From: hackbard Date: Thu, 12 Jun 2008 14:01:10 +0000 (+0200) Subject: seminar X-Git-Url: https://hackdaworld.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=4b4fe331cae31ee4043bc2aec02b8bcdca28395d;p=lectures%2Flatex.git seminar --- diff --git a/solid_state_physics/tutorial/2_04s.tex b/solid_state_physics/tutorial/2_04s.tex index 287c275..bb9024c 100644 --- a/solid_state_physics/tutorial/2_04s.tex +++ b/solid_state_physics/tutorial/2_04s.tex @@ -89,6 +89,16 @@ \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: $dH=TdS+Vdp$ \[ \frac{\partial}{\partial S} @@ -109,31 +119,40 @@ \left.-\frac{\partial S}{\partial V}\right|_T= \left.-\frac{\partial p}{\partial T}\right|_V \] - \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 $E$.