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I Proving thermodynamic relationship

  1. Nov 19, 2016 #1
    Cv = (∂U/∂T)v = T(∂S/∂T)v

    I can prove this by using the Maxwell relations, but I have trouble deriving it from the first law.

    dU = TdS - pdV
    (∂U/∂T)v = T(∂S/∂T)v + ∂S(∂T/∂T)v = T(∂S/∂T)v + ∂S

    Is there a problem with my derivation?
     
  2. jcsd
  3. Nov 19, 2016 #2
    Could you explain how you got the second term?
     
  4. Nov 19, 2016 #3
    I applied to product rule to Tds hence obtaining two terms.
     
  5. Nov 19, 2016 #4
    You don't apply the product rule here because you are not actually differentiating per se - you are not finding ##\frac{d}{dT} (dU)##, which doesn't make sense because well, what does it mean to differentiate an infinitesimal differential quantity?

    Rather what we are doing is simply "division" - the mathematicians will complain here, but what we are doing can be rigorously justified if we so wish. That is to say, we divide by ##dT## throughout and then impose the condition that ##dV = 0##.
     
  6. Nov 19, 2016 #5
    If S=S(T,V), what is dS in terms of dT and dV?
     
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