1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Diffusion coefficient Problem

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data
    A student working 3.0m away from you in a chemistry lab performs an experiment which creates a bad odor. It takes a while for the smell to get to you (assume air currents are negligible). You are able to move 2.0m further away from the source but the smell catches up with you in another 60s. What is the diffusion coefficient in the air for the molecules you smell?


    2. Relevant equations

    <R^2> = 6Dt

    Where <R^2> is the root mean squared distance, D is the diffusion coefficient and t is time.

    3. The attempt at a solution

    <R^2> = 6Dt
    <(3+2)^2> = 6Dt
    25 = 6Dt

    60s/2m = "x"s/3m
    x = 90s

    90s + 60s = 150s = t

    25/6t = D
    25/6(150) = D
    D= 0.027m^2*s^-1

    I can't think of another way to use the information I've been given. The answer in my back of my textbook says 0.044m^2*s^-1. I know for the root mean square distance you would add 2.0m and 3.0m and square them, but I do not know what to do for time. Or maybe I'm just going in the complete opposite direction that I should be, I have no idea.
     
  2. jcsd
  3. Dec 1, 2011 #2
    Never mind, I figured it out. Typical of my doing that right after I post it, huh?

    Luck be damned.

    If anyone else has trouble with this, here you go:

    <R^2> = 6Dt
    <(2+3)^2> = 6Dt
    25 = 6Dt

    Now, we know that the time it takes to travel three meters is 60s, however, this is NOT a linear rate. We can put the time "t" into two terms, though.

    25 = 6D(t1 + t2)

    We know that one of the times is 60, which will be denoted t2

    25 = 6D(t1 + 60)

    We also known that the time it takes to travel 3m can be represented by way of

    <R^2> = 6Dt
    <3^2> = 6Dt1
    9 = 6Dt1
    => D = 1.5/t1

    D is a constant, so we can use this information to do a variable cancellation in the first equation via substitution or elimination. I will use substitution.

    25 = 6(t1 +60)D
    25= 6*t1*D + 360*D
    25= 6*t1*1.5/t1 + [360(1.5)]/t1
    25 = 9 + 540/t1
    16 = 540/t1
    t1 = 33.75s


    Using this information back in the previous equation...

    9 = 6Dt1
    D = 9/6t1
    D = 9/6(33.75)
    D = 0.044m^2*s^-1

    Which is meters squared per second.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Diffusion coefficient Problem
  1. Diffusion problem (Replies: 3)

Loading...