PCRT.I.D.01

PCRT.I.D

01 Entropie des idealen Gases

A (T,V;N=const)

$(01)$	$U$	$=$	$\frac{3}{2}NkT$	$=$	$\frac{3}{2}nRT$	$U(T)$
$$	$pV$	$=$	$NkT$	$=$	$nRT$	$(pV)(T)$
$$	$dU$	$=$	$+TdS-pdV$	$=$	$+TdS-\frac{nRT}{V}dV$	$dU(S,V),dN=0$
$$	$dS$	$=$	$\frac{1}{T}dU+\frac{nR}{V}dV$	$=$	$+\frac{3}{2}\frac{nR}{T}dT+\frac{nR}{V}dV$	$dS(T,V),dN=0$
$$	${\left(\frac{\partial S}{\partial T}\right)}_{V,N}$	$=$	$+\frac{3}{2}\frac{nR}{T}$	${\displaystyle }$	${\displaystyle }$	$S(T,V,N)$
$(02)$	$T{\left(\frac{\partial S}{\partial T}\right)}_{V,N}$	$=$	$+\frac{3}{2}nR$	$=$	$S_0-nR$	$S(T,V,N)$
$$	${\left(\frac{\partial S}{\partial V}\right)}_{T,N}$	$=$	$+\frac{nR}{V}$	${\displaystyle }$	${\displaystyle }$	$S(T,V,N)$
$(03)$	$V{\left(\frac{\partial S}{\partial V}\right)}_{T,N}$	$=$	$+nR$	${\displaystyle }$	${\displaystyle }$	$S(T,V,N)$
$(04)$	$\frac{\partial }{\partial V}\frac{\partial S}{\partial T}$	$=$	$\frac{\partial }{\partial T}\frac{\partial S}{\partial V}$	$=$	$0$	$S(T,V,N)$
$(05)$	$dS$	$=$	$+\frac{3}{2}nR\frac{dT}{T}+nR\frac{dV}{V}$	$=$	$+\frac{3}{2}nR\frac{d(T/T_0)}{T/T_0}+nR\frac{d(V/V_0)}{V/V_0}$	$dS(T,V),dN=0$
$(06)$	$S$	$=$	$+\frac{3}{2}nR\ln \frac{T}{T_0}+nR\ln \frac{V}{V_0}+S_0$	$=$	$+nR\ln {\left(\frac{T}{T_0}\right)}^{3/2}\left(\frac{V}{V_0}\right)+S_0$	$S(T,V,N),S_0(T_0,V_0,N)$
$$	$G$	$=$	$U-ST+Vp$	${\displaystyle }$	${\displaystyle }$	$(S,V,N):G,U,T,p$
$$	$G$	$=$	$U-TS+Vp$	${\displaystyle }$	${\displaystyle }$	$(T,V,N):G,U,T,p$
$$	$0$	$=$	$U-TS+Vp$	${\displaystyle }$	${\displaystyle }$	$(T_0,V_0,N):U,S,p;G=0$
$(07)$	$S_0$	$=$	$\frac{U_0+V_0{p}_0}{T_0}$	$=$	$+\frac{5}{2}nR$	$(T_0,V_0,N):S,U,p;G=0$
$(08)$	$1$	$=$	${\left(\frac{T_0}{T}\right)}^{3/2}\left(\frac{V_0}{V}\right)\exp \left(\frac{S-S_0}{nR}\right)$	$=$	$\left(\frac{T_0}{T}\right){\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	$V(S,T,N),T(S,V,N)$

Übung a

Wir leiten den Wert von $S_0$ für $G=0$ her. Dazu verwenden wir die Energie und die Temperatur des idealen Gases $U=(3/2)NkT$ und $T=T_0(V_0/V{)}^{2/3}\exp ((2/3)(S-S_0)/Nk)$. 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 (T,p;N=const)

$(01)$	$H$	$=$	$U+pV$	$=$	$\frac{5}{2}nRT$	$H(T)$
$$	$dH$	$=$	$TdS+Vdp$	$=$	$+TdS+nRT\frac{dp}{p}$	$dH(S,p),dN=0$
$$	$dS$	$=$	$\frac{1}{T}dH-nR\frac{dp}{p}$	$=$	$+\frac{5}{2}\frac{nR}{T}dT-\frac{nR}{p}dp$	$dS(T,p),dN=0$
$$	${\left(\frac{\partial S}{\partial T}\right)}_{p,N}$	$=$	$+\frac{5}{2}\frac{nR}{T}$	${\displaystyle }$	${\displaystyle }$	$S(T,p,N)$
$(02)$	$T{\left(\frac{\partial S}{\partial T}\right)}_{p,N}$	$=$	$+\frac{5}{2}nR$	$=$	$S_0$	$S(T,p,N)$
$$	${\left(\frac{\partial S}{\partial p}\right)}_{T,N}$	$=$	$-\frac{nR}{p}$	${\displaystyle }$	${\displaystyle }$	$S(T,p,N)$
$(03)$	$p{\left(\frac{\partial S}{\partial p}\right)}_{T,N}$	$=$	$-nR$	${\displaystyle }$	${\displaystyle }$	$S(T,p,N)$
$(04)$	$\frac{\partial }{\partial p}\frac{\partial S}{\partial T}$	$=$	$\frac{\partial }{\partial T}\frac{\partial S}{\partial p}$	$=$	$0$	$S(T,p,N)$
$(05)$	$dS$	$=$	$+\frac{5}{2}\frac{nR}{T}dT-\frac{nR}{p}dp$	$=$	$+\frac{5}{2}nR\frac{d(T/T_0)}{T/T_0}-nR\frac{d(p/p_0)}{p/p_0}$	$dS(T,p),dN=0$
$(06)$	$S$	$=$	$+\frac{5}{2}nR\ln \frac{T}{T_0}-nR\ln \frac{p}{p_0}+S_0$	$=$	$+nR\ln {\left(\frac{T}{T_0}\right)}^{5/2}\left(\frac{p_0}{p}\right)+S_0$	$S(T,p,N),S_0(T_0,p_0,N)$
$$	$G$	$=$	$H-ST$	${\displaystyle }$	${\displaystyle }$	$(S,p,N):G,H,T,$
$$	$G$	$=$	$H-TS$	${\displaystyle }$	${\displaystyle }$	$(T,p,N):G,H,S$
$$	$0$	$=$	$H-TS$	${\displaystyle }$	${\displaystyle }$	$(T_0,p_0,N):H,S;G=0$
$(07)$	$S_0$	$=$	$\frac{H_0}{T_0}$	$=$	$\frac{5}{2}nR$	$(T_0,V_0,N):S,H;G=0$
$(08)$	$1$	$=$	${\left(\frac{T_0}{T}\right)}^{5/2}\left(\frac{p}{p_0}\right)\exp \left(\frac{S-S_0}{nR}\right)$	$=$	$\left(\frac{T_0}{T}\right){\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$p(S,T,N),T(S,p,N)$

Übung a

Wie Übung A.a, aber mit $H=(5/2)NkT$, $T=T_0(p/p_0{)}^{2/5}\exp ((2/5)(S-S_0)/Nk)$ und $G=H-S(\partial H/\partial S)$ im $(S,p,N)$-Koordinatensystem. Dann weiter wie bei Übung A.a aber im $(T,p,N)$-Koordinatensystem. Welchen Wert hat $S_0$ bei dieser Vorgehensweise?

C (V,p;N=const)

$(01)$	$\ln \frac{T}{T_0}$	$=$	$\ln \frac{Vp}{V_0{p}_0}$	$=$	$\ln \frac{V}{V_0}+\ln \frac{p}{p_0}$	$\ln T/T_0(V/V_0,p/p_0)$
$(02)$	$\frac{dT}{T}$	$=$	$\frac{1}{T}(+\frac{p}{nR}dV+\frac{V}{nR}dp)$	$=$	$+\frac{dV}{V}+\frac{dp}{p}$	$dT(V,p),dN=0$
$$	$dS$	$=$	$\frac{5}{2}nR(+\frac{dV}{V}+\frac{dp}{p})-nR\frac{dp}{p}$	$=$	$+\frac{5}{2}\frac{nR}{V}dV+\frac{3}{2}\frac{nR}{p}dp$	$dS(V,p),dN=0$
$$	${\left(\frac{\partial S}{\partial V}\right)}_{p,N}$	$=$	$+\frac{5}{2}\frac{nR}{V}$	${\displaystyle }$	${\displaystyle }$	$S(V,p,N)$
$(03)$	$V{\left(\frac{\partial S}{\partial V}\right)}_{p,N}$	$=$	$+\frac{5}{2}nR$	$=$	$S_0$	$S(V,p,N)$
$$	${\left(\frac{\partial S}{\partial p}\right)}_{V,N}$	$=$	$+\frac{3}{2}\frac{nR}{p}$	${\displaystyle }$	${\displaystyle }$	$S(V,p,N)$
$(04)$	$p{\left(\frac{\partial S}{\partial p}\right)}_{V,N}$	$=$	$+\frac{3}{2}nR$	$=$	$S_0-nR$	$S(V,p,N)$
$(05)$	$\frac{\partial }{\partial p}\frac{\partial S}{\partial V}$	$=$	$\frac{\partial }{\partial V}\frac{\partial S}{\partial p}$	$=$	$0$	$S(V,p,N)$
$(06)$	$dS$	$=$	$+\frac{5}{2}\frac{nR}{V}dV+\frac{3}{2}\frac{nR}{p}dp$	$=$	$+\frac{5}{2}nR\frac{d(V/V_0)}{(V/V_0)}+\frac{3}{2}nR\frac{d(p/p_0)}{p/p_0}$	$dS(V,p),dN=0$
$(07)$	$S$	$=$	$+\frac{5}{2}nR\ln \frac{V}{V_0}+\frac{3}{2}nR\ln \frac{p}{p_0}+S_0$	$=$	$+nR\ln {\left(\frac{V}{V_0}\right)}^{5/2}{\left(\frac{p}{p_0}\right)}^{3/2}+S_0$	$S(V,p,N),S_0(V_0,p_0,N)$
$(08)$	$S_0$	$=$	$+\frac{5}{2}nR$	${\displaystyle }$	${\displaystyle }$	$(V_0,p_0,N):S;G=0$
$(09)$	$1$	$=$	${\left(\frac{V_0}{V}\right)}^{5/2}{\left(\frac{p_0}{p}\right)}^{3/2}\exp \left(\frac{S-S_0}{nR}\right)$	${\displaystyle }$	${\displaystyle }$	$p(S,V,N),V(S,p,N)$
$$	$1$	$=$	${\left(\frac{V_0}{V}\right)}^{5/3}\left(\frac{p_0}{p}\right)\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	$=$	$\left(\frac{V_0}{V}\right){\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$p(S,V,N),V(S,p,N)$

Übung a

Wir leiten mit Hilfe von $\ln T/T_0$ Gl.(01) aus $S(V,p,N)$ Gl.(07) $S(T,V,N)$ Gl.(A.06) her.