# I Proving thermodynamic relationship

1. Nov 19, 2016

### 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?

2. Nov 19, 2016

### Fightfish

Could you explain how you got the second term?

3. Nov 19, 2016

### goggles31

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

4. Nov 19, 2016

### Fightfish

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 $dV = 0$.

5. Nov 19, 2016

### Staff: Mentor

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