bar, its logarithmic term is zero at that state. 3. Why can two gases at the same total pressure have different chemical potentials? Answer: Their μ° values differ by substance, and in a mixture their partial pressures may also differ. Equal total pressure alone does not force equal species chemical potentials.