How can the number of molecular collisio...

How can the number of molecular collisions per unit time in a gas be increased ?

Text Solution

Verified by Experts

(i) by increasing the temperature of gas (ii) by increasing the amount of gas (iii) by decreasing volume of the gas.

Topper's Solved these Questions

BEHAVIOUR OF PERFECT GAS & KINETIC THEORY
PRADEEP|Exercise Very short Answer questions|35 Videos
BEHAVIOUR OF PERFECT GAS & KINETIC THEORY
PRADEEP|Exercise Short answer questions|28 Videos
BEHAVIOUR OF PERFECT GAS & KINETIC THEORY
PRADEEP|Exercise Solved Examples|43 Videos
GRAVIATION
PRADEEP|Exercise Assertion-Reason Type Questions|19 Videos

Similar Questions

Explore conceptually related problems

What are the different ways of increasing the number of molecules collisions per unit time against the walls of the vessel containing a gas ?

The number of molecules per unit volume (n) of a gas is given by

In CsCl the number of formula units per unit cell is

If the amount of a radioactive substance is increased four times then how many times will the number of atoms disintegrating per unit time be increased ?

A sample of gas measured at 20^(@)C C and 4.0 atm is heated to 40^(@)C at constant volume. Which statement(s) is (are) true of the gas after heating relative to its initial satate? (P) The average molecular kinetic energy is increased. (Q) The average molecular speed is unchanged. (R) The pressure of the gas is increased to 8.0 atm. (S) The number of molecular collisions per second is unchanged.

Statement-1 : Mean free path of molecules of a gas decreases with increases in number density of the molecules. Statement-2 : More the molecules per unit volume, more are the chances of molecular collision. Therefore, mean free path decreases.

Collision cross-section is an area of an imaginary sphere of radius sigma around the molecule within which the centre of another molecule cannot penetrate. The volume swept by a single molecule in unit time is V=(pisigma^(2))overline(u) where overline(u) is the average speed If N^(**) is the number of molecules per unit volume, then the number of molecules within the volume V is N=VN^(**)=(pisigma^(2)overline(u))N^(**) Hence, the number of collision made by a single molecule in unit time will be Z=N=(pi sigma^(2)overline(u))N^(**) In order to account for the movements of all molecules, we must consider the average velocity along the line of centres of two coliding molecules instead of the average velocity of a single molecule . If it is assumed that, on an average, molecules collide while approaching each other perpendicularly, then the average velocity along their centres is sqrt(2)overline(u) as shown below. Number of collision made by a single molecule with other molecule per unit time is given by Z_(1)=pisigma^(2)(overline(u)_("rel"))N^(**)=sqrt(2) pisigma^(2)overline(u)N^(**) The total number of bimolecular collisions Z_(11) per unit volume per unit time is given by Z_(11)=(1)/(2)(Z_(1)N^(**))"or" Z_(11)=(1)/(2)(sqrt(2)pisigma^(2)overline(u)N^(**))N^(**)=(1)/(sqrt(2))pisigma^(2)overline(u)N^(**2) If the collsion involve two unlike molecules then the number of collisions Z_(12) per unit volume per unit time is given as Z_(12)= pisigma _(12)^(2)(sqrt((8kT)/(pimu)))N_(1)N_(2) where N_(1) and N_(2) are the number of molecules per unit volume of the two types of molecules, sigma_(12) is the average diameter of the two molecules and mu is the reduced mass. The mean free path is the average distance travelled by a molecule between two successive collisions. We can express it as follows : lambda=("Average distance travelled per unit time")/("NO. of collisions made by a single molecule per unit time")=(overline(u))/(Z_(1)) "or "lambda=(overline(u))/(sqrt(2)pisigma^(2)overline(u)N^(**))implies(1)/(sqrt(2)pisigma^(2)overline(u)N^(**)) Three ideal gas samples in separate equal volume containers are taken and following data is given : {:(" ","Pressure","Temperature","Mean free paths","Mol.wt."),("Gas A",1atm,1600K,0.16nm,20),("Gas B",2atm,200K,0.16nm,40),("Gas C",4atm,400K,0.04nm,80):} Calculate ratio of collision frequencies (Z_(11)) (A:B:C) of following for the three gases.

PRADEEP-BEHAVIOUR OF PERFECT GAS & KINETIC THEORY-Conceptual Problems

Under what conditions do the real gases obey more strictly the gas equ...
Text Solution
|
Two perfect gases at absolute temperature T(1) and T(2) are mixed. The...
Text Solution
|
A box contains equal number of molecules of hydrogen and oxygen. If th...
Text Solution
|
The graphs in Fig shown the variation of the product PV with respect t...
Text Solution
|
Two different gases at the same temperature have equal root mean squar...
Text Solution
|
Distinguish between average speed and rms speed. If three molecules ha...
Text Solution
|
A gas is filled in a cylinder fitted with a piston at a definite tempe...
Text Solution
|
A gas is filled in a cylinder fitted with a piston at a definite tempe...
Text Solution
|
On reducing the volume of the gas at constant temperature, the pressur...
Text Solution
|
Explain the phenomenon of evaporation on the basis of kinetic theory.
Text Solution
|
Why cooling is caused by evaporation?
Text Solution
|
The rms speed of oxygen molecules at a certain temperature T is v. If ...
Text Solution
|
A vessel of volume 4 litres contains a mixture of 8 grams of O(2), 14 ...
Text Solution
|
Why temperature less than OK is not possible ?
Text Solution
|
The ratio of vapour densities of two gases at the same temperature is ...
Text Solution
|
How can the number of molecular collisions per unit time in a gas be i...
Text Solution
|
On which factors does the average KE of gas molecules depend : nature ...
Text Solution
|
What happens when a compressed gas pushes a piston out and expands ?
Text Solution
|
Though the velocity of air molecules is nearly 0.5 km//s, yet the smel...
Text Solution
|
There are N molecules of a gas in a containter. If this number is incr...
Text Solution
|