Is this the true expression for isochoric enthalpy of non-ideal gases?

$\def\d{\mathrm{d}}$
$\def\D{\Delta}$


I recall seeing an equation in a thermodynamics lecture,



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> $$\tag{non-ideal gas at constant $V$} \D H = n \int\_{T\_1}^{T\_2}\! C\_{P,\mathrm{m}}\ \d T$$
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> 
> 


where $C\_{P,\mathrm{m}}$ denotes molar heat capacity at constant pressure. There was not really a derivation of this, merely a statement whose mathematical equivalent is



> 
> $$\underbrace{\D H = nC\_{P,\mathrm{m}}\D T}\_{\text{true for ideal gases}}\to\overbrace{\D H=n\int\_{T\_1}^{T\_2}\! C\_{P,\mathrm{m}}\ \d T}^{\text{the simplest generalisation}}.$$
> 
> 
> 




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I have to admit that the explanation made sense at the time. However, trying to justify the result from first principles$-$the $1^{\text{st}}$ law of thermodynamics$-$has proven unfruitful thus far.


$$\tag{$\D V=0, \text{def of}\ C\_{V,\mathrm{m}}$} \d H=nC\_{V,\mathrm{m}}\,\d T + V\,\d P$$
$$\tag{$C\_{P}-C\_{V}=\frac{\alpha^2TV}{\kappa\_T}$}\d H=n\left(C\_{P,\mathrm{m}}-\frac{\alpha^2TV}{n\kappa\_T}\right)\,\d T+V\,\d P$$
Assuming the lecture formula holds, then


$$\tag{volume constant}\frac{1}{n}\int\_{T\_1}^{T\_2}\!\frac{\alpha^2T}{\kappa\_T}\,\d T=\D P.$$
Is this the case? Why?




$\def\d{\mathrm{d}}$The general equation for the change in enthalpy of a non-ideal gas, a liquid, or a solid is $$\d H=C\_p\d T+\left[V-T\left(\frac{\partial V}{\partial T}\right)\_P\right]\d P$$ The derivation of this equation is in every thermo book. So **your lecture equation is not correct**; it omits a term related to the partial derivative of entropy with respect to pressure at constant temperature.


In terms of the coefficient of thermal expansion, this becomes$$\d H=C\_p\d T+V\left[1-T\alpha\right]\d P$$ 


Since, $V=V(T,P)$, we have:
$$\d V=\left(\frac{\partial V}{\partial T}\right)\_P\d T+\left(\frac{\partial V}{\partial P}\right)\_T\d P$$
So, for constant volume, $$\d P=-\frac{(\partial V/\partial T)\_P}{(\partial V/\partial P)\_T}\d T=\frac{\alpha}{\kappa}\d T$$
Therefore, at constant volume, $$\d H=\left[C\_p+V\frac{\alpha}{\kappa}\left(1-T\alpha\right)\right]\d T=\left[C\_p-VT\frac{\alpha^2}{\kappa}\right]\d T+V\d P$$


 3

The derivation for the first part of your question is shown below. I'm not sure what you are trying to do in the second part. 


The heat capacity is the heat required to raise the temperature of 1 mole of substance by 1 K. Suppose for the moment that heat capacity *C* is constant over a temperature range $T\_1 $ to $T\_2$, then the heat *q* required to raise 1 mole of substance over this temperature range is 
$$q=C(T\_2-T\_1) = C\Delta T$$
If the heat capacity is not constant, which is generally the case as real molecules have vibrational and rotational energy levels then, 
$$ q=\int \_{T\_1} ^{T\_2}C(T)dT$$


where in many cases the heat capacity can be approximated as $C(T)=a+bT+cT^{-2}$. For a gas the partition function can be calculated, if spectroscopic values are known, and $C(T)$ calculated that way.


The heat absorbed depends on the path taken to go between the two end points so that *C* is not well defined. To make it well defined either *T* or *P* must be constant. 


At constant volume, $q\_V=C\_V\Delta T$ and the work done as expansion against external pressure is zero, thus from the first law with internal energy as *U*,
$$ \Delta U = C\_V\Delta T $$


and a similar integral as described above is used when the temperature range is too large for $C\_V$ to be considered a constant. (This equation shows also, in the limit of small changes, that heat capacity is the slope of internal energy vs. temperature.)


In the case of constant pressure $q\_P=C\_P\Delta T$. The change in enthalpy is 
$$\Delta H = C\_P\Delta T$$
and so in general
$$ \Delta H=\int \_{T\_1} ^{T\_2}C\_P(T)dT$$


To demonstrate that $q\_P=\Delta H$ we note that in this case the volume is not constant and so the heat transferred at constant pressure $q\_P$ is related to internal energy change $\Delta U$ as 
$$ \Delta U = q\_P-P\Delta V$$
where $P\Delta V$ is the work done. Rearranging and expressing the $\Delta$'s directly gives
$$ q\_P= (U\_2+PV\_2) - (U\_1 +PV\_1)$$
from which using the definition of enthalpy $H=U+PV$ gives $q\_P=H\_2-H\_1$ and shows that $q\_P=\Delta H$.


 3

I'll try to get at the result from first principles, let's start with $U = U(S,V,N)$. I'll just assume constant mole number, so I'll write $U(S,V)$.


We get the enthalpy by taking the "[Legendre transform](https://en.wikipedia.org/wiki/Legendre_transformation)" of the internal energy.
$$H \equiv U[p]$$


which gives, $H = U + pV$ which is function of $S$ and $p$, i.e $H = H(S,p)$
taking the differential
$$\mathrm{d}H = \mathrm{d}U + p\mathrm{d}V + V\mathrm{d}p$$


$$\mathrm{d}U = \overbrace{\left(\frac{\partial U}{\partial S}\right)\_V}^T\, \mathrm{d}S + \overbrace{\left(\frac{\partial U}{\partial V}\right)\_S}^{-p}\, \mathrm{d}V$$


$$\mathrm{d}U = T\, \mathrm{d}S -p\, \mathrm{d}V$$


Using this,
$$\mathrm{d}H = T\, \mathrm{d}S + V\,\mathrm{d}p \tag{\*}$$


imposing **constant pressure** conditions, 


$$\mathrm{d}H = T\, \mathrm{d}S = \delta q\_p $$
Hmm, now we that we have identified enthalpy as the heat exchanged at constant pressure.


Now, $q\_p = n\int\_{T\_i}^{T\_f}C\_{p,m}\mathrm{d}T$ and finally, 
$$\Delta H = \int\_{i}^{f}\mathrm{d}H = q\_p = n\int\_{T\_i}^{T\_f}C\_{p,m}\mathrm{d}T $$ 


Your last equation seems fishy, and I don't see where it comes from but the other two seem fine.




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**EDIT**


So after my discussion with the OP in the comments section I was informed that we are not allowed to impose constant pressure, which I assume implicitly based on the equation he quoted from his lecture notes. If that is indeed the case, then the equation provided in his notes is incomplete. I just looked at @ChesterMiller's answer and that confirms it too.


So basically, how I would proceed is start with $(\*)$


Now, $H = (S,p)$, but $S = S(T,p)$ too, so let's derive and invoke the "2nd $T\mathrm{d}S$ equation.


$$\mathrm{d}S(T,p) = \left(\frac{\partial S}{\partial T}\right)\_p \mathrm{d}T + \left(\frac{\partial S}{\partial p}\right)\_T \mathrm{d}p
$$


multipling the LHS and RHS by $T$


$$T\,\mathrm{d}S = T \left(\frac{\partial S}{\partial T}\right)\_p \mathrm{d}T + T \left(\frac{\partial S}{\partial p}\right)\_T \mathrm{d}p
$$


Using a [Maxwell relation](https://en.wikipedia.org/wiki/Maxwell_relations)


$$T\,\mathrm{d}S = T \left(\frac{\partial S}{\partial T}\right)\_p \mathrm{d}T - T \left(\frac{\partial V}{\partial T}\right)\_p \mathrm{d}p
$$


Almost done, 
$$T\,\mathrm{d}S = C\_p \mathrm{d}T - T V \alpha \mathrm{d}p $$


Putting this in $(\*)$
$$\mathrm{d}H = C\_p \mathrm{d}T - T V \alpha \mathrm{d}p + V \mathrm{d}p$$
or $$\mathrm{d}H = C\_p \mathrm{d}T - V[1-T \alpha] \mathrm{d}p $$


which is the equation @ChesterMiller quoted. Now if you impose constant pressure, then you recover the relation I originally derived and it resembles the one you originally quoted. If you don't wish to do that, then this is the complete expression. To get the relation at constant volume, proceed as @ChesterMiller did in his answer. 


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