I have to prove the identity (∂U∂V)T=T(∂p∂T)V−p I know how to do this using Maxwell's relations, but the assignment specified that I have to use Jacobian transformations. Reading many websites, pdf's, etc. has brought me no closer to the solution.
I think that I have to start from the equation for internal energy and then differentiate it using dV - this produces almost the end result, but when I work with the Jacobian I'm always lost.
Any explanations on how to use the Jacobian would be great. I know the basic rules, but cannot solve this.
Could you help me please?
Answer
In a reversible cyclic process the heat absorbed is the work done, hence ∫CTdS=∫CpdV where C is a closed contour. But it is also true that for the enclosed area A within C we have ∫CTdS=∬ and \int_{C} pdV = \iint_{A} dpdV, so \iint_{A} dTdS = \iint_{A} dpdV, and if we go from the variables T,S to p,V one has \iint_{A} dTdS = \iint_{A} dpdV \frac{\partial(T,S)}{\partial(p,V)}. This implies that the Jacobian must be 1: \frac{\partial(T,S)}{\partial(p,V)}=1, in other words \frac{\partial T}{\partial p}\frac{\partial S}{\partial V}-\frac{\partial T}{\partial V}\frac{\partial S}{\partial p} = 1 This implies that in general \frac{\partial(T,S)}{\partial(x,y)}= \frac{\partial(p,V)}{\partial(x,y)} Now if x=T,y=V as new variables then you get \frac {\partial S}{\partial V} = \frac {\partial p}{\partial T} from which and from dU=TdS-pdV your equality follows.
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