Question

A closed vessel of known volume, containing known amount of ideal gaseous substance 'A' was observed for variation of pressure with temperature. The expected graph was to be like as in (i). However, the actual observations revealed the graph to be like (ii). The deviation was attributed to polymerisation of gas molecules as $nA\left(g\right)\rightleftharpoons A_n\left(g\right).$ If it is known that the above reaction gives only $50\%$, find the value of n to which the gas A is being polymerised into.

Answer 1

Slope $=\frac{nR}{V}$
For graph (I),
Slope $=\frac{3.2-1.54}{10}=0.166$
For graph (II)
Slope2 $=\frac{2.2875-1.25}{10}=0.10375$

$\frac{0.166}{0.10375}=\frac{1.6}{1}$
The dissociation is $50\%$.
Out of 1.6 moles, 0.8 moles will dissociate and 0.8 moles will remain undissociated.
Hence, the number of moles of $A_n$ formed will be $1-0.8=0.2$
Hence, $n=\frac{0.8}{0.2}=4$