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If the ideal gas law $PV=nRT$ is not enough, as a second approximation the van der Waals equation

$$\left(P+a\frac{n^2}{V}\right)\left(V-nb\right)=nRT$$

is applied. The constants $a$ and $b$ are van der Waals constants which take into account some of the attraction and repulsion that the ideal gas law does not.

Solving for $P$,

$$P=\frac{nRT}{V-nb}-a\frac{n^2}{V^2}$$

and plugging it into the definition formula of compression factor $Z$ yields $$Z=\frac{V_{m,real}}{(RT)/P}=\frac{V_{real}}{(nRT)/P}=\frac{V_{real}}{V-nb}-a\frac{nV_{real}}{RTV^2}.$$

This result, supposedly, helps in the measuring of constants $a$ and $b$. But how is it really done? What are the more rigorous definitions of the constants?

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    $\begingroup$ For a rigourous definition of $a$ and $b$ you can see this answer: chemistry.stackexchange.com/questions/40803/… $\endgroup$
    – user23061
    Jun 10, 2016 at 15:36
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    $\begingroup$ You don't: While educationally useful (very much so), the vdW equation is not describing the actual behaviour of any real gas. To use it, you'd have to calibrate the parameters for the relevant pressure/temperature range. There are no universal values for a and b that make vdW much better than the ideal gas law. $\endgroup$
    – Karl
    Nov 11, 2016 at 4:56

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At the critical point, the first and second derivatives of pressure with respect to volume are zero.

This provides two equations to solve for the two unknowns (a and b) in terms of the critical temperature and critical volume (or pressure).

See Critical Constants of the van der Waals Gas

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    $\begingroup$ That's correct, inside vdWaals theory, but hardly practically useful. Actual, sensible values for $a$ and $b$ are purely empiric, and vary quite a bit for different pressure/temperature ranges. $\endgroup$
    – Karl
    Nov 11, 2016 at 5:09

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