Chemical potential

Definition. In a mixture, adding an infinitesimal amount of species $i$ while holding $T$, $P$, and every other mole number fixed changes the total Gibbs energy by an amount per mole equal to $\mu_i$:

where: $G$ is total Gibbs energy, $n_i$ is moles of species $i$, and the subscript $n_{j\ne i}$ means every other component's mole number is held fixed. Because it is a ratio of an extensive change to an infinitesimal mole change, $\mu_i$ is intensive — it depends only on composition, $T$ and $P$, never on the total size of the system.

Any potential, same $\mu_i$. $G$ is the natural choice because $T,P$ are usually the controlled variables in a lab, but $\mu_i$ can equally be written as a derivative of $U$, $H$, or $A$ under their own natural variables:

All four expressions describe the same state and give the identical number — $\mu_i$ is one physical property of the system, only the bookkeeping (which variables are held fixed) changes.

The mixture rule for an ideal gas. For a pure ideal gas at pressure $p$, referenced to a standard pressure $p^\circ$ (1 bar):

where: $\mu^\circ(T)$ is the standard chemical potential (a function of $T$ alone, fixed by convention at $p=p^\circ$), $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$, and $T$ is absolute temperature. In a mixture of ideal gases, the same formula holds for each component with its own partial pressure $p_i$ in place of $p$ — total pressure never appears in this formula.

The flow rule. Starting from the fundamental relation $dG=-S\,dT+V\,dP+\sum_i\mu_i\,dn_i$, at constant $T,P$ this collapses to $dG=\sum_i\mu_i\,dn_i\le 0$ for any spontaneous change. If species $i$ can move between two regions without a net change in total moles, transfer proceeds from high $\mu_i$ to low $\mu_i$, and stops — reaching equilibrium — exactly when $\mu_i$ is equal in every region species $i$ can reach.

For an ideal-gas mixture at 298 K containing $\text{N}_2$ at partial pressure 0.8 bar and $\text{O}_2$ at partial pressure 0.2 bar (standard state $p^\circ=1$ bar), find $\mu_{N_2}-\mu_{N_2}^\circ$ and $\mu_{O_2}-\mu_{O_2}^\circ$, then use the results to decide which phase water occupies at 25°C, 1 atm in an open dish.

Step 1. Apply $\mu_i-\mu_i^\circ=RT\ln(p_i/p^\circ)$ to nitrogen. $\mu_{N_2}-\mu_{N_2}^\circ=(8.314)(298)\ln(0.8/1)=2477.6\times\ln(0.8)=2477.6\times(-0.2231)=-552.7\ \text{J mol}^{-1}$.

Step 2. Apply the same formula to oxygen, which is more dilute. $\mu_{O_2}-\mu_{O_2}^\circ=(8.314)(298)\ln(0.2/1)=2477.6\times\ln(0.2)=2477.6\times(-1.6094)=-3987.4\ \text{J mol}^{-1}$.

Step 3. Interpret the sizes. Both shifts are negative, as expected below standard pressure, but the $\text{O}_2$ shift is roughly seven times larger even though its pressure is only four times smaller than $\text{N}_2$'s — because $\mu$ depends on $P$ logarithmically, the more dilute species is pulled down disproportionately far from its reference value. This same logarithmic sensitivity is what later produces the $-RT\ln K$ relation between Gibbs energy and the equilibrium constant.

Step 4. Apply the flow rule to water in an open dish at 25°C, 1 atm. For a pure substance, $\mu^{(l)}=G_m^{(l)}$ and $\mu^{(v)}=G_m^{(v)}$. At the normal boiling point (100°C, 1 atm) the two phases coexist, so $\mu^{(l)}=\mu^{(v)}$ there by definition. At 25°C in an open dish, water vapour never accumulates to its saturation partial pressure, so using $\mu^{(v)}=\mu^{\circ(v)}+RT\ln(p_{H_2O}/p^\circ)$, the lower actual $p_{H_2O}$ makes $\mu^{(v)}$ smaller than the value at which $\mu^{(v)}=\mu^{(l)}$ would hold. Answer: hence $\mu^{(l)}>\mu^{(v)}$ under open-dish conditions, so liquid water spontaneously converts to vapour — the dish evaporates — until it empties or (in a closed container) $p_{H_2O}$ rises enough to restore equality.

• Writing $\mu_i=G/n_i$ (Gibbs energy per mole of the mixture) instead of the correct partial derivative $(\partial G/\partial n_i)_{T,P,n_j}$ — the two coincide only for a single pure substance, never for a genuine mixture where cross-terms between components matter.
• Using total pressure $P$ instead of the component's own partial pressure $p_i$ in the ideal-gas chemical potential formula for a mixture — a very common substitution error.
• Assuming $\mu_i$ has some absolute zero or must be positive — like $G$ itself, $\mu_i$ is only meaningfully defined relative to a reference/standard state $\mu_i^\circ$; only differences and equalities between phases carry physical meaning.
• Thinking equilibrium requires $\mu_i=0$ — the correct condition is $\mu_i$ equal across every phase or location species $i$ can access, not equal to zero.

Ask the tutor: If "matter flows from high mu to low mu" still feels like a hand-wavy analogy, ask the AI tutor to connect it back to $dG=\sum_i\mu_i\,dn_i\le0$ algebraically so you can see the inequality itself force the direction of flow.