Proving thermodynamic relationship

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  • Thread starter goggles31
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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
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dU = TdS - pdV
(∂U/∂T)v = T(∂S/∂T)v + ∂S(∂T/∂T)v = T(∂S/∂T)v + ∂S
Could you explain how you got the second term?
 
  • #3
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Could you explain how you got the second term?
I applied to product rule to Tds hence obtaining two terms.
 
  • #4
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I applied to product rule to Tds hence obtaining two terms.
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##.
 
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  • #5
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If S=S(T,V), what is dS in terms of dT and dV?
 
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