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

The velocity of five particles in $\mathrm{m} {\mathrm{s}}^{-1}$ are $3, 9, 4, \sqrt{15}, 2$. Calculate r.m.s. speed in $\mathrm{m} {\mathrm{s}}^{-1}$.

The root mean square velocity of a perfect gas is

The root mean square velocity of a gas molecule at $27 \mathrm{℃}$ is $1365 \mathrm{m} {\mathrm{s}}^{-1}$. The gas is

At what temperature is the root mean square speed of oxygen atom equal to the root mean square speed of helium gas atom at $-10 \mathrm{℃}$? Atomic mass of oxygen $=32$ and that of helium $=4$.

Define the following 

Mean free path

Define the following

Free path

The temperature of an ideal gas is increased from $120 \mathrm{K}$ to $480 \mathrm{K}$. If at $120 \mathrm{K}$, the RMS velocity of the gas molecules is ${\mathrm{v}}_{\mathrm{rms}}$, then at $480 \mathrm{K}$ it becomes:

$0.014 \mathrm{k}\mathrm{g}$ of nitrogen is enclosed in a vessel at a temperature of $27 °\mathrm{C}$. At which temperature the $\mathrm{r}\mathrm{m}\mathrm{s}$ velocity of nitrogen gas is twice it’s the $\mathrm{r}\mathrm{m}\mathrm{s}$ velocity at $27 °\mathrm{C}$ ?

A gas is filled in a container at pressure $P_0$. If the mass of molecules is halved and their $\mathrm{r}\mathrm{m}\mathrm{s}$ speed is doubled, then the resultant pressure would be

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.