I have a question pertaining to the following derivation:

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For the heat capacity at constant volume part, we apparently have:

$$dQ = C_v dT + P dV$$

But I find this confusing, as if we are to assume volume is constant, then $dV =0 $ so I would say that

$$dQ = C_v dT$$

Secondly, I don't understand the part how the previous thing I addressed implies

$$C_p = \frac{d Q_p}{dT}$$

As $$C_V = \left(\frac{\partial U}{\partial T}\right)_V = \frac{d U}{dT}$$

So I'd assume

$$C_P = \left(\frac{\partial U}{\partial T}\right)_P = \frac{d U}{dT}$$

but this doesn't look to be the case. What's going on here?


I call this equation ($C_V=(\partial U/\partial T)_V$) the cruelest equation in introductory thermodynamics because of how often it trips people up.

It looks like the misconception here is thinking that the heat capacity is how much the internal energy $U$ increases for a given increase in temperature $T$. This is not the case. I recommend thinking of the heat capacity as how much you have to heat the system to obtain a given increase in $T$. At constant pressure, this heating corresponds to the increase in enthalpy $H$, not the increase in internal energy $U$. Thus, $C_P=(\partial H/\partial T)_P$.


The first thing you need to do is stop thinking about heat capacity in terms of heat Q. In thermodynamics, we define heat capacity in terms of internal energy U and enthalpy H, not in terms of heat. In this way, heat capacity is a physical property of the material being processed, and not a function of the process path. So once you get rid of all the equations in your post that involve Q, things get much simpler. For an arbitrary material (liquid, solid, or gas), the two heat capacities are defined as follows: $$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$ $$C_p=\left(\frac{\partial H}{\partial T}\right)_P$$ Secondly, for an ideal gas, the internal energy and enthalpy are functions only of temperature, and do not depend on volume or pressure. So, for an ideal gas, irrespective of process path, we always have that $$dU=C_vdT$$ and $$dH=C_pdT$$irrespective of whether the volume or pressure are constant.

Thirdly, please note that the heat transferred in a constant pressure process for an ideal gas is not equal to the heat transferred in a constant volume process for an ideal gas. In a constant volume process (assuming temperature independent heat capacities), $$\Delta U=C_v\Delta T=Q$$ and $$\Delta H=C_p\Delta T=\Delta U+V\Delta P=C_v\Delta T+V\Delta P=C_v\Delta T+R\Delta T=Q+R\Delta T$$ In a constant pressure process, $$\Delta U=C_v\Delta T=Q-P\Delta V=Q-R\Delta T$$and $$\Delta H=\Delta U+P\Delta V=C_v\Delta T+R\Delta T=\Delta U+R\Delta T=Q$$ So the change in enthalpy is not equal to the change in internal energy for either of these processes (as expected, since they are different functions of temperature). The only way that $\Delta H$ and be equal to $\Delta U$ is if the temperature doesn't change, so that they are both equal to zero.

  • $\begingroup$ Thank you for this. When you state: "Secondly, for an ideal gas, the internal energy and enthalpy are functions only of temperature, and do not depend on volume or pressure. So, for an ideal gas, irrespective of process path, we always have that" (and then you present the equations) why does that statement imply the equations you then provide? I don't see where they come from, really. $\endgroup$
    – sangstar
    Jan 30 '19 at 20:08
  • $\begingroup$ Is it because the other variable is constant, so it can effectively be treated as a normal derivative, and then you can derive the result there from, say, $C_v = \frac{d U}{dT}$? $\endgroup$
    – sangstar
    Jan 30 '19 at 20:15
  • 1
    $\begingroup$ If U was a function of both T and V, the equation would read: $$dU=C_vdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$ Similarly, if H was a function of both T and P, the equation would read:$$dH=C_pdT+\left(\frac{\partial H}{\partial P}\right)_TdP$$These reduce to the equations I have written for an ideal gas, not because the other variable is constant, but because the other partial derivative is zero. $\endgroup$ Jan 30 '19 at 20:58
  • $\begingroup$ Are the other partial derivatives zero because $dV = 0$ and $dP=0$ because we are considering volume and pressure being constant respectively? $\endgroup$
    – sangstar
    Jan 31 '19 at 16:29
  • $\begingroup$ These equations apply even if dV and dP are not zero. For an ideal gas, the other partial derivatives are always zero, irrespective of dV and dP, so those terms are always zero for an ideal gas. $\endgroup$ Jan 31 '19 at 17:17

Imagine we have a container with gas in it. And you supply the heat.

$C_p$ is the heat energy required to raise the temperature of a mass by 1 K, at constant pressure. Likewise, $C_v$ is for constant volume. When you have constant volume, the energy supplied only goes to increase the internal energy $U$.

But when you do this at constant pressure, work can be done by the gas in the container as $P\Delta V$. So, you have to add more energy than that at constant volume to raise the temperature of the mass by 1 K.

$C_p\Delta T = C_v\Delta T + P\Delta V$ (we multiply $\Delta T$ as we are talking about a temperature change in general)

And the rest is clear I guess, using $P\Delta V = nR\Delta T$, you can possibly derive the relation.


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