Mean Free Path and RMS Speed

Derive the formula for time for successive collision between the walls of the container with the molecule.

For a gas with $n$ number of molecules per unit volume, Suppose the molecules of a gas are spheres of diameter $d$  the average speed of each molecule is $v$. The time between two successive collisions is on the average $\Gamma =\frac{1}{n{\mathrm{\pi vd}}^2}$

The mean free path and RMS velocity of a nitrogen molecule at a temperature $17 °\mathrm{C}$ are $1.2\times {10}^{-7}$ $\mathrm{m}$ and $5\times {10}^2 \mathrm{m} {\mathrm{s}}^{-1}$, respectively. The mean time between two successive collisions will be $x\times {10}^{-11} \mathrm{s}$, what is the value of $x$.

The mean free path of molecules of a gas (radius ‘$r$’) is inversely proportional to $r^2$.

The mean free path $\left(\mathrm{\lambda }\right)$ of a gas sample is given by:

If the temperature and pressure of a gas is doubled the mean free path of the gas molecules

Calculate the time taken between two successive collisions for a helium molecule having a diameter $7\times {10}^{-10} \mathrm{m}$ and an average velocity of ${10}^3 \mathrm{m}/\mathrm{s}$, consider ${10}^{30}$ molecules present per unit volume.

Derive the formula of the time interval between two successive collisions of molecules.

For a molecule of an ideal gas, the number density is $2\sqrt{2}\times {10}^8 {\mathrm{cm}}^{-3}$ and the mean free path is $\frac{{10}^{-2}}{\pi } \mathrm{cm}$. The diameter of the gas molecule is

Define the following 

Mean free path

Define the following

Free path

Cotyledons are also called-

At any instant, the position and velocity vectors of two particles are ${\overrightarrow{r}}_1,{\overrightarrow{r}}_{2}and{\overrightarrow{v}}_1,{\overrightarrow{v}}_2$ respectively. They will collide if :

In the following a statement of Assertion is followed by a statement of Reason.

Assertion: Mean free path of a gas molecule is inversely proportional to the density of gas.

Reason: Path of a gas molecule between two adjacent collision is straight line.

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as $V^q$ , where $V$ is the volume of the gas. The value of $q$ is $\left(\mathrm{\gamma }=\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{v}}}\right)$

Which of the following is/are correct regarding mean free path ?

Which of the following is/are correct regarding mean free path ?

Air becomes conducting when the pressure ranges between

If the mean free path for a gas is $l$ at $1atm$ pressure then its value at $5atm$ pressure will be-

The mean free path of the molecule of a certain gas at $300 \mathrm{K}$ is $2.6\times {10}^{-5} \mathrm{m}$. The collision diameter of the molecule is $0.26 \mathrm{nm}$. After calculating pressure of the gas. The number of molecules per unit volume of the gas  is $x\times 10^{20}$. The value of $x$ will be