Allgemeine Formel für spezifische Wärmen

Wir suchen einen Zusammenhang für $ C_{p}$ und $ C_{V}$ für beliebige Substanzen.

$\displaystyle C_{V}$ $\displaystyle =\left( \frac{\partial Q}{dT}\right) _{V}=T\left( \frac{\partial S}{\partial T}\right) _{V}$    
$\displaystyle C_{p}$ $\displaystyle =\left( \frac{\delta Q}{dT}\right) _{p}=T\left( \frac{\partial S}{\partial T}\right) _{p}$ (3.387)

Wir betrachten $ S\left( p,T\right) $

dann gilt:

$\displaystyle \delta Q=TdS=T\left( \left( \frac{\partial S}{\partial p}\right) _{T}dp+\left( \frac{\partial S}{\partial T}\right) _{p}dT\right)$ (3.388)

also

$\displaystyle \delta Q=C_{p}dT+T\left( \frac{\partial S}{\partial p}\right) _{T}dp$ (3.389)

Wir drücken $ dp$ als Funktion von $ T,V$ aus

$\displaystyle dp=\left( \frac{\partial p}{\partial T}\right) _{V}dT+\left( \frac{\partial p}{\partial V}\right) _{T}dV$ (3.390)

und nun $ C_{V}$

$\displaystyle C_{V}=\left( \frac{\delta Q}{\partial T}\right) _{V}=C_{p}+T\left...
...tial S}{\partial p}\right) _{T}\left( \frac{\partial p}{\partial T}\right) _{V}$ (3.391)

Maxwellrelation:

$\displaystyle \left( \frac{\partial S}{\partial p}\right) _{T}=-\left( \frac{\partial V}{\partial T}\right) _{p}$ (3.392)

also

$\displaystyle C_{V}=C_{p}-T\left( \frac{\partial V}{\partial T}\right) _{p}\left( \frac{\partial p}{\partial T}\right) _{V}$ (3.393)

Mit

$\displaystyle \alpha=\frac{1}{V}\left( \frac{\partial V}{\partial T}\right) _{p}=-\frac {1}{V}\left( \frac{\partial S}{\partial p}\right) _{T}  $   Volumenausdehnungskoeffizient (3.394)

Umschreiben von

$\displaystyle \left( \frac{\partial p}{\partial T}\right) _{V}dV =$ $\displaystyle 0=dV=\left( \frac{\partial V}{\partial p}\right) _{T}dp+\left( \frac{\partial V}{\partial T}\right) _{p}dT$    
  $\displaystyle \Rightarrow dp\left( \frac{\partial V}{\partial p}\right) _{T}=-dT\left( \frac{\partial V}{\partial T}\right) _{p}$    
$\displaystyle \left( \frac{\partial p}{\partial T}\right) _{V}$ $\displaystyle =-\frac{\left( \frac{\partial V}{\partial T}\right) _{p}}{\left( \frac{\partial V}{\partial p}\right) _{T}}$ (3.395)

Mit

$\displaystyle \kappa=-\frac{1}{V}\left( \frac{\partial V}{\partial p}\right) _{T}$ (3.396)

isotherme Kompressibilität

wird

$\displaystyle \left( \frac{\partial p}{\partial T}\right) _{V}= \frac{\alpha}{\kappa}$ (3.397)

also

$\displaystyle C_{V}=C_{p}-T\cdot V\cdot\frac{\alpha^{2}}{\kappa}$ (3.398)

oder

$\displaystyle C_{p}-C_{V}=TV\frac{\alpha^{2}}{\kappa}$ (3.399)

        für bel. Substanzen

Beispiel: Kupfer:

$\displaystyle V$ $\displaystyle =7,1\frac{cm^{3}}{mol}=\frac{7,1\cdot10^{-6}m^{3}}{mol},\alpha =5\cdot10^{-5}\frac{1}{\kappa}$    
$\displaystyle \kappa$ $\displaystyle =4,5\cdot10^{-13}\frac{cm^{2}}{dyn}=4,5\cdot10^{-13}\cdot \frac{10^{-4}m^{2}}{10^{-5}N}=4,5\cdot10^{-12}\frac{m^{2}}{N}$    
$\displaystyle C_{p}$ $\displaystyle =24,5\frac{y}{Kmol}\Rightarrow C_{p}-C_{V}=1,2\frac{J} {Kmol}\Rightarrow\gamma=\frac{C_{p}}{C_{V}}=1,05$ (3.400)

Wie ändert sich $ C_{V}$ mit dem Volumen?

Betrachte:

$\displaystyle S$ $\displaystyle =S\left( T,V\right)$    
$\displaystyle dS$ $\displaystyle =\left( \frac{\partial S}{\partial T}\right) _{V}dT+\left( \frac{\partial S}{\partial V}\right) _{T}dV$    
  $\displaystyle =\frac{1}{T}C_{V}dT+\left( \frac{\partial p}{\partial T}\right) _{V}dV$ (3.401)

Lösbar wenn die Zustandsgleichung bekannt ist:

Bsp. Ideales Gas $ \left( \frac{\partial p}{\partial T}\right) _{V}=\frac
{Nk}{V}$

$\displaystyle \left( \frac{\partial C_{V}}{\partial V}\right) _{T}$ $\displaystyle =\left( \frac{\partial}{\partial V}\right) _{T}\left[ T\left( \fr...
... S}{\partial T}\right) _{V}\right] =T\frac{\partial^{2}S}{\partial V\partial T}$    
  $\displaystyle =T\left( \frac{\partial}{\partial T}\right) _{V}\left( \frac{\par...
...rtial }{\partial T}\right) _{V}\left( \frac{\partial p}{\partial T}\right) _{V}$ (3.402)

also

$\displaystyle \left( \frac{\partial C_{V}}{\partial V}\right) _{T}=T\left( \frac {\partial^{2}p}{\partial T^{2}}\right) _{V}$ (3.403)

Wie hängt die innere Energie $ U(T,V)$ ab?

$\displaystyle dU$ $\displaystyle =TdS-pdV=\left( C_{V}dT+T\left( \frac{\partial p}{\partial T}\right) _{V}dV\right) -pdV$    
  $\displaystyle =\left( \frac{\partial U}{\partial T}\right) _{V}dT+\left( \frac{\partial U}{\partial V}\right) _{T}dV$ (3.404)

also

$\displaystyle \left( \frac{\partial U}{\partial T}\right) _{V}=C_{V}\left( \fra...
... U}{\partial V}\right) _{T}=T\left( \frac{\partial p}{\partial T}\right) _{V}-p$ (3.405)

Othmar Marti
Experimentelle Physik
Universiät Ulm