Question
$$XmL$$ of $${H_2}$$ gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is :
A.
$${\text{10 seconds : }}He$$
B.
$${\text{20 seconds :}}\,{O_2}$$
C.
$${\text{25 seconds :}}\,CO$$
D.
$${\text{55 seconds :}}C{O_2}$$
Answer :
$${\text{20 seconds :}}\,{O_2}$$
Solution :
$$\eqalign{
& {\text{Under}}\,{\text{indential}}\,{\text{conditions,}}\,\frac{{{r_1}}}{{{r_2}}} = \sqrt {\frac{{{M_2}}}{{{M_1}}}} \cr
& {\text{As rate of diffusion is also inversely proportional to}} \cr
& {\text{time,}}\,{\text{we}}\,{\text{will}}\,{\text{have}},\,\frac{{{t_2}}}{{{t_1}}} = \sqrt {\frac{{{M_2}}}{{{M_1}}}} \cr
& \left( {\text{A}} \right)\,{\text{Thus,}}\,{\text{For}}\,He,\,\,{t_2} = \sqrt {\frac{4}{2}} \left( {5s} \right) = 5\sqrt {2s} \ne 10s; \cr
& \left( {\text{B}} \right)\,{\text{For}}\,{O_2},\,\,{t_2} = \sqrt {\frac{{32}}{2}} \left( {5s} \right) = 20s \cr
& \left( {\text{C}} \right)\,{\text{For}}\,CO,\,\,{t_2} = \sqrt {\frac{{28}}{2}} \left( {5s} \right) \ne 25s\,; \cr
& \left( {\text{D}} \right)\,{\text{For}}\,C{O_2},\,\,{t_2} = \sqrt {\frac{{44}}{2}} \left( {5s} \right) \ne 55s \cr} $$