Mathematical Methods for Physics and Engineering - Matematica.NET

5.11 THERMODYNAMIC RELATIONS◮Show that (∂S/∂V) T =(∂P/∂T) V .Applying (5.45) to dS, with independent variables V and T , we find[( ) ( ) ]∂S∂SdU = TdS− PdV= T dV + dT − PdV.∂VT∂TVSimilarly applying (5.45) to dU, we find( ) ∂UdU =∂VThus, equating partial derivatives,( ) ( )∂U ∂S= T − P∂V ∂VBut, sinceT∂ 2 U∂T∂V =TdV +Tand∂2 U∂V∂T , i.e. ∂∂Tit follows that( ) ∂S∂VTThus finally we get the Maxwell relation( ) ∂S∂V( ) ∂UdT .∂TV( ) ( )∂U ∂S= T .∂TV∂TV( ) ∂U= ∂∂VT∂V( ) ∂U,∂TV( )+ T ∂2 S ∂P∂T∂V − = ∂ [ ( ) ] ∂ST= T ∂2 S∂TV∂V ∂TV T∂V∂T .T=( ) ∂P. ◭∂TVThe above derivation is rather cumbersome, however, and a useful trick thatcan simplify the working is to define a new function, called a potential. Theinternal energy U discussed above is one example of a potential but three othersare commonly defined and they are described below.◮Show that (∂S/∂V) T =(∂P/∂T) V by considering the potential U − ST.We first consider the differential d(U − ST). From (5.5), we obtaind(U − ST)=dU − SdT − TdS = −SdT − PdVwhen use is made of (5.44). We rewrite U − ST as F for convenience of notation; F iscalled the Helmholtz potential. ThusdF = −SdT − PdV,and it follows that( )( )∂F∂F= −S and= −P.∂TV∂VTUsing these results together with∂ 2 F∂T∂V =∂2 F∂V∂T ,we can see immediately that( ) ( )∂S ∂P= ,∂VT∂TVwhich is the same Maxwell relation as before. ◭177