The number of air molecules per cm^3 increase | vedclass

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(B) The collision frequency $f$ of gas molecules is given by the formula:$f = \sqrt{2} \pi d^2 v n_v$where $d$ is the molecular diameter,$v$ is the av

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(B) The collision frequency $f$ of gas molecules is given by the formula:$f = \sqrt{2} \pi d^2 v n_v$where $d$ is the molecular diameter,$v$ is the average speed,and $n_v$ is the number density (number of molecules per unit volume).From the formula,it is clear that the collision frequency is directly proportional to the number density: $f \propto n_v$.Given the initial number density $n_{v1} = 3 	imes 10^{19} 	ext{ molecules/cm}^3$ and the final number density $n_{v2} = 12 	imes 10^{19} 	ext{ molecules/cm}^3$.The ratio of collision frequencies is:$\frac{f_1}{f_2} = \frac{n_{v1}}{n_{v2}} = \frac{3 	imes 10^{19}}{12 	imes 10^{19}} = \frac{3}{12} = 0.25$.

The number of air molecules per $cm^3$ increased from $3 \times 10^{19}$ to $12 \times 10^{19}$. The ratio of collision frequency of air molecules before and after the increase in number respectively is $.........$

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