Question

In a period of 2s, 4.8E23 nitrogen molecules strike a wall area of 8.6 cm^2.

If the molecules move at 308 m/s and strike the wall head-on in a perfectly elastic collision, find the pressure exerted on the wall. The mass of one N2 molecule is 4.68E-26kg.

Answer in terms of kPa.

I found the mass of all the molecules by multiplying the total molecules by the mass given. I converted the area to .00086 m^2, and found P by dividing the Force (mass times g or 9.8) by the area. I got 255.985 Pa, and .255985 kPa. I don't think I did this right, can someone please check my work?

Answer 1

The pressure on the wall is the rate (per second) at which the momentum of colliding molecules is changed per unit area. For elastic head-on collisions, each molecular collision changes its momentum by
delta (mv) = m * 2 * 308 m/s
where m is the mass of the molecule.
At the wall, half of the molecules are heading towards the wall and half are going away

P = (1/2)*delta(mv)*(number density)
= (delta mv)*(number of molecules hitting wall per second per area)
In your case,
(number of molecules hitting wall per second per area) = (4.8*10^23)/[2s * 8.6*10^-4 m^2] = 2.79*10^26 s^-1 m^-2

Look up the mass of a nitrogen atom (in kg) and complete the calculation. The answer will be in pascals.