If I have some $H = U + pV$, where $dU = TdS - pdV$, how do I show for some varying $H$, $S$, $p$ that $$\left(\frac{\partial V}{\partial S}\right)_p = \left(\frac{\partial T}{\partial p}\right)_S$$ which will then further show that $$\left(\frac{\partial S}{\partial V}\right)_p \left(\frac{\partial T}{\partial p}\right)_V - \left(\frac{\partial S}{\partial p}\right)_V \left(\frac{\partial T}{\partial V}\right)_p = 1$$ by considering $U$ as a function of $p$ and $V$ and considering two expressions for $\partial^2U/\partial p \partial V$.
What I have done:
Looking at the question at first, I'm thinking that I should first integrate $dU$ to get $U = Ts - pV$, which I then plug back into the first equation concerning $H$.
Then I would have $H = TS$. But this doesn't contain $p$ in it, so it is hard to find the partial derivatives as required.
- Looking at $dU = TdS - pdV$, I can make an expression of $\frac{dV}{dS}$ but this is not a partial derivative.
For the first part.
We have that the enthalpy $H = U + pV$, so $$\mathrm dH=\mathrm dU+p\,\mathrm dV+V\,\mathrm dp\quad\Longrightarrow\quad\mathrm d H= T\,\mathrm d S+V\,\mathrm d p$$
Now $H=H(S,p)$ so that $$\mathrm d H=\left(\frac{\partial H}{\partial S}\right)_p\,\mathrm d S+\left(\frac{\partial H}{\partial p}\right)_S\,\mathrm d p\quad\Longrightarrow\quad \left(\frac{\partial H}{\partial S}\right)_p=T\quad\text{and}\quad \left(\frac{\partial H}{\partial p}\right)_S=V$$ For the differential form of $H$ to be exact the second derivatives of both terms in the differential are equal, i.e. $$\left(\frac{\partial }{\partial S}\left(\frac{\partial H}{\partial p}\right)_S\right)_p=\left(\frac{\partial }{\partial p}\left(\frac{\partial H}{\partial S}\right)_p\right)_S $$ that is $$\left(\frac{\partial V}{\partial S}\right)_p = \left(\frac{\partial T}{\partial p}\right)_S$$
Now you can prove the second part.