# A Thermodynamics Pathria Eq.4.5.9

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1. Oct 9, 2018

### IcedCoffee

I'm confused about the mathematics that led to the equation 4.5.9.

Specifically, I'm confused about what the variables that describe U are.

From the equation

I think temperature T(through beta), chemical potential (through alpha), V (through E_s) and N (through... restriction on the summation?) defines the (average) internal energy U,

but how should I do the differentiation? What I ended up was that the chemical potential dependence can be changed to the dependence on the fugacity z so that

but it doesn't seem to be exactly eq.4.5.9 to me.

2. Oct 9, 2018

I have the answer. It's going to take me a few minutes to write out the partial derivatives in Latex... $\\$ $z=z(T, N,V)$, so that $U(T,z,V)=U(T,N,V)$ . $\\$ Three parameters are need to specify $U$, not 4 . Any 3 of the 4, ($z,T,N,V$), can be chosen.$\\$ Now using the first expression for $U$, $\\$ $dU=(\frac{\partial{U}}{\partial{T}})_{z,V} \, dT+(\frac{\partial{U}}{\partial{z}})_{V,T} \, dz+(\frac{\partial{U}}{\partial{V}})_{z,T} \, dV$. $\\$ Taking the second expression: $\\$ $dU=(\frac{\partial{U}}{\partial{T}})_{N,V} \, dT+(\frac{\partial{U}}{\partial{N}})_{V,T} \, dN+(\frac{\partial{U}}{\partial{V}})_{N,T} \, dV$. $\\$ Now, set $dz=0$ and $dV=0$, and set the two $dU's$ equal. Take a partial w.r.t. $T$ at constant $z,V$, (we set $dz$ and $dV=0$ ), =divide through by $dT$ everywhere, and the result is the equation 4.5.9. $\\$ @Chestermiller This is a good thermodynamics problem that the OP has. Might you have anything to add?