# (Gases) How to get dU=(PdV+VdP)/(gamma-1) from U=PV/

1. Feb 15, 2015

### InvalidState

Hello I'm reading Feynman here: http://www.feynmanlectures.caltech.edu/I_39.html

I'm having problems understanding how he turns equation 39.11 into 39.12.

He does some U -> dU trick and he turns:
$$U=PV/(\gamma-1)$$ into $$dU=(PdV+VdP)/(\gamma-1)$$

I don't understand this intuitively. Any help? Specificaly, how he expands PV into that (PdV+VdP) thing.

2. Feb 15, 2015

### zoki85

Without reading the link, seems to me U is function in 2 variables: U=f(P,V)= P⋅V⋅const ⇒ dU=(PdV+VdP)⋅const.

3. Feb 15, 2015

### Staff: Mentor

It looks like he is using the product rule for differentiation.

Chet

4. Feb 15, 2015

### InvalidState

Ah yes of course! Thanks