Estimating Diameter of an Oil Slick with One Liter of Spilled Oil on a Lake

  • Thread starter Thread starter mlowery
  • Start date Start date
AI Thread Summary
One liter of oil, when spilled on a lake, spreads uniformly to form an oil slick one molecule thick, with oil molecules assumed to have a diameter of 2 x 10^-10 m. The volume of the oil is converted to cubic meters, and using the formula for the volume of a cylinder, the radius is calculated to be approximately 1.3 x 10^3 m, leading to a diameter of 2.5 x 10^3 m. A participant suggests verifying the answer by considering the physical dimensions of the oil slick and highlights the importance of checking calculations through alternative methods. Concerns about the assumption of oil molecules being perfect spheres and significant figures are raised, leading to a revised diameter estimate of 3 x 10^3 m. The discussion emphasizes the importance of confidence in calculations and the need for rough checks to validate results.
mlowery
Messages
23
Reaction score
0
The Problem
One liter (1000 cm^3) of oil is spilled onto a smooth lake. If the oil spreads out uniformly until it makes an oil slick just one molecule thick, with adjacent molecules just touching, estimate the diameter of the oil slick. Assume the oil molecules have a diameter of 2 x 10^-10 m.

---------------------------------
My Work
---------------------------------

1. I will first change liters from cm^3 to m^3 so that the units will match up.
1 liter = 1000 cm^3 = 1.0 x 10^-3 m^3

2. I will use the formula for volume of a cylinder (V = h π r^2) where:

V = 1.0 x 10^-3 m^3
h = 2 x 10^-10 m
r = unknown

3. Simply solve for the unknown radius:
(1.0 x 10^-3 m^3) = (2 x 10^-10 m) π r^2
(1.0 x 10^-3 m^3) / (2 x 10^-10 m) = π r^2
[(1.0 x 10^-3 m^3) / (2 x 10^-10 m)] / π = r^2
{[(1.0 x 10^-3 m^3) / (2 x 10^-10 m)] ^ (1/2)} = r = 1.3 x 10^3m

4. Calculate the diameter using the radius:
D = 2r
D = 2(1.3 x 10^3)
D = 2.5 x 10^3m

---------------------------------
Final Answer
---------------------------------

D = 2.5 x 10^3m

I have no idea if this is right or not, but it is as close as I have been able to get. This question has been giving me problems for hours. Can somebody confirm my calculation, or inform me where I went wrong?

Thanks,
Mitch
 
Physics news on Phys.org
Good job - but -
How come you don't know if it is right? Do you have zero confidence?
Why do you think "went wrong"? Here's how to "check" your answer:

If you sliced a 1-liter bottle into disks (10^9 of them)
they'd fill a rectangle 10^5 disks x 10^4 disks , or 10^4 m x 10^3 m

How big an oil slick *** 1 - atom thick *** did you expect 1 liter to make?

Always ry to check answers ROUGHLY, using a different approach
 
Thanks.

I was having trouble grasping the concept that we assume the oil molecules are perfect spheres. I was thinking, you know, complex branch-chain alkanes and stuff. So even when I did get the correct calculation, I was unsure whether my assumption was valid.

Additionally, I found one flaw in my answer. Significant figures. On the diameter of the molecule, I cannot assume that

2 x 10^-10 m == 2.0 x 10^-10 m.

Therefore, my answer should be D = 3 x 10^3m

Thanks for the response lightgrav,
Mitch
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top