Ideal gas: Energy: Difference between revisions

Revision as of 18:13, 7 June 2007

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

This energy is all kinetic energy, $1/2.kT$ per degree of freedom. This is because there are no intermolecular forces, thus no potential energy.

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

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 The energy of the [[ideal gas]] is given by (Hill Eq. 4-16)
-:<math>E = 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</math>
+:<math>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</math>
 This energy is all ''kinetic energy'', <math>1/2.kT</math> per degree of freedom. This is because there are no intermolecular forces, thus no potential energy.