Larger value of b signifies that the size of the molecules is larger and hence the forces acting should become larger due to increased surface area.This should ensure easier liquefication. But my book says that smaller value of b corresponds to easier liquefication. Where am going wrong?

The book says "Chlorine is more easily liquefied than Ethane because 'a' for chlorine is greater than 'a' for ethane but 'b' for fluorine is less than 'b' for ethane. Van der Waals constant 'a' is due to force of attraction and 'b' due to finite size of molecules. Thus, greater the value of 'a' and smaller the value of b, larger the liquefaction"


The $b$ in the Van der Waals equation of state signifies the volume that is effectively taken out per molecule by intermolecular repulsive interactions. You can imagine that a small $b$ will help the gas to liquefy, since in most ordinary liquids molecules are packed closer together than in a gas.

As written in Atkins Physical Chemistry:

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    $\begingroup$ For an ideal gas b = 0, does that mean it is very easy to liquify an ideal gas? But, ideal gases cannot be liquified, right? $\endgroup$ – Ayush Pateria Mar 21 '15 at 16:29
  • $\begingroup$ @AyushPateria I don't think you can resume assuming ideal gas when the gas is about to/in the process of liquify/liquefaction. $\endgroup$ – Jori Mar 21 '15 at 18:00
  • $\begingroup$ What do you meam by "taken out per mole by intermolecular repulsive interactions?" $\endgroup$ – the_random_guy42 Mar 22 '15 at 11:20
  • $\begingroup$ That it is the volume that cannot be occupied by other molecules due to repulsive interaction between them. It is how the Van der Waal equation is derived. I updated my answer with an excerpt from Atkins. $\endgroup$ – Jori Mar 22 '15 at 11:36
  • $\begingroup$ I'll point out that the a and b constants are not really independent of temperature or pressure. The constants are "best fit" over some range of experimental data, say 0.100 moles of gas at some various temperatures and pressures. So if you change the range of fit, then the constants would change. Thus a and b provide temperature and pressure corrections to ideal gas law. The a term is mostly a pressure correction, and the b term mostly a volume correction. $\endgroup$ – MaxW Nov 3 '15 at 7:49

The virial equation of state reads:

$$Z = 1 + \left(b - \frac{a}{RT}\right)\frac{1}{V_\mathrm{m}} + \left(\frac{b}{V_\mathrm{m}}\right)^2 \cdots$$

Going by this equation, it seems like increasing the value of $b$ will increase $Z$, which means repulsive forces will increase, making the gas liquefaction difficult. Also for $a = 0$ and $b = 0$, the gas becomes an ideal gas whose liquefaction is not possible, so I think what you are saying is incorrect.

The question you are talking about, that why chlorine is more easily liquefied than ethane? It is because $a$ of $\ce{Cl2}$ is greater than $a$ of $\ce{C2H6}$, which means $\ce{Cl2}$ is easier to liquify. $\ce{C2H6}$, being greater in size than $\ce{Cl2}$, has a higher value of $b$. It's wrong to say "the smaller the value of $b$, the larger the liquefaction".

  • $\begingroup$ Increase in the value of z would simply imply increase in the volume of the gas. It does not mean that liquefication would become difficult due to increas in volume. $\endgroup$ – the_random_guy42 Mar 22 '15 at 11:27
  • $\begingroup$ And does Vm represent ideal volume or real volume? $\endgroup$ – the_random_guy42 Mar 22 '15 at 11:28
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    $\begingroup$ Z > 1 signifies positive deviation. Which means the gas will occupy more volume than what an ideal gas would have. But why is that? It's because of repulsive forces that molecules are not able to come closer and hence have larger volume. Thus, the gas is more difficult to compress than an ideal gas. $\endgroup$ – Ayush Pateria Mar 22 '15 at 11:50
  • $\begingroup$ Vm represent the volume measured, or simply the volume which an ideal gas would have occupied. $\endgroup$ – Ayush Pateria Mar 22 '15 at 11:51
  • $\begingroup$ This isn't the Van der Waals equation. There are a zillion gas equations with various fudge factors. Each different equation has experimentally derived coefficients which are based on some rationalization. See Real Gas en.wikipedia.org/wiki/Real_gas at Wikipedia. $\endgroup$ – MaxW Nov 3 '15 at 7:39

b value has no contribution to the liqufication/critical temperature of a gas. Whereas ‘a’ value has implication over the critical temp.. Higher the ‘a’ value easier liquification / high critical temp. because ‘a’ value is a direct measure of intermolecular force. Krypton and Nitrogen has nearly equal ‘b’ value. Whereas ‘a’ value of Krypton is higher than the Nitrogen. Hence the critical temp of Krypton is higher than the Nitrogen. Intermolecular force decides the critical temperature, not the volume factor. Neon has much lower ‘b’ value than Krypton and Nitrogen, but lower ‘a’ value keep its critical temp at 44 kelvin much much lower than critical temp of Nitrogen 126K and Krypton 209K.

  • $\begingroup$ I'm not sure I would call $a$ a "direct Measure". $\endgroup$ – A.K. Sep 13 '18 at 13:51

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