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

  1. Feb 15, 2015 #1
    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.
     
  2. jcsd
  3. Feb 15, 2015 #2
    Without reading the link, seems to me U is function in 2 variables: U=f(P,V)= P⋅V⋅const ⇒ dU=(PdV+VdP)⋅const.
     
  4. Feb 15, 2015 #3
    It looks like he is using the product rule for differentiation.

    Chet
     
  5. Feb 15, 2015 #4
    Ah yes of course! Thanks
     
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