The graph of compressibility factor (Z) vs 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 (dZdp) is 110atm−1, the volume of one molecules of real gas in cm3 is: [Given: R=22.4273 Latm K-1 mol-1 and NA=6×1023]

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The graph of compressibility factor (Z) vs 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 $\frac{1}{10} {atm}^{- 1},$ the volume of one molecules of real gas in ${cm}^{3}$ is: [Given: $R= \frac{22.4}{273} {Latm K}^{-1} {mol}^{-1}$ and $N_{A} = 6 \times 10^{23}$ ]

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$9.3 \times 10^{- 22}$

$9.3 \times 10^{- 23}$

$3.7 \times 10^{- 20}$

$5.6 \times 10^{- 20}$

answer is C.

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Detailed Solution

At very high pressure $Z = 1 + \frac{P b}{R T}$

$\begin{array}{l} \frac{d Z}{d p} = \frac{b}{R T} = \frac{1}{10} {atm}^{- 1} \\ b = \frac{22.4}{273} \times \frac{273}{10} = 2.24 litre/mole \\ b=4 \times V \times N_{A} \\ V = \frac{2.24 \times 10^{3}}{4 \times 6 \times 10^{23}} \Rightarrow 9.3 \times 10^{- 22} cc \end{array}$