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Pathria: Specific Heat in a gravitational field

  • Thread starter howin
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1. Homework Statement

Pathria 6.8: Evaluate the partition function of a classical ideal gas consisting of N molecules of mass m confined to a cylinder of vertical height L, which is in a state of thermal equilibrium at constant T in a uniform gravitational field of acceleration g. Calculate the specific heat and explain why it is larger than the free space value. (The last part is my problem).

2. Homework Equations

PV=nrT
partition function Z = sum (e^(-BE))

3. The Attempt at a Solution

I made it pretty far fr the first part, although I think I made a mistake somewhere. I'm really looking for help on why the specific heat is larger than the free space value. I am wondering if it has to do with pressure due to gravity.

I used the barometric Boltzmann distribution, taking an arbitrary cross-sectional area A.

So a cross-section of the column has a volume Adh and mass m=ro(h) Adh

So the cross-section exerts a force mg ro(h) n the gass below it

So the excess pressure at height h over the pressure at h+dh is: P(h) -P(h+dh) = -dp = mg ro(h) dh

pV=nRT
ro = N/V
n=N/N(0)
so P=ro k t
dro(h)/dh = -(mg/kt) ro(h)
ro(h)= ro(h(0)) exp[(-mg(h-h0)/kT]

epsilon=mgh
sum(n eps) = E
sum (n) = N

general partition function is: Q=sum [exp(-BE)]

E=sum(n(sub h) mgh)
assume h(0)=0

Z=sum [exp(-B n(sub h) mgh)] = exp(-BmgNL^2/2) --> is this right??

to get the specific heat:

U = N partial ln Q / partial B = N exp (-BmgNL^2/2) (-mgNL^2/2)
= -(mgN^2L^2/2) exp(-BmgNL^2/2)

Cv = dU/dT = (-mgN^2L^2/2T)^2 (1/k) exp(-BmgNL^2/2) --> somehow I think this is wrong, it looks like it has the wrong units

now how do I explain why the specific heat with the gravitational field (this answer) is larger than the free space value?
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

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