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:

Why do gases conduct electricity at low pressure?

An ideal gas is enclosed in a cylinder at $4 \text{atm}$ pressure and $500 \text{K}$ temperature. The mean time between two successive collisions is $0.0002 \text{ms}$. If the pressure is doubled and temperature is made one fourth, determine the time between two successive collisions in $\text{ms}$.

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 $1.0 \mathrm{Å}$)

Give the mathematical expression for the time of successive collisions between the walls of the container with.

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.

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 $\left(\mathrm{\lambda }\right)$ 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.

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

Define the following 

Mean free path

Define the following

Free path

A container is divided into two equal part I and II by a partition with small hole of diameter d. The two partitions are filled with same ideal gas, but held at temperature $T_I=150K$ and $T_{II}=300K$ by connecting to heat reservoirs. Let ${\lambda }_I$ and ${\lambda }_{II}$ be the mean free paths of the gas particles in the two parts such that $d>>{\lambda }_I$ and $d>>{\lambda }_{II}.$ Then $\frac{{\lambda }_I}{{\lambda }_{II}}$ is close to.

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 $\left(\mathrm{\gamma }=\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{v}}}\right)$