Thermodynamic/Partial Differentiation Quesiton

Question

I have a question which is somewhat physics-y but it's still a maths question at heart, I hope someone here can help.

Basically, here is the problem, I get to this intermediate stage in a thermo proof:

$$ dH = T \bigg( \frac{\partial S}{\partial T} \bigg)_P dT $$

The next stage just skips to $$ \bigg( \frac{\partial H}{\partial T} \bigg)_P = T \bigg( \frac{\partial S}{\partial T} \bigg) _P $$

Did they just divide by dT? I thought that, that was not something that is even possible (in terms of proper maths). I've tried applying the product rule but then I end up in the precarious position of trying to to figure out what the derivative of dT is wrt to T, can someone please guide me through how you get from the first equation to the second in formal math terms please!

Thank you.

Answer 1

Let $H = H(T,p)$ just for an example. Then we say that the total differential of H is given by

$$\mathrm dH = \frac{\partial H}{\partial T}\mathrm dT + \frac{\partial H}{\partial p}\mathrm dp $$

Now let $p$ be constant then you don't have a variation in pressure and it becomes

$$\mathrm dH = \frac{\partial H}{\partial T}\mathrm dT$$

To be precise in order to others can understand what is constant physicians wright

$$\mathrm dH = \left(\frac{\partial H}{\partial T}\right)_{p}\mathrm dT\tag{1}$$

Your first equation says that

$$\mathrm dH = T \left( \frac{\partial S}{\partial T} \right)_P \mathrm dT\tag{2}$$

See $(1)$ and $(2)$ together then you'll have the equality you want. An important question is: Why the $H$ have a total differential well defined in the first place? This is a really important physical and mathematical question.