1. Limited time only! Sign up for a free 30min personal 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!

Kinetic Theory of Gases Derivation

  1. Feb 28, 2015 #1
    Hi, I am struggling with a particular point on the derivation of the kinetic theory of gases. Between 5mins and 6mins 30seconds of the video below they discuss how to get a value for the average of the squared velocities of the molecules in the gas. The bit I don't get is why they divide by 'N' on BOTH sides of the equation (i.e. why he divides the force by N). At first I thought he did it to make sure he did it to both sides of the equation but then that doesn't make sense to me. If you add up all the squared velocities and divide through the total number of molecules then you get the average of the squared velocities. So instead of the force being equal to the squared velocity of a single particle it is equal to an average for them all which is fine. But then why would you go and divide the left side by N as well?

  2. jcsd
  3. Feb 28, 2015 #2


    User Avatar
    Homework Helper
    Gold Member

    If F = k ( sum of the square of each molecule's velocity ) that is the total force for ALL the molecules
    So when you divide the RHS by N to get the average velocity, you now have the velocity of ONE typical molecule.
    F is much too big for that. It was the force for ALL the molecules combined. So we have to divide that also by N.
    Now ##\frac{F}{N}## = k ( average squared velocity of one molecule )

    As you said, it is mathematically required that you divide both sides of an equation by the same thing.

    It is also what averages are about.
    Say F = the total number of chocolates in N boxes

    Then ##\frac{F}{N}## = ##\frac{the\ total\ number\ of\ chocolates\ in\ N\ boxes}{N}## = the average number of chocolates in ONE box
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook