Question

Use the concepts of the elementary kinetic theory of gases to derive an expression for the root mean square velocity of the molecules of a gas(Nitrogen) in terms of the pressure and density of the nitrogen gas.

Answer 1

The collisions being perfectly clastic, the molecule after striking the wall with velocity $x$ , returns with exactly the same velocity but in opposite direction. The momentum of the molecule just when it strikes is $= mx$ , and when it rebounds, the momentum is $= -mx$ ,. The change of momentum due to a collision is therefore,

$mx-(-mx) = 2mx$

walls in unite time would be $= (x/l)$

Therefore, the total momentum imparted to the two opposite walls G and H by this molecule per second

$= (x/l)×2mx$

$= 2mx²/l$

If we now consider the directions along Y-axis and Z-axis, the momentum imparted on the corresponding pairs of walls by the collisions of this molecule would be $2my²/l$ and $2mz²/l$ respectively.

Hence, the net momentum imparted on all the six walls by the collisions of this molecule per second is

$= 2m(x² + y² +z²)/l$

$= 2mc²/l$

Let there be $n1,n2,n3,...$ molecules having different, resultant velocities $c1, c2, c3....$ etc. The change of momentum per second from collisions of all the molecules on all the walls,

$= (2mn1c1²/l) + (2mn2c2² /l)+ (2mn3c3²/l)+...$

$= 2mNC²/l$

$C²= (n1c1²+n2c2²+n3c3²+...)$

where is $C²$ the mean square velocity.

(Hence, $C$ = root mean square velocity)

This change of momentum per unit time is the force $F$ exerted by the molecules (Newton's Law) on all the walls

$F= 2mNC²/l$

The pressure of the gas P is the force per unit area. of area $P$ There are six walls each of areaHence the total force on all the walls,

$F = 6l² x P$

$Or, P = F/6l²$

$Or, P = 2mNC²/6l³$

$Or, P = mNC²/3V$

($l³ = V$ , volume of the container)

$Or, PV = mNC²/3$

$Or, P = [(mN/V)C²]/3$

$Or, P = DC²/3$

($mN/V = D$ , density of the gas)

$Or, C² = 3P/D$

$Or, C = √(3P/D)$

$C = √(3P/D)$ is the required expression of root mean square velocity.