Answer:
[tex]dU=C_{v} dT+(T(\frac{\beta }{\kappa }) -P)dV[/tex]
[tex]dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV[/tex]
Explanation:
The internal energy is equal to:
[tex]dU=C_{v} dT+(T(\frac{\delta P}{\delta T} )_{v} -P)dV[/tex]
The entropy is equal to:
[tex]dS=C_{v} \frac{dT}{T} +(\frac{\delta P}{\delta T} )_{v} dV[/tex]
If we write the pressure derivative in terms of isothermal compresibility and volume expansivity, we have
[tex]\frac{\delta P}{\delta T}=\frac{\beta }{\kappa }[/tex]
Replacing:
[tex]dU=C_{v} dT+(T(\frac{\beta }{\kappa }) -P)dV[/tex]
[tex]dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV[/tex]
