# Thermodynamics: DT/DV at constant entropy? (last maxwell relation I haven't figured)

by tsuwal
Tags: maxwell, thermodynamics
 PF Patron P: 103 So, until now I know: (DV/DS)p=(DT/Dp)s=a*T/cp*(rho) (enthalpy) (Dp/DT)v=(DS/DV)t=-a/k (helmoltz) (DS/Dp)t=-(DV/DT)p=-Va (gibbs) a=expansion coefficient k=isothermal compression coefficent cp=heat capacity at constante pressure I want to deduce DT/DV at constant entropy=(DT/DV)s. BUT HOW? Let me try to write S(T,V), then, dS=Cv/T*dT-a/k*dV putting S=0, i get, a/k*dV=Cv/T*dT <=> (DT/DV)s=a*T/Cv*k am I right?
 P: 5,462 Is this the one you want? $$\begin{array}{l} T = {\left( {\frac{{\partial U}}{{\partial S}}} \right)_V} \\ {\left( {\frac{{\partial T}}{{\partial V}}} \right)_S} = \left[ {\frac{\partial }{{\partial V}}{{\left( {\frac{{\partial U}}{{\partial S}}} \right)}_V}} \right] = \frac{{{\partial ^2}U}}{{\partial V\partial S}} \\ \end{array}$$ and $$\begin{array}{l} P = - {\left( {\frac{{\partial U}}{{\partial V}}} \right)_S} \\ {\left( {\frac{{\partial P}}{{\partial S}}} \right)_V} = - \left[ {\frac{\partial }{{\partial S}}{{\left( {\frac{{\partial U}}{{\partial S}}} \right)}_V}} \right] = - \frac{{{\partial ^2}U}}{{\partial S\partial V}} \\ \end{array}$$ Therefore $${\left( {\frac{{\partial T}}{{\partial V}}} \right)_S} = - {\left( {\frac{{\partial P}}{{\partial S}}} \right)_V}$$
 PF Patron P: 103 Hey, thanks for worring so much, but until there I knew... I want to evaluate that derivative further and write in terms of a,k,Cv,Cp,T,p,... as I did (∂T/∂p)s=a*T/cp*(rho)

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