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Grade upto college level Physical Chemistry

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.
(a) Determine
(i) Molecular weight,
(ii) molar volume,
(iii) Compression factor (Z) of the vapour and
(iv) Which forces among the gas molecules are domination, the attractive or the repulsive?
(b) If the vapour behaves ideally at 1000 K, determine the average translational kinetic energy f a molecule.

Profile image of Shane Macguire
12 Years agoGrade upto college level
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1 Answer

Profile image of Deepak Patra
12 Years ago
(a) d = 0.36 kg m-3 = 0.36 g/L
(i) From Graham’s Law of diffusion
\frac{r_{v}}{r_{O_{2}}} = \sqrt{\frac{M_{O_{2}}}{M_{v}}} ; 1.33 = \sqrt{\frac{32}{M_{v}}}
∴ Mv = 32/(1.33)2 = 18.09;
Where Mv = MW of the vapour
(ii) Thus, 0.36g = 0.36/18.09 mol
0.36/18.09 mol occupies 1 L volume, so 1 mol occupies
18.09/0.36 L = 50.25L
Thus, molar volume of vapour = 50.25 L
Assuming ideal behavior the volume of the vapour can be calculated by
V1/T1 = V2/T2 => 22.4 * 500/273 = 41.025L
(iii) Compressibility factor (Z)
= (PV)obs/(PV)ideal = 1 *50.25/1 *41.025 = 1.224
(iv) Z is greater than unity, hence it is the short range repulsive force that would dominate.
(∵ actual density is less than given density)
(b) E = 3/2 KT = 3/2 * 8.31/6.02 *1023 *100
= 2.07 *10-20 J per molecule
(∵ K, Boltzmann constant = R/N)