f the slope of $Z$ (compressibility factor) vs $P$ curve be constant (slope $= π / 492.6 {a t m}^{- 1}$ ) at a particular temperature $300 K$ and very high pressure. Calculate the diameters of molecules (given $N_{A} = 6 \times 10^{23}, R = 0.0821 L a t m . {m o l}^{- 1} K^{- 1}$

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Solution

$Z = 1 + \frac{P b}{R T}$ at high pressure.
Here, slope $\frac{b}{R T}$ which is $\frac{π}{492.6}$

$b = \frac{π}{492.6} (0.0821) (300)$
But $b = \frac{4}{3} π r^{3} \times 4 N_{A}$
Hence, $\frac{π}{492.6} (0.0821) (300)$ $= \frac{4}{3} π r^{3} \times 4 \times 6.022 \times 10^{23}$
$\Rightarrow r = 2.5_{\circ}$
Hence the value of d is $5_{\circ}$

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