The chemical potential is the partial molar Gibbs function:
\[ \mu = \left( \frac{\partial G} {\partial n} \right)_{T,p} \]
Since the chemical potential is a partial molar Gibbs function, it is natural to ask how it changes with pressure.
To answer this question, begin with the total differential of the Gibbs function:
\[ dG = V\,dp - S\,dT \]
Dividing by the number of moles converts the extensive quantities into molar quantities:
\[ d\mu = \bar V\,dp - \bar S\,dT \]
where \(\bar V\) and \(\bar S\) are the molar volume and molar entropy, respectively.
Most measurements of chemical potential are made at constant temperature.
Under these conditions,
\[ dT = 0 \]
and therefore
\[ \boxed{ d\mu = \bar V\,dp } \]
This result shows that the pressure dependence of the chemical potential is controlled by the molar volume.
For most liquids and solids, the molar volume changes very little with pressure.
In this case, \(\bar V\) may be treated as a constant and removed from the integral:
\[ \int_{\mu^\circ}^{\mu} d\mu = \bar V \int_{p^\circ}^{p} dp \]
yielding
\[ \boxed{ \mu = \mu^\circ + \bar V (p-p^\circ) } \]
where \(p^\circ\) is a reference pressure, usually \(1\ \mathrm{atm}\) or \(1\ \mathrm{bar}\).
This expression predicts that the chemical potential increases linearly with pressure.
For gases, the molar volume changes dramatically with pressure and cannot be treated as a constant.
For an ideal gas,
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\[ \bar V = \frac{RT}{p} \]
Substituting into
\[ d\mu = \bar V\,dp \]
gives
\[ d\mu = \frac{RT}{p}\,dp \]
Integrating from the standard pressure \(p^\circ\) to the pressure \(p\):
\[ \int_{\mu^\circ}^{\mu} d\mu = RT \int_{p^\circ}^{p} \frac{dp}{p} \]
yields
\[ \boxed{ \mu = \mu^\circ + RT \ln \left( \frac{p}{p^\circ} \right) } \]
This is the pressure dependence of the chemical potential for an ideal gas.
Increasing the pressure increases the chemical potential.
Physically, compressing a substance makes it thermodynamically less stable, increasing its tendency to move to a region of lower pressure.
The effect is modest for liquids and solids because their molar volumes are nearly constant, but much larger for gases because their volumes change dramatically with pressure.
Big picture: The pressure dependence of the chemical potential provides a quantitative connection between composition and pressure. For condensed phases the dependence is approximately linear, while for ideal gases it follows a logarithmic relationship that will become central to the thermodynamics of mixtures and chemical equilibrium.
Calculate the change in chemical potential when an ideal gas is compressed isothermally from \(1.00\ \mathrm{atm}\) to \(5.00\ \mathrm{atm}\) at \(298\ \mathrm{K}\).
For an ideal gas, the pressure dependence of chemical potential is
\[ \Delta \mu = RT \ln\left( \frac{p_2}{p_1} \right) \]
Substitute the values:
\[ \Delta \mu = (8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}) (298\ \mathrm{K}) \ln\left( \frac{5.00\ \mathrm{atm}} {1.00\ \mathrm{atm}} \right) \]
\[ \Delta \mu = 3.99\times10^3\ \mathrm{J\,mol^{-1}} \]
\[ \boxed{ \Delta \mu = +3.99\ \mathrm{kJ\,mol^{-1}} } \]
Interpretation: Compressing the gas increases its chemical potential. The compressed gas is thermodynamically less stable than it was at the lower pressure, so it has a greater tendency to expand or move toward a region of lower pressure.
The molar volume of liquid water at \(25^\circ\mathrm{C}\) is
\[ \bar V = 18.0\ \mathrm{cm^3\,mol^{-1}} \]
Calculate the change in chemical potential when the pressure is increased from \(1.00\ \mathrm{atm}\) to \(100.0\ \mathrm{atm}\) at constant temperature.
For a liquid, the molar volume is nearly independent of pressure, so
\[ \Delta\mu = \bar V (p_2-p_1) \]
First convert the molar volume to SI units:
\[ \bar V = 18.0\ \mathrm{cm^3\,mol^{-1}} \left( \frac{1\ \mathrm{m^3}} {10^6\ \mathrm{cm^3}} \right) \]
\[ \bar V = 1.80\times10^{-5} \ \mathrm{m^3\,mol^{-1}} \]
Next calculate the pressure change:
\[ \Delta p = 100.0\ \mathrm{atm} - 1.00\ \mathrm{atm} \]
\[ \Delta p = 99.0\ \mathrm{atm} \]
Converting to pascals,
\[ \Delta p = (99.0\ \mathrm{atm}) (101325\ \mathrm{Pa\,atm^{-1}}) \]
\[ \Delta p = 1.00\times10^7 \ \mathrm{Pa} \]
Therefore,
\[ \Delta\mu = (1.80\times10^{-5}\ \mathrm{m^3\,mol^{-1}}) (1.00\times10^7\ \mathrm{Pa}) \]
\[ \Delta\mu = 1.82\times10^2 \ \mathrm{J\,mol^{-1}} \]
\[ \boxed{ \Delta\mu = +0.182\ \mathrm{kJ\,mol^{-1}} } \]
Interpretation: Increasing the pressure by nearly a factor of 100 changes the chemical potential by only \(0.182\ \mathrm{kJ\,mol^{-1}}\). This is much smaller than the \(3.99\ \mathrm{kJ\,mol^{-1}}\) change calculated for an ideal gas under a much smaller pressure increase. The reason is that liquids are nearly incompressible and have very small molar volumes compared with gases.
This example illustrates why the pressure dependence of chemical potential is often negligible for liquids and solids, but extremely important for gases.
An ideal gas will be generated at random. Calculate the change in chemical potential resulting from an isothermal pressure change.
Big picture: The chemical potential increases with pressure because compressing a substance makes it thermodynamically less stable. For condensed phases the dependence is approximately linear, while for ideal gases it is logarithmic. This pressure dependence provides the foundation for understanding how chemical potential varies in mixtures and why composition affects equilibrium.