Consider the following graph of $Z vs \frac{1}{V_{m}}$ drawn at a low pressure and at inversion temperature, using the Virial form of van der Waal's equation:

$Z = 1 + (b - \frac{a}{R T}) \frac{1}{V_{m}} + \frac{b^{2}}{V_{m}^{2}} + \frac{b^{3}}{V_{m}^{3}} + \dots \dots$ Find the critical molar volume $(V_{c})$ in ${cm}^{3}$ ?

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Solution

$Z vs \frac{1}{V_{m}}$ equation is :
$\frac{Z - 1.01}{1.02 - 1.01} = \frac{\frac{1}{V_{m}} - 0.001}{0.011 - 0.001}$
or $\frac{Z - 1.01}{0.01} = \frac{\frac{1}{V_{m}} - 0.001}{0.01}$
$\Rightarrow Z = 1.009 + \frac{1}{V_{m}} \Rightarrow s l o p e = 1$
Now, $T_{i} = 2 T_{b} = \frac{2 a}{R b}$
at $low P, Z = 1 + (b - \frac{9}{R T}) \frac{1}{V_{m}}$
Putting $T = T_{i} = \frac{2 a}{R b}, z = 1 + \frac{b}{2 V_{m}}$
$\Rightarrow Slope = \frac{b}{2} = 1 \Rightarrow b = 2$
$∴ V_{C} = 3 b = 6 c m^{3}$

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