Question

Using Maxwell’s relations, deduce first and second TdS equations

Answer 1

Answer

According to thermodynamics

We can express entropy in terms of any two of PVT. Let us first express entropy as a function of V and T

$d S=\left(\frac{\partial S}{\partial V}\right)_{T} d V+\left(\frac{\partial S}{\partial T}\right)_{V} d T.$

As

$T d S=T\left(\frac{\partial S}{\partial V}\right)_{T} d V+T\left(\frac{\partial S}{\partial T}\right)_{V} d T.$

Now using maxwell equation

$\left(\frac{\partial S}{\partial V}\right)_{T}=\left(\frac{\partial P}{\partial T}\right)_{V}$ Using Tds=dU

$T\left(\frac{\partial S}{\partial T}\right)_{V}=\left(\frac{\partial U}{\partial T}\right)_{V}=C_{V}$

This is the first of the TdS equations.

Now let us express entropy as a function of P and T

$d S=\left(\frac{\partial S}{\partial P}\right)_{T} d P+\left(\frac{\partial S}{\partial T}\right)_{P} d T.$

$T d S=T\left(\frac{\partial S}{\partial P}\right)_{T} d P+T\left(\frac{\partial S}{\partial T}\right)_{P} d T.$

Using maxwell equation

$\left(\frac{\partial S}{\partial P}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{P}$

We know Tds=dU

$T\left(\frac{\partial S}{\partial T}\right)_{P}=\left(\frac{\partial H}{\partial T}\right)_{P}=C_{P}$

So

$T d S=-T\left(\frac{\partial V}{\partial T}\right)_{P} d P+C_{P} d T.$

This is the first of the TdS equations.