Ideal gas: Energy: Difference between revisions

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(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 17:13, 7 June 2007

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

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 = -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, 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. This is because there are no intermolecular forces, thus no potential energy.

References

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