Proving thermodynamic relationship

[goggles31]

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?

 

[Fightfish]

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?

 

[goggles31]

Could you explain how you got the second term?

I applied to product rule to Tds hence obtaining two terms.

 

[Fightfish]

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 $dV = 0$.

 

[Chestermiller]

If S=S(T,V), what is dS in terms of dT and dV?