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)\u00d72mx"
"= 2mx\u00b2\/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 "2my\u00b2\/l" and "2mz\u00b2\/l" respectively.
Hence, the net momentum imparted on all the six walls by the collisions of this molecule per second is
"= 2m(x\u00b2 + y\u00b2 +z\u00b2)\/l"
"= 2mc\u00b2\/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,
"= (2mn1c1\u00b2\/l) + (2mn2c2\u00b2 \/l)+ (2mn3c3\u00b2\/l)+..."
"= 2mNC\u00b2\/l"
"C\u00b2= (n1c1\u00b2+n2c2\u00b2+n3c3\u00b2+...)"
where is "C\u00b2" 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= 2mNC\u00b2\/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 = 6l\u00b2 x P"
"Or, P = F\/6l\u00b2"
"Or, P = 2mNC\u00b2\/6l\u00b3"
"Or, P = mNC\u00b2\/3V"
("l\u00b3 = V" , volume of the container)
"Or, PV = mNC\u00b2\/3"
"Or, P = [(mN\/V)C\u00b2]\/3"
"Or, P = DC\u00b2\/3"
("mN\/V = D" , density of the gas)
"Or, C\u00b2 = 3P\/D"
"Or, C = \u221a(3P\/D)"
"C = \u221a(3P\/D)" is the required expression of root mean square velocity.
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