Heat and Thermodynamics 4 Question 4
4. A $25 \times 10^{-3} m^{3}$ volume cylinder is filled with 1 mole of $O _2$ gas at room temperature $(300 K)$. The molecular diameter of $O _2$ and its root mean square speed are found to be $0.3 nm$ and $200 m / s$, respectively. What is the average collision rate (per second) for an $O _2$ molecule?
(2019 Main, 10 April I)
(a) $\sim 10^{10}$
(b) $\sim 10^{12}$
(c) $\sim 10^{11}$
(d) $\sim 10^{13}$
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Solution:
- Given,
Volume, $V=25 \times 10^{-3} m^{3}$
$N=1$ mole of $O _2=6.023 \times 10^{23}$ atoms of $O _2$,
$T=300 K$
Root mean square velocity of a gas molecule of $O _2, v _{rms}=200 m / s$
Radius, $r=\frac{0.3}{2} nm=\frac{0.3}{2} \times 10^{-9} m$
Now, average time, $\frac{1}{\tau}=\frac{\nu _{av}}{\lambda}$ where,
$$ \begin{aligned} \lambda & =\frac{R T}{\sqrt{2} N \pi r^{2} p} \\ p & =\frac{R T}{V} \Rightarrow \lambda=\frac{V}{\sqrt{2} N \pi r^{2}} \end{aligned} $$
$\therefore$ Average collision per second,
$$ \begin{aligned} & \frac{1}{\tau}=\frac{v _{av}}{\lambda}=\frac{\sqrt{\frac{8}{3 \pi}} \times v _{rms}}{\lambda} \\ & \frac{1}{\tau}=\sqrt{\frac{8}{3 \pi}} \times \frac{200 \times \sqrt{2} \times 6.023 \times 0^{23} \times \pi \times \frac{0.09}{4} \times 10^{-18}}{25 \times 10^{-3}} \\ & \Rightarrow \frac{1}{\tau}=4.4 \times 10^{8} \text { per second } \approx 10^{8} \end{aligned} $$
$\therefore$ No option given is correct.
