States of Matter 1 Question 69
69. The density of the vapour of a substance at $1 \mathrm{~atm}$ pressure and $500 \mathrm{~K}$ is $0.36 \mathrm{~kg} \mathrm{~m}^{-3}$. 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 $1000 \mathrm{~K}$, determine the average translational kinetic energy of a molecule.
(2002, 5M)
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Answer:
Correct Answer: 69. (1025)
Solution:
- $\frac{r_{\text {gas }}}{r_{\mathrm{O}{2}}}=1.33=\sqrt{\frac{32}{M{\text {gas }}}}$
(i) (a) $M_{\text {gas }}=18 \mathrm{~g} \mathrm{~mol}^{-1}$
(b) $V_{m}=\frac{18}{0.36}=50 \mathrm{~L} \mathrm{~mol}^{-1}$ (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) $\bar{E}{k}=\frac{3}{2} k{B} T=\frac{3}{2} \times 1.38 \times 10^{-23} \times 1000 \mathrm{~J}=2.07 \times 10^{-20} \mathrm{~J}$
