Ideal gas: Energy

The energy of the ideal gas is given by (Hill Eq. 4-16)

$E=-T^{2}\left.{\frac {\partial (A/T)}{\partial T}}\right\vert _{V,N}=kT^{2}\left.{\frac {\partial \ln Q}{\partial T}}\right\vert _{V,N}=NkT^{2}{\frac {d\ln T^{3/2}}{dT}}={\frac {3}{2}}NkT\equiv {\frac {3}{2}}RT$

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} is the molar gas constant. This energy is all kinetic energy, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1/2 kT} per degree of freedom, by equipartition. This is because there are no intermolecular forces, thus no potential energy. This result is valid only for a monoatomic ideal gas. The general expression would be

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E = \frac{n}{2} NkT = \frac{n}{2} RT, }

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} is the number of degrees of freedom. This number is 3 for atoms; if would be 6 in principle for diatomic molecules, but in normal conditions 5 is a very good approximation since vibrations are "frozen" (as explained in the entry about degrees of freedom.)

References

1. Terrell L. Hill "An Introduction to Statistical Thermodynamics" 2nd Ed. Dover (1962)