The density of the vapour of a substance at $1 a t m$ pressure and $500 k$ is $0.36 k g m^{- 3}$ . The vapour effuse through a small hole at a rate of $1.33$ times faster than oxygen under the same condition.

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Solution

Using Graham's law of diffusion,
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \Rightarrow \frac{1.33}{1} = \sqrt{\frac{32 g m o l^{- 1}}{M_{1}}} \Rightarrow M_{1} = 18.09 g m o l^{- 1}$
The molar volume of the vapour, $V_{m} = \frac{M_{1}}{α}$
$= \frac{18.9 g m o l^{- 1}}{0.36 K g m^{- 3}}$
$= \frac{18.09 g m o l^{- 1}}{0.36 g d m^{- 3}}$
$= 50.25 d m^{3} m o l^{- 1}$
The compressibilty factor, $Z = \frac{V_{m}, r e a l}{V_{m}, i d e a l} = \frac{V_{m}, r e a l}{R T / α}$
$= \frac{101.325 K P_{a} \times 50.25 d m^{3} m o l^{- 1}}{8.314 J K^{- 1} m o l^{- 1} \times 500 K}$
$= 1.22$
Since $Z > 1$ the repulsive forces dominate among the gaseous molecules.
Average $K . E = \frac{3}{2} K T = \frac{3}{2} \times 3.8 \times 10^{- 23} J K^{- 1} \times 1000 K$
$= 2.07 \times 10^{- 20} J$ .

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The density of the vapour of a substance at 1atm pressure and 500k is 0.36kgm−3. The vapour effuse through a small hole at a rate of 1.33 times faster than oxygen under the same condition.

The density of the vapour of a substance at $1 a t m$ pressure and $500 k$ is $0.36 k g m^{- 3}$ . The vapour effuse through a small hole at a rate of $1.33$ times faster than oxygen under the same condition.