# 24.10: The Gibbs Free Energy for One Mole of An Ideal Gas

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In our discussion of ensembles, we find that the thermodynamic functions for a system can be expressed as functions of the system’s partition function. Now that we have found the molecular partition function for a diatomic ideal gas molecule, we can find the partition function, $$Z_{IG}$$, for a gas of $$N$$ such molecules. From this system partition function, we can find all of the thermodynamic functions for this $$N$$-molecule ideal-gas system. The system entropy, energy, and partition function are related to each other by the equation

$S=\frac{E}{T}+k{ \ln Z\ }_{IG} \nonumber$

Rearranging, and adding $$\left(PV\right)_{\mathrm{system}}$$ to both sides, we find the Gibbs free energy

$G=E-TS+\left(PV\right)_{\mathrm{system}}= \left(PV\right)_{\mathrm{system}}-kT \ln Z_{IG} \nonumber$

For a system of one mole of an ideal gas, we have $$\left(PV\right)_{\mathrm{system}}=\overline{N}kT$$. If the ideal gas is diatomic, we can substitute the molecular partition functions developed above to find

\begin{align*} G_{IG}&=\overline{N}kT-kT \ln Z_{IG} \\[4pt] &=\overline{N}kT-\mathrm{kT ln} \left[\frac{\left(z_t\right)^{\overline{N}}}{\overline{N}!}\right] -\overline{N}kT \ln z_r -\overline{N}kT \ln z_v -\overline{N}kT \ln z_e \\[4pt] &=\overline{N}kT-\overline{N}kT-\overline{N}kT \ln \left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\frac{kT}{P}\right] -\overline{N}kT \ln \left(\frac{8{\pi }^2IkT}{\sigma h^2}\right) -\overline{N}kT \ln \left(\frac{\mathrm{exp} \left(-h\nu/2kT\right)}{1-\mathrm{exp} \left(-h\nu /kT\right)}\right) -\overline{N}kT \ln \left(\frac{D_0}{RT}+\frac{h\nu }{2kT}\right) \end{align*} \nonumber

For the standard Gibbs free energy of an ideal gas, we define the pressure to be one bar. Introduction of this condition $$\left(P=P^o=1\ \mathrm{bar}={10}^5\ \mathrm{Pa}\right)$$ and further simplification gives

$G^o_{IG}=-RT \ln \left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\frac{kT}{P^o}\right] -RT \ln \left(\frac{8{\pi }^2IkT}{\sigma h^2}\right) -RT \ln \left(\frac{\mathrm{exp} \left(-h\nu/2kT\right)}{1-\mathrm{exp} \left(-h\nu /kT\right)}\right)-RT\left(\frac{D_0}{RT}+\frac{h\nu }{2kT}\right) \nonumber$

In this form, the successive terms represent, respectively, the translational, rotational, vibrational, and electronic contributions to the Gibbs free energy. Further simplification results because vibrational and electronic contributions from terms involving $$h\nu /2kT$$ cancel. This is a computational convenience. Factoring out $$RT$$,

$G^o_{IG}=-RT\left\{ \ln \left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\frac{kT}{P^o}\right] + \ln \left(\frac{8{\pi }^2IkT}{\sigma h^2}\right) - \ln \left(1-\mathrm{exp} \left(-h\nu/kT \right) \right) +\frac{D_0}{RT}\right\} \nonumber$

This page titled 24.10: The Gibbs Free Energy for One Mole 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 the style and standards of the LibreTexts platform.