Editing Ideal gas partition function

Jump to navigation Jump to search
Warning: You are not logged in. Your IP address will be publicly visible if you make any edits. If you log in or create an account, your edits will be attributed to your username, along with other benefits.

The edit can be undone. Please check the comparison below to verify that this is what you want to do, and then publish the changes below to finish undoing the edit.

Latest revision Your text
Line 2: Line 2:
for a system of ''N'' identical particles each of mass ''m'' is given by
for a system of ''N'' identical particles each of mass ''m'' is given by


:<math>Q_{NVT}=\frac{1}{N!}\frac{1}{h^{3N}}\int\int d{\mathbf p}^N d{\mathbf r}^N \exp \left[ - \frac{H({\mathbf p}^N,{\mathbf r}^N)}{k_B T}\right]</math>
:<math>Q_{NVT}=\frac{1}{N!}\frac{1}{h^{3N}}\int\int dp^N dr^N \exp \left[ - \frac{H(p^N, r^N)}{k_B T}\right]</math>


where ''h'' is [[Planck constant |Planck's constant]], ''T'' is the [[temperature]] and <math>k_B</math> is the [[Boltzmann constant]]. When the particles are distinguishable then the factor ''N!'' disappears. <math>H(p^N, r^N)</math> is the [[Hamiltonian]]
where ''h'' is [[Planck constant |Planck's constant]], ''T'' is the [[temperature]] and <math>k_B</math> is the [[Boltzmann constant]]. When the particles are distinguishable then the factor ''N!'' disappears. <math>H(p^N, r^N)</math> is the [[Hamiltonian]]
Line 9: Line 9:
The Hamiltonian can be written as the sum of the kinetic and the potential energies of the system as follows
The Hamiltonian can be written as the sum of the kinetic and the potential energies of the system as follows


:<math>H({\mathbf p}^N, {\mathbf r}^N)= \sum_{i=1}^N \frac{|{\mathbf p}_i |^2}{2m} + {\mathcal V}({\mathbf r}^N)</math>
:<math>H(p^N, r^N)= \sum_{i=1}^N \frac{|p_i |^2}{2m} + \Phi(r^N)</math>


Thus we have  
Thus we have  


:<math>Q_{NVT}=\frac{1}{N!}\frac{1}{h^{3N}}\int d{\mathbf p}^N \exp \left[ - \frac{|{\mathbf p}_i |^2}{2mk_B T}\right]
:<math>Q_{NVT}=\frac{1}{N!}\frac{1}{h^{3N}}\int dp^N \exp \left[ - \frac{|p_i |^2}{2mk_B T}\right]
\int  d{\mathbf r}^N  \exp \left[ - \frac{{\mathcal V}({\mathbf r}^N)} {k_B T}\right]</math>
\int  dr^N  \exp \left[ - \frac{\Phi(r^N)} {k_B T}\right]</math>


This separation is only possible if <math>{\mathcal V}({\mathbf r}^N)</math> is independent of velocity (as is generally the case).
This separation is only possible if <math>V(r^N)</math> is independent of velocity (as is generally the case).
The momentum integral can be solved analytically:
The momentum integral can be solved analytically:


:<math>\int d{\mathbf p}^N \exp \left[ - \frac{|{\mathbf p} |^2}{2mk_B T}\right]=(2 \pi m k_B T)^{3N/2}</math>
:<math>\int dp^N \exp \left[ - \frac{|p |^2}{2mk_B T}\right]=(2 \pi m k_b T)^{3N/2}</math>


Thus we have  
Thus we have  


:<math>Q_{NVT}=\frac{1}{N!} \frac{1}{h^{3N}} \left( 2 \pi m k_B T\right)^{3N/2}
:<math>Q_{NVT}=\frac{1}{N!} \frac{1}{h^{3N}} \left( 2 \pi m k_B T\right)^{3N/2}
\int  d{\mathbf r}^N  \exp \left[ - \frac{{\mathcal V}({\mathbf r}^N)} {k_B T}\right]</math>
\int  dr^N  \exp \left[ - \frac{\Phi(r^N)} {k_B T}\right]</math>




Line 31: Line 31:
<math>Z_{NVT}</math> (from the German ''Zustandssumme'' meaning "sum over states")
<math>Z_{NVT}</math> (from the German ''Zustandssumme'' meaning "sum over states")


:<math>Z_{NVT}= \int  d{\mathbf r}^N  \exp \left[ - \frac{{\mathcal V}({\mathbf r}^N)} {k_B T}\right]</math>
:<math>Z_{NVT}= \int  dr^N  \exp \left[ - \frac{\Phi(r^N)} {k_B T}\right]</math>


In an [[ideal gas]] there are no interactions between particles so <math>{\mathcal V}({\mathbf r}^N)=0</math>.
In an [[ideal gas]] there are no interactions between particles so <math>\Phi(r^N)=0</math>
Thus <math>\exp(-{\mathcal V}({\mathbf r}^N)/k_B T)=1</math> for every gas particle.
Thus <math>\exp(-\Phi(r^N)/k_B T)=1</math> for every gas particle.
The integral of 1 over the coordinates of each atom is equal to the volume so for ''N'' particles
The integral of 1 over the coordinates of each atom is equal to the volume so for ''N'' particles
the ''configuration integral'' is given by <math>V^N</math> where ''V'' is the volume.
the ''configuration integral'' is given by <math>V^N</math> where ''V'' is the volume.
Please note that all contributions to SklogWiki are considered to be released under the Creative Commons Attribution Non-Commercial Share Alike (see SklogWiki:Copyrights for details). If you do not want your writing to be edited mercilessly and redistributed at will, then do not submit it here.
You are also promising us that you wrote this yourself, or copied it from a public domain or similar free resource. Do not submit copyrighted work without permission!

To edit this page, please answer the question that appears below (more info):

Cancel Editing help (opens in new window)