# 13.1: The Gibbs Free Energy of an Ideal Gas

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In Chapter 11, we find a general equation for the molar Gibbs free energy of a pure gas. We adopt the Gibbs free energy of formation of the hypothetical ideal gas, in its standard state at 1 bar, $$P^o$$, as the reference state for the Gibbs free energy of the gas at other pressures and the same temperature. Then, the molar Gibbs free energy of pure gas $$A$$, at pressure $$P$$, is

${\overline{G}}_A\left(P\right)={\Delta }_fG^o\left(A,{HIG}^o\right)+RT{ \ln \left(\frac{P}{P^o}\right)\ }+RT\int^P_0{\left(\frac{\overline{V}}{RT}-\frac{1}{P}\right)dP}$ (any pure gas)

$${\overline{G}}_A\left(P\right)$$ is the difference between the Gibbs free energy of the gas at pressure $$P$$ and that of its constituent elements at 1 bar and the same temperature. If gas $$A$$ is an ideal gas, the integral is zero, and the standard-state Gibbs free energy of formation is that of an “actual” ideal gas, not a “hypothetical state” of a real gas. To recognize this distinction, let us write $${\Delta }_fG^o\left(A,P^o\right)$$, rather than $${\Delta }_fG^o\left(A,{HIG}^o\right)$$, when the gas behaves ideally. In a mixture of ideal gases, the partial pressure of gas $$A$$ is given by $$P_A=x_AP$$, where $$x_A$$ is the mole fraction of $$A$$ and $$P$$ is the pressure of the mixture. In §3, we find that the Gibbs free energy of one mole of pure ideal gas $$A$$ at pressure $$P_A$$ has the same Gibbs free energy as one mole of gas $$A$$ in a gaseous mixture in which the partial pressure of $$A$$ is $$P_A=x_AP$$. Recognizing these properties of an ideal gas, we can express the molar Gibbs free energy of an ideal gas—pure or in a mixture—as

$\overline{G}_A\left(P_A\right)=\Delta_fG^o\left(A,P^o\right)+RT \ln \left(\frac{P_A}{P^o}\right)\$

(ideal gas)

Note that we can obtain this result for pure gas $$A$$ directly from $${\left({\partial \overline{G}}/{\partial P}\right)}_T=\overline{V}={RT}/{P}$$ by evaluating the definite integrals

$\int^{\overline{G}_A\left(P_A\right)}_{\Delta_fG^o\left(A,P^o\right)}{d\overline{G}}=\int^{P_A}_{P^o}{\frac{RT}{P}dP}$

Including the constant, $$P^o$$, in these relationships is a useful reminder that $$RT{ \ln \left({P_A}/{P^o}\right)\ }$$ represents a Gibbs free energy difference. Including $$P^o$$ makes the argument of the natural-log function dimensionless; if we express $$P$$ in bars, including$${\ P}^o=1\ \mathrm{bar}$$ leaves the numerical value of the argument unchanged. If we express $$P$$ in other units, $$P^o$$ becomes the conversion factor for converting those units to bars; if we express $$P$$ in atmospheres, we have $${\ P}^o=1\ \mathrm{bar}=0.986923\ \mathrm{atm}$$.

However, including the “$$P^o$$” is frequently a typographical nuisance. Therefore, let us introduce another bit of notation; we use a lower-case “$$p$$” to denote the ratio “$${P}/{P^0}$$”. That is, $$p_A$$ is a dimensionless quantity whose numerical value is that of the partial pressure of $$A$$, expressed in bars. The molar Gibbs free energy becomes

${\overline{G}}_A\left(P_A\right)={\Delta }_fG^o\left(A,P^o\right)+RT{ \ln p_A\ }$ (ideal gas)

13.1: The Gibbs Free Energy of an Ideal Gas is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Paul Ellgen via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.