6.2: Other Relations
- Last updated
- Mar 16, 2021
- Save as PDF
( \newcommand{\kernel}{\mathrm{null}\,}\)
The TdS Equations
The entropy S is a function of the state of the system. We can take it to be a function of any two of the three variables (p, T, V). Taking S to be a function of p and T, we write
TdS=T(∂S∂T)pdT+T(∂S∂p)Tdp
For the first term on the right hand side, we can use
T(∂S∂T)pdT=(∂Q∂T)pdT=Cp
where Cp is the specific heat at constant pressure. Further, using the last of the Maxwell relations, we can now write Equation ??? as
Tds=CpdT−T(∂V∂T)pdp
The coefficient of volumetric expansion (due to heating) is defined by
α=1V(∂V∂T)p
Equation ??? can thus be rewritten as
TdS=CpdT−αTVdp
If we take S to be a function of V and T,
TdS=T(∂S∂T)VdT+T(∂S∂V)TdV
Again the first term on the right hand side can be expressed in terms of Cv, the specific heat at constant volume, using
T(∂S∂T)V=Cv
Further using the Maxwell relations, we get
TdS=CvdT+T(∂p∂T)VdV
Equations ??? (or ??? and ???) are known as the TdS equations.
Equations for Specific Heats
Equating the two expressions for TdS, we get
(Cp−Cv)dT=T[(∂p∂T)VdV+(∂V∂T)pdp]
By the equation of state, we can write p as a function of V and T, so that
dp=(∂p∂T)VdT+(∂p∂V)TdV
Using this in Equation ???, we find
(Cp−Cv)dT=T[(∂p∂T)V+(∂V∂T)p(∂p∂V)T]dV+T(∂V∂T)p(∂p∂T)VdT
However, using Equation 6.1.9, taking X=V, Y=p and Z=T, we have
(∂p∂T)V+(∂V∂T)p(∂p∂V)T=0
Thus the coefficient of dV in Equation ??? vanishes and we can simplify it as
Cp−Cv=T(∂V∂T)p(∂p∂T)V=−T[(∂V∂T)p]2(∂p∂V)T
where we have used Equation ??? again. We have already defined the coefficient of volumetric expansion α. The isothermal compressibility κT is defined by
1κT=−V(∂p∂V)T
In terms of these we can express Cp−Cv as
Cp−Cv=Vα2TκT
This equation is very useful in calculating Cv from measurements of Cp and α and κT. Further, for all substances, κT>0. Thus, we see from this equation that Cp≥Cv. (The result κT>0 can be proved in statistical mechanics.)
In the TdS equations, if dT, dV and dp are related adiabatically, dS=0 and we get
Cp=T(∂V∂T)p(∂p∂T)S,Cv=−T(∂p∂T)V(∂V∂T)S
This gives
CpCv=−(∂V∂T)p(∂p∂T)S[(∂p∂T)V(∂V∂T)S]−1
We have the following relations among the terms involved in this expression,
(∂p∂T)S=(∂V∂T)S(∂p∂V)S=(∂V∂T)S1VκS(∂V∂T)p=(∂p∂T)V(∂T∂p)V(∂V∂T)p=−(∂p∂T)V1(∂p∂V)T=(∂p∂T)VVκT
Using these we find
CpCv=κTκS
Going back to the Maxwell relations and using the expressions for TdS, we find
dU=CvdT+[T(∂p∂T)V−p]dVdH=CpdT+[V−T(∂V∂T)p]dp
these immediately yield the relations
Cv=(∂U∂T)V,Cp=(∂H∂T)p(∂U∂V)T=T(∂p∂T)V−p(∂H∂p)T=V−T(∂V∂T)p
Gibbs-Helmholtz Relation
Since the Helmholtz free energy is defined as F=U−TS,
dF=dU−TdS−SdT=−SdT−pdV
This gives immediately
S=−(∂F∂T)V,p=−(∂F∂V)T
Using this equation for entropy, we find
U=F+TS=F−T(∂F∂T)V
This is known as the Gibbs-Helmholtz relation. If F is known as a function of T and V, we can use these to obtain S, p and U. Thus, all thermodynamic variables can be obtained from F as a function of T and V.
