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I Another "Partial Derivatives in Thermodynamics" Question

  1. Jun 16, 2016 #1
    Hi all,

    It seems I haven't completely grasped the use of Partial Derivatives in general; I have seen many discussions here dealing broadly with the same topic, but can't find the answer to my doubt. So, any help would be most welcome:

    In Pathria's book (3rd ed.), equation (1.3.11) says:
    [tex] P = \frac{\left( \frac{\partial S}{\partial V}\right )_{N,E} } {\left (\frac{ \partial S}{\partial E} \right)_{N,V}} = - \left( \frac{\partial E}{\partial V} \right)_{N,S} [/tex]
    My question is 2 fold:

    1. How is he writing the first equality in the above equation?
    2. What properties of partial derivatives are being used here to figure out the correct subscripts on the extreme right in the equation, given the subscripts in [itex] \frac{\left( \frac{\partial S}{\partial V}\right )_{N,E} } {\left (\frac{ \partial S}{\partial E} \right)_{N,V}} [/itex]?
  2. jcsd
  3. Jun 16, 2016 #2
    He's using the triple product rule
    [tex]\left(\frac{\partial x}{\partial y}\right)_{z} \left(\frac{\partial y}{\partial z}\right)_{x} \left(\frac{\partial z}{\partial x}\right)_{y} = -1 [/tex]
  4. Jun 16, 2016 #3
    The equality ##P = - \left( \frac{\partial E}{\partial V} \right)_{N,S} ## comes from ##dE = TdS-PdV+\mu dN## by setting dS and dN equal to zero.

    The equality ## \frac{\left( \frac{\partial S}{\partial V}\right )_{N,E} } {\left (\frac{ \partial S}{\partial E} \right)_{N,V}} = - \left( \frac{\partial E}{\partial V} \right)_{N,S} ## comes from
    $$dS=\left(\frac{\partial S}{\partial V}\right)_{E,N}dV+\left(\frac{\partial S}{\partial E}\right)_{V,N}dE+\left(\frac{\partial S}{\partial N}\right)_{E,V}dN$$ by setting dN and dS equal to zero.
  5. Jun 17, 2016 #4
    Thanks @Fightfish , that helps.
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