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(D) The collision frequency $Z_{11}$ is given by the formula: $Z_{11} = \frac{1}{2} (\pi \sigma^{2}) (\sqrt{2} v_{avg}) (N^{*})^{2}$.Since $N^{*} = \f

Vedclass · Accepted Answer

On increasing volume four times,$Z_{11}$ will become $\frac{1}{8}$ times of its initial value.

Vedclass · Answer

(D) The collision frequency $Z_{11}$ is given by the formula: $Z_{11} = \frac{1}{2} (\pi \sigma^{2}) (\sqrt{2} v_{avg}) (N^{*})^{2}$.Since $N^{*} = \frac{N}{V} = \frac{P}{kT}$ and $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$,we substitute these into the expression:$Z_{11} = \frac{1}{2} (\pi \sigma^{2}) \sqrt{2} \sqrt{\frac{8RT}{\pi M}} (\frac{P}{kT})^{2}$.Assuming constant pressure $P$,we find the relationship: $Z_{11} \propto \sqrt{T} 	imes \frac{1}{T^{2}} = T^{-3/2}$.For an isobaric process,$V \propto T$. If the volume increases $4$ times,the temperature $T$ also increases $4$ times.Therefore,the new collision frequency $Z_{11}'$ is related to the initial value $Z_{11}$ as: $Z_{11}' = Z_{11} 	imes (4)^{-3/2} = Z_{11} 	imes \frac{1}{8}$.Thus,$Z_{11}$ becomes $\frac{1}{8}$ times its initial value.

Which of the following is correct about collision frequency $(Z_{11})$ of an ideal gas participating in an isobaric process?

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