I'm looking at a textbook question (Engel & Reid, Thermodynamics, Statistical Thermodynamics & Kinetics, 4 ed, Q3.1) which states the following:

The heat capacity $C_{\mathrm m,p}$ is less than $C_{\mathrm m,V}$ for $\ce{H2O(l)}$ near $\pu{4 °C}$. Explain this result.

Their answer is that water contracts as you heat it in this regime, thus (at constant $p$) the surroundings are doing work on the system, hence $C_{\mathrm m,p}\lt C_{\mathrm m,V}$.

But the textbook also contains the following equation (which is correct, and whose only restrictions are that the system is of fixed phase and composition, and $\text{đ}w = –pdV$, i.e., pV-work only):

$C_{\mathrm m,p}-C_{\mathrm m,V} = T V_\mathrm m \beta^2/\kappa$

where $\beta$ is the isobaric thermal expansivity, and $\kappa$ is the isothermal compressibility.

Based on this, for a substance of fixed phase and composition, $C_{\mathrm m,p} \ge C_{\mathrm m,V}$, always, because, while $\beta$ can be $\le0$* (water near $\pu{4 °C}$ being a notable example), $\beta^2$, as well as $T, V_\mathrm m$ and $\kappa$, are always positive. [*Edited from $\lt 0$ to $\le 0$ based on Night Writer's answer.]

So is the statement in the textbook's question just a flat-out mistake, or am I missing something here?

It appears their error is in assuming the difference between Cp and Cv is due to pV expansion alone, when in fact there are two terms:

$$C_p - C_V = p \left(\partial V\over\partial T\right)_p+ \left(\partial U\over\partial V\right)_T \left(\partial V\over\partial T\right)_p,$$

where the first term on the RHS is the PV work per unit change in T, while the second term is the change in internal energy with respect to volume (which results from changing the intermolecular distance between interacting particles) times the rate at which the volume changes with temperature.

It would be nice to have direct reference-quality experimental values for $C_{\mathrm m,p}$ and $C_{\mathrm m,V}$ for $\ce{H2O(l)}$ near $\pu{4 °C}$, but I've not been able to locate them, nor do I expect to: because of the difficulty of accurately measuring $C_{\mathrm m,V}$ for a liquid, measurements are typically made at constant pressure, giving $C_{\mathrm m,p}$, and then $C_{\mathrm m,V}$ is calculated using the above equation.

  • $\begingroup$ In addition to the answers below, the data do show that $C_p \gt C_V$, albeit by a very small amount near zero $^\circ$C ($\approx 0.05$ J/mol/K) at temperatures from just above zero to 10 $^\circ$C. So the question is factually wrong. $\endgroup$
    – porphyrin
    Apr 16 '19 at 10:38
  • $\begingroup$ @porphyrin Can you provide a reference for that data? $\endgroup$
    – theorist
    Apr 16 '19 at 15:04
  • $\begingroup$ I just looked on the web: engineeringtoolbox.com/specific-heat-capacity-water-d_660.html . $\endgroup$
    – porphyrin
    Apr 16 '19 at 19:53
  • $\begingroup$ @porphyrin I suspect the only experimental data there is for Cp, and that Cv was calculated from the Cp values using the above formula. There's no way to tell without a reference to the original source. $\endgroup$
    – theorist
    Apr 16 '19 at 20:13
  • $\begingroup$ Ok, I had'n't realised that. You will have to search for some original data, but I can see no reason why $C_v$ should be greater than $C_p$. That $\beta \to 0$ seems to be the best explanation as shown by @ Night Writer in that answer. $\endgroup$
    – porphyrin
    Apr 17 '19 at 6:27

Repeating your line of reasoning, if $C_{\mathrm m,p}<C_{\mathrm m,V}$ you would conclude based on the expression you provide ($C_{\mathrm m,p}-C_{\mathrm m,V} = T V_\mathrm m \alpha^2/\kappa$) that $\kappa<0$ which in turn would mean that

$$\left(\frac{\partial V_m}{\partial P}\right)_T > 0,\label{eqn:1}\tag{1}$$

which is not physically reasonable except presumably in some very unusual circumstances. Therefore, either the question statement contains a fallacy, or the above analysis contains a fallacy, i.e. $\eqref{eqn:1}$ is reasonable. My bet is on the first scenario, as you suggest.

What seems to happen is that near $\pu{4 °C}$

$$C_{\mathrm m,p} = C_{\mathrm m,V}\label{eqn:2}\tag{2}$$

because the change in the density or molar volume with $T$ (and therefore also the thermal expansion coefficient $\beta$ - usually designated with the symbol $\alpha$) becomes zero, that is there is a (well known) minimum in the dependence of the density or molar volume on $T$ at constant $P$, illustrated in the following figure (source: wikipedia)

enter image description here

In agreement with this, if you inspect available heat capacity data you will see that the isochoric and isobaric heat capacities converge as the temperature of liquid water nears $\pu{0 °C}$.

The International Association for the Properties of Water and Steam has released a document (IAPWS R7-97(2012)) containing expressions for the heat capacities of water parameterized based on experimental data (see Table 3).

  • $\begingroup$ That's a nice point about $\beta$ for water going to zero at 4C. $\endgroup$
    – theorist
    Apr 16 '19 at 20:23
  • $\begingroup$ @theorist I think the answer is a combination of my explanation and the alternate answer: that answer explains some of the flaws in the book's reasoning, mine explains others (well, I echo your explanation - good catch - and follow it to it's logical conclusion based on the available data). I did find some standard data but have not crunched the numbers. $\endgroup$
    – Buck Thorn
    Apr 20 '19 at 19:43

The textbook's question is flawed because, as you state, its assertion that $c_p < c_v$ is false.

The textbook's answer messes up the math a little bit. Their logic is based on the First Law:

\begin{align} \Delta u &= q - P \Delta v \\ q &= \Delta u + P \Delta v \end{align} Considering an isobaric process, we have \begin{align} \overbrace{c_p \Delta T}^q &= \overbrace{\left(\frac{\partial u}{\partial T}\right)_P \Delta T}^{\Delta u} + P \overbrace{\left(\frac{\partial v}{\partial T}\right)_P \Delta T}^{\Delta v} \\ c_p &= \underbrace{\left(\frac{\partial u}{\partial T}\right)_P}_X + \underbrace{\left(\frac{\partial v}{\partial T}\right)_P}_{<0} \end{align} If term $X$ were the same as $c_v$ then it would follow that $c_p < c_v$, but the wrong property is held constant: $c_v$ would be $\left(\frac{\partial u}{\partial T}\right)_v$. Evidently $\left(\frac{\partial u}{\partial T}\right)_P - \left(\frac{\partial u}{\partial T}\right)_v$ is more positive than $P \left(\frac{\partial v}{\partial T}\right)_P$ is negative. The textbook seems to be incorrectly assuming that $X = c_v$.

Note that the corner case $c_p = c_v$ is allowed; this would occur when $\beta = 0$, $T = 0$, $v = 0$, or $\kappa = \infty$ (the first case being the one that can actually occur in everyday life).

Original take-down of the textbook answer:

Based on your summary, the textbook's answer fails to account for the fact that switching from an isobaric path to an isothermal path changes the end state. The (flawed) logic seems to be: "if I go from A to B isochorically then I need to provide some amount of heat, but if I go from A to B isobarically then the surroundings provide some work so I don't need to provide as much heat." The flaw is that if A-B is an isochor then an isobar beginning at A doesn't go through B. An apples-to-oranges comparison would consider the heat required to reach a third point C which is at the same pressure as A and temperature as B, and would find that this heat is never lower than that required to reach B.

  • $\begingroup$ You seem to be saying the textbook's answer is flawed because you can't compare Cp and Cv, the reason being that the difference in constraints implicit to Cp vs. Cv brings you to different final states (making it apples to oranges). However, even with this, you certainly can compare Cp to Cv (their difference is $TV\beta^2/\kappa$). Thus (and forgive me if I'm not understanding you correctly) I don't think the specific issue you raise (the difference in final states) is relevant here. I.e., I think the textbook is wrong, but not for the reason you give. $\endgroup$
    – theorist
    Apr 16 '19 at 22:05
  • $\begingroup$ I'm saying that their argument that the isobaric process involves less work than the isochoric process is overly simplistic (because it assumes the two processes have the same endpoint). You can certainly compare $c_p$ and $c_v$, for example using the formula you provided, but their attempt to provide an explanation based directly on the First Law/intuition is flawed. $\endgroup$ Apr 16 '19 at 22:52
  • $\begingroup$ I think their argument is flawed, but not for the reason you give. The problem isn't that they don't go to the same endpoint; that doesn't preclude the comparison. For instance, it's not a problem to say the reason Cp = Cv +nR for an ideal gas is that, during constant-P heating, you have to input additional thermal energy for the PV work associated with expansion. By your argument, you couldn't make a universal statement like that for ideal gases simply because they don't go to the same final state in a constant-V vs a constant-P heating. But that's simply not the case. [continued....] $\endgroup$
    – theorist
    Apr 17 '19 at 0:55
  • $\begingroup$ ...Hence the disparity of final states is not the logical issue per se. Rather, I think the reason their argument is simplistic/flawed is that it doesn't account for the difference in internal energy due to changes in volume in real materials (which have attractive and repulsive interactions). I.e., it accounts only for the difference in the work done on the surroundings. I.e., you're saying that the problem is that they're comparing processes with different final states. I'm saying it's OK for the final states to be different, so long as you account for the effect of those differences. $\endgroup$
    – theorist
    Apr 17 '19 at 0:57
  • $\begingroup$ I suppose we are arguing over what is the "right" way is to do something wrong, haha. You have the advantage of actually having read the textbook's answer in its entirety ;) I have updated my answer to reflect the flaw that you see but also retained the flaw that I saw so that our comment chain still makes sense. $\endgroup$ Apr 17 '19 at 1:50

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