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Specific Heat Capacity

One mole tria...

Question

One mole triatomic vapours of an unknown susbtance effuses 4/3 times faster than 1 mole $O_{2}$ under same conditions. If the density of unknown vapours at pressure P and temperature T is d, which of the following holds true for the unknown substance:

A

d_{N, T, P} = 0.8035 g / L

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B

Z (a t o m i c n u m b e r) = 6

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C

z (c o m p r e s s i b i l i t y f a c t o r) = \frac{18 P}{d R T}

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D

Vapour density = 9

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Solution

The correct options are
A $d_{N, T, P} = 0.8035 g / L$
C $z (c o m p r e s s i b i l i t y f a c t o r) = \frac{18 P}{d R T}$
D Vapour density = 9
a)We know, $\frac{r_{v a p o u r s}}{r_{O_{2}}} = \sqrt{\frac{M_{O_{2}}}{M_{v a p o u r s}}}$
$\Rightarrow \frac{4}{3} = \sqrt{\frac{32}{M}}$
$\Rightarrow M = 18$
Again, $D e n s i t y = \frac{m a s s o f 1 m o l e}{V o l u m e o f 1 m o l e} = \frac{18}{22.4} = 0.8035 g / L$
b) Because the vapour is triatomic ,
$∴$ atomic weight = $\frac{18}{3} = 6$
So, atomic number cannot be 6.
c) Compressibility factor $z = \frac{P V}{n R T}$
$\Rightarrow z = \frac{P V}{\frac{w}{M} R T} = \frac{P V M}{w R T} = \frac{P M}{d R T} = \frac{18 P}{d R T}$
d) vapour density= $\frac{M}{2} = \frac{18}{2} = 9$

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