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.

  • 1
    $\begingroup$ It came from chain rule. $\endgroup$ Oct 23, 2017 at 21:13
  • 1
    $\begingroup$ it is assumed that the internal energy depends on the temperature and volume, $E=E(T,V)$ $\endgroup$
    – hyportnex
    Oct 23, 2017 at 21:14
  • 1
    $\begingroup$ Actually, $$C_p=\left(\frac{\partial H}{\partial T}\right)_p$$ $\endgroup$ Oct 23, 2017 at 21:57

1 Answer 1


If $$E = E(T,V),$$ then $$dE = \frac{\partial E}{\partial T}dT + \frac{\partial E}{\partial V}dV.$$

Or, generally if

$$ f = f(x^i), \quad \text{for }i=1,\dots, n, $$

then $$ df = \partial_1 f dx^1 + \partial_2 f dx^2 + \dots \partial_n f dx^n, $$

where $\partial_1 \equiv \partial/\partial x^1$.

  • $\begingroup$ thanks so much. so does that mean for an ideal gas $E=E(T)$ we would get $dE = \dfrac{\partial E}{\partial T}dT$? $\endgroup$ Oct 23, 2017 at 22:08
  • $\begingroup$ @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. $\endgroup$ Oct 23, 2017 at 22:43

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