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

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    152808
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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

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

    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; a detailed edit history is available upon request.