Canonical ensemble, entropy of a classical gas

  • Thread starter hansbahia
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  • #1
hansbahia
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Homework Statement



I have the equation
Z=1/N!h3N∫∫d3qid3pie-βH(q,p)

How can I get the entropy from this equation assuming a classical gas of N identical, noninteracting atoms inside a volume V in equilibrium at T where it has an internal degree of freedom with energies 0 and ε
What about the specific heat at constant volume Cv?
Can anyone explain the equation?

Homework Equations



Z=1/N!h3N∫∫d3qid3pie-βH(q,p)

The Attempt at a Solution



Well I integrated the momentum and the volume separately. At the end I did get PV=NRT where I'm supposed to show that from this equation I can derive to the ideal gas law equation
∫d3q=VN=Vn/N!
∫d3p=0

However by using Z equation I can derive the entropy for this problem, how?
what about specific heat?
 

Answers and Replies

  • #2
clamtrox
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Lets say the partition function is Z = Ʃ e-βH where the sum runs over all possible configurations (it's a path integral or whatever). Then the probability of a single state is P = e-βH / Z. The entropy is defined as usual, S = -kB Ʃ P ln P where sum is again over all states. Using this knowledge, you should be able to show that S = -β2 ∂/∂β (1/β ln Z).

As for the actual calculation, remember that the Hamiltonian of classical gas molecules is Hi = pi2/2m. Also take into account the extra degree of freedom by adding an extra factor into the Hamiltonian (and sum over all possibilities again). Finally, remember that since your particles are not interacting, the partition function should simplify into a direct product of N single particle systems
[tex] Z_1 = \int d^3p d^3q \sum_{\epsilon} \exp(-\beta H_1(p,q,\epsilon)) [/tex]
[tex] Z_N = \frac{1}{N!} Z_1^N [/tex]
 

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