Why do we use the ratio of Critical Pressure and Critical Temperature to calculate size of a gas molecule?

Our teacher explained that using the ratio of Critical Pressure and Critical Temperature is preferred over Critical Volume to calculate the size of a gas molecule when all these quantities are taken in terms of van der Waals constants. He told us that it was due to increased accuracy.


How exactly does using the ratio increase the accuracy?


Reference:


Critical Temperature :
$\frac{8a}{27Rb}$


Critical Pressure : $\frac{a}{27b^2}$


Critical Volume : $3b$


where a and b are van der Waals constants.




Critical temperature and pressure are easily measurable with good accuracy. OTOH $\frac{\mathrm{d}V}{\mathrm{d}p}$ or $\frac{\mathrm{d}V}{\mathrm{d}T}$ reach extremely high values near the critical point.


By other words, the ratio of the critical temperature and pressure has much smaller experimental relative error than the critical volume. Therefore obtaining $b$ from the former is more accurate.


Note that b is 4 times the true volume of 1 mol of gaseous (supposed spherical) particles in the [van der Waals model](https://en.wikipedia.org/wiki/Van_der_Waals_equation). Therefore, according to the model, $V\_\mathrm{molecule} = \frac{b}{4N\_\mathrm{A}}=\frac{V\_\mathrm{mol, c}}{12N\_\mathrm{A}}$


See e.g. [chem.libretexts.org: Real\_Gases\_and\_Critical\_Phenomena](https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chem1_(Lower)/06%3A_Properties_of_Gases/6.06%3A_Real_Gases_and_Critical_Phenomena)


[![chem.libretexts.org: Real_Gases_and_Critical_Phenomena - isopentane](https://i.stack.imgur.com/1se96.png)](https://i.stack.imgur.com/1se96.png)


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