Thermodynamics / micro/macro connection problem

In summary, the wall of a 10 cm*10 cm*10 cm box contains .01 mol of nitrogen at 20 degrees C. The rate of collisions (collisions/s) on one wall of the box is 1.54*10^25.
  • #1
greenskyy
17
0

Homework Statement


A 10 cm * 10 cm * 10 cm box contains .01 mol of nitrogen at 20 degrees C.
What is the rate of collisions (collisions/s) on one wall of the box?

Homework Equations


[tex]\frac{N_{coll}}{t}=\frac{NAv_{x}}{2V}[/tex]

[tex]v_{rms}=\sqrt{\frac{3k_{B}T}{m}}[/tex]

The Attempt at a Solution


I'm really banging my head on the desk with this one. So far, I've calculated the following information.

T = 20 degrees C = 293 degrees K

N = .01 mol * avogadros constant = [tex]6.02*10^{21}[/tex] particles.

Nitrogen is diatomic, so the molecular mass is 28u.

m = molecular mass / avogadros constant = [tex]4.65*10^{-26} kg[/tex]

[tex]v_{rms}=\sqrt{\frac{3k_{B}T}{m}} = 511 m/s[/tex]

Plugging this into the first equation given

[tex]\frac{N_{coll}}{t}=\frac{NAv_{x}}{2V}[/tex] = [tex]\frac{Nv_{x}}{2*.1} = 1.54*10^{25}[/tex]

This answer is incorrect according to masteringphysics, and I have no ideas from here. If anyone could offer some suggestions I would be very grateful.
 
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  • #2
Thanks for the help everyone... =\

I figured it out after talking to some other students today. I was lacking one equation:

[tex](v_{x}^{2})_{avg}=\frac{v_{rms}^{2}}{3}[/tex]

Good luck on this problem for whoever needs this :D
 
  • #3
I happen to also be working on this problem. In your last equation it is possible to get:

[tex](v_{x})_{avg}=sqrt(\frac{v_{rms}^{2}}{3})[/tex]

Do you plug this into this equation:

[tex]
\frac{N_{coll}}{t}=\frac{NAv_{x}}{2V}
[/tex]

for [tex](v_{x})[/tex]? I tried doing this and came up with the wrong answer. I assume the number of collisions per second would be an average of all of the molecules hitting the wall, however Mastering Physics would have me believe that I am wrong in this regard.
 
  • #4
Yes, that's what you do. The mastering physics system has the incorrect answer stored as well, as if the number of moles were .1 rather than .01. The "true" answer should be to the magnitude of 10^24, while the one accepted by mastering physics is 10^25.
 
  • #5
Alright, thanks. That problem was giving me many problems as well, and I am glad that forums such as these exist!
 

FAQ: Thermodynamics / micro/macro connection problem

What is thermodynamics?

Thermodynamics is a branch of physics that deals with the behavior of energy and its transformation from one form to another.

What is the micro/macro connection problem in thermodynamics?

The micro/macro connection problem in thermodynamics refers to the challenge of bridging the gap between the behavior of individual particles at the microscopic level and the overall behavior of a system at the macroscopic level.

What are the laws of thermodynamics?

The laws of thermodynamics are fundamental principles that govern the behavior of energy in a system. They include the first law, which states that energy cannot be created or destroyed, only transformed, and the second law, which states that the total entropy of a system and its surroundings always increases.

What is the difference between micro and macro in thermodynamics?

In thermodynamics, the micro level refers to the behavior of individual particles, such as atoms or molecules, while the macro level refers to the overall behavior of a system. The micro level is concerned with the kinetic and potential energy of individual particles, while the macro level is concerned with the total energy of the system.

How is the micro/macro connection problem addressed in thermodynamics?

The micro/macro connection problem is addressed through the use of statistical mechanics, which uses probability and statistical methods to describe the behavior of large systems of particles. It allows for the connection between the macroscopic properties of a system and the microscopic behavior of its constituent particles.

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