Thermodynamics: DT/DV at constant entropy? (last maxwell relation I haven't figured)by tsuwal Tags: maxwell, thermodynamics 

#1
Jan2213, 10:31 AM

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*dTa/k*dV putting S=0, i get, a/k*dV=Cv/T*dT <=> (DT/DV)s=a*T/Cv*k am I right? 



#2
Jan2213, 12:35 PM

P: 5,462

Is this the one you want?
[tex]\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}[/tex] and [tex]\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}[/tex] Therefore [tex]{\left( {\frac{{\partial T}}{{\partial V}}} \right)_S} =  {\left( {\frac{{\partial P}}{{\partial S}}} \right)_V}[/tex] 


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