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

Click For Summary
SUMMARY

The diffusion coefficient for bad odor molecules in the air was calculated to be 0.044 m²/s based on the root mean squared distance formula = 6Dt. The experiment involved a student creating an odor 3.0 meters away, with the smell taking 60 seconds to reach an additional 2.0 meters further away. The calculations involved substituting time values into the diffusion equation and solving for D, confirming the textbook answer of 0.044 m²/s.

PREREQUISITES
  • Understanding of the diffusion equation = 6Dt
  • Knowledge of root mean squared distance calculations
  • Familiarity with basic algebra for solving equations
  • Concept of time as a variable in diffusion processes
NEXT STEPS
  • Research advanced diffusion models in gases
  • Learn about factors affecting diffusion coefficients in different mediums
  • Explore applications of diffusion coefficients in environmental science
  • Study the impact of temperature on diffusion rates
USEFUL FOR

Chemistry students, environmental scientists, and anyone interested in the principles of diffusion and odor dispersion in gases.

Laserbeam
Messages
4
Reaction score
0

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.
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 1 ·
Replies
1
Views
13K
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K