Question

particles undergoing diffusion move a few millimeters per hour. In reality, particles are traveling around 500 m/s. The slow overall movement is because they collide with other particles very frequently. (density of water =1000 kg/m^3, Avogadro's # = 6.022x10^23 particles/mole, mass of the earth =5.97 x10^24 kg) using Xrms = sqrt(2Dt)

a. Estimate how far a water molecule travels before colliding with another water molecule.

b. How much time is in between each collision?

c. How many collisions occur for a molecule that diffuses 1 mm in 1 hour? d. Use your estimates from a, b, and c to figure out how many times a water molecule moves away from its starting point and how many times it moves towards it to end up 1 mm away after 1 hour.

Answer 1

a) distance a water molecule travels before colliding with another water molecule=2*D/U

D= diffusion coefficient of water molecules=2 *10^-9

so distance =2*2*10^-9/500=8*10^-12 m

b) time between each collisions =distance /velocity=8*10^-12/500=1.6*10^-14 seconds

no of collisions =1/t=10^14/1.6=6.25*10^13 collisions per second

c) diffusion coefficient=d^2/t=10^-6/3600=2.78*10^-10

one "gram molecular weight" of water, , weighs 18 grams. One gram molecular weight of something contains one Avogradro's number, of molecules of the stuff. So there are molecules of water in 18 grams of water, or molecules in one gram of water. Now 1 gram of water at room temperature occupies of volume. So of water contains molecules, giving a density of . The distance between collisions is therefore

v0=d/l=2.78*10^-10/3.1*10^-10=0.896 m/s.

no of collision=0.896/3.1*10^-10=2.89*10^9 collisions per second.

exactly half times towards and away from starting point.