Ideal gas: Energy

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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 R is the molar gas constant. This energy is all kinetic energy, 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

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.)


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