Ideal gas: Energy

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

$E=-T^2{\frac{\partial (A/T)}{\partial T}|}_{V,N}=k{T}^2{\frac{\partial \ln Q}{\partial T}|}_{V,N}=Nk{T}^2\frac{d\ln T^{3/2}}{dT}=\frac{3}{2}NkT\equiv \frac{3}{2}RT$

where $R$ is the molar gas constant. This energy is all kinetic energy, $1/2kT$ 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

$E=\frac{n}{2}NkT=\frac{n}{2}RT,$

where $n$ 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)