C_p for monoatomic gas mixture of same volume = $${5 \over 2}R$$

$$ \therefore $$ C_V = $${3 \over 2}R$$

$$ \Rightarrow {{{C_P}} \over {{C_V}}} = {{{5 \over 2}R} \over {{3 \over 2}R}} = {5 \over 3} = 1.67$$

Van der Waal gas constant '$$a$$' represent intermolecular force of attraction of gaseous molecules and Van der Waal gas constant 'b' represent effective size of molecules . Therefore order should be

(I) H₂ < He < O₂ < CO₂   (II) CH₄ > O₂ > H₂

According to Graham's law of diffusion

$$r \propto {1 \over {\sqrt d }} \propto {1 \over {\sqrt M }}$$

$$ \Rightarrow {{{r_1}} \over {{r_2}}} = \sqrt {{{{M_2}} \over {{M_1}}}} $$

Rate of diffusion = $${{Volume\,of\,gas\,diffused\,(V)} \over {Times\,taken\,(t)}}$$

$$ \therefore $$ $${{{V_1}/{t_1}} \over {{V_2}/{t_2}}} = \sqrt {{{{M_2}} \over {{M_1}}}} $$

If same volume of two gases diffuse then V₁ = V₂

$$ \Rightarrow $$ $${{{t_1}} \over {{t_2}}} = \sqrt {{{{M_2}} \over {{M_1}}}} $$

Here t₂ = 3t, M₁ = 4 u, M₂ = ?

$$ \therefore {{3{t_1}} \over {{t_1}}} = \sqrt {{{{M_2}} \over 4}} \Rightarrow 3 = \sqrt {{{{M_2}} \over 4}} $$

$$ \Rightarrow 9 = {{{M_2}} \over 4} \Rightarrow {M_2} = 36\,u$$

According to Graham's law of diffusion,

$${{{r_1}} \over {{r_2}}} = \sqrt {{{{d_2}} \over {{d_1}}}} = \sqrt {{{{M_2}} \over {{M_1}}}} $$

$${r_A} = {{{V_A}} \over {{T_A}}},\,{r_B} = {{{V_B}} \over {{T_B}}}$$

$${{{V_A}/{T_A}} \over {{V_B}/{T_B}}} = \sqrt {{{{M_B}} \over {{M_A}}}} $$

$${V_A} = {V_B},\,{T_A} = 150\,\sec $$, T_B = 200 sec, M_B = 36, M_A = ?

$${{{T_B}} \over {{T_A}}} = \sqrt {{{{M_B}} \over {{M_A}}}} \Rightarrow {{200} \over {150}} = \sqrt {{{36} \over {{M_A}}}} $$

$$ \Rightarrow {4 \over 3} = \sqrt {{{36} \over {{M_A}}}} \,\,or\,\,{{4 \times 4} \over {3 \times 3}} = {{36} \over {{M_A}}}$$

$$ \Rightarrow $$ M_A = $${{36} \over {4 \times 4}} \times 3 \times 3 = 20.25$$