Difference between revisions of "Ideal gas: Energy"

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:<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>
 
:<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.
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This energy is all ''kinetic energy'', <math>1/2 kT</math> per [[degree of freedom]], by [[equipartition]]. This is because there are no intermolecular forces, thus no potential energy.
 
==References==
 
==References==
 
#Terrell L. Hill "An Introduction to Statistical Thermodynamics"  2nd Ed. Dover (1962)  
 
#Terrell L. Hill "An Introduction to Statistical Thermodynamics"  2nd Ed. Dover (1962)  
 
[[category: ideal gas]]
 
[[category: ideal gas]]

Revision as of 14:46, 9 May 2008

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, by equipartition. 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)