Ideal gas: Energy: Difference between revisions
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Carl McBride (talk | contribs) (New page: 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</m...) |
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The energy of the [[ideal gas]] is given by (Hill Eq. 4-16) | 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. | 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. | ||
Revision as of 16:13, 7 June 2007
The energy of the ideal gas is given by (Hill Eq. 4-16)
This energy is all kinetic energy, per degree of freedom. This is because there are no intermolecular forces, thus no potential energy.
References
- Terrell L. Hill "An Introduction to Statistical Thermodynamics" 2nd Ed. Dover (1962)