Cp vs Cv for liquid water @ < 4 °C

Question

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.

Answer 1

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)

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).

Answer 2

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.