01 Die Entropie des idealen Gases bei konstanter Stoffmenge

A Entropie bei (T,V;N=const)

$(3.9.0.1.1)\qquad$	$U$	$=$	${\frac {3}{2}}N\,k\,T$	$=$	${\frac {3}{2}}n\,R\,T$	$\quad U(T)\qquad$
$\qquad$	$p\,V$	$=$	$N\,k\,T$	$=$	$n\,R\,T$	$\quad (p\,V)(T)\qquad$
$\qquad$	$dU$	$=$	$+\,T\,dS-p\,dV$	$=$	$+\,T\,dS-{\frac {n\,R\,T}{V}}\,dV$	$\quad dU(S,V),\,dN=0\qquad$
$\qquad$	$dS$	$=$	${\frac {1}{T}}dU+{\frac {n\,R}{V}}\,dV$	$=$	$+\,{\frac {3}{2}}\,{\frac {n\,R}{T}}\,dT+{\frac {n\,R}{V}}\,dV$	$\quad dS(T,V),\,dN=0\qquad$
$\qquad$	$\left({\frac {\partial S}{\partial T}}\right)_{V,N}$	$=$	$+\,{\frac {3}{2}}\,{\frac {n\,R}{T}}$			$\quad S(T,V,N)\qquad$
$(3.9.0.1.2)\qquad$	$T\,\left({\frac {\partial S}{\partial T}}\right)_{V,N}$	$=$	$+\,{\frac {3}{2}}\,n\,R$	$=$	$S_{0}-\,n\,R$	$\quad S(T,V,N)\qquad$
$\qquad$	$\left({\frac {\partial S}{\partial V}}\right)_{T,N}$	$=$	$+\,{\frac {n\,R}{V}}$			$\quad S(T,V,N)\qquad$
$(3.9.0.1.3)\qquad$	$V\,\left({\frac {\partial S}{\partial V}}\right)_{T,N}$	$=$	$+\,n\,R$			$\quad S(T,V,N)\qquad$
$(3.9.0.1.4)\qquad$	${\frac {\partial }{\partial V}}{\frac {\partial S}{\partial T}}$	$=$	${\frac {\partial }{\partial T}}{\frac {\partial S}{\partial V}}$	$=$	$0$	$\quad S(T,V,N)\qquad$
$(3.9.0.1.5)\qquad$	$dS$	$=$	$+\,{\frac {3}{2}}\,n\,R\,{\frac {dT}{T}}+n\,R\,{\frac {dV}{V}}$	$=$	$+\,{\frac {3}{2}}n\,R\,{\frac {d(T/T_{0})}{T/T_{0}}}+n\,R{\frac {d(V/V_{0})}{V/V_{0}}}$	$\quad dS(T,V),\,dN=0\qquad$
$(3.9.0.1.6)\qquad$	$S$	$=$	$+\,{\frac {3}{2}}n\,R\,\ln {\frac {T}{T_{0}}}+n\,R\,\ln {\frac {V}{V_{0}}}+\,S_{0}$	$=$	$+\,n\,R\,\ln \left({\frac {T}{T_{0}}}\right)^{3/2}\left({\frac {V}{V_{0}}}\right)+\,S_{0}$	$\quad S(T,V,N),\,S_{0}(T_{0},V_{0},N)\qquad$
$\qquad$	$G$	$=$	$U-\,S\,T+\,V\,p$			$\quad (S,V,N):\,G,\,U,\,T,\,p\qquad$
$\qquad$	$G$	$=$	$U-\,T\,S+\,V\,p$			$\quad (T,V,N):\,G,\,U,\,T,\,p\qquad$
$\qquad$	$0$	$=$	$U-\,T\,S+\,V\,p$			$(T_{0},V_{0},N)\,:\,U,\,S,\,p\,;\,G=0\qquad$
$(3.9.0.1.7)\qquad$	$S_{0}$	$=$	${\frac {U_{0}+\,V_{0}\,p_{0}}{T_{0}}}$	$=$	$+\,{\frac {5}{2}}\,n\,R$	$(T_{0},V_{0},N)\,:\,S,\,U,\,p\,;\,G=0\qquad$
$(3.9.0.1.8)\qquad$	$1$	$=$	$\left({\frac {T_{0}}{T}}\right)^{3/2}\left({\frac {V_{0}}{V}}\right)\exp \left({\frac {S-S_{0}}{n\,R}}\right)$	$=$	$\left({\frac {T_{0}}{T}}\right)\left({\frac {V_{0}}{V}}\right)^{2/3}\exp \left({\frac {2}{3}}{\frac {S-S_{0}}{n\,R}}\right)$	$\quad V(S,T,N),T(S,V,N)\qquad$

Übung a

Wir leiten den Wert von $S_{0}$ für $G=0$ her. Dazu verwenden wir die Energie und die Temperatur des idealen Gases $U=(3/2)\,N\,k\,T$ und $T=T_{0}\,(V_{0}/V)^{2/3}\exp((2/3)(S-S_{0})/N\,k)$ . Mit Hilfe der Formel $G=U-\,S\,(\partial U/\partial S)-\,V\,(\partial U/\partial V)$ im $(S,V,N)$ -Koordinatensystem berechnen wir $G(S,V,N)$ . Dann drücken wir die freie Enthalpie $G$ im $(T,V,N)$ -Koordinatensystem aus. Für $G=0$ ergibt sich $S_{0}$ . Welchen Wert hat $S_{0}$ bei dieser Vorgehensweise?

B Entropie bei (T,p;N=const)

$(3.9.0.2.1)\qquad$	$H$	$=$	$U+p\,V$	$=$	${\frac {5}{2}}n\,R\,T$	$\quad H(T)\qquad$
$\qquad$	$dH$	$=$	$T\,dS+V\,dp$	$=$	$+\,T\,dS+\,n\,R\,T\,{\frac {dp}{p}}$	$\quad dH(S,p),dN=0\qquad$
$\qquad$	$dS$	$=$	${\frac {1}{T}}dH-n\,R{\frac {dp}{p}}$	$=$	$+\,{\frac {5}{2}}{\frac {n\,R}{T}}\,dT-\,{\frac {n\,R}{p}}\,dp$	$\quad dS(T,p),dN=0\qquad$
$\qquad$	$\left({\frac {\partial S}{\partial T}}\right)_{p,N}$	$=$	$+\,{\frac {5}{2}}\,{\frac {n\,R}{T}}$			$\quad S(T,p,N)\qquad$
$(3.9.0.2.2)\qquad$	$T\,\left({\frac {\partial S}{\partial T}}\right)_{p,N}$	$=$	$+\,{\frac {5}{2}}\,n\,R$	$=$	$S_{0}$	$\quad S(T,p,N)\qquad$
$\qquad$	$\left({\frac {\partial S}{\partial p}}\right)_{T,N}$	$=$	$-\,{\frac {n\,R}{p}}$			$\quad S(T,p,N)\qquad$
$(3.9.0.2.3)\qquad$	$p\,\left({\frac {\partial S}{\partial p}}\right)_{T,N}$	$=$	$-\,n\,R$			$\quad S(T,p,N)\qquad$
$(3.9.0.2.4)\qquad$	${\frac {\partial }{\partial p}}{\frac {\partial S}{\partial T}}$	$=$	${\frac {\partial }{\partial T}}{\frac {\partial S}{\partial p}}$	$=$	$0$	$\quad S(T,p,N)\qquad$
$(3.9.0.2.5)\qquad$	$dS$	$=$	$+\,{\frac {5}{2}}{\frac {n\,R}{T}}\,dT-\,{\frac {n\,R}{p}}\,dp$	$=$	$+\,{\frac {5}{2}}n\,R\,{\frac {d(T/T_{0})}{T/T_{0}}}-\,n\,R{\frac {d(p/p_{0})}{p/p_{0}}}$	$\quad dS(T,p),dN=0\qquad$
$(3.9.0.2.6)\qquad$	$S$	$=$	$+\,{\frac {5}{2}}n\,R\,\ln \,{\frac {T}{T_{0}}}-n\,R\,\ln \,{\frac {p}{p_{0}}}+\,S_{0}$	$=$	$+\,n\,R\,\ln \left({\frac {T}{T_{0}}}\right)^{5/2}\left({\frac {p_{0}}{p}}\right)+\,S_{0}$	$\quad S(T,p,N)\,,\,S_{0}(T_{0},p_{0},N)\qquad$
$\qquad$	$G$	$=$	$H-\,S\,T$			$\quad (S,p,N):\,G,\,H,\,T,\qquad$
$\qquad$	$G$	$=$	$H-\,T\,S$			$\quad (T,p,N):\,G,\,H,\,S\qquad$
$\qquad$	$0$	$=$	$H-\,T\,S$			$\quad (T_{0},p_{0},N)\,:\,H,\,S\,;\,G=0\qquad$
$(3.9.0.2.7)\qquad$	$S_{0}$	$=$	${\frac {H_{0}}{T_{0}}}$	$=$	${\frac {5}{2}}\,n\,R$	$\quad (T_{0},V_{0},N)\,:\,S,\,H;\,G=0\qquad$
$(3.9.0.2.8)\qquad$	$1$	$=$	$\left({\frac {T_{0}}{T}}\right)^{5/2}\left({\frac {p}{p_{0}}}\right)\exp \left({\frac {S-S_{0}}{n\,R}}\right)$	$=$	$\left({\frac {T_{0}}{T}}\right)\left({\frac {p}{p_{0}}}\right)^{2/5}\exp \left({\frac {2}{5}}{\frac {S-S_{0}}{n\,R}}\right)$	$\quad p(S,T,N),T(S,p,N)\qquad$

Übung a

Wie Übung 3.9.0.1a, aber mit $H=(5/2)\,N\,k\,T$ , $T=T_{0}\,(p/p_{0})^{2/5}\exp((2/5)(S-S_{0})/N\,k)$ und $G=H-\,S\,(\partial H/\partial S)$ im $(S,p,N)$ -Koordinatensystem. Dann weiter wie bei Übung 3.9.0.1a aber im $(T,p,N)$ -Koordinatensystem. Welchen Wert hat $S_{0}$ bei dieser Vorgehensweise?

C Entropie bei (V,p;N=const)

$(3.9.0.3.1)\qquad$	$\ln {\frac {T}{T_{0}}}$	$=$	$\ln {\frac {V\,p}{V_{0}\,p_{0}}}$	$=$	$\ln {\frac {V}{V_{0}}}+\ln {\frac {p}{p_{0}}}$	$\quad \ln \,T/T_{0}(V/V_{0},p/p_{0})\qquad$
$(3.9.0.3.2)\qquad$	${\frac {dT}{T}}$	$=$	${\frac {1}{T}}\left(+\,{\frac {p}{n\,R}}\,dV+\,{\frac {V}{n\,R}}\,dp\right)$	$=$	$+\,{\frac {dV}{V}}+\,{\frac {dp}{p}}$	$\quad dT(V,p),\,dN=0\qquad$
$\qquad$	$dS$	$=$	${\frac {5}{2}}n\,R\,\left(+\,{\frac {dV}{V}}+{\frac {dp}{p}}\right)-\,n\,R{\frac {dp}{p}}$	$=$	$+\,{\frac {5}{2}}{\frac {n\,R}{V}}\,dV+{\frac {3}{2}}{\frac {n\,R}{p}}\,dp$	$\quad dS(V,p),\,dN=0\qquad$
$\qquad$	$\left({\frac {\partial S}{\partial V}}\right)_{p,N}$	$=$	$+\,{\frac {5}{2}}\,{\frac {n\,R}{V}}$			$\quad S(V,p,N)\qquad$
$(3.9.0.3.3)\qquad$	$V\,\left({\frac {\partial S}{\partial V}}\right)_{p,N}$	$=$	$+\,{\frac {5}{2}}\,n\,R$	$=$	$S_{0}$	$\quad S(V,p,N)\qquad$
$\qquad$	$\left({\frac {\partial S}{\partial p}}\right)_{V,N}$	$=$	$+\,{\frac {3}{2}}\,{\frac {n\,R}{p}}$			$\quad S(V,p,N)\qquad$
$(3.9.0.3.4)\qquad$	$p\,\left({\frac {\partial S}{\partial p}}\right)_{V,N}$	$=$	$+\,{\frac {3}{2}}\,n\,R$	$=$	$S_{0}-\,n\,R$	$\quad S(V,p,N)\qquad$
$(3.9.0.3.5)\qquad$	${\frac {\partial }{\partial p}}{\frac {\partial S}{\partial V}}$	$=$	${\frac {\partial }{\partial V}}{\frac {\partial S}{\partial p}}$	$=$	$0$	$\quad S(V,p,N)\qquad$
$(3.9.0.3.6)\qquad$	$dS$	$=$	$+\,{\frac {5}{2}}{\frac {n\,R}{V}}\,dV+{\frac {3}{2}}{\frac {n\,R}{p}}\,dp$	$=$	$+\,{\frac {5}{2}}n\,R\,{\frac {d(V/V_{0})}{(V/V_{0})}}+\,{\frac {3}{2}}n\,R\,{\frac {d(p/p_{0})}{p/p_{0}}}$	$\quad dS(V,p),\,dN=0\qquad$
$(3.9.0.3.7)\qquad$	$S$	$=$	$+\,{\frac {5}{2}}n\,R\,\ln \,{\frac {V}{V_{0}}}+\,{\frac {3}{2}}n\,R\,\ln \,{\frac {p}{p_{0}}}+\,S_{0}$	$=$	$+\,n\,R\,\ln \left({\frac {V}{V_{0}}}\right)^{5/2}\left({\frac {p}{p_{0}}}\right)^{3/2}+\,S_{0}$	$S(V,p,N),\,S_{0}(V_{0},p_{0},N)\qquad$
$(3.9.0.3.8)\qquad$	$S_{0}$	$=$	$+\,{\frac {5}{2}}n\,R$			$\quad (V_{0},p_{0},N):\,S;\,G=0\qquad$
$(3.9.0.3.9)\qquad$	$1$	$=$	$\left({\frac {V_{0}}{V}}\right)^{5/2}\left({\frac {p_{0}}{p}}\right)^{3/2}\exp \left({\frac {S-S_{0}}{n\,R}}\right)$			$\quad p(S,V,N),V(S,p,N)\qquad$
$\qquad$	$1$	$=$	$\left({\frac {V_{0}}{V}}\right)^{5/3}\left({\frac {p_{0}}{p}}\right)\exp \left({\frac {2}{3}}{\frac {S-S_{0}}{n\,R}}\right)$	$=$	$\left({\frac {V_{0}}{V}}\right)\left({\frac {p_{0}}{p}}\right)^{3/5}\exp \left({\frac {2}{5}}{\frac {S-S_{0}}{n\,R}}\right)$	$\quad p(S,V,N),V(S,p,N)\qquad$

Übung a

Wir leiten mit Hilfe von $\ln \,T/T_{0}$ Gl.(3.9.0.3.1) aus $S(p,V,N)$ Gl.(3.9.0.3.3) $S(T,V,N)$ Gl.(3.9.0.1.3) her und bestimmen so indirekt $S_{0}$ Gl.(3.9.0.1.4).