What is the diffusion coefficient for the bad odor molecules in the air?

In summary, the root mean squared distance of the molecules causing a bad odor in a chemistry lab is 5 meters. Using the equation <R^2> = 6Dt, and given that the time it takes for the smell to travel 3 meters is 60 seconds, the diffusion coefficient for the molecules can be calculated to be 0.044m^2*s^-1. This is in line with the given answer of 0.044m^2*s^-1 in the textbook.
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Homework Statement


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?


Homework Equations



<R^2> = 6Dt

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

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.
 
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  • #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 = 33.75sUsing 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.
 

1. What is diffusion coefficient problem?

Diffusion coefficient problem is a mathematical concept that describes the rate at which particles spread or diffuse in a given substance or medium.

2. How is diffusion coefficient calculated?

The diffusion coefficient is calculated by dividing the diffusion flux (the number of particles diffusing per unit area per unit time) by the concentration gradient (the change in concentration over a given distance).

3. What factors affect diffusion coefficient?

The diffusion coefficient can be affected by various factors such as temperature, molecular weight, size and shape of particles, and the nature of the medium in which diffusion occurs.

4. What are the applications of diffusion coefficient problem?

The diffusion coefficient problem has numerous applications in various fields such as chemistry, physics, biology, engineering, and materials science. It is used to understand processes such as mass transport, heat transfer, and chemical reactions.

5. How does diffusion coefficient relate to Fick's laws of diffusion?

Fick's laws of diffusion describe the relationship between the diffusion flux, concentration gradient, and diffusion coefficient. The first law states that the diffusion flux is directly proportional to the concentration gradient, while the second law states that the diffusion coefficient is directly proportional to the rate of change of concentration with respect to time.

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