Theoretical comparison of critical constants of real gases

Question

We know that for a real gas, its critical constants are: $T_\mathrm c=\frac{8a}{27Rb}$, attempt to $P_\mathrm c=\frac{a}{27b^2}$ and $V_\mathrm c=3b$. Now, let's compare the value of these critical constants for three real gases, say, $\ce{CO2}$, $\ce{NH3}$ and $\ce{H2O}$.

Knowing that $V_\mathrm c=\frac38\cdot\frac{RT_\mathrm c}{P_\mathrm c}$ (for one mole), experimental data from Wikipedia is:

$$ \begin{array}{|c|c|c|c|}\hline &T_\mathrm c&P_\mathrm c&V_\mathrm c\\\hline \text{Carbon dioxide}&\pu{304K}&\pu{7380 kPa}&\pu{0.1284L}\\\hline \text{Ammonia}&\pu{406K}&\pu{11280 kPa}&\pu{0.1122L}\\\hline \text{Water}&\pu{647K}&\pu{22060kPa}&\pu{0.0914L}\\\hline \end{array} $$

We will now attempt to justify this regular trend in the values qualitatively.

For $T_\mathrm c$, we know that it is the maximum temperature above which a real gas cannot be liquefied. If a gas has stronger intermolecular forces (IMFs), its molecules possess greater attractive forces to be able to overcome separatist forces that might separate them, and thus they'll still be able to liquefy. Thus, if a gas has higher IMFs, its $T_\mathrm c$ would be higher.
We know that H-bonds formed by water are much effective than those by ammonia (read), and carbon dioxide is a non-polar molecule with only London dispersion forces. Hence, we would expect $T_\mathrm c(\text{water})>T_\mathrm c(\text{ammonia})>T_\mathrm c(\text{carbon dioxide})$ by our qualitative reasoning, which agrees with the experimental order.

By definition, the critical pressure is the pressure needed to liquefy a real gas at its critical temperature. Now, since water molecules have the highest attractive forces, one would expect that they require the least external "support" (aka force per unit area) to be able to liquefy themselves (by bringing the molecules closer). Hence, we would expect that $P_\mathrm c(\text{water})<P_\mathrm c(\text{ammonia})<P_\mathrm c(\text{carbon dioxide})$. However, this does not match with the experimental trend! (which happens to actually be in the reverse order)

Moreover, I am unable to come up with a way to predict the trend in the critical volume at all.

Hence, my questions are:

1. Since $V_\mathrm c$ depends on the ratio of $T_\mathrm c$ to $P_\mathrm c$, am I correct in saying that its trend cannot be determined without actually being given the values for the other two constants? Or is there some clever way that I am overlooking?
2. Is there any other way to justify the trend in $T_\mathrm c$?
3. What is the fault in my reasoning for the trend in $P_\mathrm c$? And what is the correct reasoning?

Answer 1

In essence regarding this problem
$$T_\mathrm c \propto \frac{a}{b}$$ $$P_\mathrm c \propto \frac{a}{b^2}$$ $$V_\mathrm c \propto {b}$$

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Values of $a$:

$\ce{NH3} \to 4.225\ \mathrm{l^2\ bar/mol^2}$
$\ce{CO2} \to 3.640\ \mathrm{l^2\ bar/mol^2}$
$\ce{H2O} \to 5.536\ \mathrm{l^2\ bar/mol^2}$.

This trend as justified by the OP through hydrogen bonding is absolutely correct.

Values of $b$:

$\ce{NH3} \to 0.0371\ \mathrm{l/mol}$
$\ce{CO2} \to 0.04267\ \mathrm{l/mol}$
$\ce{H2O} \to 0.03049\ \mathrm{l/mol}$.

$b$ or effective molecular size is essentially calculated by multiplying the size of a molecule by four. The trend over here is obviously determined by molecular size only.

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Now:

1. The critical temperature and critical pressure are in direct /inverse proportion with $a$ as well as $b$ .An interplay of two factors. Predicting the orders of both of these cases become extremely difficult without data. The opposing orders of $P_\mathrm c$ and $T_\mathrm c$ are seen as the order of proportion changes i.e $P_\mathrm c$ is inversely proportional to the square of $b$ and $T_\mathrm c$ is inversely proportional to $b$.
2. Trends in $V_\mathrm c$ on the other hand can be safely predicted by estimating the size of molecules as it only dependent upon one variable that is $b$.

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^{All data is from this Wikipedia page}