On the basis of kinetic theory derive an expression for the pressure exerted by an ideal gas

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On the basis of kinetic theory of gases, derive an expression for the pressure exerted by gas.

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Consider a cubical vessel (of side l) with walls perfectly elastic as in the figure.  Let the vessel have one gram molecule of a gas with its molecules in random motion.

Consider a molecule (mass m) moving with a velocity c, that can be resolved into three components u,v, and w in the direction of the edges of the cube along x,y,z axes.

∴c 2 =u 2 +v 2 +w 2 ........(1)

Let us consider two faces of the cube, say P and Q, normal to x−axis.When the molecule collides with the side Q with a velocity u, it rebounds with −u, while its other components remain unchanged.

Change in momentum of the molecule due to this collision =−2mu, which is imparted to the wall per collision.

The time taken by molecule to cover distance l=

u l ​

∴, after every interval of time

u 2l ​

, the molecule will again collide with the wall Q, and the number of collisions per unit time with the wall Q is equal to

2l u ​ .

∴, Momentum imparted to the wall per unit time =2mu×

2l u ​ = l mu 2 ​

(ignoring the negative sign).

The pressure exerted on the wall Q due to one molecule =

l mu 2 ​ × l 2 1 ​ = l 3 mu 2 ​

If the vessel contains N molecules with velocities u

1 ​ ,u 2 ​ ,u 3 ​ ,....u n ​

along x−axis, then the pressure exerted by them on Q, say

P= l 3 m ​ (u 1 2 ​ +u 2 2 ​ +u 3 2 ​ +...+u n 2 ​ )= v mN ​ ( u ˉ ) 2

where V is the volume of the vessel and (

u ˉ ) 2

is the average value of u

2 of all N molecules. ( u ˉ ) 2 = N u 1 2 ​ +u 2 2 ​ +u 3 2 ​ +...+u n 2 ​ ​

As the molecules move randomly,

( u ˉ ) 2 =( v ˉ ) 2 =( w ˉ ) 2 = 3 ( c ˉ ) 2 = ​ where c ˉ = N c 1 2 ​ +c 2 2 ​ +c 3 2 ​ +...+c n 2 ​ ​ ​ .........(2) c ˉ

is called the root mean square velocity of the molecules.

∴P= 3 1 ​ V mN( c ˉ ) 2 ​ .......(3)

This equation gives the pressure exerted by the gas on the walls of the vessel.It has the same value in any direction because the molecules have no preference for direction.

We have ρ= V mN ​ ∴P= 3 1 ​ ρ( c ˉ ) 2 or P= 3 2 ​ E where E= 2 1 ​ ρ( c ˉ ) 2

is called the mean kinetic energy per unit volume of the gas.

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Using the kinetic theory derive an expression for the pressure exe

Using the kinetic theory derive an expression for the pressure exerted by an ideal gas, stating clearly the assumption which you make.

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Properties Of Matter

Using the kinetic theory deriv...

Using the kinetic theory derive an expression for the pressure exerted by an ideal gas, stating clearly the assumption which you make.

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00:00 - 00:59 in this question using the Kinetic Theory derive an expression for pressure exerted by an ideal gas stating clearly does Amazon which you make 20 stand this concept firstly we will consider 1qq latest consider one cube now when I joined this two terms this will become one of the face of the cube this will become another face of the cube hear this will become another face of the cube after seeing what let's take a letter status quarters like x-axis will take this path as my y-axis and will take this part is my z-axis if I am considering here one particular

01:00 - 01:59 molecule or it's better if we consider a molecule here and this molecule is moving towards its side with a velocity V check if it is moving in the direction we can take velocity v y if it is moving in the direction we can take velocity visit here so what exactly happened this molecule when touches to the walls of the container and the pressure is exerted physically the pressure of a gas is due to the conditions of the molecules with respect to each other and with respect to the walls of the container so here in this case the momentum is transferred to the basic concept of momentum can be taken as admin to be if it is moving along x-axis I can take react when it is going in this direction I can take plus V accent is coming back and take - of MBA Momentum if I'm taking + MBA and while it is coming back minus of MBA I can say whatever the collision is this particular molecule when it collides with the walls of the container the collision is perfectly elastic if it is

02:00 - 02:59 elastic at exhibit both kinetic energy as well as the momentum so we will find out the change in momentum firstly the change in momentum is equals to change in momentum is equals to like a while it is coming back rest API 2 - 31 in that case I can write your - MBA - of MBA so I can write the change in momentum -2 MBA I can directly to this is the change in momentum now if I would like to find out the magnitude of change in momentum like to which indicates plus two MB now would like to find out the time taken for this molecule to complete this complete what to complete this total parts we can see the time taken is equals to time taken is equals to a physically time is nothing but we can say velocity is equal to displacement by time to time can be taken as displacement by velocity

03:00 - 03:59 now just time T is equals to the displacement is that if I consider the side of the cube as well while it is going it is l i am coming back to the wall will take it as well as well as well as other Total distance year will be to hell hole / it is moving with a velocity v at this time taken is to make only one impact is to make only one impact only one impact sunao if I want to find out if I want to find out number of impacts per second so I can say for a number of impact and number of impact first second impact for second we can represent the spot as En dash and dash can be taken as one by two lbx this goes to the numerator and we will be getting a number of impact but S

04:00 - 04:59 suggest will be written as and dash is equals to B at y12 so this is the number of impacts per second now I want to find out the change in momentum per second so I can write a change in momentum for second change in momentum for second is equals to the magnitude of the momentum into number of impact as the magnitude of momentum already we got it here as to MBA and the impact also we have you later substitute their respective values so I can write here to mvx and this can be written as we act by 12 suggest to address to will get cancel I can write directly MBA x square by health services change in momentum for S so once we got the change in momentum this change from Newton's second law cancel change in momentum is directly proportional to force work and right here change in momentum change in momentum is directly proportional to the

05:00 - 05:59 force exerted by the wall of the container in this case we can say that the force will be equal to MV act square whole divided by l so we got the force right over here now I would like to find out the pressure so this pressure is along a direction so I can take a p x is equals to force per unit area such as TX can be taken as the force is MBA x square whole divided by l whole divided by the area area of a cube is elsewhere this will be multiplied with this terms we can write p x is equals to x square whole divided by the pressure along x-axis but this is only for individual molecule if we have a number of molecules so for the total for that purpose we can write about total pressure as admin Jhoom vx1 square + m into vx2 square Plus I'm into vx3 I just got to hear

06:00 - 06:59 vx3 square and so on less m into v x 10 square whole divided by LG this is the pressure complete along x-axis S ap is equal to hear I can take out like MBL commons will be getting like vx1 square + x square + 3 X 3 square and so on up to 50 X and Square now I can represent this water Saviour x square is equals to 3 X square + bx to square + 3 X 3 square and so on up to reaction square whole divided by capital and that means I can send this capital and this site so that I can replace this quarter Saifai x square into an is equals to vx1 square + b square + b square and so on up to

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Derive the expression of pressure exerted by the gas on the walls of the container.

Derive the expression of pressure exerted by the gas on the walls of the container.

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Derive the expression of pressure exerted by the gas on the walls of the container.

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SOLUTION

The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy, but a change in momentum occurs.

The molecules of the gas exert pressure on the walls of the container due to the collision on it. During each collision, the molecules impart certain momentum to the wall. Due to the transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time.

A molecule of mass m moving with a velocity

v v→

having components (vx, vy, vz) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with the same speed and its x-component are reversed. This is shown in the figure. The components of velocity of the molecule after the collision are (- vx, vy, vz).

The x-component of momentum of the molecule before collision = mvx

The x-component of momentum of the molecule after collision = – mvx

The change in momentum of the molecule in x-direction = Final momentum – initial momentum

= – mvx – mvx = – 2mvx

According to the law of conservation of linear momentum, the change in momentum of the wall = 2mvx

The number of molecules hitting the right side wall in a small interval of time At.

The molecules within the distance of vx∆t from the right side wall and moving towards the right will hit the wall in the time interval ∆t. The number of molecules that will hit the right side wall in a time interval At is equal to the product of volume (Avx∆t) and number density of the molecules (n). Here A is an area of the wall and n is the number of molecules per unit volume

N V NV .

We have assumed that the number density is the same throughout the cube.

Not all the n molecules will move to the right, therefore on average only half of the n molecules move to the right and the other half moves towards the left side.

The number of molecules that hit the right side wall in a time interval ∆t

= n n2

Avx∆t ..............(1)

In the same interval of time At, the total momentum transferred by the molecules

∆P = n Av x tmv x Av x mnt

n2AvxΔt×2mvx=Avx2mnΔt

.......(2)

From Newton’s second law, the change in momentum in a small interval of time gives rise to force.

The force exerted by the molecules on the wall (in magnitude)

F = P t nmAv x ΔPΔt=nmAvx2 ...........(3)

Pressure, P = force divided by the area of the wall

P = F A nmv x FA=nmvx2

..................(4)

Since all the molecules are moving completely in a random manner, they do not have the same speed. So we can replace the term v

x vx2 with the average v x vx2¯ in equation (4). P = nm v x nmvx2¯ ........(5)

Since the gas is assumed to move in a random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has the same average speed in all three directions. So,

v x v y v z vx2¯=vy2¯=vz2¯

. The mean square speed is written as

v v x v y v z v x

v2¯=vx2¯+vy2¯+vz2¯=3vx2¯

v x v vx2¯=13v2¯

Using this in equation (5), we get

P = nm v 13nmv2¯ or P = N V m v 13NVmv2¯ ...........(6) as n N V [n=NV]

Concept: Pressure Exerted by a Gas

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Chapter 9: Kinetic Theory of Gases - Evaluation [Page 185]

Q III. 2. Q III. 1. Q III. 3.

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Tamil Nadu Board Samacheer Kalvi Class 11th Physics Volume 1 and 2 Answers Guide

Chapter 9 Kinetic Theory of Gases

Evaluation | Q III. 2. | Page 185

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