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Q.

The graph of compressibility factor (Z) vs P for one mole of a real gas is plotted at constant temperature 273 K. If the slope of the graph at very high pressure $(\frac{d Z}{d p})$ is $\frac{1}{10} a t m^{- 1},$ the volume of one molecule of real gas in $c m^{3}$ is $x \times 10^{- 22} .$ Find ‘x’.[ Take $N_{A} = 6 \times 10^{23}$ ].

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answer is 9.33.

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Detailed Solution

$z = 1 + \frac{P b}{R T} \frac{d z}{d p} = \frac{1}{10} = \frac{b}{R T} b = \frac{R T}{10} = \frac{22.4}{10} = 2.24 4 \times V \times N_{o} = 2.24 V = \frac{2.24}{4 \times 6} \times 10^{- 23} l = \frac{2.24 \times 1000}{24} \times 10^{- 23} c m^{3} = 9.33 \times 10^{- 22} c m^{3}$

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The graph of compressibility factor (Z) vs P for one mole of a real gas is plotted at constant temperature 273 K. If the slope of the graph at very high pressure (dZdp) is 110 atm−1, the volume of one molecule of real gas in cm3 is x×10−22. Find ‘x’.[ Take NA=6×1023 ].