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 a 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: $(i n L / m o l)$ $(G i v e n, N_{A} = 6 \times 10^{23}$ and $R = \frac{22.4}{273} L a t m K^{- 1} m o l^{- 1})$

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Solution

At high pressures,
$Z = 1 + P b / R T$
So,
$(\frac{d Z}{d P}) = (\frac{1}{2.8}) a t m^{- 1} = b / R T$
$b = R T / 2.8 = 4 V$
Hence,
$V = \frac{22.4}{4 \times 2.8}$
$V (v o l u m e o f 1 m o l e o f g a s) = 2 L$

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At high pressures, Z=1+Pb/RT So, (dZdP)=(12.8)atm−1=b/RT b=RT/2.8=4V Hence, V=22.44×2.8 V(volume of 1 mole of gas)=2 L

At high pressures,
$Z = 1 + P b / R T$
So,
$(\frac{d Z}{d P}) = (\frac{1}{2.8}) a t m^{- 1} = b / R T$
$b = R T / 2.8 = 4 V$
Hence,
$V = \frac{22.4}{4 \times 2.8}$
$V (v o l u m e o f 1 m o l e o f g a s) = 2 L$