Question

Derivation of pressure of an ideal gas

Answer 1

Dude your answer is here Imagine an ideal gas contained in a container which is cubical in shape. Let one corner of the cube be the origin O, and let the x, y, z-axes along the edges. Let A1and A2 be the parallel faces perpendicular to the x-axis. Consider a molecule moving with velocity v in the container. The components of velocity along the axes are vx, vy, and vz. Now, when the molecule is colliding with face A1, the x component of velocity is reversed while the y and z component of velocity remains unchanged (as per our assumption that the collisions are elastic).



The change in momentum of the molecule is:
ΔP = (−mvx)–(mvx) = −2mvx ……….. (1)

Since the momentum remains conserved, the change in momentum of the wall is 2mvx

After the collision, the molecule travels towards the face A2 with x component of the velocity equal to −vx.

Distance traveled from A1 to A2 = L
Therefore time = Lvx

After collision with A2, the molecule again travels to A1. Thus the time between two collisions is 2Lvx

Therefore the number of collisions of the molecule per unit time:

n = vx2L ………………… (2)

Using equations 1 and 2,

The momentum imparted per unit time to the wall by the molecule:

ΔF = nΔp

= mLv2x

Therefore, total force on the wall A1 due to all the molecules is

F = ƩmLvx2

= mLƩvx2

Now, Ʃvx2 = Ʃvy2 = Ʃvz2 (symmetry)

= 13Ʃv2

Therefore, F = 13mLƩv2

Pressure is force per unit area so that

P = FL2

= 13ML3∑v2N

= 13ρ∑v2N

Where

M= total mass of gas

ρ = density of the gas

Now, ∑v2N is written as v2 and is called the mean square speed.