Question

The molecular diameter of A & B are 4 nm and 8 nm respectively. The pressure of A and B are 10 atm and 5 atm respectively. The temperature of A and B are 300 K and 400 K respectively. The ratio of molecular mass of A and B are 30 : 40. What will be the ratio of binary collisions among A molecules to that of B molecules ?

• A. $\frac{50}{57}$
• B. $\frac{29}{21}$
• C. $\frac{16}{9}$
• D. $\frac{49}{36}$

Answer 1

Answer: (C)

$\frac{16}{9}$

Answer 2

Z₁₁ = $\frac{1}{\sqrt{2}}$ p s² u_avg N^*2 where Z₁₁ = total no. of binary collisions

or Z₁₁ = $\frac{1}{\sqrt{2}}$ p s² × $\sqrt{\frac{8RT}{\pi M}}$ × $\frac{P^{2}N_{A}^{2}}{R^{2}T^{2}}$

PV = $\frac{N}{N_{A}}$ RT

or $\frac{N}{V}$ = $\frac{PN_{A}}{RT}$ = N^*

or Z₁₁ µ $\frac{\sigma^{2} \times P^{2}}{\sqrt{M} \times T^{3/2}}$

\$\frac{(Z_{11})_{A}}{(Z_{11})_{B}}$ = $\frac { \sigma _ { \mathrm { A } } ^ { 2 } } { \sigma _ { \mathrm { B } } ^ { 2 } }$ × $\frac { \mathrm { P } _ { \mathrm { A } } ^ { 2 } } { \mathrm { P } _ { \mathrm { B } } ^ { 2 } }$ × $\sqrt{\frac{M_{B}}{M_{A}}}$ × $\left( \frac{T_{B}}{T_{A}} \right)^{3/2}$

or $\frac{(Z_{11})_{A}}{(Z_{11})_{B}}$ = $\frac{1}{4}$ × 4 × $\sqrt{\frac{4}{3}}$ × $\left( \frac{4}{3} \right)^{3/2}$

= $\left( \frac{4}{3} \right)^{2}$= $\frac{16}{9}$