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The density of the vapour of a substance...

The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 kg `m^(-3)`. The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition.
Determine, (a) molecular weight (b) molar volume (c) compression factor (Z) of the vapour and (d) which forces among the gas molecules are dominating, the attractive or the repulsive?

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To solve the problem step by step, we will determine the molecular weight, molar volume, compressibility factor (Z), and analyze the dominating forces among the gas molecules. ### Step 1: Determine the Molecular Weight (M) We know that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Given that the vapor effuses 1.33 times faster than oxygen (O₂), we can set up the following equation: \[ \frac{\text{Rate of effusion of gas}}{\text{Rate of effusion of O}_2} = \frac{1}{\sqrt{\frac{M_{\text{O}_2}}{M_{\text{gas}}}}} \] ...
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