Change of Independent Variables in the First Law of Thermodynamics

Question

In Fermi's book "Thermodynamics", in page 20, Fermi shows the first law of thermodynamics:

$dU + pdV = dQ$ (equation 21 in the book).

He then goes on and shows how, if we were to take $T$ and $p$ as independent variables, we'd have:

$\left[ \left( \frac{\partial U}{\partial T} \right)_p + p \left( \frac{\partial V}{\partial T} \right)_p \right] dT + \left[ \left( \frac{\partial U}{\partial p} \right)_T + p \left( \frac{\partial V}{\partial p} \right)_T \right] dp = dQ$ (equation 23 in the book).

I do understand how $dU$ transformed, what I don't understand is how $pdV$ transformed under constant $T$

What I would expect is:

$pdV = \left( \frac{p \partial V}{\partial T} \right)_p + \left( \frac{p \partial V}{\partial p} \right)_T = \left( p \frac{\partial V}{\partial T} \right)_p + \left( \frac{dp}{dp} dV \right)_T + p \left( \frac{\partial V}{\partial p} \right)_T = \left( p \frac{\partial V}{\partial T} \right)_p + (1+p) \left( \frac{\partial V}{\partial p} \right)_T$

Why does Fermi's equation have a factor of $(p)$ instead of $(1+p)$ in the term of the derivative with respect to $p$ with constant $T$?

Thanks!

Answer 1

Your final result should alert you that your calculus is wrong: (i) the LHS is a differential but the RHS is not (we can't have that); and (ii) you are adding a unitless number, 1, to a quantity with dimensions, $p$ (we can't have that either). It all starts with your first step which is not proper calculus.

Simply divide $pdV$ by $dp$ keeping constant $T$: $$ \frac{pdV}{dp}\Big|_\text{const. $T$} = p\left(\frac{\partial V}{\partial p}\right)_T \Rightarrow pdV = p\left(\frac{\partial V}{\partial p}\right)_T dp $$ which is the result we want.