Pressure Dependence of the Gibbs Energy

Squeeze a bottle of water and almost nothing happens to its Gibbs energy. Squeeze a balloon of gas and the Gibbs energy climbs steeply. This contrast is not an accident. It follows from one compact thermodynamic relation. The Gibbs energy is the natural potential for tracking chemical change at constant temperature and pressure. Its pressure dependence tells you how phase stability, reaction spontaneity, and equilibrium constants shift when you change the pressure on a system. Because liquids and solids barely compress, their Gibbs energy responds only weakly to pressure. Gases, being highly compressible, respond strongly. This article derives the relation between Gibbs energy and pressure from the fundamental equation of $G$ . It then integrates that relation for condensed phases, where the molar volume is nearly constant. You will see the exact differential result first. Then you will see the practical integrated formula. Finally you will see two worked numerical examples that show why deep-sea pressure barely dents a protein's stability, while it dramatically shifts gas-phase equilibria.

Key takeaways

The master relation is $(\partial G/\partial p)_T = V$ , derived directly from $dG = -SdT + Vdp$ .
For a pure substance the molar version reads $(\partial G_m/\partial p)_T = V_m$ .
Condensed phases (liquids, solids) have nearly pressure-independent $V_m$ , so integration gives $\Delta G_m \approx V_m \Delta p$ .
Gases have $p$ -dependent $V_m$ , so the same integration instead produces a logarithmic term, foreshadowed here and completed with the ideal gas law.
Because $V_m$ for condensed phases is small (tens of cm³ mol⁻¹), enormous pressure changes are needed to shift $G_m$ by a chemically significant amount.
Key idea: the slope of Gibbs energy against pressure, at fixed temperature, is exactly the molar volume — so how compressible a phase is dictates how sensitive its stability is to pressure.

The core insight

Because $G$ is defined through $dG = -SdT + Vdp$ , its partial derivative with respect to pressure at constant temperature is nothing more exotic than the system's volume — a fact that turns a differential definition into a directly measurable, and directly integrable, thermodynamic rule.

The physical picture

Every thermodynamic potential has "natural variables" — the pair of variables in which its total differential takes the simplest, most information-rich form. For the Gibbs energy those variables are temperature and pressure, so it is natural to ask how $G$ changes as you dial the pressure up or down while holding $T$ fixed. Picture a piston compressing a fixed amount of substance at constant temperature. As the piston moves in, the system does negative work on its surroundings (the surroundings do positive work on it), and the Gibbs energy increases if the substance has any positive volume at all — which every real substance does. The rate at which $G$ rises with pressure is exactly the volume the substance occupies at that pressure. A mole of liquid water occupies about 18 cm³; a mole of an ideal gas at 1 bar and 298 K occupies roughly 24,790 cm³ — nearly 1400 times more. That volume ratio is the entire reason gases are so much more pressure-sensitive than liquids or solids. For a condensed phase the molar volume changes only slightly as pressure rises, because liquids and solids are nearly incompressible. Treating $V_m$ as constant across the pressure range of interest converts the exact differential relation into a simple linear formula: the change in molar Gibbs energy is just the (nearly constant) molar volume multiplied by the pressure change. This linear approximation is the workhorse used in geochemistry (mineral stability at depth), biophysics (pressure effects on protein folding), and industrial process design (compressing liquids in pumps).

Figure

Molar Gibbs energy versus pressure at fixed T, comparing a condensed phase (nearly straight line, small slope) to an ideal gas (steep, curving line)

The core relations

\left(\frac{\partial G}{\partial p}\right)_T = V

where: $G$ is Gibbs energy, $p$ is pressure, $T$ is temperature held constant, $V$ is the system's total volume.

\left(\frac{\partial G_m}{\partial p}\right)_T = V_m

where: $G_m$ is molar Gibbs energy, $V_m$ is molar volume — the per-mole version used for pure substances.

\Delta G_m = \int_{p_1}^{p_2} V_m\, dp \;\approx\; V_m\,(p_2 - p_1) \quad \text{(condensed phase, } V_m \text{ constant)}

where: $\Delta G_m$ is the change in molar Gibbs energy between pressures $p_1$ and $p_2$ , valid when $V_m$ barely varies over the interval.

G_m(p_2) = G_m(p_1) + V_m\,(p_2 - p_1)

where: this is the practical form used to shift a tabulated standard-state Gibbs energy to any other pressure, for a liquid or solid.

Derivation AFrom dG = −SdT + Vdp to the integrated condensed-phase formula

Goal. Starting only from the fundamental equation for $G$ , obtain $(\partial G/\partial p)_T = V$ , then integrate it for a phase whose molar volume is essentially pressure-independent.

Step 1. Write the fundamental equation of Gibbs energy. Gibbs energy is defined as $G = H - TS = U + pV - TS$ . Differentiating totally and substituting the combined first-and-second-law expression $dU = TdS - pdV$ (reversible, no non-expansion work) gives, after the $pV$ and $TS$ product-rule terms partially cancel,

dG = -S\,dT + V\,dp .

This is the fundamental equation of $G$ in its natural variables $T$ and $p$ ; every term on the right is exact because $G$ , being built from state functions, is itself a state function.

Step 2. Read off the pressure derivative. Because $dG = -SdT + Vdp$ is an exact differential of the form $dG = (\partial G/\partial T)_p dT + (\partial G/\partial p)_T dp$ , matching coefficients term by term gives immediately

\left(\frac{\partial G}{\partial p}\right)_T = V, \qquad \left(\frac{\partial G}{\partial T}\right)_p = -S .

No approximation has been made yet — this is exact for any closed system doing only $pV$ work, at any pressure, for any phase.

Step 3. Restrict to one mole of a pure substance and integrate formally. Divide through by the amount of substance to work in molar quantities, then integrate the exact relation at constant $T$ between two pressures $p_1$ and $p_2$ :

G_m(p_2) - G_m(p_1) = \int_{p_1}^{p_2} V_m(p)\, dp .

This integral is completely general; what changes between a gas and a condensed phase is only how strongly $V_m$ depends on $p$ inside it.

Step 4. Invoke near-incompressibility of condensed phases. For a liquid or solid, the isothermal compressibility $\kappa_T = -\frac{1}{V_m}\left(\partial V_m/\partial p\right)_T$ is typically of order $10^{-4}$ to $10^{-5}$ bar⁻¹. Over pressure changes of tens or even hundreds of bar, $V_m$ therefore changes by a fraction of a percent — negligible for most chemical purposes. Treating $V_m$ as a constant equal to its value at $p_1$ pulls it out of the integral:

G_m(p_2) - G_m(p_1) \approx V_m \int_{p_1}^{p_2} dp = V_m (p_2 - p_1).

Step 5. State the boxed result.

\boxed{\,\Delta G_m \approx V_m \,\Delta p \quad \text{(condensed phase, constant } T\text{)}\,}

This linear approximation is accurate to within experimental precision for most laboratory-scale pressure changes on liquids and solids. It is the condensed-phase counterpart of the logarithmic pressure dependence that emerges for an ideal gas once $V_m = RT/p$ is substituted into Step 3 — a derivation taken up in the ideal-gas Gibbs energy discussion.

Worked examples

Example 1 — Compressing liquid water (foundational). Take one mole of liquid water at 298 K, with molar volume $V_m = 18.07\ \text{cm}^3\,\text{mol}^{-1}$ (density 0.997 g cm⁻³, molar mass 18.02 g mol⁻¹). Find $\Delta G_m$ on raising the pressure from 1 bar to 101 bar (a 100 bar increase), assuming $V_m$ stays constant. Convert units first: $V_m = 18.07\ \text{cm}^3\,\text{mol}^{-1} = 18.07\times10^{-6}\ \text{m}^3\,\text{mol}^{-1}$ , and $\Delta p = 100\ \text{bar} = 100\times10^5\ \text{Pa} = 1.00\times10^7\ \text{Pa}$ .

\Delta G_m = V_m\,\Delta p = (18.07\times10^{-6}\ \text{m}^3\,\text{mol}^{-1})(1.00\times10^7\ \text{Pa})

\Delta G_m = 180.7\ \text{J mol}^{-1} \approx 0.181\ \text{kJ mol}^{-1}.

Sanity check: $1\ \text{Pa}\cdot\text{m}^3 = 1\ \text{J}$ , units are consistent. Compare this to $RT = 8.314 \times 298 = 2478\ \text{J mol}^{-1}$ — the Gibbs energy change is only about 7% of $RT$ , confirming that even a 100-fold increase in pressure over atmospheric barely perturbs a liquid's thermodynamic state. This is why liquid-phase equilibrium constants are treated as pressure-independent in ordinary bench chemistry.

Example 2 — Deep-ocean pressure on a folded protein (applied, harder). Marine biophysicists ask whether hydrostatic pressure at oceanic trench depths can destabilize a folded protein. Model the folded state as a condensed phase with an effective molar volume change on unfolding of $\Delta V_m^{\ddagger} = -20\ \text{cm}^3\,\text{mol}^{-1}$ (folding is usually accompanied by a small volume decrease relative to the unfolded, more solvent-exposed state; take the unfolding volume change as $+20\ \text{cm}^3\,\text{mol}^{-1}$ for concreteness, a typical experimental magnitude). Estimate the change in the Gibbs energy of unfolding, $\Delta G_{unfold}$ , on going from the surface (1 bar) to the bottom of the Mariana Trench (about 1086 bar, corresponding to roughly 10,900 m depth), at constant $T = 298\ \text{K}$ . Step 1: identify the relevant volume change. The pressure dependence of the unfolding Gibbs energy itself is governed by the volume change of unfolding, since

\left(\frac{\partial \Delta G_{unfold}}{\partial p}\right)_T = \Delta V_{unfold}.

This follows by applying $(\partial G_m/\partial p)_T = V_m$ separately to folded and unfolded states and subtracting. Step 2: convert units. $\Delta V_{unfold} = +20\ \text{cm}^3\,\text{mol}^{-1} = 20\times10^{-6}\ \text{m}^3\,\text{mol}^{-1}$ . $\Delta p = 1086 - 1 = 1085\ \text{bar} = 1.085\times10^8\ \text{Pa}$ . Step 3: integrate assuming $\Delta V_{unfold}$ stays roughly constant over this range (a common approximation, since the compressibility difference between folded and unfolded states is itself a second-order effect):

\Delta(\Delta G_{unfold}) = \Delta V_{unfold}\,\Delta p = (20\times10^{-6}\ \text{m}^3\,\text{mol}^{-1})(1.085\times10^{8}\ \text{Pa})

\Delta(\Delta G_{unfold}) = 2170\ \text{J mol}^{-1} \approx 2.17\ \text{kJ mol}^{-1}.

Interpretation: pressurizing the protein to trench depth raises the Gibbs energy of unfolding by about 2.2 kJ mol⁻¹, meaning unfolding becomes less favorable (folded state is stabilized) — consistent with the smaller molar volume assigned to the folded state. Sanity check against thermal energy: $RT \ln 2 \approx 1.72\ \text{kJ mol}^{-1}$ at 298 K, so this pressure effect is comparable to about a factor of 2 change in the folding/unfolding equilibrium constant, a modest but measurable shift — matching the real experimental observation that most proteins are only mildly perturbed by ocean-floor pressures, while much larger pressures (kilobar range, used in high-pressure NMR) are needed to fully unfold them.

Limiting cases & common traps

Situation	What happens to $(\partial G_m/\partial p)_T$	Practical consequence
Liquid or solid, modest $\Delta p$ (< few hundred bar)	Essentially constant, equal to $V_m$	$\Delta G_m \approx V_m\Delta p$ , usually a few tenths of a kJ mol⁻¹
Liquid or solid, huge $\Delta p$ (kilobar range, e.g. geophysics)	$V_m$ itself drifts measurably; need compressibility correction	Use $V_m(p) = V_m^\circ(1 - \kappa_T \Delta p)$ or tabulated equations of state
Ideal gas	$V_m = RT/p$ , strongly $p$ -dependent	Integration gives $\Delta G_m = RT\ln(p_2/p_1)$ , a logarithmic — not linear — law
Phase transition pressure (e.g. melting point shift)	$V_m$ differs discontinuously between phases	Governs the slope of coexistence lines via the Clapeyron equation

The chemist behind it

The formalism connecting Gibbs energy, volume, and pressure traces back to Josiah Willard Gibbs, whose 1870s memoirs "On the Equilibrium of Heterogeneous Substances" established the mathematical machinery of chemical thermodynamics — chemical potential, phase rule, and the free energy function that bears his name. Gibbs worked largely in isolation at Yale, publishing in an obscure regional journal, and his results were not widely appreciated by chemists until James Clerk Maxwell and later Wilhelm Ostwald championed and popularized them in Europe. The pressure-derivative relation used throughout this article is a direct, almost immediate consequence of the differential form Gibbs wrote down for what he called the "available energy" of a system — recognized today as one of the most quietly powerful equations in physical chemistry, because it turns a single measured molar volume into a full prediction for how stability shifts with pressure.

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 Gibbs energy and chemical potential

Ask the tutor: How does this same relation change when we integrate it for a real (non-ideal) gas instead of a condensed phase?

Is ΔGm = VmΔp exact, or just an approximation?

It is an approximation, though an excellent one for condensed phases under typical laboratory pressure changes. The exact statement is the integral $\Delta G_m = \int V_m\,dp$ ; replacing $V_m(p)$ with a single constant value is valid only because liquids and solids have tiny isothermal compressibility, so $V_m$ barely changes even over hundreds of bar. At geophysical pressures (kilobars), the approximation breaks down and a pressure-dependent $V_m$ , or an equation of state, is needed instead.

Why is the gas-phase result logarithmic instead of linear?

Because an ideal gas's molar volume is $V_m = RT/p$ , which is strongly pressure-dependent, unlike a condensed phase's nearly constant $V_m$ . Substituting this into $\Delta G_m = \int V_m\,dp$ and integrating from $p_1$ to $p_2$ gives $\Delta G_m = RT\ln(p_2/p_1)$ . The same parent relation, $(\partial G_m/\partial p)_T = V_m$ , produces a linear law for condensed phases and a logarithmic law for gases purely because of how differently $V_m$ behaves in each case.

Does this relation apply to mixtures, or only pure substances?

The total-system relation $(\partial G/\partial p)_T = V$ holds for mixtures exactly as it does for pure substances, since it follows only from the fundamental equation of $G$ and requires no assumption about composition. For a mixture, the corresponding per-component quantity is the partial molar volume, and the pressure dependence of each component's chemical potential is governed by its own partial molar volume rather than by a simple pure-substance molar volume.

How does this relation connect to the Clapeyron equation for phase boundaries?

At a phase transition, the molar Gibbs energies of both phases are equal along the coexistence curve. Differentiating that equality with respect to pressure and temperature, and using $(\partial G_m/\partial p)_T = V_m$ together with $(\partial G_m/\partial T)_p = -S_m$ , yields the Clapeyron equation $dp/dT = \Delta_{trs}S_m/\Delta_{trs}V_m$ . The volume-pressure relation derived here is thus one of the two ingredients needed to derive that famous phase-boundary slope formula.

Why does pressure barely affect equilibrium constants for reactions among liquids and solids?

Equilibrium constants are governed by $\Delta_r G^\circ$ , whose pressure dependence is set by the reaction's molar volume change, $\Delta_r V_m$ . For reactions involving only liquids and solids, $\Delta_r V_m$ is typically just a few cm³ mol⁻¹, so even hundreds of bar produce Gibbs energy shifts of only a fraction of a kJ mol⁻¹ — far smaller than the thermal energy scale $RT \approx 2.5\ \text{kJ mol}^{-1}$ at room temperature. Gas-phase reactions, in contrast, can have $\Delta_r V_m$ of tens of liters per mole, making them far more pressure-sensitive.