Problem 3-4

Ebtehal Alrewaily 28 Nov 2013 04:02

Sloution:

p = kT ln Ξ V

From equation (1-56)

dG=−SdT+Vdp+ μjdN

In The Grand Canonical Ensemble

G = μN =E + pV − T S

Thus we have

pV =−E+TS−μN

p=1/V(−E+TS−μN)→ 1

From equation (3-12)

S=E ̄ -N ̄μ +klnΞ

We can rewrite (1) in term of Entropy in equation (3-12)

p=1/V [-E+T(E ̄ -N ̄μ +k lnΞ )-μN ]

p=1/V [-E+ E − Nμ + TklnΞ − μN]

p = 1/V [ KT ln Ξ ]

p = kT ln Ξ/V

Apply Euler’s theorem

nF( x1 , x2,…, xN )= x1 ∂F + …..+ x2 ∂F

￼￼￼￼V lnΞ (V,T,μ)= v ∂lnΞ =v[∂(pV)]= v p\kT

Therefor

V lnΞ(V,T,μ)= V p/kT

We can rewrit it as

p = kT ln Ξ/V

￼￼￼(∂lnΞ)μ,T= lnΞ = p/kT

p = kT ( ∂ ln Ξ )

And we can use the same method to state that

(∂lnΞ)μ,T=lnΞ = p/kT

p = kT ( ∂ ln Ξ )/ V