Derive an equation for the pressure due to an ideal gas during kinetic theory of gases?
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Let us see how to do this for a single gas molecule having a velocity v. The three velocity components of the molecule along the three axes X, Y and Z are given as vx, vy, and vz. The sum of these velocity components is made equivalent to the velocity v as
V2 = vx2 +, vy2, + vz2
Now, consider a molecule moving along with x-axis with a velocity of vx.
The momentum of the molecule before striking the face A is given as mvx.
Whenever a molecule strikes the walls of a container, it will bounce with same speed but in opposite direction i.e. - vx
Thus, the momentum after striking the surface is given as - mvx
The change in momentum before and after striking is given as mvx- (-mvx) = 2mvx
For the molecule to strike the same face again it has to go to the opposite face and come back. Thus, the molecule has to travel a distance of 2Lcm. (as the length of each side is Lcm).
The time required for colliding on the same face again is given as
Distance 2L
---------- = ----- sec.
Velocity vx
Number of collisions made by the molecule on the face A per sec is given as
Vx
------ per sec.
2L
As seen above the change in momentum for one collision is given as 2mvx
vx 2mvx.x vx m vx2
For ----- collisions the change in momentum is given as --------------- = ----------
2L 2L L
Likewise, the change in momentum per second due to collision on other surface is given as
m vx2
= ------
L
Now, the total change in momentum per second due to collision of one molecule on two faces along the x-axis is given as
m vx2 m vx2
---------- = -----
L L 2m vx2
= ------
L
Similarly, the change in momentum due to collisions of the molecule on the surfaces y and z is given as
2 m vy2 2 m vz2
= ---------- and ----- respectively.
L L
Thus, the total change in momentum per second on all the six faces along the three axes is given as
2m vx2 m vy2 2 m vz2
------- + ------- + -------
L L L
2m
= --------(vX2 + vY2 + vz2)
L
2mv2
= --------as ((vX2 + v Y2 + v z2 = v2)
L
As there are N molecules of a gas each moving with a different velocity, the velocity of each molecule contributes to change in momentum per second on all the six faces along the three axes.
Thus, the total change in momentum due to all molecules on all the six faces along the three axes is given as
2m
= --------(v12 + v 22 + v 32 +-------------+ v n2)
L
By multiplying the numerator and denominator with N, the total number of molecules gives
2mN
= --------((v12 + v 22 + v 32 +-------------+ v n2) / N )
L
But, ((v12 + v 22 + v 32+-------------+ v n2) / N is equal to u2 which is root mean square velocity.
Thus, the total change of momentum per second due to N molecules is given as
2m Nu2
------
L
The change momentum per second is termed as force F.
Hence,
2m Nu2
F = ------
L
Total force
Now, pressure P is given as force per unit area. = --------------
Total area
2m Nu2
= ----------
L x area
The area of six faces is given as 6L2
2m Nu2 2m Nu2
Thus P is given as --------------- = ----------
L x 6L2 L3
We know that L3 = volume, V.
mNu2 1
Thus P = ------- or --- mNu2 which is the kinetic gas equation.
3L 3
V2 = vx2 +, vy2, + vz2
Now, consider a molecule moving along with x-axis with a velocity of vx.
The momentum of the molecule before striking the face A is given as mvx.
Whenever a molecule strikes the walls of a container, it will bounce with same speed but in opposite direction i.e. - vx
Thus, the momentum after striking the surface is given as - mvx
The change in momentum before and after striking is given as mvx- (-mvx) = 2mvx
For the molecule to strike the same face again it has to go to the opposite face and come back. Thus, the molecule has to travel a distance of 2Lcm. (as the length of each side is Lcm).
The time required for colliding on the same face again is given as
Distance 2L
---------- = ----- sec.
Velocity vx
Number of collisions made by the molecule on the face A per sec is given as
Vx
------ per sec.
2L
As seen above the change in momentum for one collision is given as 2mvx
vx 2mvx.x vx m vx2
For ----- collisions the change in momentum is given as --------------- = ----------
2L 2L L
Likewise, the change in momentum per second due to collision on other surface is given as
m vx2
= ------
L
Now, the total change in momentum per second due to collision of one molecule on two faces along the x-axis is given as
m vx2 m vx2
---------- = -----
L L 2m vx2
= ------
L
Similarly, the change in momentum due to collisions of the molecule on the surfaces y and z is given as
2 m vy2 2 m vz2
= ---------- and ----- respectively.
L L
Thus, the total change in momentum per second on all the six faces along the three axes is given as
2m vx2 m vy2 2 m vz2
------- + ------- + -------
L L L
2m
= --------(vX2 + vY2 + vz2)
L
2mv2
= --------as ((vX2 + v Y2 + v z2 = v2)
L
As there are N molecules of a gas each moving with a different velocity, the velocity of each molecule contributes to change in momentum per second on all the six faces along the three axes.
Thus, the total change in momentum due to all molecules on all the six faces along the three axes is given as
2m
= --------(v12 + v 22 + v 32 +-------------+ v n2)
L
By multiplying the numerator and denominator with N, the total number of molecules gives
2mN
= --------((v12 + v 22 + v 32 +-------------+ v n2) / N )
L
But, ((v12 + v 22 + v 32+-------------+ v n2) / N is equal to u2 which is root mean square velocity.
Thus, the total change of momentum per second due to N molecules is given as
2m Nu2
------
L
The change momentum per second is termed as force F.
Hence,
2m Nu2
F = ------
L
Total force
Now, pressure P is given as force per unit area. = --------------
Total area
2m Nu2
= ----------
L x area
The area of six faces is given as 6L2
2m Nu2 2m Nu2
Thus P is given as --------------- = ----------
L x 6L2 L3
We know that L3 = volume, V.
mNu2 1
Thus P = ------- or --- mNu2 which is the kinetic gas equation.
3L 3
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