\left.\frac{\partial G}{\partial T}\right|_p=-S
\]
\item Maxwell relations:\\
+ Internal energy: $dE=TdS-pdV$
+ \[
+ \frac{\partial}{\partial S}
+ \left(\left.\frac{\partial E}{\partial V}\right|_S\right)_V=
+ \frac{\partial}{\partial V}
+ \left(\left.\frac{\partial E}{\partial S}\right|_V\right)_S
+ \Rightarrow
+ \left.-\frac{\partial p}{\partial S}\right|_V=
+ \left.\frac{\partial T}{\partial V}\right|_S
+ \]
Enthalpy: $dH=TdS+Vdp$
\[
\frac{\partial}{\partial S}
\left.-\frac{\partial S}{\partial V}\right|_T=
\left.-\frac{\partial p}{\partial T}\right|_V
\]
- \item For a thermodynamic potential $\Phi(X,Y)$ the following identity
- expressing the permutability of derivatives holds:
+ Gibbs free energy: $dG=Vdp-SdT$
\[
- \frac{\partial^2 \Phi}{\partial X \partial Y} =
- \frac{\partial^2 \Phi}{\partial Y \partial X}
+ \frac{\partial}{\partial p}
+ \left(\left.\frac{\partial G}{\partial T}\right|_p\right)_T=
+ \frac{\partial}{\partial T}
+ \left(\left.\frac{\partial G}{\partial p}\right|_T\right)_p
+ \Rightarrow
+ \left.-\frac{\partial S}{\partial p}\right|_T=
+ \left.\frac{\partial V}{\partial T}\right|_p
\]
- Derive the Maxwell relations by taking the mixed derivatives of the
- potentials in (b) with respect to the variables they depend on.
- Exchange the sequence of derivation and use the identities gained in (b).
\end{enumerate}
\section{Thermal expansion of solids}
-It is well known that solids change their length $L$ and volume $V$ respectively
-if there is a change in temperature $T$ or in pressure $p$ of the system.
-The following exercise shows that
-thermal expansion cannot be described by rigorously harmonic crystals.
-
\begin{enumerate}
- \item The coefficient of thermal expansion of a solid is given by
- $\alpha_L=\frac{1}{L}\left.\frac{\partial L}{\partial T}\right|_p$.
- Show that the coefficient of thermal expansion of the volume
- $\alpha_V=\frac{1}{V}\left.\frac{\partial V}{\partial T}\right|_p$
- equals $3\alpha_L$ for isotropic materials.
- \item Find an expression for the pressure as a function of the free energy
+ \item Coefficients of thermal expansion:\\
+ Consider a cube with side lengthes $L_1,L_2,L_3$.
+ Isotropic material: $\frac{1}{L_1}\frac{\partial L_1}{\partial T}=
+ \frac{1}{L_2}\frac{\partial L_2}{\partial T}=
+ \frac{1}{L_3}\frac{\partial L_3}{\partial T}=
+ \alpha_L$.
+ \begin{eqnarray}
+ \alpha_V&=&\frac{1}{V}\frac{\partial V}{\partial T}=
+ \frac{1}{L_1L_2L_3}\frac{\partial}{\partial T}(L_1L_2L_3)=
+ \frac{1}{L_1L_2L_3}\left(L_2L_3\frac{\partial L_1}{\partial T}+
+ L_1L_3\frac{\partial L_2}{\partial T}+
+ L_1L_2\frac{\partial L_3}{\partial T}\right)
+ \nonumber\\
+ &=&\frac{1}{L_1}\frac{\partial L_1}{\partial T}+
+ \frac{1}{L_2}\frac{\partial L_2}{\partial T}+
+ \frac{1}{L_3}\frac{\partial L_3}{\partial T}=3\alpha_L\nonumber
+ \end{eqnarray}
+ \item
+ Find an expression for the pressure as a function of the free energy
$F=E-TS$.
Rewrite this equation to express the pressure entirely in terms of
the internal energy $E$.