Question

A 14cm×14cm×14cm box contains 0.010 mol of nitrogen at 20∘C. What is the rate of collisions (collisions/s) on one wall of the box?

Answer 1

T = 293 degrees K

N = .01 mol * avogadros constant = 6.02*10^{21} particles.

Nitrogen is diatomic, so the molecular mass is 28u.

m = molecular mass / avogadros constant = 4.65*10^{-26} kg

so you have to solve for v_rms

v_{rms}=sqrt{3k_{B}T}/m} = 511 m/s

So you can figure out v_x in one direction because this is for v_rms

so you use

v_x= sqrt[(v_rms)^2/3] = 295 m/s

then to find rate of collisions/s so N_coll/t you use

(N*v_x)/(2 *0.1) = 8.86×10^25 collisions/s