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

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Discussion Overview

The discussion revolves around the derivation of the differential form of internal energy, specifically how to transition from the equation U=PV/(\gamma-1) to dU=(PdV+VdP)/(\gamma-1). Participants are exploring the mathematical manipulation involved in this transformation, focusing on the application of differentiation techniques.

Discussion Character

  • Technical explanation

Main Points Raised

  • One participant references Feynman's work and expresses confusion about the transition between the two equations, specifically the intuitive understanding of expanding PV into (PdV + VdP).
  • Another participant suggests that U is a function of two variables, proposing that dU can be expressed as a function of changes in P and V, indicating a constant factor.
  • A third participant notes that the product rule for differentiation is likely being applied in this context.

Areas of Agreement / Disagreement

There is no clear consensus on the intuitive understanding of the derivation, as participants are offering different perspectives and interpretations of the mathematical manipulation involved.

Contextual Notes

Participants have not fully resolved the assumptions or steps involved in the differentiation process, and the discussion reflects varying levels of familiarity with the underlying mathematical principles.

InvalidState
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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:
[tex]U=PV/(\gamma-1)[/tex] into [tex]dU=(PdV+VdP)/(\gamma-1)[/tex]

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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