# $c_p - c_v$ proof

My professor has done the proof for us in his lecture notes and I'm having trouble understanding one line.

\begin{align} E &= Q-W \\ dQ & = dE+dW \\ & = \Bigg{(}\dfrac{\partial E}{\partial V}\Bigg{)}_T dV + \Bigg{(}\dfrac{\partial E}{\partial T}\Bigg{)}_V dT + pdV \end{align}

I understand that $c_p = \Bigg{(}\dfrac{\partial Q}{\partial T}\Bigg{)}_p$ and $c_v = \Bigg{(}\dfrac{\partial E}{\partial T}\Bigg{)}_v$ But where did $\Bigg{(}\dfrac{\partial E}{\partial V}\Bigg{)}_T dV$ come from. I've been looking for ages for relationships and I can't find where that term has come from. One conclusion I have is that

$$-P = \Bigg{(}\dfrac{\partial E}{\partial V}\Bigg{)}_T$$

Could someone help me understand where it came from. P.s I don't need the proof I just need what's asked thanks.

• It came from chain rule. Oct 23, 2017 at 21:13
• it is assumed that the internal energy depends on the temperature and volume, $E=E(T,V)$ Oct 23, 2017 at 21:14
• Actually, $$C_p=\left(\frac{\partial H}{\partial T}\right)_p$$ Oct 23, 2017 at 21:57

If $$E = E(T,V),$$ then $$dE = \frac{\partial E}{\partial T}dT + \frac{\partial E}{\partial V}dV.$$
$$f = f(x^i), \quad \text{for }i=1,\dots, n,$$
then $$df = \partial_1 f dx^1 + \partial_2 f dx^2 + \dots \partial_n f dx^n,$$
where $\partial_1 \equiv \partial/\partial x^1$.
• thanks so much. so does that mean for an ideal gas $E=E(T)$ we would get $dE = \dfrac{\partial E}{\partial T}dT$? Oct 23, 2017 at 22:08
• @PatrickMoloney well, if that were the case it would not use partial derivatives as it is a function of only one variable. To prove it to yourself. Consider a trivial example: For an ideal gas $E(T) = 3T^2 + C$, where $C$ is some constant. Then $dE(T) = 6T dT$. does that make it a bit more clear? The partial derivatives in the above answer is due to the fact that it is a multivariable function. Oct 23, 2017 at 22:43