States of Matter - Result Question 69
####69. 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.
(i) 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?
(ii) If the vapour behaves ideally at $1000 K$, determine the average translational kinetic energy of a molecule.
(2002, 5M)
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Answer:
Correct Answer: 69. (1025)
Solution:
- $\frac{r _{\text {gas }}}{r _{O _2}}=1.33=\sqrt{\frac{32}{M _{\text {gas }}}}$
(i) (a) $M _{\text {gas }}=18 g mol^{-1}$
(b) $V _m=\frac{18}{0.36}=50 L mol^{-1}$ (c) $Z=\frac{p V}{R T}=\frac{1 \times 50}{0.082 \times 500}=1.22$
(d) $\because \quad Z>1$, repulsive force is dominating.
(ii) $\bar{E} _k=\frac{3}{2} k _B T=\frac{3}{2} \times 1.38 \times 10^{-23} \times 1000 J=2.07 \times 10^{-20} J$
