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\begin{document}
\centerline{\textbf{Lösungen zum Übungsblatt 10}  }
\vspace{0,5cm} 
Johannes Dörr, Jan Schumann-Bischoff \hfill Übungsgruppe 2, Jürgen Lampe\\
\mbox{}\hrulefill\\
\vspace {1cm}

\textbf{Problem 1: Thermodynamic Relations and Clausius-Clapeyron equation:} \\

\textbf{Tasks:}
\begin{itemize}
	\item[a)] Show the following thermodynamic relation:
		\begin{equation}
			\pd{T}{p}{S} = \frac{T}{C_p}\pd{V}{T}{p}
		\label{eq:relation1}
		\end{equation}
	\item[b)] Show the following thermodynamic relation:
		\begin{equation}
			\pd{V}{p}{S} = \pd{V}{p}{T} + \frac{T}{C_p} \pd{V}{T}{p}^2
		\label{eq:relation2}
		\end{equation}
\end{itemize}

\textbf{Methods:}\\
\begin{itemize}
	\item[a)] We'll show the Maxwell-relation (\textsl{proof 1}):
		\begin{equation}
			\pd{T}{p}{S} = \pd{V}{S}{P}
		\label{eq:maxwell1}
		\end{equation}
		and after a few transformations we'll insert it in the definition of the spesific heat by a constant 		pressure:
		\begin{equation}
			C_p :=	T \pd{S}{T}{p}
		\label{eq:def_pressure}
		\end{equation}
	\item[b)] We concider the functions $f(u,v)$ and $g(u,v)$ and want to prove the following assertions:
		\begin{itemize}
			\item[\textsl{2)}] The Jacobians can be simplified as followed
				\begin{equation}
					\frac{\partial (f,g)}{\partial (u,g)} = \pd{f}{u}{g} \quad \frac{\partial (f,g)}{\partial  						(f,v)} = \pd{g}{v}{f}
					\label{eq:jacobi1}
				\end{equation}
			\item[\textsl{3)}]
				\begin{equation}
					\pd{S}{p}{T} = -\pd{V}{T}{p}
					\label{eq:maxwell2}
				\end{equation}
			\item[\textsl{4)}] 
				\begin{equation}
					\pd{f}{u}{v} = -\frac{\pd{f}{v}{u}}{\pd{u}{v}{f}}
					\label{eq:jacobi2}
				\end{equation}
			\item[\textsl{5)}]
				\begin{equation}
					\frac{C_p}{C_V} = \frac{\kappa_T}{\kappa_S}
					\label{eq:cpv}
				\end{equation}
		\end{itemize}
\end{itemize}

\textbf{Results:}\\

\textbf{Appendix:}\\
a) If we multiply \eqref{eq:def_pressure} with $\partial V$ we obtain:
\begin{align}
	\partial V C_p				&=	T \partial V \pd{S}{T}{p} \notag \\
	\Rightarrow \pd{V}{S}{p}	&=	\frac{T}{C_p} \pd{V}{T}{p} \notag
\end{align}
With the Maxwell-relation \eqref{eq:maxwell1} prooved in \textsl{proof 1} we obtain our relation \eqref{eq:relation1}:
\begin{displaymath}
	\pd{T}{p}{S} = \frac{T}{C_p}\pd{V}{T}{p}
\end{displaymath}

\underline{\textsl{proof 1}}: 
We start with the differential of the enthalpy $H(S,p)$:
\begin{displaymath}
	\diff{H} = T\diff{S} + V \diff{p}
\end{displaymath}
with the obvios equations:
\begin{displaymath}
	T=\pd{H}{S}{p} \qquad \qquad V=\pd{H}{p}{S}
\end{displaymath}
Now we use the permutability of differentations:
\begin{displaymath}
	\pd{T}{p}{S}	= \frac{\partial^2 H}{\partial p \, \partial S}
					= \frac{\partial^2 H}{\partial S \, \partial p}
					= \pd{V}{S}{p}  \qed
\end{displaymath}

b) We start with the definition of the specific heat by a constant temperature and use the relations \eqref{eq:jacobi1}:
\begin{align}
	C_V	&:=	T \pd{S}{T}{V} = T \cdot \frac{\partial (S,V)}{\partial (T,V)} \notag \\
		&=	T \cdot \frac{\partial (S,V)}{\partial (T,p)} \frac{\partial (T,p)}{\partial (T,V)} \notag \\
		&=	T \pd{p}{V}{T}\frac{\partial (S,V)}{\partial (T,p)} \notag \\
		&=	T \pd{p}{V}{T} \left[ \pd{S}{T}{p}\pd{V}{p}{T} - \pd{S}{p}{T}\pd{V}{T}{p} \right] \notag \\
		&=	T\pd{S}{T}{p} - T\pd{p}{V}{T}\pd{S}{p}{T}\pd{V}{T}{p} \notag
\end{align}
With the Maxwell-relation \eqref{eq:maxwell2}, $\alpha := \frac{1}{V} \pd{V}{T}{p}$ and the definition of the isothermal compressibility $\kappa_T := -\frac{1}{V} \pd{V}{p}{T}$ we obtain:
\begin{align}
	C_V	&=	C_p + T\frac{\pd{V}{T}{p}^2}{\pd{V}{p}{T}} \notag \\
	\Rightarrow	C_p - C_V	&= \frac{T\alpha^2}{\kappa_T} \notag
\end{align}
With \eqref{eq:cpv} we obtain:
\begin{align}
	\kappa_S								&=	\kappa_T - \frac{TV\alpha^2}{C_p} \notag \\
	\Rightarrow -\frac{1}{V}\pd{V}{p}{S}	&=	-\frac{1}{V}\pd{V}{p}{T} - \frac{TV\pd{V}{T}{p}^2}{V^2C_p} 			\notag	\\
	\Rightarrow \pd{V}{p}{S}	&=	\pd{V}{p}{T} + \frac{T}{C_p}\pd{V}{T}{p}^2 \notag					
\end{align}

\underline{\textsl{proof 2:}} We also concider the functions $f(u,v)$ and $g(u,v)$. The Jacobian is defined by:
\begin{displaymath}
	\frac{\partial (f,g)}{\partial (u,v)} = \pd{f}{u}{v}\pd{g}{v}{u} - \pd{f}{v}{u}\pd{g}{u}{v}
\end{displaymath}
Now we set $g=v$. $\pd{v}{u}{v}=0$ and  $\pd{g}{v}{u}=1$ is trivial and we obtain:
\begin{displaymath}
	\frac{\partial (f,v)}{\partial (u,v)} = \pd{f}{u}{v}
\end{displaymath}
With the same argumentation you can find with $f=u$:
\begin{displaymath}
	\frac{\partial (u,g)}{\partial (u,v)} = \pd{g}{v}{u}	\quad \qed
\end{displaymath}

\underline{\textsl{proof 3:}} We concider the potential of the free enthalpy $G(T,p)$
\begin{displaymath}
	\diff{G} = -S\diff{T} + V \diff{p}
\end{displaymath}
and find the relations:
\begin{displaymath}
	S=-\pd{G}{T}{p} \qquad \qquad V=\pd{G}{p}{T}
\end{displaymath}
With the same argumentation as in \textsl{proof 1} we obtain the Maxwell-relation:
\begin{displaymath}
	-\pd{S}{p}{T} = \frac{\partial^2 G}{\partial p \partial T}=\frac{\partial^2 G}{\partial T \partial p} = \pd{V}{T}{p} \qquad \qquad \qed
\end{displaymath}

\underline{\textsl{proof 4:}} With the use of \textsl{proof 2} we obtain:
\begin{displaymath}
	\pd{f}{u}{v} = \frac{\partial (f,v)}{\partial (u,v)} = \frac{\partial (f,v)}{\partial (f,u)}\frac{\partial (f,u)}{\partial (u,v)} = -\frac{\pd{f}{v}{u}}{\pd{u}{v}{f}} \qquad \qed
\end{displaymath}

\underline{\textsl{proof 5:}} We start with the entropy $S(V,T)$ and \eqref{eq:jacobi2}:
\begin{align}
	\pd{S}{V}{T}	&= \pd{T}{V}{S}\cdot \frac{\pd{S}{V}{T}}{\pd{T}{V}{S}} \notag \\
					&= -\pd{T}{V}{S}\pd{S}{T}{V} \notag \\
	\Rightarrow -\pd{S}{p}{T} \pd{p}{V}{T}	&= \pd{T}{V}{S}\pd{S}{T}{V} \notag \\
	\Rightarrow -\frac{\pd{S}{p}{T}}{\pd{T}{p}{S}} \pd{p}{V}{T}	&= \pd{T}{V}{S}\pd{p}{T}{S}\pd{S}{T}{V} \notag \\
	\Rightarrow \pd{S}{T}{p} \pd{p}{V}{T}	&= \pd{p}{V}{S} \pd{S}{T}{V} \notag \\
	\Rightarrow -\frac{T}{V}\pd{S}{T}{p} \pd{V}{p}{S}	&= -\frac{T}{V}\pd{S}{T}{V} \pd{V}{p}{T} \notag 
\end{align}
With the definitions of the specific heat and the compressibility we obtain:
\begin{displaymath}
	C_p\kappa_S = C_V\kappa_T \qquad \qed
\end{displaymath}

\newpage
\textbf{Problem 2: Solar Cells and Geothermal Energy} 
	\begin{itemize}
		\item[(a)] The solar constant specifies the average energy, that reaches the surface of the earth averaged 
							 per square meter per second, is $1.367\tx{kJm}^{-2}\tx{s}^{-1}$. Since about one third of the 
							 arriving energy is reflected, only $I = 0.96\tx{kJm}^{-2}\tx{s}^{-1} \approx 1\tx{kJm}^{-2}\tx{s}^{-1}$
							 actually reach the surface.
							 
							 Knowing that the radiation's
							 temperature is $T_2 = 6000\tx{K}$ and estimating $T_1 = 290\tx{K}$ on earth, we obtain the Carnot efficiency:
 							 \begin{align*}
								 \eta_{\tx{C}} &= \frac{T_2 - T_1}{T_2} = 0.95 \ \ .
							 \end{align*}
							 This means that the maximum power $P_{\tx{max}}$, a $1\tx{m}^2$ solar cell can obtain hypothetically, is
 							 \begin{align*}
								 P_{\tx{max}} = \eta_{\tx{C}} I \cdot 1\tx{m}^2 = 0.91\tx{kJs}^{-1} = 910\tx{W} \ \ .
							 \end{align*}
							 This calculation ignores, that the solar cell's temperature probably will increase to higher
							 values than $290\tx{K}$.
		
		\item[(b)] We assume a real efficiency $\eta_{\tx{r}} = 0.05$ of a solar cell. This leads to:
 							 \begin{align*}
								 P_{\tx{r}} = \eta_{\tx{r}} I \cdot 1\tx{m}^2 = 48\tx{W} \ \ .
							 \end{align*}				
							 We regard a cell, that does its job for $30$ years, $5$ hours a day. The "produced" electrical energy 
							 during this time is
 							 \begin{align*}
								 E = \underbrace{30 \cdot 365 \cdot 5 \cdot 60 \cdot 60}_{=: \ t \ [\tx{s}]} \cdot 
								 		 48 \tx{W} = 9461 \cdot 10^6 \tx{J} = 2628 \tx{kWh}\ \ ,
							 \end{align*}
							 since $[\tx{kWh}] = 1000 \cdot [\tx{W}] \cdot 60^2$. 
							 
							 We can compute the adequate price $p$ of the solar cell
							 as follows, assuming $0.15\$/\tx{kWh}$:
 							 \begin{align*}
								 p = 2628 \tx{kWh} \cdot 0.15\$ / \tx{kWh} = 394.2\$ \ \ .
							 \end{align*}
							 After $30$ years, the solar cell is profitable.
		\item[(c)] With 
		 					 \begin{align*}
								 \diff{Q} = c \cdot \diff{T} \ \ ,
							 \end{align*}
							 where $c = 1000 \tx{J}\tx{kg}^{-1}\tx{K}^{-1} \cdot 10^{14}\tx{kg}$ is the specific
							 heat of the rock, we get the change of energy during a change of temperature. To
							 compute the energy $\diff{W}$, the power plant "produces", we need its efficiency.
					 					 						 
							 Similar to task (a), the efficiency is computed by:
		 					 \begin{align*}
								 \eta_{\tx{C}} = \eta_{\tx{C}}(T) &= \frac{T - T_W}{T} \ \ \ \ \ (T_W=20°C=293\tx{K}) \ \ ,
							 \end{align*}
							 with the difference, that one of both temperatures ($T$) is not a constant.
							 
							 Now we get:
		 					 \begin{align*}
								 \diff{W} = c \cdot \eta_{\tx{C}}(T) \diff{T} \ \ .
							 \end{align*}
							 The total amount of energy is:
		 					 \begin{align*}
								 W &= \int \diff{W} = c \cdot \int\limits_{T_i}^{T_f} \eta_{\tx{C}}(T) \diff{T} \\
								   &= c \cdot \int\limits_{T_i}^{T_f} \frac{T - T_W}{T} \diff{T} 
								    = c \cdot \int\limits_{T_i}^{T_f} \left(1 - \frac{T_W}{T} \right) \diff{T} \\
								   &= c \cdot \left[T - T_W \cdot \ln T \right]_{T_i}^{T_f} \\
								   &= c \cdot (T_f - T_W \cdot \ln T_f - T_i + T_W \cdot \ln T_i) \\
								   &= -2.49 \cdot 10^{19} \ \tx{J} \\
								   &= -6.91 \cdot 10^{12} \ \tx{kWh} \ \ ,
							 \end{align*}
							 with $T_i = 600°C = 873K$ and $T_f = 110°C = 383K$. The negative value indicates,
							 that the energy leaves the system (the power plant).
	\end{itemize}
	
	
	
\end{document}