# Solve this

Question:

A $25 \times 10^{-3} \mathrm{~m}^{3}$ volume cylinder is filled with $1 \mathrm{~mol}$ of $\mathrm{O}_{2}$ gas at room temperature $(300 \mathrm{~K})$. The molecular diameter of $\mathrm{O}_{2}$, and its root mean square speed, are found to be $0.3 \mathrm{~nm}$ and $200 \mathrm{~m} / \mathrm{s}$, respectively. What is the average

1. (1) $\sim 10^{12}$

2. (2) $\sim 10^{11}$

3. (3) $\sim 10^{10}$

4. (4) $\sim 10^{13}$

Correct Option: , 3

Solution:

(3) $\mathrm{V}=25 \times 10^{-3} \mathrm{~m}^{3}, \mathrm{~N}=1$ mole of $\mathrm{O}_{2}$

$\mathrm{T}=300 \mathrm{~K}$

$\mathrm{V}_{\mathrm{rms}}=200 \mathrm{~m} / \mathrm{s}$

$\therefore \lambda=\frac{1}{\sqrt{2} \mathrm{~N} \pi \mathrm{r}^{2}}$

Average time $\frac{1}{\tau}=\frac{\langle\mathrm{V}\rangle}{\lambda}=200 \cdot \mathrm{N} \pi \mathrm{r}^{2} \cdot \sqrt{2}$

$=\frac{\sqrt{2} \times 200 \times 6.023 \times 10^{23}}{25 \times 10^{-3}} . \pi \times 10^{-18} \times 0.09$

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

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