Mean free path - class-XI
Description: mean free path | |
Number of Questions: 31 | |
Created by: Chandra Bhatti | |
Tags: behaviour of perfect gas and kinetic theory of gases behavior of perfect gas and kinetic theory kinetic theory of gases kinetic theory physics kinetic theory of matter |
A molecule of gas in a container hits one wall (1) normally and rebounds back. It suffers no collision and hits the opposite wall (2) which is at an angle of $30^o$ with wall 1.
Assuming the collisions to be elastic and the small collision time to be the same for both the walls, the magnitude of average force by wall 2. $(F _2)$ provided the molecule during collision satisfy
Mean free path depends on which of the following?
A gas has an average speed of 10 m/s and a collision frequency of 10 $s^{-1}$. What is its mean free path?
If the pressure in a closed vessel is reduced by drawing out some gas, the mean-free path of molecules :
If the pressure of a gas is increased then its mean free path becomes :
The mean free path of a gas varies with absolute temperature as :
A gas has a molecular diameter of 0.1 m. It also has a mean free path of 2.25 m. What is its density?
In physics, the mean free path is the average distance traveled by a moving particle (such as an atom , a molecule, a photon) between successive impacts (collisions), which modify its direction or energy or other particle properties. In which of the following mean free path is used ?
A satellite sent into space samples the density of matter within the solar system and gets a value $2.5$ hydrogen atoms per cubic centimeter. What is the mean free path of the hydrogen atoms? Take the diameter of a hydrogen atoms as $d=0.24\ nm$.
Estimate the mean free path of nitrogen molecule in a cylinder containing nitrogen at 2.0atm pressure and temperature ${17^o}C$.(take the radius of nitrogen molecule to be 1.0A, Molecular mass=28gm
The mean free path of a molecule of He gas is $\alpha $. Its mean free path along any arbitrary coordinate axis will be
The mean free path and rms velocity of a nitrogen molecule at a temperature 17C are $1.2 \times 10^{-7}$ m and $5 \times 10^2$ m/s respectively.The time between two successive collisions
Modern vacuum pumps can evacuate a vessel down to a pressure of $4.0\times { 10 }^{ -15 }atm$. At room temperature $(300K)$, taking $R=8.3J{ K }^{ -1 }\quad { mole }^{ -1 },1\quad atm={ 10 }^{ 5 }Pa\quad \quad $ and ${ N } _{ Avagadro }=6\times { 10 }^{ 23 }{ mole }^{ -1 }$, the mean distance between the molecules of gas in an evacuated vessel will be of the order of :
The mean free path of the molecules of a gas depends on
If the pressure in a closed vessle is reduced by drawing out some gas the mean-free path of molecules
Mean free path does not depend on
State whether true or false:
$He > H _2 > O _2 > N _2 > CO _2$
The mean free path of the molecule of a certain gas at 300 K is $2.6\times10^{-5}:m$. The collision diameter of the molecule is 0.26 nm. Calculate
(a) pressure of the gas, and
(b) number of molecules per unit volume of the gas.
A gas has an average speed of $10 m/s$ and a collision frequency of $10$ $s^{-1}$. What is its mean free path?
A gas has an average speed of $10 m/s$ and an average time of $0.1 s$ between collisions. What is its mean free path?
A gas has a density of $10$ particles$/m^3$ and a molecular diameter of $0.1 $m. What is its mean free path?
Consider the following statements for air molecules in an air tight container.
(I) The average speed of molecules is larger than root mean square speed.
(II) Mean free path of molecules is larger than the mean distance between molecules.
(III) Mean free path of molecules increases with temperature.
(IV) The rms speed of nitrogen molecule is smaller than oxygen molecule. The true statements are.
Mean free path in kinetic theory can be written as: ( every symbol has standard meaning)
A gas in a 1 $m^3$ container has a molecular diameter of 0.1 m. There are 10 molecules. What is its mean free path?
Estimate the mean free path of a cosmic ray proton in the atmosphere at sea level. Given $\sigma = 10 ^{-26} cm^2$
Estimate the average number of collisions per second that each $N _2$ molecule undergoes in air at room temperature and at atmospheric pressure. The diameter of a $N _2$ molecule is $0.3\ mm$.
A container is divided into two equal parts I and II by a partition with a small hole of diameter d. The two partitions are filled with same ideal gas, but held at temperatures $T _I=150$K and $T _{II}=300$K 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 $\lambda _I/\lambda _{II}$ is close to.
Calculate the means free path of nitrogen molecule at $27^o$C when pressure is $1.0$ atm. Given, diameter of nitrogen molecule $=1.5\overset{o}{A}$, $k _B=1.38\times 10^{-23}$J $K^{-1}$. If the average speed of nitrogen molecule is $675$ $ms^{-1}$. The time taken by the molecule between two successive collisions is?
Ten small planes are flying at a speed of $150$km $h^{-1}$ in total darkness in an air space that is $20\times 20\times 1.5km^3$ in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius $10$m.
Estimate the mean free path for a water molecule in water vapor at $373K$,the diameter of the molecule is $2\ \times 10^{-10}\ m$ and at $STP$ number of molecular per unit volume is $2.7\ \times 10^{25}\ m^{-3}$ :
There are two vessels of same consisting same no of moles of two different gases at same temperature . One of the gas is $CH _{4}$ & the other is unknown X. Assuming that all the molecules of X are under random motion whereas in $CH _{4}$ except one all are stationary. Calculate $Z _{1}$ for X in terms of $Z _{1}$ of $CH _{4}$. Given that the collision diameter for both gases are same & $\displaystyle (U _{rms}) _{x}=\frac{1}{\sqrt{6}}(Uav) _{CH _{4}}$.