Question

A $25\times {10}^{-3}{\text{m}}^3$ volume cylinder is filled with $1$ mol of $O_2$ gas at room temperature $\left(300\text{K}\right)$. The molecular diameter of $O_2$, and its root mean square speed, are found to be $0.3\text{nm}$ and $200\text{m/s}$, respectively. what is the average collision rate (per second) for an $O_2$ molecule ?

• A. $\sim {10}^{11}$
• B. $\sim {10}^{12}$
• C. $\sim {10}^{13}$
• D. $\sim {10}^{10}$

Answer 1

The correct option is D $\sim {10}^{10}$

Given:
Volume of cylinder $V=25\times {10}^{-3}{\text{m}}^3,$
$N=1\text{mole of}{O}_2$
$T=300\text{K}$
Root mean square speed of gas$V_{\text{rms}}=200\text{m/s}$
Average speed of gas is $V_{\text{avg}}=\sqrt{\frac{8}{3\pi }}200\text{m/s}$

Molecular diameter $=0.3\text{nm}$

We know that
$\lambda =\frac{1}{\sqrt{2}{N}_V\pi d^2}$
Average time of collision
$\frac{1}{\tau }=\frac{V_{\text{avg}}}{\lambda }=V_{\text{avg}}.\sqrt{2}\text{N}\pi d^2.=\sqrt{\frac{8}{3\pi }}\frac{200\times \sqrt{2}\times 6.023\times {10}^{23}}{25\times {10}^{-3}}.\pi \times (0.3\times {10}^{-9}{)}^2\approx {10}^{10}$

The closest value in the given option is $={10}^{10}$

Hence option $\left(C\right)$ is correct.