We know that $${{{r_A}} \over {{r_B}}} = {{v/{t_A}} \over {v/{t_B}}} = \sqrt {{{{M_B}} \over {{M_A}}}} $$

$$ \Rightarrow {{{t_A}} \over {{t_B}}} = \sqrt {{{{M_B}} \over {{M_A}}}} \Rightarrow {{10} \over {20}} = \sqrt {{{{M_B}} \over {49}}} $$

$$ \Rightarrow {\left( {{{10} \over {20}}} \right)^2} = {{{M_B}} \over {49}} \Rightarrow {{100} \over {400}} = \,{{{M_B}} \over {49}}$$

$$ \Rightarrow {M_B} = {{49 \times 100} \over {400}} = 12.25$$ u

Given

P₁ = 1.5 bar, T₁ = 273 + 15 = 288 K, V₁ = V

P₂ = 1.0 bar, T₁ = 273 + 25 = 298K, V₂ = ?

$${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}$$

$$ \Rightarrow $$ $${{1.5 \times V} \over {288}} = {{1 \times {V_2}} \over {298}}$$

$$ \Rightarrow $$ V₂ = 1.55V

$$ \therefore $$ Volume of bubble will be almost 1.6 time to initial volume of bubble.

Number of moles $$ \Rightarrow $$ n_CO = n_N2

Volume of container is same. So, V_CO = V_N2 .

Also, temperature is same for both the gases thus, T_CO = T_N2

According to ideal gas equation, PV = nRT

Now, V, n, R and T for both gases are same. So, P_CO = P_N2

Now, total pressure is 1 atm and according to Dalton’s sum of partial pressure,

P_CO + P_N2 = 1 atm

$$ \Rightarrow $$ 2P_N2 = 1 atm {$$ \because $$ P_CO = P_N2 }

$$ \Rightarrow $$ P_N2 = 0.5 atm

Average velocity (v_AV) = $$\sqrt {{{8RT} \over {\pi M}}} $$

$$ \Rightarrow $$ v_AV $$ \propto $$ $$\sqrt T $$

$$ \therefore $$ $${{{{\left( {{v_{AV}}} \right)}_2}} \over {{{\left( {{v_{AV}}} \right)}_1}}} = \sqrt {{{2T} \over T}} $$ = 1.4