Question

derivation of pressure of gas on the basis of the kenitic theroy
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Answer 1

Consider a cubic vessel, having each side of length l, contains n number of gas molecules moving in x, y and z directions. So the no. of molecules moving in x direction is equal to that of molecules moving in y direction, which is equal to that of the molecules moving in z direction.

Hence we can say that among n molecules, n/3 molecules move in x direction, n/3 molecules move in y direction, or n/3 molecules move in z direction.

Consider one molecule of the gas moving in x direction. Here the gas molecule is assumed to move from one side of the cubic vessel, collide with the opposite wall and return to its initial position. So the gas has to travel a distance of 2l in total.

Before colliding with the opposite side, the linear momentum of the gas molecule is mv, where m and v are its mass and velocity respectively.

After collision, the direction of its velocity changes but not magnitude. Hence the final linear momentum is -mv.

Now, change in momentum = - mv - mv = - 2mv.

This is the change in momentum of the particle. The change in momentum of the wall undergoing collision, so, is given by 2mv.

By Newton's Second Law of Motion, we know that force is the rate of change of linear momentum. Thus the force exerted on the wall by the particle is,

$F=\dfrac {2mv}{\Delta t}\quad\longrightarrow\quad (1)$

where ∆t is the time taken for the particle to travel the distance 2l.

But, we know that time is the distance travelled by speed. Thus

$\Delta t=\dfrac {2l}{v}$

Then (1) becomes,

$F=\dfrac {2mv}{\left (\dfrac {2l}{v}\right)}\\\\\\F=\dfrac {mv^2}{l}$

This is for one gas molecule. For n/3 gas molecules,

$F=\dfrac {mnv^2}{3l}\quad\longrightarrow\quad(2)$

But the exception is that, by kinetic energy, the velocity of each particle is not same but constant. Then we have to take the mean of all these velocities.

Here we take the mean square velocity of the particles,

$\displaystyle\overline{v^2}=\dfrac {1}{n}\sum_{i=1}^n(v_i)^2$

Hence (2) is actually,

$F=\dfrac {mn\overline{v^2}}{3l}$

Then, pressure,

$P=\dfrac {F}{A}$

where A is the area of one side of the cubical vessel, which is nothing but l² since length of each side of the vessel is l. Then,

$P=\dfrac {F}{l^2}\\\\\\P=\dfrac {mn\overline{v^2}}{3l^3}$

Let l³ = V, the volume of the vessel. Then,

$P=\dfrac {mn\overline{v^2}}{3V}\quad\longrightarrow\quad (3)$

But, mass,

$m=\rho V$

where $\rho$ is the density of the gas. Then (3) becomes,

$P=\dfrac {\rho Vn\overline{v^2}}{3V}\\\\\\\boxed {P=\dfrac {1}{3}\rho n\overline{v^2}}$

For one gas molecule, n = 1. Therefore,

$\boxed {\mathbf{P=\dfrac {1}{3}\rho \overline{v^2}}}$

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