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Issue with notation

  1. Apr 8, 2010 #1

    danago

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    Gold Member

    I am currently doing a class on chemical thermodynamics which involves a fair amount of calculus. So far it is going well, however i have a very quick question about the notation being used for partial derivatives.

    If there is some property of a mixture, K=K(T,P), then the differential change in that propety is given by:

    [tex]
    dK = \left(\frac{\partial K}{\partial T}\right)_P dT + \left(\frac{\partial K}{\partial P}\right)_T dP
    [/tex]

    Where the subscripts T and P imply that they are being held constant. My question is -- Does the partial derivative not already imply everything except for one variable is held constant? Would [tex]\frac{\partial K}{\partial T}[/tex], by definition, be the change in K when ONLY T changes, without having to specift that P is held constant?

    I guess what i am asking is -- is there is a difference between [tex]\frac{\partial K}{\partial T}[/tex] and [tex]\left(\frac{\partial K}{\partial T}\right)_P[/tex] that i was not made aware of in my first year calculus courses?
     
  2. jcsd
  3. Apr 8, 2010 #2

    Mark44

    Staff: Mentor

    I agree with you. The P and T subscripts seem redundant to me, for exactly the same reason you gave.
     
  4. Apr 8, 2010 #3

    danago

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    Thanks Mark for clearing that up. I find it a bit strange that the book does it, because it really just makes equations look messier than they should.
     
  5. Apr 8, 2010 #4

    Mark44

    Staff: Mentor

    I'm open to someone who can give a justification for those subscripts.
     
  6. Apr 8, 2010 #5

    D H

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    As a justification, suppose [itex]u=y/x[/itex] and [itex]v=ux[/itex]. Is [itex]\partial v/\partial x=u[/itex] just because you only see that one occurrence of [itex]x[/itex] directly in the equation for [itex]v[/itex]? Of course not. That [itex]u[/itex] in the equation for [itex]v[/itex] is not truly an independent variable -- and neither are most of the variables involved in statistical physics.
     
  7. Apr 9, 2010 #6

    danago

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    So are you saying that [tex]\partial v/\partial x \ne u[/tex], but [tex]\left(\partial v/\partial x\right)_u = u[/tex]? Have i understood you correctly?
     
  8. Apr 9, 2010 #7

    D H

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    Exactly.
     
  9. Apr 9, 2010 #8

    danago

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    Alright :smile:

    Thanks for shedding some light on that
     
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