Gibbs Energy of an Ideal Gas with Pressure

Squeeze a mole of gas into a smaller container and its Gibbs energy rises. This is not exotic chemistry — it is the thermodynamic reason gases expand, effuse through leaks, and drive turbines. Understanding how $G$ depends on pressure is the single most useful piece of bookkeeping in physical chemistry. The result matters because it is the direct ancestor of chemical potential, reaction quotients, and the equilibrium constant $K_p$ . Every equilibrium calculation you will ever do with gases traces back to one integral. This article integrates $dG = V\,dp$ at constant temperature for an ideal gas. You will see the derivation in full. Then two worked numerical examples. Finally the traps that catch students who forget the standard-state pressure.

Key takeaways

At constant $T$ , $dG = V\,dp$ ; for an ideal gas $V = nRT/p$ , so $dG = nRT\,dp/p$ .
Integrating from $p^\circ$ to $p$ gives $G(p) = G^\circ + nRT\ln(p/p^\circ)$ .
$G^\circ$ is the molar Gibbs energy at the standard pressure $p^\circ = 1\ \text{bar}$ , at the same temperature.
The logarithm means $G$ rises steeply at low pressure and only gently at high pressure.
This equation is the direct route to the chemical potential $\mu(p) = \mu^\circ + RT\ln(p/p^\circ)$ of a pure ideal gas.
Key idea: because molar volume scales as $1/p$ for an ideal gas, compressing it costs Gibbs energy logarithmically in pressure, not linearly.

The core insight

Because $V = nRT/p$ for an ideal gas, the exact differential $dG = V\,dp$ integrates to a logarithm rather than a linear term, giving the universal form $G = G^\circ + nRT\ln(p/p^\circ)$ that underlies every gas-phase equilibrium expression in chemistry.

The physical picture

Start from the general thermodynamic relation for a closed system at constant composition, $dG = V\,dp - S\,dT$ . At constant temperature the $S\,dT$ term vanishes, leaving $dG = V\,dp$ . This is exact for any substance — solid, liquid, or gas — and was established in the earlier discussion of the pressure dependence of the Gibbs energy. The only thing that changes from substance to substance is how $V$ depends on $p$ . For a solid or liquid, $V$ is nearly constant, so $G$ rises almost linearly with pressure and the effect is tiny (a few J mol⁻¹ for a pressure change of 1 bar). For an ideal gas, $V$ is inversely proportional to $p$ through $V = nRT/p$ . As pressure increases, the volume shrinks, so the "cost" in Gibbs energy per unit pressure increase — the slope $dG/dp = V$ — shrinks too. The result of integrating a $1/p$ dependence is a natural logarithm, not a straight line. That single feature — logarithmic pressure dependence — is what makes gas equilibria behave the way they do: doubling a pressure never doubles $G$ , but multiplying it by $e$ adds a fixed increment $RT$ . Because absolute Gibbs energies cannot be measured, we always report $G$ relative to a reference. Chemists chose a standard pressure $p^\circ = 1\ \text{bar}$ (exactly $10^5$ Pa) at which $G$ equals a reference value $G^\circ$ , the standard molar Gibbs energy, at the temperature of interest. Every real pressure is then compared to that fixed benchmark through the ratio $p/p^\circ$ , which is dimensionless — essential, since you cannot take the logarithm of a dimensional quantity.

Figure

Molar Gibbs energy of an ideal gas versus pressure at 298 K, showing the logarithmic rise relative to G° at 1 bar

The core relations

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

where: $V$ is the molar volume and $dp$ an infinitesimal pressure change at fixed temperature and composition.

V = \frac{nRT}{p}

where: $n$ = amount of gas (mol), $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ , $T$ = temperature (K), $p$ = pressure.

G(p) = G^\circ + nRT\ln\!\left(\frac{p}{p^\circ}\right)

where: $G^\circ$ is the Gibbs energy at the standard pressure $p^\circ = 1\ \text{bar}$ , same $T$ ; the ratio $p/p^\circ$ is dimensionless.

G_\text{m}(p) = G_\text{m}^\circ + RT\ln\!\left(\frac{p}{p^\circ}\right)

where: molar form (per mole, $n=1$ ); this is the version usually quoted and used directly for chemical potential.

Derivation AIntegrating dG = V dp for an ideal gas

Goal. Find $G(p)$ for $n$ mol of ideal gas at fixed temperature $T$ , starting from the exact relation $dG = V\,dp$ .

Step 1 — Write the differential at constant T. Because $dG = V\,dp - S\,dT$ in general, holding $T$ fixed kills the entropy term:

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

Step 2 — Insert the ideal-gas equation of state. For $n$ moles of ideal gas, $pV = nRT$ , so

V = \frac{nRT}{p}.

Substituting,

dG = \frac{nRT}{p}\,dp.

Step 3 — Integrate between two pressures. Choose the lower limit to be the standard pressure $p^\circ$ , where $G = G^\circ$ , and the upper limit to be the pressure of interest $p$ , where $G = G(p)$ . Since $n$ , $R$ , $T$ are constants at fixed temperature and amount, they leave the integral:

\int_{G^\circ}^{G(p)} dG = nRT \int_{p^\circ}^{p} \frac{dp'}{p'}.

Step 4 — Evaluate both integrals. The left side is simply $G(p) - G^\circ$ . The right side is a standard logarithmic integral:

\int_{p^\circ}^{p}\frac{dp'}{p'} = \ln p - \ln p^\circ = \ln\!\left(\frac{p}{p^\circ}\right).

G(p) - G^\circ = nRT \ln\!\left(\frac{p}{p^\circ}\right).

Step 5 — Rearrange to the final form.

\boxed{G(p) = G^\circ + nRT\ln\!\left(\frac{p}{p^\circ}\right)}

Dividing by $n$ gives the molar form $G_\text{m} = G_\text{m}^\circ + RT\ln(p/p^\circ)$ , which is the version that feeds directly into the definition of chemical potential for a pure ideal gas.

Worked examples

Example 1 — Foundational: compressing nitrogen isothermally. One mole of N₂(g), treated as ideal, is compressed at $298\ \text{K}$ from the standard pressure $p^\circ = 1\ \text{bar}$ to $p = 20\ \text{bar}$ . Find $\Delta G$ . Step 1. Identify quantities: $n = 1\ \text{mol}$ , $T = 298\ \text{K}$ , $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ , $p/p^\circ = 20$ . Step 2. Apply the boxed result directly, since the initial state is the standard state ( $G_i = G^\circ$ ):

\Delta G = G(p) - G^\circ = nRT\ln\!\left(\frac{p}{p^\circ}\right).

Step 3. Compute the logarithm: $\ln 20 = 2.996$ . Step 4. Compute $RT$ : $8.314 \times 298 = 2477.6\ \text{J mol}^{-1}$ . Step 5. Multiply:

\Delta G = (1\ \text{mol})(2477.6\ \text{J mol}^{-1})(2.996) = 7423\ \text{J} = 7.42\ \text{kJ}.

Sanity check: compression increases $G$ , and $\Delta G$ came out positive — consistent with the sign of $\ln(p/p^\circ) > 0$ for $p > p^\circ$ . The magnitude, a few kJ per mole for a twenty-fold pressure change, is typical for gas-phase pressure effects (compare: a 1 bar change in a liquid's pressure moves $G$ by only ~ $10^{-5}$ kJ mol⁻¹, since $V_\text{m,liquid} \ll V_\text{m,gas}$ ).

Example 2 — Applied: Gibbs energy change on mixing/expansion relevant to Kp. Consider 2.50 mol of an ideal gas at $350\ \text{K}$ expanding isothermally from an initial pressure $p_1 = 12.0\ \text{bar}$ to a final pressure $p_2 = 0.800\ \text{bar}$ (e.g., a gas released from a cylinder into a larger reactor volume, relevant to later $K_p$ calculations). Find $\Delta G$ and comment on its sign. Step 1. Because $G$ depends on pressure through the same logarithmic law regardless of reference point, the difference between two arbitrary pressures is:

\Delta G = G(p_2) - G(p_1) = nRT\ln\!\left(\frac{p_2}{p^\circ}\right) - nRT\ln\!\left(\frac{p_1}{p^\circ}\right) = nRT\ln\!\left(\frac{p_2}{p_1}\right).

Note the standard pressure cancels — it must, since $\Delta G$ between two real states cannot depend on an arbitrary reference choice. Step 2. Compute the pressure ratio: $p_2/p_1 = 0.800/12.0 = 0.06667$ . Step 3. Take the logarithm: $\ln(0.06667) = -2.708$ . Step 4. Compute $nRT$ : $2.50\ \text{mol} \times 8.314\ \text{J K}^{-1}\text{mol}^{-1} \times 350\ \text{K} = 7274.75\ \text{J}$ . Step 5. Multiply:

\Delta G = 7274.75\ \text{J} \times (-2.708) = -19701\ \text{J} = -19.7\ \text{kJ}.

Sanity check: expansion to lower pressure should decrease $G$ (a gas expanding into vacuum-like conditions releases Gibbs energy at constant $T$ ), and indeed $\Delta G < 0$ . The size, about $-20\ \text{kJ}$ for a fifteen-fold pressure drop with 2.5 mol, scales sensibly compared to Example 1's $+7.4\ \text{kJ}$ for a twenty-fold increase with only 1 mol — same order of magnitude per mole once the ratio and mole number are accounted for.

Limiting cases & common traps

Situation	Behavior	Why
$p = p^\circ$	$G = G^\circ$ exactly	$\ln(1) = 0$
$p \to 0$	$G \to -\infty$	ideal-gas law breaks down physically first; log diverges mathematically
$p \gg p^\circ$	$G$ grows only logarithmically	consistent with $V \propto 1/p$ shrinking the "cost" per unit $dp$
Solids/liquids	$G$ roughly linear in $p$ , tiny slope	$V$ nearly pressure-independent, so $dG = V\,dp$ integrates to $V\Delta p$
Real gases at high $p$	Formula becomes approximate	must replace $p$ with fugacity $f$

The chemist behind it

The modern formalism of chemical potential and standard states, built on exactly this pressure-logarithm relation, owes its systematic development to Gilbert N. Lewis, who in the early twentieth century unified activity, fugacity, and standard states into the coherent framework still taught today. Lewis's insight was to recognize that real gases and solutions could be handled with the same mathematical form as the ideal case, simply by replacing $p$ with an effective "fugacity," making the ideal-gas logarithmic law the template for all of chemical thermodynamics.

Origins · 1 minGilbert N. Lewis and how this was discovered

Gilbert N. Lewis

1875–1946 · American physical chemist who introduced fugacity, activity, and the modern standard-state convention underlying G(p) = G° + RT ln(p/p°)

Ask the tutor: How does this pressure-dependence formula turn into the chemical potential of a gas mixture, and eventually into the equilibrium constant Kp?

Why does G increase with pressure for a gas but the everyday intuition says compressed gas has more "stored energy," not more Gibbs energy specifically

Compressed gas indeed stores more Gibbs energy at constant $T$ , and this is exactly what drives it to expand and do work when released. The rise in $G$ with $p$ reflects reduced entropy per mole (less accessible volume) at fixed enthalpy, so $G = H - TS$ increases as $S$ decreases. That increase is a genuine thermodynamic potential difference, not just "energy" in the everyday sense — it quantifies the maximum non-expansion work extractable at constant $T,p$ .

Does this equation apply to gas mixtures, not just pure gases

The derivation as shown is for one pure ideal gas, but it generalizes directly to a component in an ideal gas mixture by replacing total pressure $p$ with the partial pressure $p_i$ of that component. This gives $\mu_i = \mu_i^\circ + RT\ln(p_i/p^\circ)$ , the chemical potential of a gas in a mixture, which is the direct next step toward the reaction quotient and $K_p$ .

Why is the standard pressure defined as exactly 1 bar and not 1 atm

IUPAC redefined the standard pressure as $1\ \text{bar} = 10^5\ \text{Pa}$ in 1982, replacing the older convention of $1\ \text{atm} = 101{,}325\ \text{Pa}$ . The change was made for a rounder, SI-consistent reference value. The numerical difference between the two standards is small, changing $G^\circ$ by about $0.034\ \text{kJ mol}^{-1}$ at 298 K, but you should check which convention a data table uses.

What happens to this equation for a real, non-ideal gas

For real gases, $V \neq nRT/p$ exactly, so the clean integration breaks down. The practical fix, due to G. N. Lewis, is to define a fugacity $f$ such that the same functional form holds, $G = G^\circ + nRT\ln(f/p^\circ)$ , with $f \to p$ in the low-pressure ideal limit. All the nonideality is absorbed into the fugacity coefficient $\gamma = f/p$ , so the elegant logarithmic law survives even when the ideal gas law does not.

How does this connect to chemical potential and equilibrium constants

Dividing $G(p) = G^\circ + nRT\ln(p/p^\circ)$ by $n$ gives the molar Gibbs energy, which for a pure substance equals the chemical potential $\mu$ . This $\mu(p) = \mu^\circ + RT\ln(p/p^\circ)$ is the starting point for writing the Gibbs energy of a reacting mixture, from which $\Delta_rG = \Delta_rG^\circ + RT\ln Q_p$ and, at equilibrium, $\Delta_rG^\circ = -RT\ln K_p$ are derived in the following topics.