err... isn't this in the notes?
Anyway for deduction, you have to imagine a cubic volume of
sides d. And we consider only 1 side of the wall.
for a molecule of mass m, it's change in momentum. So, that's mc
- (-mc) = 2mc
time taken by a molecule between 2 sides of a wall = 2d / c
(distance / speed)
hence, num of collisions per unit time = (N/3)/t = Nc/6d
We take N/3 because there are 3 directions, and we are only
considering one direction. It's an average value.
multiply above, you get Nmc²/3d for total change in momentum per
unit time.
Remember Newton's 2nd law, force is d(mv)/dt, which means change
in momentum per unit time. This means Nmc²/3d is the force on one
side of the wall. Thus, pressure = force/area
p = ⅓(Nmc²/d) / A
p = ⅓(Nmc²/Ad) ============> A*d = V (volume)
p = ⅓(Nmc²/V)
(deduced)
btw, c is the root-mean-squared speed... You should use
<c> in your workings. I was lazy to type out...
