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Derivation for Kinetic Theory of Gases

  1. Aug 3, 2007 #1
    My question is with regard to the derivation for the kinetic theory of gases that allows us to relate temperature to the motion of the particles.

    I've looked at several introductory Physics texts and the same derivation is given (derivation in italics, my question in red, regular font):

    To proceed we need an expression for the force

    (1) Consider an ideal gas of N identical particles in a cubical container of length L.

    (2)These particles do not interact except for elastic collisions.

    Let’s just concentrate on one of these particles of mass m moving towards a wall with velocity +v and momentum +mv. Since the wall essentially has infinite mass, the particle bounces off with a velocity –v and momentum –mv. Now the time between collisions with the right hand wall is the round trip distance 2L divided by the particle speed or t=2L/v.

    We now have enough information to calculate the average force on the wall given by the particle's change in momentum. Using the impulse-momentum theorem,

    Fave(on particle) = (Final p – Initial p) / (Time between Collisions with the wall)

    Substituting our information

    Fave(on particle) = (-mv – mv) / (2L/v) = -2mv/2L/v = - mv2/L

    I'm okay with the rest of it (based on the above), but I'm unsure why the "time between collisions" is being used for the time in the above equation? In that equation, shouldn't the time of the collision be used?

    Now this is the average force exerted on the particle by the wall. But according to Newton’s 3rd Law the opposite but equal force will be exerted by the particle on the wall,

    Fave(on wall) = +mv2/L.

    The total force exerted on the right wall is equal to the force of each particle striking the wall. Since there are N particles and three dimensions, on average one-third will strike the right wall during the time t with an average value of the squared speed,

    Ftotal(on wall) = [N/3]*[m(v2)ave/L]

    Note we are using the average value of the squared velocity. The average must be used since the particles have a Maxwell distribution. Often one uses the definition: the square root of (v2)ave equals the root-mean-square speed or rms speed, vrms = ((v2)ave)1/2. So we can re-express the total force on the wall as

    Ftotal(on wall) = [N/3]*[mv2rms/L]

    So let’s convert to the pressure on the wall by dividing by the wall’s area L2,

    P = Ftotal(on wall)/L2 = [N/3]* [mv2rms/L3]

    But not, and this is nice, that the length cubed is just the volume of the box,so

    PV = [N/3]* [mv2rms].

    This is starting to look a lot like the ideal gas law! Note how we’ve related quantitatively macroscopic properties of the gas on the left to microscopic properties on the right. A little more manipulation and

    PV = 2/3 * N *[mv2rms /2]

    Since [mv2rms /2] is the average translational kinetic energy of an individual particle we can also state,

    PV = 2/3 * N * KEave

    Let’s compare this last result with the second version of the ideal gas law

    PV = NkT

    This shows that

    2/3 *KEave = kT


    KEave = ½*mv2rms = 3/2*kT
  2. jcsd
  3. Aug 3, 2007 #2

    Doc Al

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    Staff: Mentor

    You want the average force exerted by the particle, not the instantaneous force. For every collision, you have a certain change in momentum. To find the average force, consider the number of collisions per unit time--which is equivalent to considering a single collision every 2L/v seconds since that's the average time between collisions with the wall. Make sense?
  4. Aug 3, 2007 #3
    I think so. Thank you.

    So the 2L/v establishes a unit of time, and then we go on to assume that 1/3 of the particles hit the wall during that time.

    So, we're looking at the average force of all particles (N/3) being applied during the to-and-from travel time of only one particle?
  5. Aug 3, 2007 #4

    Doc Al

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    Staff: Mentor

    Now that I've read the rest of your quoted proof more carefully, I'd say that assuming 1/3 of particles hit during that time is a bit lame. I prefer thinking in terms of one dimension at a time, thus considering the x-component of the velocity ([itex]v_x[/itex]) for calculating the change in momentum and the travel time for all particles along the x-axis. Thus your force term will equal [itex]m v_x^2/L[/itex]. Since, by symmetry, there's nothing special about the x-direction, you can write the average force in terms of the average value of [itex](1/3) m v^2/L[/itex]. (The average values of [itex]v_x^2[/itex], [itex]v_y^2[/itex], and [itex]v_z^2[/itex] are all equal and add up to [itex]v^2[/itex].) This gets you to the same place, with a bit less handwaving.
  6. Aug 4, 2007 #5
    So all particles hit the wall within that time increment of 2L/v, with some x-component of velocity?
  7. Aug 4, 2007 #6

    Doc Al

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    Staff: Mentor

    Imagine a cube aligned with the axes. Every particle has some component of velocity along the x-axis, [itex]v_x[/itex]. Thus its travel time (between the walls perpendicular to the x-axis) must be [itex]2L/v_x[/itex]. (The actual value will be different for each particle--that's why we end up taking the average value.)
  8. Aug 4, 2007 #7
    That makes sense. Thank you.

    Sorry it took me so long.
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