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gbertoli
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Homework Statement
Variables: N (number of particles), μ (chemical potential), P (pression), V (volume).
k is Boltzmann's constant. I often use β=1/kT.
The (isothermal) compressibility is given by
\kappa_{T} = -\frac{1}{V}\left (\frac{\partial V}{\partial P}\right )_{N,T}
The following is also true (this was the first part of the assignment but I've already done it):
\left ( \frac{\partial P}{\partial V} \right )_{\mu ,T} = \left ( \frac{\partial P}{\partial V} \right )_{N,T} + \left ( \frac{\partial P}{\partial N} \right )_{V,T}\left ( \frac{\partial N}{\partial V} \right )_{\mu,T}
The questions are:
a) Show that the left-hand side of this equation vanishes
b) Show that
\kappa_{T} = \frac{1}{V}\left ( \frac{\partial \mu}{\partial P} \right )_{V,T} \left ( \frac{\partial N}{\partial \mu} \right )_{V,T}\left ( \frac{\partial V}{\partial N} \right )_{\mu,T}
c) Show that
\left ( \frac{\partial \mu}{\partial P} \right )_{V,T} = \left ( \frac{\partial V}{\partial N} \right )_{\mu,T} = \frac{V}{N}
d) Show that
\kappa_{T} = \frac{V^{2}}{N^{2}} \frac{g_{0}}{2} \int_{0}^{\infty }\frac{\mathrm{d}\varepsilon}{\sqrt{\varepsilon}(e^{\beta\varepsilon-\beta\mu}-1)}
e) The Bose-Einstein condensation can be approached by isothermally compressing the
gas. Describe qualitatively the behavior of κ in that process. What do you consider
an essential difference with the behavior of the ideal gas at low temperatures?
f) Show that in the high temperature the ideal gas result of \kappa_{T} = \frac{1}{P} = \frac{V}{NkT} is recovered.
Homework Equations
To keep the V dependence explicit, the density of states can be written as
g(\varepsilon) = V g_{0} \sqrt{\varepsilon }
Here is g_{0} = 2\pi (2m)^{\frac{3}{2}}h^{-3}
An expression for N is:
N = \int_{0}^{\infty }\mathrm{d}\varepsilon g(\varepsilon) \frac{1}{e^{\beta \varepsilon - \beta \mu} -1} = V g_{0}\int_{0}^{\infty }\mathrm{d}\varepsilon \sqrt{\varepsilon }\frac{1}{e^{\beta \varepsilon - \beta \mu}-1}
The pressure can be found from \beta PV = \ln Z, with Z the gibbs sum, leading to the equation
P = kTg_{0}\int_{0}^{\infty }\mathrm{d}\varepsilon \sqrt{\varepsilon }\log(1-e^{\beta \mu - \beta \varepsilon})
The Attempt at a Solution
My guesses are:
a) The expression for P does not explicitely depend on V and μ and T are held constant. This gives that \left ( \frac{\partial P}{\partial V} \right )_{\mu ,T} is equal to 0.
b) If what I said by a) is true, I can just use the equality
\left ( \frac{\partial P}{\partial V} \right )_{N,T} = - \left ( \frac{\partial P}{\partial N} \right )_{V,T}\left ( \frac{\partial N}{\partial V} \right )_{\mu,T}
and "turn it upside down" (sorry for poor english) to get
\left ( \frac{\partial V}{\partial P} \right )_{N,T} = - \left ( \frac{\partial N}{\partial P} \right )_{V,T}\left ( \frac{\partial V}{\partial N} \right )_{\mu,T}
Then I think I can use (chain rule)
\left ( \frac{\partial N}{\partial P} \right )_{V,T} = \left ( \frac{\partial N}{\partial \mu} \right )_{V,T}\left ( \frac{\partial \mu}{\partial P} \right )_{V,T}
Combining all this should give the right answer:
\kappa_{T} = -\frac{1}{V}\left (\frac{\partial V}{\partial P}\right )_{N,T} = -\frac{1}{V} \times - \left ( \frac{\partial N}{\partial P} \right )_{V,T}\left ( \frac{\partial V}{\partial N} \right )_{\mu,T} =\frac{1}{V} \left ( \frac{\partial N}{\partial \mu} \right )_{V,T}\left ( \frac{\partial \mu}{\partial P} \right )_{V,T}\left ( \frac{\partial V}{\partial N} \right )_{\mu,T}
c) I really don't know, maybe I could get an expression for μ but that seems unlikely.
d)Assuming I already have the results from c), the equation found in b) becomes
\kappa_{T} = \frac{1}{V} \frac{V^{2}}{N^{2}}\left ( \frac{\partial N}{\partial \mu} \right )_{V,T}
so I just need to take the derivative of N with respect to μ. I can use integration by parts but it gets quite messy because you integrate in ε but derive in μ. Anyway, this is what I have:
\frac{\partial }{\partial \mu}V g_{0}\int_{0}^{\infty }\mathrm{d}\varepsilon \sqrt{\varepsilon }\frac{1}{e^{\beta \varepsilon - \beta \mu}-1} = V g_{0}\int_{0}^{\infty }\mathrm{d}\varepsilon \sqrt{\varepsilon }\frac{\partial }{\partial \mu}\frac{1}{e^{\beta \varepsilon - \beta \mu}-1} = \left [ V g_{0} \sqrt{\varepsilon}\frac{1}{e^{\beta \varepsilon - \beta \mu}-1} \right ]_{0}^{\infty} - V g_{0} \int_{0}^{\infty }\mathrm{d}\varepsilon \frac{1}{2\sqrt{\varepsilon }}\frac{1}{e^{\beta \varepsilon - \beta \mu}-1}
I put the derivative inside the integral and derive only the Bose-Einstein distribution because the square root is not μ-dependent. The get the correct answer, the evaluated term should go to zero but i still get a minus sign. Help?
e) I don't really know. Probably κ would diverge but I can only say that from the fact that if T goes to 0 then β goes to infinity. I can't quite grasp the physical insight of the question.
f) No very good ideas. For a small β the exponential goes to 1 but that doesn't really help.
This is my first post so go easy on me.
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