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I've been learning for my Thermodynamics exam when something puzzled me. For an ideal gas, $E=E(T)$, so especially $E(T,V)=E(T)$ and $E(T,P)=E(T)$. But then $dE=(\frac{\partial E}{\partial T})_V dT + (\frac{\partial E}{\partial V})_T dV=(\frac{\partial E}{\partial T})_P dT + (\frac{\partial E}{\partial P})_T dP=\frac{\partial E}{\partial T}dT$ and therefore $C_V=(\frac{\partial E}{\partial T})_V=\frac{\partial E}{\partial T}=(\frac{\partial E}{\partial T})_P=C_P$, and that is clearly not the case, so I must have understood something wrong. But where is my mistake?

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As stated by Chester Miller, you have used an incorrect definition of the heat capacity at constant pressure $C_p$. We define $$C_p := \bigg(\frac{\delta Q}{dT} \bigg)_p = \bigg(\frac{dH}{dT}\bigg)_p,$$ so that $C_p$ gives the amount of heat $\delta Q$ needed to raise the temperature of a system by $dT$ at constant pressure $p$. The second equality follows from the definition of the enthalpy $H = E + pV$. Hence, $$dH = dE + p\,dV +V\,dp= \delta Q - p\,dV +p\,dV +V\,dp = \delta Q +V\,dp.$$ We have here used $dE = \delta Q - p\,dV$. It now immediately follows that $\big(\frac{\delta Q}{dT} \big)_p = \big(\frac{dH}{dT}\big)_p$ since $dp = 0$ when $p$ is held constant.

Now let's move to the monoatomic ideal gas which obeys $pV = Nk_b T$ and $E=E(T)=\frac{3}{2}Nk_b T$. Then the enthalpy $H = E +pV =\frac{3}{2}Nk_b T + Nk_b T = \frac{5}{2}Nk_bT = H(T)$. Therefore, $$C_p = \bigg(\frac{dH}{dT}\bigg)_p = \frac{5}{2}Nk_b = C_V + Nk_b$$ exactly as we might expect since the heat capacity at constant volume of an ideal gas $C_V = \frac{3}{2}Nk_b$.

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  • $\begingroup$ I see, I just assumed that the expression for $C_P$ would look the same as the one for $C_V$. So does the quantity $(\frac{\partial E}{\partial T})_P$ even have any bigger significance/name, just like $(\frac{\partial E}{\partial T})_V$? $\endgroup$ – Intergalakti Jul 19 '18 at 9:54
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For an ideal gas at constant temperature, if V is constant then P is constant, and if P is constant, then V is constant. So, either way, E is a function only of T. But, your equation $$\left(\frac{\partial E}{\partial T}\right)_P=C_p$$ is not the correct relationship (definition) for Cp. The correct definition for Cp is $$C_p\equiv\left(\frac{\partial H}{\partial T}\right)_p$$.

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