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(A) The mean free path $\lambda$ of a gas molecule is given by the formula $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$,where $k_B$ is the Boltzmann c

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Directly proportional to temperature

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(A) The mean free path $\lambda$ of a gas molecule is given by the formula $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$,where $k_B$ is the Boltzmann constant,$T$ is the absolute temperature,$d$ is the molecular diameter,and $P$ is the pressure.Assuming the pressure $P$ remains constant,the mean free path $\lambda$ is directly proportional to the absolute temperature $T$ (i.e.,$\lambda \propto T$).

How does the mean free path of a gas depend on temperature?

Similar Questions

How does the mean free path vary with the number density of a gas?

For the given concentration,if the ratio of the diameters of the molecules of two gases is $1: 2$,then the ratio of their mean free paths is

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For a van der Waals gas,if $P_c, V_c$,and $T_c$ are the critical pressure,volume,and temperature respectively,then the value of $P_cV_c/T_c$ is:

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