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

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The discussion centers on understanding the transformation of equation 39.11 into 39.12 in Feynman's lectures. The key point is the application of the product rule in calculus to differentiate the product of pressure (P) and volume (V), leading to the expression dU = (PdV + VdP)/(\gamma-1). The user seeks clarity on how Feynman expands PV into the differential form (PdV + VdP). The explanation confirms that U is a function of both P and V, reinforcing the use of the product rule for differentiation. This highlights the importance of understanding multivariable functions in thermodynamics.
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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.
 
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Without reading the link, seems to me U is function in 2 variables: U=f(P,V)= P⋅V⋅const ⇒ dU=(PdV+VdP)⋅const.
 
It looks like he is using the product rule for differentiation.

Chet
 
Ah yes of course! Thanks
 
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