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

by tsuwal
Tags: maxwell, thermodynamics
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tsuwal
#1
Jan22-13, 10:31 AM
tsuwal's Avatar
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?
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Studiot
#2
Jan22-13, 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]
tsuwal
#3
Jan22-13, 02:21 PM
tsuwal's Avatar
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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