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Pressue due to Ideal Gas

  1. Sep 3, 2004 #1
    I posted this in the College Level Help forum, but no one responded. I really need some explanation on this, so I'm moving this post from that forum to this. Thanks in advance:

    Can someone explain why the pressure due to an ideal gas in a container is given by [tex]P = \frac{1}{3}\rho<c^2>[/tex]? (rho is the density of the gas and <c^2> is the mean square speed)
    I happen to have the derivation of the eqn at hand, but somehow, there is this particular step in it I didn't understand; it's the step in which they calculated the change of momentum for a single gas molecule to be 2mv. Since force exerted on container walls is given by [tex]F = \frac {\triangle p}{\triangle t}[/tex] which means the change of momentum (p) divided by the time taken for the momentum to change.

    However, in the case of the gas particle, [tex]\triangle t[/tex] was taken to be [tex]\frac {2l}{v}[/tex]! (l is the length of the cuboid container, and v is velocity of the gas molecule). Why?
  2. jcsd
  3. Sep 4, 2004 #2
    Its actually a manipulation of the ideal gas law combined with the equation for the rms of a molecule.

    [tex] PV = nRT[/tex]

    [tex] PV = \frac{mRT}{M}[/tex] where m = mass of molecule and M = moler mass.

    [tex] P = \frac{(m/V)RT}{M} = \frac{\rho RT}{M}[/tex]

    [tex] c_r_m_s = \sqrt{\frac{3RT}{M}}[/tex]

    [tex] c_r_m_s^2 = \frac{3RT}{M}[/tex]

    [tex] M = \frac{3RT}{c^2}[/tex]

    Now you substitute this value for M into the manipulated Ideal Gas law...

    [tex] P = \frac{\rho RT}{3RT/c^2}[/tex]

    [tex] P = \frac{1}{3}\rho c^2[/tex]

    And there's your answer, :smile:
  4. Sep 4, 2004 #3


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    In an ideal gas the "atoms" do not interact with each other so the [itex]\triangle t[/itex] in your derivation represents the interval of time between collisions of a given particle with a wall. The velocity you used is really the component of velocity normal to the wall.
  5. Sep 4, 2004 #4
    Can you show me how to derive this?

    But I thought the [tex]\triangle t[/tex] should the time taken for the momentum of the gas particle to change, which is the time it spends in contact with the wall, since it is the case that the force exerted on the wall is defined as the rate of change of momentum with respect to the time taken for linear momentum to change.

    P.S. For some reason I can't edit some of my posts. Anyone know the reason why?
  6. Sep 4, 2004 #5


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    No, you have no way of calculating the "time taken for the momentum of the gas particles to change" (i.e. the time a molecule is in contact with the wall). Since your equation is an "average" (it doesn't apply only while a molecule is in contact with a wall, you are actually calculating the "average" force on the molecule during "lap" (from the time it bounces against a particular wall until it comes back to that wall). You can calculate the average force by finding the total change in momentum (from mc in one direction to mc in the other is mc-(-mc)= 2mc) and then divide by the total time between "bounces". [tex]\triangle t[/tex] is the time between collisions of a particular molecule with a particular wall. If the distance between walls is l, the the molecule bounces of the wall, travels distance l to the other wall, bounces of that and comes back- a total distance of 2l. Given that its speed is c (in both directions), [tex]\triangle t[/tex]t= 2l/c.
  7. Sep 4, 2004 #6
    Thanks Halls, but where did this come from:
    [tex] c_r_m_s = \sqrt{\frac{3RT}{M}}[/tex]
  8. Sep 4, 2004 #7


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    I think it comes from the equipartition theorem. Every active degree of freedom of a thermodynamic system at temperature T has an average energy of 1/2*k*T, where k is boltzmann's constant.

    An ideal gas, by definition, has only three degrees of freedom - motion in the x direction, motion in the y direction, and motion in the z direction. So we can write

    .5*m*vx_avg^2 = .5*m*vy_avg^2 = .5*m*vz_avg^2 = .5*k*T

    Multiplying each side by 2/m and adding we get

    vx_avg^2+vy_avg^2+vz_avg^2 = 3*k*T/m
  9. Sep 4, 2004 #8


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    That doesn't come from the ideal gas law. It comes out of an assumption about how the velocities are distributed. If the atomic velocities are normally distributed (gaussian aka the Maxwell-Boltzmann distribution) then that expression for the "average" speed follows.
  10. Sep 4, 2004 #9
    This is a well known and often used equation, if u want to see derivations just google it.
    I found some by googling "derivation root mean square velocity".
    http://itl.chem.ufl.edu/2041_f97/kin_thr/kin_thr.html [Broken]
    http://pitons.uwaterloo.ca/chem350/Resources/pressurederiv.pdf [Broken]
    Last edited by a moderator: May 1, 2017
  11. Sep 5, 2004 #10


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    Gyod. sometimes I slap my head here. That is the formula and that is the way it is. It is a very simple derivative.
  12. Sep 5, 2004 #11
    Thanks for all your replies, but why does [tex]\frac {2mu}{2l/c}[/tex] give you the average force on the wall?
  13. Sep 5, 2004 #12


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    [itex]2mu[/itex] is equal to the change of momentum of the particle bouncing off the wall if the collision is elastic.
  14. Feb 16, 2010 #13
    I'm having the same problem. I can't understand why the change in momentum has to be divided by the time for taken for one 'lap'...
  15. Feb 16, 2010 #14
    Old thread.

    You want the force on the wall. But you're too slow to perceive the individual impulses from each time a particular particle bounces against the wall again. (And when there is some astounding number of particles in total, you certainly can't identify the gaps between impulses from different individual particles bouncing against the wall - in fact they'll tend to smooth the sum impulse out anyway.) But you can work out the average force (or impulse) on the wall, since you can find how much momentum, per particle, per lap, is transferred from the gas to the wall (and recall force is just rate of momentum change) - thus determining how much force (on average at least) must be applied in the other direction to prevent the wall accelerating out.
    Last edited: Feb 16, 2010
  16. Feb 16, 2010 #15
    You might prefer to use the time that the molecule is actually in contact with the wall but for an elastic collision this time is taken to be zero.Since force=momentum change/t you might assume that the force is infinite and then you would have an infinite force acting for zero time????What!:surprised Let's use the time between laps.
  17. Feb 16, 2010 #16
    From what i learnt was,
    there are three degrees of freedom for monoatomic molecules/atoms
    up down, left right, forward back
    for that reason, its 1/3 when u consider particles moving at one direction we are considering

    now, its pressure we are dealing with ,, the force in a certain area

    so now we know where the 1/3 comes in.. the root mean squared speed is well the average lets say, and will be striking the wall with a certain force with that speed,
    also, the greater the number of molecules, the greater the force, hence, the density of rho
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