Mean Free Path and RMS Speed

The mean free path of molecules of a certain gas at STP is $1500d$, where $d$is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at $373 K$ is approximately:

In a gas at STP, if $n$ is the number density of the molecules and $r$ is the radius of the molecule, then the mean free path of the molecule is inversely proportional to

The collision frequency of nitrogen molecule in a cylinder containing at $2.0 \mathrm{atm}$ pressure and temperature $17°\mathrm{C}$ is approximately: (Take radius of a nitrogen molecule is $1.0 \mathrm{Å}$)

Time for successive collisions between walls :-

(ii) Time for two successive collision between the walls is.

Time for successive collisions between walls ;-

(i) When there is collision between two walls then charge in momentum per second is $\left(\text{along} X- \text{axis}\right)$.

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

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 :- $\left(\gamma =\frac{C_{\mathrm{p}}}{C_{\mathrm{v}}}\right)$

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