# #((delS)/(delV))_T = alpha/beta#. Prove this relation please?

##

#S = "entropy"#

#V = "volume"#

#alpha = "coefficient of isobaric expansion"#

#beta = "coefficient of isothermal compression"#

##### 1 Answer

From the definitions of

#alpha = 1/V((delV)/(delT))_P#

#beta = -1/V((delV)/(delP))_T#

Now, since

#dV = ((delV)/(delT))_PdT + ((delV)/(delP))_TdP#

Notice how this contains both

#cancel(((delV)/(delT))_V)^(0) = ((delV)/(delT))_Pcancel(((delT)/(delT))_V)^(1) + ((delV)/(delP))_T((delP)/(delT))_V#

#-((delV)/(delT))_P = ((delV)/(delP))_T((delP)/(delT))_V#

#((delP)/(delT))_V = -[((delV)/(delT))_P]/[((delV)/(delP))_T]#

This looks quite familiar. Considering what

#((delP)/(delT))_V = -[1/V((delV)/(delT))_P]/[1/V((delV)/(delP))_T]#

#= -[alpha]/[-beta] = alpha/beta#

Now we just have to show that

Consider the *Helmholtz free energy*, the thermodynamic state function that contains the natural variables **total derivative** is:

#dA = ((delA)/(delT))_VdT + ((delA)/(delV))_TdV#

Just like the Gibbs' free energy, where we had

#bb(dA) = dU - d(TS)#

#= dU - SdT - TdS#

From the first law of thermodynamics for a *reversible process in a closed system*,

#=> cancel(TdS) - PdV - SdT - cancel(TdS)#

#= bb(-SdT - PdV)#

which gives the **Maxwell relation** for the Helmholtz free energy. Depending on your professor, you may simply be able to use this relation without deriving it. Compare back to the total derivative to find:

#S = -((delA)/(delT))_V#

#P = -((delA)/(delV))_T#

Since state functions have exact differentials, it follows that ** the second-order cross-derivatives are equal**:

#((del^2A)/(delVdelT))_(V,T) = ((del^2A)/(delTdelV))_(T,V)#

So, let us take the cross-derivatives of the above expressions of

#((delS)/(delV))_T = -((del^2A)/(delVdelT))_(V,T)#

#((delP)/(delT))_V = -((del^2A)/(delTdelV))_(T,V)#

We now see that since the second-order cross-derivatives are equal from the definition of a state function, we've established that **cross-derivative relationship**, and you should be allowed to invoke this any time.

Therefore:

#color(blue)(((delS)/(delV))_T = (alpha)/(beta))#