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Confused on statistical mechanics problem

  • #1

Homework Statement


A dilute gas of N non-interacting atoms of mass m is contained in a volume V and in equilibrium with the surroundings at a temperature T. Each atom has two (active) intrinsic states of energies ε = 0 and ∆, respectively. Find the total partition function of the gas.


Homework Equations


Partition function.

The Attempt at a Solution


For discrete energy levels we normally write ##Z = \sum_{i}e^{-\beta \epsilon_{i}},## and the volume of the system appears in the energy spectrum e.g. particle in a box energy spectrum. For a classical system we write $$Z=\frac{1}{h^{3}}\int e^{H(q,p)}d^{3}qd^{3}p.$$ We can write

$$Z = \frac{V}{h^{3}}\int e^{-\beta p^{2}/2m}d^{3}p,$$
##\frac{p^{2}}{2m} = \epsilon## implies ##d\epsilon = \frac{p}{m}dp = \frac{\sqrt{2m\epsilon}}{m}dp## which means we can write ##dp = \sqrt{\frac{m}{2\epsilon}}d\epsilon## and the partition function becomes

$$Z = (\frac{m}{2})^{3/2}\frac{V}{h^{3}}\int e^{-\epsilon}\epsilon^{-3/2}d^{3}\epsilon.$$

Should I be including delta functions in order to pick off the two discrete energy levels? or is there an easier way in going about this problem?

Thank you!
 

Answers and Replies

  • #2
265
52
You said yourself; “for discrete energy levels we use the discrete sum over energy levels,” but then you use the continuous energy formulation??

Finding a discrete sum by integrating over Dirac-delta spikes seems pretty silly!


(Sorry if I’m missing something; I’m also learning the subject.)
 
  • #3
I am just confused on how to incorporate the volume ##V## into the picture by evaluating ##Z = \sum_{i} e^{-\beta \epsilon_{i}} = (1+e^{-\beta \Delta})^{N}.## Maybe it is as simple as multiplying this by ##\frac{V}{h^{3}}.##
 
  • #4
265
52
Just because they give the volume V doesn’t mean we have to use it!

I don’t know quantum mechanics, but I would think V would appear inside Δ. But since some authority tells us what Δ is, then I don’t think we need to use V.

Anyway I’ll shut up now. Let’s wait for someone who actually knows what they’re talking about to chime in! o0)
 

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