Partition function for a gas in a cylinder -

T-7
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Partition function for a gas in a cylinder -- urgent!

Hi,

Here's the problem -- it's supposed to be a specimen of what I can expect in my exam, but it isn't much like the tutorial questions I've been doing. I'd really appreciate some help -- fast!

Homework Statement



An ideal gas consisting of N particles of mass m is enclosed in an infinitely-tall cylindrical container of cross-section area A placed vertically in a uniform gravitational field with gravitational acceleration g. The conditions are such that the problem can be treated classically, and the system is in equilibrium at a temperature T.

a) Explain under what circumstances the partition function of a gas can be factored into single particle partition functions.
(b) Obtain an expression for the partition function for a single particle in the cylinder.
(c) Obtain an expression for the total partition function for all gas particles in the cylinder.
(d) Derive the Helmholtz free energy for the gas

The Attempt at a Solution



Clearly g is supposed to be important here. But I would have thought the energy of a single particle could be written

E = p_{i}^2 / 2m

Then, for a single particle,

Z_{c}_{sp} = \sum_{i} e^{-\beta\epsilon_{i}} = Z_{c}_{sp} = \sum_{i} e^{-\beta p_{i}^2/2m}

Then, since \epsilon_{i} can take a large number of values, approximate the sum by an integral:

Z_{c}^{sp} = \int \rho(p) dp e^{-\beta p_{i}^2/2m}

I'll need to derive a density function (presumably the usual 3D one derived in the first octant of a sphere is inappropriate... though the system is surely still 3D).

I would then form the total partition function by raising the single one to the power N and dividing by N! to compensate for the indistinguishability of particles.

However, my partition function is almost certainly wrong... do I just add mgh to it, and if so, how do I compute the partition function integral?

Thanks in advance!
 
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T-7 said:
Then, since \epsilon_{i} can take a large number of values, approximate the sum by an integral:

Z_{c}^{sp} = \int \rho(p) dp e^{-\beta p_{i}^2/2m}

I'll need to derive a density function (presumably the usual 3D one derived in the first octant of a sphere is inappropriate... though the system is surely still 3D).

I don't know why you need a density function or a single particle. The energy of the particle is

E=\frac{p^2}{2\,m}+m\,g\,z

thus

Z_{c}^{sp} = \int e^{ \frac{-\beta\,p^2}{2\,m}}\,d^3\,p \int e^{ -\beta\,m\,g\,z}\,d^3\,x

For the first integral, with the aid of shperical coordinates we have

I_1=\sqrt{8\,\pi^3\,m^3}\,\beta^{-3/2}

while the 2nd one yields

I_2=\frac{A}{\beta\,m\,g}

Finally
Z_{c}^{sp} =\sqrt{\frac{8\,A^2\,\pi^3\,m}{g^2}}\,\beta^{-5/2}

if my calculations are correct! :smile:
 
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Rainbow Child said:
I don't know why you need a density function or a single particle. The energy of the particle is

E=\frac{p^2}{2\,m}+m\,g\,z

thus

Z_{c}^{sp} = \int e^{ \frac{-\beta\,p^2}{2\,m}}\,d^3\,p \int e^{ -\beta\,m\,g\,z}\,d^3\,x

For the first integral, with the aid of shperical coordinates we have

I_1=\sqrt{8\,\pi^3\,m^3}\,\beta^{-3/2}

while the 2nd one yields

I_2=\frac{A}{\beta\,m\,g}

Finally
Z_{c}^{sp} =\sqrt{\frac{8\,A^2\,\pi^3\,m}{g^2}}\,\beta^{-5/2}

if my calculations are correct! :smile:

Thanks for your reply :-)

I believe we are supposed to use a density function, however -- at least, it says later in the question to show that

F \approx NkT log \left(\frac{AkT}{Nh^{3}mg}(2\pi mkT)^{3/2}\right)

h is Plank's constant.

This *looks* to me like an expression has been derived for the k-density, turned into a p-density (p = \hbar k ) and introduced somewhere down the line.
 
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If you need to use a density function then the first integral I_1 is to be evaluated for the magnititude of the momentum.
So the number of the states in the interval (p,d\,p) is

g(p)\,d\,p=\frac{4\,\pi\,p^2\,V}{h^3}\,d\,p

i.e., the density function you mentioned in the 1st post.
 
how would you obtain the total partition function for this question?
 
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