At a constant pressure,the collision frequenc | vedclass

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(A) The collision frequency $(Z)$ is given by the formula: $Z = \sqrt{2} \pi d^2 \bar{v} \frac{N}{V}$.Here,$\bar{v}$ is the average speed of gas molec

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Decrease with an increase in temperature.

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(A) The collision frequency $(Z)$ is given by the formula: $Z = \sqrt{2} \pi d^2 \bar{v} \frac{N}{V}$.Here,$\bar{v}$ is the average speed of gas molecules,which is directly proportional to the square root of temperature $(\bar{v} \propto \sqrt{T})$.According to the ideal gas law,$PV = nRT$,so $\frac{N}{V} = \frac{P}{kT}$.Substituting these into the expression,$Z \propto \frac{\sqrt{T}}{T} = \frac{1}{\sqrt{T}}$.However,for a fixed mass of gas at constant pressure,$V \propto T$. The number density $\frac{N}{V}$ decreases as $T$ increases.Specifically,$Z \propto \frac{\sqrt{T}}{V} \propto \frac{\sqrt{T}}{T} = \frac{1}{\sqrt{T}}$.Wait,let us re-evaluate: For a fixed amount of gas at constant pressure,$V \propto T$. The collision frequency $Z$ is proportional to $\frac{\bar{v}}{V}$. Since $\bar{v} \propto T^{1/2}$ and $V \propto T$,then $Z \propto T^{1/2} / T = T^{-1/2}$.Thus,as temperature increases,the collision frequency decreases.

At a constant pressure,the collision frequency $(Z)$ of a gas will.......

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