Question

The molecular diameters of $A$ and $B$ are $4nm$ and $8nm$ respectively. The pressure of $A$ and $B$ are $10atm$ and $5atm$ respectively. The temperature of $A$ and $B$ are $300K$ and $400K$ respectively. The ratio of molecular mass of $A$ and $B$ is $30:40$. 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

The correct option is C $\frac{16}{9}$

$Z_{11}=\frac{1}{\sqrt{2}}\pi {\sigma }^2{u}_{avg}{N}^{\ast 2}$
where $Z_{11}=\text{total number of binary collisions},u_{avg}=\text{the average speed of molecules},N^{\ast }=\text{number of molecules per unit volume}.$

We know, $PV=\frac{N}{N_A}RT\text{or}\frac{N}{V}=\frac{P{N}_A}{RT}=N^{\ast }\therefore Z_{11}=\frac{1}{\sqrt{2}}\pi {\sigma }^2\times \sqrt{\frac{8RT}{\pi M}}\times \frac{P^2{N}_A^2}{R^2{T}^2}\therefore Z_{11}\propto \frac{{\sigma }^2\times P^2}{\sqrt{M}\times T^{3/2}}$
Hence, the ratio of binary collision,
$\frac{(Z_{11}{)}_A}{(Z_{11}{)}_B}={\left(\frac{{\sigma }_A}{{\sigma }_B}\right)}^2\times {\left(\frac{P_A}{P_B}\right)}^2\times \sqrt{\frac{M_B}{M_A}}\times {\left(\frac{T_B}{T_A}\right)}^{3/2}\frac{(Z_{11}{)}_A}{(Z_{11}{)}_B}=\frac{1}{4}\times 4\times \sqrt{\frac{4}{3}}\times {\left(\frac{4}{3}\right)}^{3/2}$
$={\left(\frac{4}{3}\right)}^2$
$=\frac{16}{9}$