PCRT/I/f9b2

16 Molare freie Energie

$\qquad \qquad \qquad \qquad \qquad$	$dF_{N,T,V}-\mu \,dN+S\,dT-p\,d(-V)$	$=$	$0$
$\qquad \qquad \qquad \qquad \qquad$	$dF_{N,T,V}-\mu \,dN+S\,dT+p\,dV$	$=$	$0$
$\qquad \qquad \qquad \qquad \qquad$	$d(N\ f_{T,v})-\mu \,d(N\,{\it {1}})+(N\,s)\,dT+p\,d(N\,v)$	$=$	$0$
$\qquad \qquad \qquad \qquad \qquad$	$f_{T,v}\,dN+N\,df_{T,v}-\mu \,dN-\mu \,d{\it {1}}+N\,s\,dT+p\,v\,dN+N\,p\,dv$	$=$	$0$
$\qquad \qquad \qquad \qquad \qquad$	$f_{T,v}\,dN+N\,df_{T,v}-\mu \,dN+N\,s\,dT+p\,v\,dN+N\,p\,dv$	$=$	$0$
$\qquad \qquad \qquad \qquad \qquad$	$N(df_{T,v}+s\,dT+p\,dv)+dN(f_{T,v}-\mu +p\,v)$	$=$	$0$
$\qquad \qquad \qquad \qquad \qquad$	$N{\big (}df_{T,v}+s\,dT-p\,d(-v){\big )}+dN{\big (}f_{T,v}-\mu -p\,(-v){\big )}$	$=$	$0$

Die Klammern müssen wieder Null sein und die molare freie Energie und deren Differential ergibt sich zu

$(01)\qquad \qquad \qquad \qquad$	$df_{T,v}=-\,s\,dT-p\,dv$	$\qquad \qquad$	$f_{T,v}=+\,\mu +p\,v$	$\qquad \qquad$	$g_{T,v}=+\mu$
$(01)\qquad \qquad \qquad \qquad$	$df_{T,v}=-s\,dT+p\,d(-v)$	$\qquad \qquad$	$f_{T,v}=+\,\mu \,{\it {1}}+p\,(-v)$	$\qquad \qquad$	$g_{T,v}=+\,\mu \,{\it {1}}$