CPRT.I.C.02

CPRT.I.C

02 Zustandsänderungen bei (S,V;N=const)

A Energie

$$	$dU$	$=$	$+TdS-pdV$	$dU(S,V),dN=0;(S,V,N):T,p$
$(3.9.1.1.1)$	$dU$	$=$	$+nRT\frac{dS}{nR}-pV\frac{dV}{V}$	$dU(S,V),dN=0,T(S,V,N)$
$$	$T$	$=$	$T_0{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	$T(S,V,N)$
$$	$p$	$=$	$p_0{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	$p(S,V,N)$
$(3.9.1.1.2)$	$dU$	$=$	${\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)[+nR{T}_0\frac{dS}{nR}-p_0{V}_0\frac{dV}{V}]$	$dU(S,V),dN=0$
$(3.9.1.1.3)$	${\left(\frac{\partial U}{\partial S}\right)}_{V,N}$	$=$	$+T_0{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)=+T$	${\partial }_{S}U(S,V,N)$
$(3.9.1.1.4)$	${\left(\frac{\partial U}{\partial V}\right)}_{S,N}$	$=$	$-p_0{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)=-p$	${\partial }_{V}U(S,V,N)$
$$	$\frac{\partial }{\partial V}\frac{\partial U}{\partial S}$	$=$	$-\frac{2}{3}\frac{T_0}{V}{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)=-\frac{2}{3}\frac{T}{V}$	${\partial }_V{\partial }_{S}U(S,V,N),T(S,V,N)$
$$	$\frac{\partial }{\partial S}\frac{\partial U}{\partial V}$	$=$	$-\frac{2}{3}\frac{p_0}{nR}{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)=-\frac{2}{3}\frac{p}{nR}$	${\partial }_S{\partial }_{V}U(S,V,N),p(S,V,N)$
$(3.9.1.1.5)$	$\frac{\partial }{\partial V}\frac{\partial U}{\partial S}$	$=$	$\frac{\partial }{\partial S}\frac{\partial U}{\partial V}$	${\partial }_V{\partial }_{S}U(S,V,N),{\partial }_S{\partial }_{V}U(S,V,N)$

B Enthalpie

$$	$dH$	$=$	$TdS+Vdp$	$dH(S,p),dN=0;(S,p,N):T,V$
$$	$dH$	$=$	$[T-V\frac{\partial T}{\partial V}]dS-(0-V\frac{\partial p}{\partial V})dV$	$dH(S,V),dN=0;(S,V,N):T,p$
$$	$T$	$=$	$T_0{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	$T(S,V,N)$
$$	$-V\frac{\partial T}{\partial V}$	$=$	$-V(-\frac{2}{3})\frac{T}{V}$	${\partial }_{V}T(S,V,N)$
$(3.9.1.2.1)$	$-V\frac{\partial T}{\partial V}$	$=$	$+\frac{2}{3}T$	${\partial }_{V}T(S,V,N)$
$(3.9.1.2.2)$	$-V\frac{\partial T}{\partial V}$	$=$	$+\frac{2}{3}{T}_0{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	${\partial }_{V}T(S,V,N)$
$$	$p$	$=$	$p_0{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	$p(S,V,N)$
$$	$-V\frac{\partial p}{\partial V}$	$=$	$-V(-\frac{5}{3})\frac{p}{V}$	${\partial }_{V}p(S,V,N)$
$(3.9.1.2.3)$	$-V\frac{\partial p}{\partial V}$	$=$	$+\frac{5}{3}p$	${\partial }_{V}p(S,V,N)$
$(3.9.1.2.4)$	$-V\frac{\partial p}{\partial V}$	$=$	$+\frac{5}{3}{p}_0{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	${\partial }_{V}p(S,V,N)$
$(3.9.1.2.5)$	$dH$	$=$	$\frac{5}{3}TdS-\frac{5}{3}pdV$	$dH(S,V),dN=0;(S,V,N):T,p$
$(3.9.1.2.6)$	$dH$	$=$	$\frac{5}{3}dU$	$dH(U)$
$(3.9.1.2.7)$	$dH$	$=$	$\frac{5}{3}{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)[+nR{T}_0\frac{dS}{nR}-p_0{V}_0\frac{dV}{V}]$	$dH(S,V),dN=0$
$(3.9.1.2.8)$	${\left(\frac{\partial H}{\partial S}\right)}_{V,N}$	$=$	$+\frac{5}{3}{T}_0{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)=+\frac{5}{3}T$	${\partial }_{S}H(S,V,N)\ne T,T(S,V,N)$
$(3.9.1.2.9)$	${\left(\frac{\partial H}{\partial V}\right)}_{S,N}$	$=$	$-\frac{5}{3}{p}_0{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)=-\frac{5}{3}p$	${\partial }_{V}H(S,V,N),p(S,V,N)$
$$	$\frac{\partial }{\partial V}\frac{\partial H}{\partial S}$	$=$	$-\frac{10}{9}\frac{T_0}{V}{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)=-\frac{10}{9}\frac{T}{V}$	${\partial }_V{\partial }_{S}H(S,V,N),T(S,V,N)$
$$	$\frac{\partial }{\partial S}\frac{\partial H}{\partial V}$	$=$	$-\frac{10}{9}\frac{p_0}{nR}{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)=-\frac{10}{9}\frac{p}{nR}$	${\partial }_S{\partial }_{V}H(S,V,N),p(S,V,N)$
$(3.9.1.2.10)$	$\frac{\partial }{\partial V}\frac{\partial H}{\partial S}$	$=$	$\frac{\partial }{\partial S}\frac{\partial H}{\partial V}$	${\partial }_V{\partial }_{S}H(S,V,N),{\partial }_S{\partial }_{V}H(S,V,N)$

Übung a

Beim Übergang von der inneren Energie $U(S,V,N)$ zur Enthalpie $H(S,p,N)$ wird $(V\to p)$ ersetzt. Wir berechnen $(T-V\partial T/\partial V)$ und $(p-V\partial p/\partial V)$ für ein ideales Gas im (S,V,N)-Koordinatensystem und nehmen dabei $\mu N=0$ an. Sind beide Ausdrücke zueinander proportional? Welche Ausdrücke sind zueinander proportional und was bedeutet das für $dH$ eines idealen Gases in den unabhängigen Variablen (S,V,N)?

C Freie Energie

$$	$dF$	$=$	$-SdT-pdV$	$dF(T,V),dN=0;(T,V,N):S,p$
$$	$dF$	$=$	$(0-S\frac{\partial T}{\partial S})dS-[p-S\frac{\partial p}{\partial S}]dV$	$dF(S,V),dN=0;(S,V,N):T,p$
$$	$T$	$=$	$T_0{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	$T(S,V,N)$
$$	$-S\frac{\partial T}{\partial S}$	$=$	$-S\left(\frac{2}{3}\frac{1}{nR}\right)T$	${\partial }_{S}T(S,V,N),T(S,V,N)$
$(3.9.1.3.1)$	$-S\frac{\partial T}{\partial S}$	$=$	$-\frac{2}{3}\frac{S}{nR}T$	${\partial }_{S}T(S,V,N),T(S,V,N)$
$(3.9.1.3.2)$	$-S\frac{\partial T}{\partial S}$	$=$	$-\frac{2}{3}\frac{S}{nR}{T}_0{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	${\partial }_{S}T(S,V,N)$
$$	$p$	$=$	$p_0{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	$p(S,V,N)$
$$	$-S\frac{\partial p}{\partial S}$	$=$	$-S\left(\frac{2}{3}\frac{1}{nR}\right)p$	${\partial }_{S}p(S,V,N),p(S,V,N)$
$(3.9.1.3.3)$	$-S\frac{\partial p}{\partial S}$	$=$	$-\frac{2}{3}\frac{S}{nR}p$	${\partial }_{S}p(S,V,N),p(S,V,N)$
$(3.9.1.3.4)$	$-S\frac{\partial p}{\partial S}$	$=$	$-\frac{2}{3}\frac{S}{nR}{p}_0{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	${\partial }_{S}p(S,V,N)$
$(3.9.1.3.5)$	$dF$	$=$	$(-\frac{2}{3}\frac{S}{nR})TdS-(1-\frac{2}{3}\frac{S}{nR})pdV$	$dF(S,V),dN=0;(S,V,N):T,p$
$(3.9.1.3.6)$	$dF$	$=$	$(-\frac{2}{3}\frac{S}{nR})dU-pdV$	$dF(U,V),dN=0;(U,V,N):S,p$
$(3.9.1.3.7)$	$dF$	$=$	${\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	$$
$$	${\displaystyle }$	${\displaystyle }$	$[(-\frac{2}{3}\frac{S}{nR})nR{T}_0\frac{dS}{nR}-(1-\frac{2}{3}\frac{S}{nR})p_0{V}_0\frac{dV}{V}]$	$dF(S,V),dN=0$
$(3.9.1.3.8)$	${\left(\frac{\partial F}{\partial S}\right)}_{V,N}$	$=$	$(-\frac{2}{3}\frac{S}{nR})T_0{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	${\partial }_{S}F(S,V,N)$
$$	${\displaystyle }$	$=$	$(-\frac{2}{3}\frac{S}{nR})T$	$T(S,V,N)$
$(3.9.1.3.9)$	${\left(\frac{\partial F}{\partial V}\right)}_{S,N}$	$=$	$-(1-\frac{2}{3}\frac{S}{nR})p_0{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	${\partial }_{V}F(S,V,N)\ne -p$
$$	${\displaystyle }$	$=$	$-(1-\frac{2}{3}\frac{S}{nR})p$	$p(S,V,N)$
$$	$\frac{\partial }{\partial V}\frac{\partial F}{\partial S}$	$=$	$+\frac{4}{9}\frac{S}{nR}\frac{T_0}{V}{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)=+\frac{4}{9}\frac{S}{nR}\frac{T}{V}$	${\partial }_V{\partial }_{S}F(S,V,N)$
$$	$\frac{\partial }{\partial S}\frac{\partial F}{\partial V}$	$=$	$[+\frac{2}{3}\frac{1}{nR}-(1-\frac{2}{3}\frac{S}{nR})\frac{2}{3}\frac{1}{nR}]$	${\partial }_S{\partial }_{V}F(S,V,N)$
$$	${\displaystyle }$	${\displaystyle }$	$p_0{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	$$
$$	${\displaystyle }$	$=$	$+\frac{4}{9}\frac{S}{nR}\frac{p_0}{nR}{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)=+\frac{4}{9}\frac{S}{nR}\frac{p}{nR}$	$$
$(3.9.1.3.10)$	$\frac{\partial }{\partial V}\frac{\partial F}{\partial S}$	$=$	$\frac{\partial }{\partial S}\frac{\partial F}{\partial V}$	${\partial }_V{\partial }_{S}F(S,V,N),{\partial }_S{\partial }_{V}F(S,V,N)$

Übung a

Beim Übergang von der inneren Energie $U(S,V,N)$ zur freien Energie $F(T,V,N)$ wird $(S\to T)$ ersetzt. Wir berechnen $(T-S\partial T/\partial S)$ und $(p-S\partial p/\partial S)$ für ein ideales Gas im (S,V,N)-Koordinatensystem und nehmen dabei $\mu N=0$ an. Sind beide Ausdrücke zueinander proportional? Welche Ausdrücke sind zueinander proportional und was bedeutet das für $dF$ eines idealen Gases in den unabhängigen Variablen (S,V,N)?

D Freie Enthalpie

$$	$dG$	$=$	$-SdT+Vdp$	$(T,p,N):dG,S,V,dN=0$
$$	$dG$	$=$	$(0-S\frac{\partial T}{\partial S}-V\frac{\partial T}{\partial V})dS-(0-S\frac{\partial p}{\partial S}-V\frac{\partial p}{\partial V})dV$	$(S,V,N):dG,T,p,dN=0$
$dH:(3.9.1.2.1)$	$-V\frac{\partial T}{\partial V}$	$=$	$+\frac{2}{3}T$	${\partial }_{V}T(S,V,N)$
$dH:(3.9.1.2.3)$	$-V\frac{\partial p}{\partial V}$	$=$	$+\frac{5}{3}p$	${\partial }_{V}p(S,V,N)$
$dF:(3.9.1.3.1)$	$-S\frac{\partial T}{\partial S}$	$=$	$-\frac{2}{3}\frac{S}{nR}T$	${\partial }_{S}T(S,V,N)$
$dF:(3.9.1.3.3)$	$-S\frac{\partial p}{\partial S}$	$=$	$-\frac{2}{3}\frac{S}{nR}p$	${\partial }_{S}p(S,V,N)$
$(3.9.1.4.1)$	$dG$	$=$	$(+\frac{2}{3}-\frac{2}{3}\frac{S}{nR})TdS-(+\frac{5}{3}-\frac{2}{3}\frac{S}{nR})pdV$	$(S,V,N):dG,T,p,dN=0$
$(3.9.1.4.2)$	$dG$	$=$	$(+\frac{2}{3}-\frac{2}{3}\frac{S}{nR})dU-pdV$	$(U,V,N):dG,S,p,dN=0$
$(3.9.1.4.3)$	$dG$	$=$	${\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	$$
$$	${\displaystyle }$	${\displaystyle }$	$[(+\frac{2}{3}-\frac{2}{3}\frac{S}{nR})nR{T}_0\frac{dS}{nR}-(+\frac{5}{3}-\frac{2}{3}\frac{S}{nR})p_0{V}_0\frac{dV}{V}]$	$dG(S,V),dN=0$
$(3.9.1.4.4)$	${\left(\frac{\partial G}{\partial S}\right)}_{V,N}$	$=$	$(+\frac{2}{3}-\frac{2}{3}\frac{S}{nR})T_0{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	${\partial }_{S}G(S,V,N)$
$$	${\displaystyle }$	$=$	$(+\frac{2}{3}-\frac{2}{3}\frac{S}{nR})T$	$T(S,V,N)$
$(3.9.1.4.5)$	${\left(\frac{\partial G}{\partial V}\right)}_{S,N}$	$=$	$-(+\frac{5}{3}-\frac{2}{3}\frac{S}{nR})p_0{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	${\partial }_{V}G(S,V,N)$
$$	${\displaystyle }$	$=$	$-(+\frac{5}{3}-\frac{2}{3}\frac{S}{nR})p$	$p(S,V,N)$
$$	$\frac{\partial }{\partial V}\frac{\partial G}{\partial S}$	$=$	$(-\frac{4}{9}+\frac{4}{9}\frac{S}{nR})\frac{T_0}{V}{\left(\frac{V_0}{V}\right)}^{2/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	${\partial }_V{\partial }_{S}G(S,V,N)$
$$	${\displaystyle }$	$=$	$(-\frac{4}{9}+\frac{4}{9}\frac{S}{nR})\frac{T}{V}$	$T(S,V,N)$
$$	$\frac{\partial }{\partial S}\frac{\partial G}{\partial V}$	$=$	$[+\frac{6}{9}\frac{1}{nR}-(+\frac{5}{3}-\frac{2}{3}\frac{S}{nR})\frac{2}{3}\frac{1}{nR}]$	${\partial }_S{\partial }_{V}G(S,V,N)$
$$	${\displaystyle }$	${\displaystyle }$	$p_0{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	$$
$$	${\displaystyle }$	$=$	$(-\frac{4}{9}+\frac{4}{9}\frac{S}{nR})\frac{p_0}{nR}{\left(\frac{V_0}{V}\right)}^{5/3}\exp \left(\frac{2}{3}\frac{S-S_0}{nR}\right)$	$$
$$	${\displaystyle }$	$=$	$(-\frac{4}{9}+\frac{4}{9}\frac{S}{nR})\frac{p}{nR}$	$p(S,V,N)$
$(3.9.1.4.6)$	$\frac{\partial }{\partial V}\frac{\partial G}{\partial S}$	$=$	$\frac{\partial }{\partial S}\frac{\partial G}{\partial V}$	${\partial }_V{\partial }_{S}G(S,V,N),{\partial }_S{\partial }_{V}G(S,V,N)$

Übung a

Beim Übergang von der inneren Energie $U(S,V,N)$ zur freien Enthalpie $G(T,p,N)$ wird $(V\to p)$ und $(S\to T)$ ersetzt. Wir berechnen $(T-S\partial T/\partial S-V\partial T/\partial V)$ und $(p-S\partial p/\partial S-V\partial p/\partial V)$ für ein ideales Gas im (S,V,N)-Koordinatensystem und nehmen dabei $\mu N=0$ an. Sind beide Ausdrücke zueinander proportional? Welche Ausdrücke sind zueinander proportional und was bedeutet das für $dG$ eines idealen Gases in den unabhängigen Variablen (S,V,N)?