States of Matter - Result Question 69
69. The density of the vapour of a substance at $1 \hspace {1mm} atm$ pressure and $500 K$ is $0.36 kg \hspace {1mm} m^{-3}$. 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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Solution: 69
$\dfrac{r _{\text {gas }}}{r _{O _2}}=1.33=\sqrt{\dfrac{32}{M _{\text {gas }}}}$
(i) (a) $M _{\text {gas }}=18 \hspace {1mm} g \hspace {1mm} mol^{-1}$
(b) $V _m=\dfrac{18}{0.36}=50 L mol^{-1}$
(c) $Z=\dfrac{p V}{R T}=\dfrac{1 \times 50}{0.082 \times 500}=1.22$
(d) $\because \quad Z>1$, repulsive force is dominating.
(ii) $\bar{E} _k=\dfrac{3}{2} k _B T=\dfrac{3}{2} \times 1.38 \times 10^{-23} \times 1000 J=2.07 \times 10^{-20} J$
BETA
