Chemical Potential of an Ideal Gas

Pump up a bicycle tire and the air inside gains more than pressure. It gains a thermodynamic "eagerness" to escape, expand, or react. That eagerness is the chemical potential, and for gases it has one of the cleanest formulas in all of thermodynamics. This relationship matters because chemical potential is the master variable of equilibrium. Reaction direction, phase stability, and equilibrium constants all reduce to comparing chemical potentials. For ideal gases the dependence on pressure is exactly logarithmic, which makes every downstream calculation tractable. This article derives the pressure dependence of $\mu$ from the Gibbs energy of an ideal gas. It extends the result to gas mixtures using partial pressures. It closes with two worked numerical examples and a preview of activities and equilibrium constants.

Key takeaways

The chemical potential of a pure ideal gas is $\mu = \mu^\circ + RT\ln(p/p^\circ)$ , with $p^\circ = 1\,\text{bar}$ .
This comes directly from $dG = V\,dp$ at constant $T$ , integrated using $V = nRT/p$ .
In a mixture of ideal gases, each component obeys the same law using its own partial pressure $p_i$ .
Doubling pressure raises $\mu$ by $RT\ln 2$ , about 1.7 kJ/mol at 298 K.
Chemical potential differences, not absolute values, drive spontaneous change and equilibrium.
Key idea: for an ideal gas, chemical potential rises logarithmically with pressure or partial pressure, each species behaving as if the others were not there.

The core insight

Because the molar Gibbs energy of an ideal gas increases as $RT\ln p$ (relative to a standard state), and because chemical potential is just molar Gibbs energy for a pure substance, the chemical potential of an ideal gas—pure or in a mixture, using its partial pressure—is $\mu = \mu^\circ + RT\ln(p/p^\circ)$ .

The physical picture

Chemical potential, $\mu$ , measures how much the Gibbs energy of a system changes when you add one more mole of a substance, at constant temperature, pressure, and composition of everything else. For a pure substance it equals the molar Gibbs energy, $G_m$ . Gases are the simplest system in which to see how $\mu$ depends on a state variable, because the equation of state is simple: $pV = nRT$ . Picture a fixed amount of ideal gas at fixed temperature. As you compress it, the molecules are squeezed into a smaller volume. They collide with walls more often and more energetically in an entropic sense: the number of accessible microstates per particle shrinks. Gibbs energy rises. Since $G = n\mu$ for a pure gas, the chemical potential rises too. The relationship is not linear in $p$ ; it is logarithmic, because the entropy of an ideal gas depends on the logarithm of volume, and $V \propto 1/p$ at fixed $n, T$ . Now consider a mixture: nitrogen and oxygen in air. Each gas, being ideal, does not "feel" the other. Ideal gas molecules occupy space and exchange energy without intermolecular forces, so each component's chemical potential depends only on its own partial pressure, $p_i = x_i p_\text{total}$ , exactly as if it alone filled the container at pressure $p_i$ . This is the microscopic reason the mixture formula is simply the pure-gas formula with $p$ replaced by $p_i$ .

Figure

Chemical potential of an ideal gas versus pressure at 298 K, showing the logarithmic rise through the standard state at 1 bar

The core relations

\mu(T,p) = \mu^\circ(T) + RT\ln\left(\frac{p}{p^\circ}\right)

where $\mu^\circ(T)$ is the standard chemical potential at $p^\circ = 1\ \text{bar}$ and the same temperature $T$ .

\mu_i(T,p) = \mu_i^\circ(T) + RT\ln\left(\frac{p_i}{p^\circ}\right)

where $p_i = x_i\,p$ is the partial pressure of component $i$ in an ideal gas mixture, $x_i$ its mole fraction.

G = \sum_i n_i \mu_i

where the total Gibbs energy of the mixture is the mole-weighted sum of the partial chemical potentials.

Derivation AFrom dG = V dp to μ = μ° + RT ln(p/p°)

Goal. Show that the molar Gibbs energy—and hence the chemical potential—of a pure ideal gas depends on pressure as $\mu^\circ + RT\ln(p/p^\circ)$ , starting from the fundamental relation for $G$ .

Step 1. Start from the pressure dependence of Gibbs energy. For a closed system of fixed composition, the combined first and second law gives $dG = -S\,dT + V\,dp$ . At constant temperature, $dT = 0$ , so

dG = V\,dp \quad (\text{constant } T).

Step 2. Convert to molar quantities. Divide through by the (fixed) amount of substance $n$ . Since $G/n = G_m = \mu$ for a pure substance,

d\mu = V_m\,dp \quad (\text{constant } T),

where $V_m$ is the molar volume.

Step 3. Insert the ideal gas equation of state. For one mole of ideal gas, $pV_m = RT$ , so $V_m = RT/p$ . Substituting,

d\mu = \frac{RT}{p}\,dp \quad (\text{constant } T).

Step 4. Integrate from the standard pressure to the pressure of interest. Integrate both sides from a reference state $(p^\circ, \mu^\circ)$ to the current state $(p, \mu)$ , holding $T$ fixed so $RT$ leaves the integral:

\int_{\mu^\circ}^{\mu} d\mu = RT\int_{p^\circ}^{p}\frac{dp'}{p'}.

Step 5. Evaluate both integrals. The left side gives $\mu - \mu^\circ$ . The right side gives $RT\ln(p/p^\circ)$ , since $\int dp'/p' = \ln p'$ . So

\mu - \mu^\circ = RT\ln\left(\frac{p}{p^\circ}\right).

Step 6. Extend to a mixture component. In an ideal gas mixture, each species behaves as though it alone occupied the container, exerting only its partial pressure $p_i$ and contributing independently to the total pressure ( $p = \sum_i p_i$ , Dalton's law). Because there are no intermolecular forces distinguishing species in the ideal limit, the same integration applies to each component using $p_i$ in place of $p$ :

\mu_i - \mu_i^\circ = RT\ln\left(\frac{p_i}{p^\circ}\right).

Boxed result.

\boxed{\mu_i(T,p) = \mu_i^\circ(T) + RT\ln\left(\frac{p_i}{p^\circ}\right)}

reducing to $\mu = \mu^\circ + RT\ln(p/p^\circ)$ for a pure gas, where $p_i = p$ .

Worked examples

Example 1 — Pure nitrogen gas compressed. A cylinder holds pure N₂(g) at 298.15 K. Find the change in chemical potential when the pressure is raised from 1.00 bar (the standard pressure) to 50.0 bar, assuming ideal behavior. Use

\Delta\mu = \mu(50\ \text{bar}) - \mu^\circ = RT\ln\left(\frac{p}{p^\circ}\right).

With $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ , $T = 298.15\ \text{K}$ , $p/p^\circ = 50.0$ :

RT = 8.314 \times 298.15 = 2478.8\ \text{J mol}^{-1}

\ln(50.0) = 3.912

\Delta\mu = 2478.8 \times 3.912 = 9698\ \text{J mol}^{-1} = 9.70\ \text{kJ mol}^{-1}

Sanity check: pressure rose 50-fold, well above 1 bar, so $\mu$ should rise substantially and positively—consistent with compression driving the gas toward higher Gibbs energy, favoring expansion back toward 1 bar. The size (about four times the "one $RT$ " scale of 2.48 kJ/mol) is reasonable for a factor of 50 inside a logarithm.

Example 2 — Partial pressures in a reacting mixture (ammonia synthesis feed). A gas mixture at 400 K and total pressure 200 bar contains N₂, H₂, and NH₃ in mole fractions $x_{\text{N}_2} = 0.225$ , $x_{\text{H}_2} = 0.675$ , $x_{\text{NH}_3} = 0.100$ . Treating the mixture as ideal, find the chemical potential of each species relative to its own standard state, $\mu_i - \mu_i^\circ$ , at this composition and pressure.

Step 1: partial pressures.

p_{\text{N}_2} = x_{\text{N}_2}\,p = 0.225 \times 200\ \text{bar} = 45.0\ \text{bar}

p_{\text{H}_2} = 0.675 \times 200\ \text{bar} = 135.0\ \text{bar}

p_{\text{NH}_3} = 0.100 \times 200\ \text{bar} = 20.0\ \text{bar}

Check: $45.0 + 135.0 + 20.0 = 200.0$ bar. ✓

Step 2: RT at 400 K.

RT = 8.314 \times 400 = 3325.6\ \text{J mol}^{-1}

Step 3: apply $\mu_i - \mu_i^\circ = RT\ln(p_i/p^\circ)$ to each gas (with $p^\circ = 1$ bar, so $p_i/p^\circ$ is numerically the pressure in bar): For N₂: $\ln(45.0) = 3.807$

\mu_{\text{N}_2} - \mu_{\text{N}_2}^\circ = 3325.6 \times 3.807 = 12{,}660\ \text{J mol}^{-1} = 12.66\ \text{kJ mol}^{-1}

For H₂: $\ln(135.0) = 4.905$

\mu_{\text{H}_2} - \mu_{\text{H}_2}^\circ = 3325.6 \times 4.905 = 16{,}310\ \text{J mol}^{-1} = 16.31\ \text{kJ mol}^{-1}

For NH₃: $\ln(20.0) = 2.996$

\mu_{\text{NH}_3} - \mu_{\text{NH}_3}^\circ = 3325.6 \times 2.996 = 9963\ \text{J mol}^{-1} = 9.96\ \text{kJ mol}^{-1}

Sanity check: all three values are positive because every partial pressure exceeds 1 bar, correctly raising each $\mu_i$ above its standard value. The ordering tracks the partial pressures: H₂ has the highest partial pressure and the largest boost, NH₃ the smallest. These numbers are exactly the pieces needed later to build $\Delta_rG$ and the reaction quotient $Q_p$ for the Haber process.

Limiting cases & common traps

Situation	What happens	Why
$p \to p^\circ$	$\mu \to \mu^\circ$	$\ln 1 = 0$
$p \to 0$	$\mu \to -\infty$	Gas becomes infinitely dilute, entropy diverges
$p \to \infty$ (ideal model)	$\mu \to +\infty$	Formula still logarithmic, no physical divergence at real high $p$
Real gas at high pressure	Formula breaks down	Must replace $p$ with fugacity $f$ (foreshadowing activity and fugacity)
Mixture component	Use $p_i$ , not total $p$	Each ideal gas responds only to its own partial pressure

The chemist behind it

The mathematical machinery here—Gibbs energy, chemical potential, and their pressure and composition dependence—traces to Josiah Willard Gibbs, whose 1876 and 1878 papers "On the Equilibrium of Heterogeneous Substances" essentially invented chemical thermodynamics from scratch. Gibbs defined chemical potential as a partial molar quantity and showed it governs phase and reaction equilibrium, decades before physical chemists had the vocabulary of partial pressures and standard states to apply it neatly to gases. The logarithmic pressure dependence used constantly by chemists today, from vapor pressures to equilibrium constants, is a direct descendant of Gibbs's abstract formalism.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American mathematical physicist who founded chemical thermodynamics and defined the chemical potential

Ask the tutor: How does replacing pressure with fugacity fix this formula for real, non-ideal gases?

Why does chemical potential depend on ln p rather than p itself?

The logarithm traces back to the ideal gas law entering through $V_m = RT/p$ in $d\mu = V_m\,dp$ . Integrating $dp/p$ produces a natural logarithm. Physically, molar volume shrinks as $1/p$ , so each successive increment of pressure squeezes the gas proportionally less in absolute volume terms but represents the same relative compression. Entropy, and hence Gibbs energy, tracks that relative, multiplicative change, which is exactly what a logarithm captures.

What exactly is the standard state for a gas, and why 1 bar?

The standard state is the pure ideal gas at pressure $p^\circ = 1$ bar exactly, at whatever temperature $T$ is specified. IUPAC adopted 1 bar (rather than the older 1 atm = 1.01325 bar) as the standard pressure in 1982. The gas is treated as ideal even if the real gas is not, meaning tabulated standard values already extrapolate to ideal behavior at that reference pressure.

Does this formula apply to real gases like CO2 at high pressure?

Not accurately. Real gases deviate from ideality especially at high pressure or low temperature, where intermolecular attractions and finite molecular volume matter. The fix is to replace pressure with fugacity, $f$ , an effective pressure defined so that $\mu = \mu^\circ + RT\ln(f/p^\circ)$ remains exactly true. Fugacity equals pressure in the ideal limit and corrects for nonideality otherwise—the subject of the next topic in this sequence.

How does this connect to equilibrium constants like Kp?

At equilibrium, the sum of $\nu_i\mu_i$ over products minus reactants equals zero. Substituting $\mu_i = \mu_i^\circ + RT\ln(p_i/p^\circ)$ for every species and collecting terms yields $\Delta_rG^\circ = -RT\ln K_p$ , where $K_p$ is built from equilibrium partial pressures raised to their stoichiometric powers. The logarithmic pressure dependence derived here is the direct ancestor of that equilibrium expression.

Is μ the same as Gibbs energy for a gas?

Only for a pure substance, where $\mu = G_m$ , the molar Gibbs energy. In a mixture, $\mu_i$ is a partial molar Gibbs energy: the slope of total $G$ with respect to moles of species $i$ , holding $T$ , $p$ , and all other mole numbers fixed. Total $G$ for the mixture is then $\sum_i n_i\mu_i$ , not simply a molar average.