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

I recall seeing an equation in a thermodynamics lecture,

$$\tag{non-ideal gas at constant $V$} \D H = n \int_{T_1}^{T_2}\! C_{P,\mathrm{m}}\ \d T$$

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

$$\underbrace{\D H = nC_{P,\mathrm{m}}\D T}_{\text{true for ideal gases}}\to\overbrace{\D H=n\int_{T_1}^{T_2}\! C_{P,\mathrm{m}}\ \d T}^{\text{the simplest generalisation}}.$$

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?

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    $\begingroup$ The equation in the lecture notes is incorrect. It omits a term involving the partial derivative of entropy with respect to pressure at constant temperature. See the fleshed-out version of my Answer below. $\endgroup$ – Chet Miller Dec 24 '16 at 12:48
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$\def\d{\mathrm{d}}$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, $$\d H=\left[C_p+V\frac{\alpha}{\kappa}\left(1-T\alpha\right)\right]\d T=\left[C_p-VT\frac{\alpha^2}{\kappa}\right]\d T+V\d P$$

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  • $\begingroup$ Great answer! I wouldn't say that the derivation of the first equation is in every thermodynamics book, though; I've looked over four and it wasn't in any of them ;) then again during a lab test I was given seven dysfunctional pipette squeeze bulbs in a row $\endgroup$ – Linear Christmas Dec 24 '16 at 20:02
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    $\begingroup$ @LinearChristmas If you can't find the derivation, I have provided one $\endgroup$ – getafix Dec 25 '16 at 0:37
  • $\begingroup$ All the books I have in my library have the derivation, including Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness and Fundamentals of Engineering Thermodynamics by Moran et al. $\endgroup$ – Chet Miller Dec 25 '16 at 1:59
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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 $q_P=\Delta H$.

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  • $\begingroup$ I have no issue with $$\Delta H = \int_{T_1}^{T_2}C_P\ \mathrm{d}T$$ when pressure is constant. The question is about constant volume. What I am attempting to do is 'prove' the same formula holds for isochoric processes, using only the first law of thermodynamics, definition of entalphy and the condition of constant volume. It is easy to show that $$\tag{constant $V$, ideal gas} \Delta H = C_P \Delta T.$$ But it is not mathematically rigorous to conclude from this the general form. In other words, induction ought to be justified. $\endgroup$ – Linear Christmas Dec 23 '16 at 18:04
  • $\begingroup$ my second equation serves both to produce $C_V$ and $C_P$. I can see where you get your penultimate and antepenultimate equations from but not the last. $\endgroup$ – porphyrin Dec 23 '16 at 18:26
  • $\begingroup$ $\def\d{\mathrm{d}} \def\D{\Delta}$Yes, I understand that $q$ itself is the integral you have given. But to show that $q$ is equal to entalphy when only volume (not pressure) is constant would still need demostrating. $$ $$ In the final step, $$ H=n\left(C_{P,\ m}-\frac{\alpha^2TV}{n\kappa_T}\right)\d T+V\d P,$$assume that $$\D H = \int_{T_1}^{T_2}C_P\ \d T$$ is true and substitute. Therefore, the other terms must amount to $0$, or equivalently $$\frac{V}{n}\int_{T_1}^{T_2}\frac{\alpha^2T}{\kappa_T}\d T=V\D P.$$ Divide both sides by $V$ to get my final equation. $\endgroup$ – Linear Christmas Dec 23 '16 at 18:42
  • $\begingroup$ $\def\d{\mathrm{d}} \def\D{\Delta}$ PS. It is highly suspicious that something like $$\tag{volume constant}\frac{1}{n}\int_{T_1}^{T_2}\frac{\alpha^2T}{\kappa_T}\d T=\D P$$ would hold, and this is why I am not sure that $$\D H = \int_{T_1}^{T_2}C_P\ \d T$$ is true for constant volume. Then again, record shows I might have gone horribly wrong somewhere ;). $\endgroup$ – Linear Christmas Dec 23 '16 at 18:49
  • $\begingroup$ It follows from the first law that $\Delta U = C_V\Delta T$ since V is constant and thats that; no more is needed as the work done is zero. Similarly for $\Delta H =C_P\Delta T$ when volume is allowed to change because pressure is constant. These eqns are quite general. The $\Delta H$ eqn above is only true at constant pressure, which was assumed during its derivation. You can check if your eqns in $\alpha$ etc are correct by assuming ideal gas law and working out derivatives. $\endgroup$ – porphyrin Dec 24 '16 at 9:12
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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" 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.

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

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

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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  • $\begingroup$ You have applied the condition of constant pressure which is not allowed. $+1$ for using \mathrm and deltas tho :) $\endgroup$ – Linear Christmas Dec 24 '16 at 17:46
  • $\begingroup$ @LinearChristmas I thought it was implicit, the equation you quote from your notes has no pressure term ？ $\endgroup$ – getafix Dec 24 '16 at 20:02
  • $\begingroup$ $\def\d{\mathrm{d}}$ $\def\D{\Delta}$No indeed. It only states that $$\tag{non-ideal gas at constant $V$} \D H = n \int_{T_1}^{T_2}\! C_{P,\mathrm{m}}\ \d T$$ 'because then $$\D H = nC_{P,\mathrm{m}}\D T$$ follows for ideal gases.' $\endgroup$ – Linear Christmas Dec 24 '16 at 20:10
  • $\begingroup$ I might assume the integral form to be a good approximation for solids and liquids (at constant $V$) in most cases. Not entirely sure, though. $\endgroup$ – Linear Christmas Dec 24 '16 at 20:13
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    $\begingroup$ Thanks for your personalised update and the derivation! For the time being, I will accept Chester Miller's answer because he was first. However, I might bounty (+50) both your and Chester's answer in the future ;) $\endgroup$ – Linear Christmas Dec 27 '16 at 15:01

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