The graph of compressibility factor $(Z) v s P$ for one mole of a real gas is shown in following diagram. The graph is plotted at constant temperature $273 K$ . If the slope of graph at very high pressure $(\frac{d Z}{d P})$ is $(\frac{1}{2.8}) a t m^{- 1}$ , then the volume of one mole of real gas molecules is: (in $L / m o l$ ) [Given, $N_{A} = 6 \times 10^{23}$ and $R = \frac{22.4}{273} L a t m K^{- 1} m o l^{- 1}$ ]

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Solution

As we know,
At very high pressure,
$(\frac{d Z}{d P}) = (\frac{1}{2.8}) a t m^{- 1}$
$V / R T = 1 / 2.8, V = 8$ .

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