PCRT.I.D.03

PCRT.I.D

03 Zustandsänderungen des idealen Gases bei (S,p;N=const)

A Enthalpie

$$	$dH$	$=$	$TdS+Vdp$	$dH(S,p),dN=0;(S,p,N):T,V$
$(01)$	$dH$	$=$	$nRT\frac{dS}{nR}+pV\frac{dp}{p}$	$dH(S,p),dN=0;(S,p,N):T,V$
$$	$T$	$=$	$T_0{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$T(S,p,N)$
$$	$V$	$=$	$V_0{\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$V(S,p,N)$
$$	$\left(\frac{p_0}{p}\right){\left(\frac{p}{p_0}\right)}^{2/5}$	$=$	${\left(\frac{p_0}{p}\right)}^{3/5}$	$$
$(02)$	$dH$	$=$	${\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)[+nR{T}_0\frac{dS}{nR}+p_0{V}_0\frac{dp}{p}]$	$dH(S,p),dN=0$
$(03)$	${\left(\frac{\partial H}{\partial S}\right)}_{p,N}$	$=$	$+T_0{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)=+T$	${\partial }_{S}H(S,p,N)=T,T(S,p,N)$
$(04)$	${\left(\frac{\partial H}{\partial p}\right)}_{S,N}$	$=$	$+V_0{\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)=+V$	${\partial }_{p}H(S,p,N)=V,V(S,p,N)$
$$	$\frac{\partial }{\partial p}\frac{\partial H}{\partial S}$	$=$	$+\frac{2}{5}\frac{T_0}{p}{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)=+\frac{2}{5}\frac{T}{p}$	${\partial }_p{\partial }_{S}H(S,p,N),T(S,p,N)$
$$	$\frac{\partial }{\partial S}\frac{\partial H}{\partial p}$	$=$	$+\frac{2}{5}\frac{V_0}{nR}{\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)=+\frac{2}{5}\frac{V}{nR}$	${\partial }_S{\partial }_{p}H(S,p,N)$
$(05)$	$\frac{\partial }{\partial p}\frac{\partial H}{\partial S}$	$=$	$\frac{\partial }{\partial S}\frac{\partial H}{\partial p}$	${\partial }_p{\partial }_{S}H(S,p,N),{\partial }_S{\partial }_{p}H(S,p,N)$

B Innere Energie

$$	$dU$	$=$	$TdS-pdV$	$dU(S,V),T(S,V,N),p(S,V,N),dN=0$
$$	$dU$	$=$	$[T-p\frac{\partial T}{\partial p}]dS+(0-p\frac{\partial V}{\partial p})dp$	$dU(S,p),T(S,p,N),V(S,p,N),dN=0$
$$	$T$	$=$	$T_0{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$T(S,p,N)$
$$	$-p\frac{\partial T}{\partial p}$	$=$	$-p(+\frac{2}{5})\frac{T}{p}$	${\partial }_{p}T(S,p,N)$
$(01)$	$-p\frac{\partial T}{\partial p}$	$=$	$-\frac{2}{5}T$	${\partial }_{p}T(S,p,N)$
$(02)$	$-p\frac{\partial T}{\partial p}$	$=$	$-\frac{2}{5}{T}_0{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	${\partial }_{p}T(S,p,N)$
$$	$V$	$=$	$V_0{\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$V(S,p,N)$
$$	$-p\frac{\partial V}{\partial p}$	$=$	$-p(-\frac{3}{5})\frac{V}{p}$	${\partial }_{p}V(S,p,N)$
$(03)$	$-p\frac{\partial V}{\partial p}$	$=$	$+\frac{3}{5}V$	${\partial }_{p}V(S,p,N)$
$(04)$	$-p\frac{\partial V}{\partial p}$	$=$	$+\frac{3}{5}{V}_0{\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	${\partial }_{p}V(S,p,N)$
$(05)$	$dU$	$=$	$\frac{3}{5}TdS+\frac{3}{5}Vdp$	$dU(S,p),T(S,p,N),V(S,p,N),dN=0$
$(06)$	$dU$	$=$	$\frac{3}{5}dH$	$dU(H)$
$(07)$	$dU$	$=$	$\frac{3}{5}{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)[+nR{T}_0\frac{dS}{nR}+p_0{V}_0\frac{dp}{p}]$	$dU(S,p),dN=0$
$(08)$	${\left(\frac{\partial U}{\partial S}\right)}_{p,N}$	$=$	$+\frac{3}{5}{T}_0{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)=+\frac{3}{5}T$	${\partial }_{S}U(S,p,N)\ne T,T(S,p,N)$
$(09)$	${\left(\frac{\partial U}{\partial p}\right)}_{S,N}$	$=$	$+\frac{3}{5}{V}_0{\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)=+\frac{3}{5}V$	${\partial }_{p}U(S,p,N),V(S,p,N)$
$$	$\frac{\partial }{\partial p}\frac{\partial U}{\partial S}$	$=$	$+\frac{6}{25}\frac{T_0}{p}{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)=+\frac{6}{25}\frac{T}{p}$	${\partial }_p{\partial }_{S}U(S,p,N),T(S,p,N)$
$$	$\frac{\partial }{\partial S}\frac{\partial U}{\partial p}$	$=$	$+\frac{6}{25}\frac{V_0}{nR}{\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)=+\frac{6}{25}\frac{V}{nR}$	${\partial }_S{\partial }_{p}U(S,p,N),V(S,p,N)$
$(10)$	$\frac{\partial }{\partial p}\frac{\partial U}{\partial S}$	$=$	$\frac{\partial }{\partial S}\frac{\partial U}{\partial p}$	${\partial }_p{\partial }_{S}U(S,p,N),{\partial }_S{\partial }_{p}U(S,p,N)$

Übung a

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

C Freie Enthalpie

$$	$dG$	$=$	$-SdT+Vdp$	$dG(T,p),dN=0;(T,P,N):S,V$
$$	$dG$	$=$	$(0-S\frac{\partial T}{\partial S})dS+[V-S\frac{\partial V}{\partial S}]dp$	$dG(S,p),dN=0;(S,p,N):T,V$
$$	$T$	$=$	$T_0{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$T(S,p,N)$
$$	$-S\frac{\partial T}{\partial S}$	$=$	$-S\left(\frac{2}{5}\frac{1}{nR}\right)T$	${\partial }_{S}T(S,p,N)$
$(01)$	$-S\frac{\partial T}{\partial S}$	$=$	$-\frac{2}{5}\frac{S}{nR}T$	${\partial }_{S}T(S,p,N)$
$(02)$	$-S\frac{\partial T}{\partial S}$	$=$	$-\frac{2}{5}\frac{S}{nR}{T}_0{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	${\partial }_{S}T(S,p,N)$
$$	$V$	$=$	${\left(\frac{p_0}{p}\right)}^{3/5}{V}_0\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$V(S,p,N)$
$$	$-S\frac{\partial V}{\partial S}$	$=$	$-S\left(\frac{2}{5}\frac{1}{nR}\right)V$	${\partial }_{S}V(S,p,N)$
$(03)$	$-S\frac{\partial V}{\partial S}$	$=$	$-\frac{2}{5}\frac{S}{nR}V$	${\partial }_{S}V(S,p,N)$
$(04)$	$-S\frac{\partial V}{\partial S}$	$=$	$-\frac{2}{5}\frac{S}{nR}{\left(\frac{p_0}{p}\right)}^{3/5}{V}_0\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	${\partial }_{S}V(S,p,N)$
$(05)$	$dG$	$=$	$+(-\frac{2}{5}\frac{S}{nR})TdS+(1-\frac{2}{5}\frac{S}{nR})Vdp$	$dG(S,p),dN=0;(S,p,N):T,V$
$(06)$	$dG$	$=$	$+(-\frac{2}{5}\frac{S}{nR})dH+Vdp$	$dG(H,p),V(H,p),dN=0$
$(07)$	$dG$	$=$	${\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$$
$$	${\displaystyle }$	${\displaystyle }$	$[+(-\frac{2}{5}\frac{S}{nR})nR{T}_0\frac{dS}{nR}+(1-\frac{2}{5}\frac{S}{nR})V_0{p}_0\frac{dp}{p}]$	$dG(S,p),dN=0$
$(08)$	${\left(\frac{\partial G}{\partial S}\right)}_{p,N}$	$=$	$+(-\frac{2}{5}\frac{S}{nR})T_0{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	${\partial }_{S}G(S,p,N)$
$$	${\displaystyle }$	$=$	$+(-\frac{2}{5}\frac{S}{nR})T$	$T(S,p,N)$
$(09)$	${\left(\frac{\partial G}{\partial p}\right)}_{S,N}$	$=$	$+(1-\frac{2}{5}\frac{S}{nR})V_0{\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	${\partial }_{p}G(S,p,N)\ne V$
$$	${\displaystyle }$	$=$	$+(1-\frac{2}{5}\frac{S}{nR})V$	$V(S,p,N)$
$$	$\frac{\partial }{\partial p}\frac{\partial G}{\partial S}$	$=$	$+(-\frac{4}{25}\frac{S}{nR})\frac{T_0}{p}{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	${\partial }_p{\partial }_{S}G(S,p,N)$
$$	${\displaystyle }$	$=$	$+(-\frac{4}{25}\frac{S}{nR})\frac{T}{p}$	$T(S,p,N)$
$$	$\frac{\partial }{\partial S}\frac{\partial G}{\partial p}$	$=$	$+[(-\frac{2}{5}\frac{1}{nR})+(1-\frac{2}{5}\frac{S}{nR})\left(\frac{2}{5}\frac{1}{nR}\right)]$	${\partial }_S{\partial }_{p}G(S,p,N)$
$$	${\displaystyle }$	${\displaystyle }$	$V_0{\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$$
$$	${\displaystyle }$	$=$	$+(-\frac{4}{25}\frac{S}{nR})\frac{V_0}{nR}{\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$$
$$	${\displaystyle }$	$=$	$+(-\frac{4}{25}\frac{S}{nR})\frac{V}{nR}$	$V(S,p,N)$
$(10)$	$\frac{\partial }{\partial p}\frac{\partial G}{\partial S}$	$=$	$\frac{\partial }{\partial S}\frac{\partial G}{\partial p}$	${\partial }_p{\partial }_{S}G(S,p,N),{\partial }_S{\partial }_{p}G(S,p,N)$

Übung a

Beim Übergang von der Enthalpie $H(S,p,N)$ zur freien Enthalpie $G(T,p,N)$ wird $(S\to T)$ ersetzt. Wir berechnen $(T-S\partial T/\partial S)$ und $(V-S\partial V/\partial S)$ für ein ideales Gas im (S,p,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,p,N)?

D Freie Energie

$$	$dF$	$=$	$-SdT-pdV$	$(T,V,N):dF,S,p,dN=0$
$$	$dF$	$=$	$(0-S\frac{\partial T}{\partial S}-p\frac{\partial T}{\partial p})dS+(0-S\frac{\partial V}{\partial S}-p\frac{\partial V}{\partial p})dp$	$(S,p,N):dF,T,V,dN=0$
$dU:(\text{B}.01)$	$-p\frac{\partial T}{\partial p}$	$=$	$-\frac{2}{5}T$	${\partial }_{p}T(S,p,N)$
$dU:(\text{B}.03)$	$-p\frac{\partial V}{\partial p}$	$=$	$+\frac{3}{5}V$	${\partial }_{p}V(S,p,N)$
$dG:(\text{C}.01)$	$-S\frac{\partial T}{\partial S}$	$=$	$-\frac{2}{5}\frac{S}{nR}T$	${\partial }_{S}T(S,p,N)$
$dG:(\text{C}.03)$	$-S\frac{\partial V}{\partial S}$	$=$	$-\frac{2}{5}\frac{S}{nR}V$	${\partial }_{S}V(S,p,N)$
$(01)$	$dF$	$=$	$(-\frac{2}{5}-\frac{2}{5}\frac{S}{nR})TdS+(+\frac{3}{5}-\frac{2}{5}\frac{S}{nR})Vdp$	$(S,p,N):dF,T,V,dN=0$
$(02)$	$dF$	$=$	$(-\frac{2}{5}-\frac{2}{5}\frac{S}{nR})dH+Vdp$	$(H,p,N):dF,S,V,dN=0$
$(03)$	$dF$	$=$	${\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$$
$$	${\displaystyle }$	${\displaystyle }$	$[+(-\frac{2}{5}-\frac{2}{5}\frac{S}{nR})nR{T}_0\frac{dS}{nR}+(+\frac{3}{5}-\frac{2}{5}\frac{S}{nR})V_0{p}_0\frac{dp}{p}]$	$dF(S,p),dN=0$
$(04)$	${\left(\frac{\partial F}{\partial S}\right)}_{p,N}$	$=$	$+(-\frac{2}{5}-\frac{2}{5}\frac{S}{nR})T_0{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	${\partial }_{S}F(S,p,N)$
$$	${\displaystyle }$	$=$	$+(-\frac{2}{5}-\frac{2}{5}\frac{S}{nR})T$	$T(S,p,N)$
$(05)$	${\left(\frac{\partial F}{\partial p}\right)}_{S,N}$	$=$	$+(+\frac{3}{5}-\frac{2}{5}\frac{S}{nR})V_0{\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	${\partial }_{p}F(S,p,N)$
$$	${\displaystyle }$	$=$	$+(+\frac{3}{5}-\frac{2}{5}\frac{S}{nR})V$	$V(S,p,N)$
$$	$\frac{\partial }{\partial p}\frac{\partial F}{\partial S}$	$=$	$+(-\frac{4}{25}-\frac{4}{25}\frac{S}{nR})\frac{T_0}{p}{\left(\frac{p}{p_0}\right)}^{2/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	${\partial }_p{\partial }_{S}F(S,p,N)$
$$	${\displaystyle }$	$=$	$+(-\frac{4}{25}-\frac{4}{25}\frac{S}{nR})\frac{T}{p}$	$T(S,p,N)$
$$	$\frac{\partial }{\partial S}\frac{\partial F}{\partial p}$	$=$	$+[(-\frac{10}{25}\frac{1}{nR})+(+\frac{3}{5}-\frac{2}{5}\frac{S}{nR})\left(\frac{2}{5}\frac{1}{nR}\right)]$	${\partial }_S{\partial }_{p}F(S,p,N)$
$$	${\displaystyle }$	${\displaystyle }$	$V_0{\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$$
$$	${\displaystyle }$	$=$	$+(-\frac{4}{25}-\frac{4}{25}\frac{S}{nR})\frac{V_0}{nR}{\left(\frac{p_0}{p}\right)}^{3/5}\exp \left(\frac{2}{5}\frac{S-S_0}{nR}\right)$	$$
$$	${\displaystyle }$	$=$	$+(-\frac{4}{25}-\frac{4}{25}\frac{S}{nR})\frac{V}{nR}$	$V(S,p,N)$
$(06)$	$\frac{\partial }{\partial p}\frac{\partial F}{\partial S}$	$=$	$\frac{\partial }{\partial S}\frac{\partial F}{\partial p}$	${\partial }_p{\partial }_{S}F(S,p,N),{\partial }_S{\partial }_{p}F(S,p,N)$

Übung a

Beim Übergang von der Enthalpie $H(S,p,N)$ zur freien Energie $F(T,V,N)$ wird $(p\to V)$ und $(S\to T)$ ersetzt. Wir berechnen $(T-S\partial T/\partial S-p\partial T/\partial p)$ und $(V-S\partial V/\partial S-p\partial V/\partial p)$ für ein ideales Gas im (S,p,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,p,N)?