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RMS speed of an ideal gas

  1. Nov 30, 2013 #1
    Hello everyone,

    When I was studying about the way to find the RMS speed of an ideal gas by using classical mechanics, I wondered that why the time interval of the collisions can be approached as the time between two collisions, instead of contact-time between particle and wall.

    remember? Δt=2l/v,

    Thankyou
     
  2. jcsd
  3. Nov 30, 2013 #2

    mathman

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    Off hand guess. It might depend on the size of the container. Wall collisions would be relatively rare compared to collisions between particles.
     
  4. Nov 30, 2013 #3
    For a cubic meter container of humid air at a temperature of 25°C, a total pressure of 1,000 hectopascals, and a vapor pressure of 100 pascals, I get a collision rate with a wall of some 1.79 X 1028 impacts per square meter per second.

    The number of intermolecular collisions would approximate some 6.21 X 1034 collisions per cubic meter per second. This makes intermolecular collisions some 112 thousand times more common than collisions with a wall [there are six walls].

    By the by, this is only a rough approximation because intermolecular collisions depend upon estimates of effective molecular "size"--a subject for which precise answers are lacking.
     
  5. Dec 1, 2013 #4
    Thanks for your helps,
    I know that collisions between particles and walls is relatively rare compare to collisions between particles themselves. But, when we're trying to find the pressure at the wall using classical mechanics law, we're beginning with F=Δp/Δt right?

    where Δp= -2mvx (If x is perpendicular to the wall)
    and why should Δt=2l/vx , instead of contact-time between particle and wall?
    This breaks newton's 2nd law isn't it?
    or do you have some explanations for that?

    Thankyou
     
  6. Dec 1, 2013 #5
    In Feynman's lecture, I found his derivation different from common derivation.


    [itex]P=\frac{N}{A_{collision}}[/itex]

    [itex]N=F_{net,acted\:by\:wall\:eek:n\:particles}=\frac{dp_{colliding\:particles}}{Δt_{collision}}[/itex]

    [itex]dp_{colliding\:particles}=N_{colliding\:particles}\:\times\:dp_{one\:particle}[/itex]

    [itex]N_{colliding\:particles}=\frac{1}{2}ρA_{collision}V_{rms,x}\:Δt_{collision}[/itex]

    [itex]ρ=density=\frac{N_{total}}{V_{total}}[/itex]

    The reason why there is [itex]\frac{1}{2}[/itex] in front of ρ is that the center velocity is zero so that there is only half particles colliding with wall in that space.

    [itex]dp_{one\:particle}=m\times(2V_{rms,x})[/itex]

    [itex]∴P=\frac{dp_{colliding\:particles}}{A_{collision}Δt_{collision}}[/itex]

    [itex]∴P=N_{colliding\:particles}\times\frac{dp_{one\:particle}}{A_{collision}Δt_{collision}}[/itex]

    [itex]∴P=\frac{1}{2}ρA_{collision}V_{rms,x}\:Δt_{collision}\frac{2mV_{rms,x}}{A_{collision}Δt_{collision}}---------------(*)[/itex]

    [itex]∴P=ρmV^{2}_{rms,x}=\frac{N_{total}}{V_{total}}mV^{2}_{rms,x}[/itex]

    And...

    [itex]V^{2}_{rms,x}=\frac{V^{2}_{rms}}{3}[/itex]

    [itex]∴P=\frac{N_{total}}{3V_{total}}mV^{2}_{rms}[/itex]

    [itex]∴N_{total}\frac{1}{2}mV^{2}_{rms}=E_{k,total}=\frac{3}{2}PV_{total}[/itex]


    In my opinion, the only problem is.... why can we use [itex]V_{rms,x}[/itex] in (*)?...
     
  7. Dec 1, 2013 #6
    I don't know if this will be of any help, but in statistical mechanics the root-mean-square axial speed is easily defined as:

    σ=(kT/m)1/2

    Here, sigma (σ) is the root-mean-square axial speed in meters per second, k is Boltzmann's Constant in joules per molecule per Kelvin, T is the temperature in Kelvins, and m is the unique molecular mass in kilograms. σ is used instead of vrms, because this quantity is also the standard deviation of the axial velocity distribution.
     
  8. Dec 3, 2013 #7
    I think this derivation is good, in terms to avoid misrepresentation (so far) of Δt terms in common derivation.

    I guess we should use V rms because it contains the average and variance. It means rms is arbitrary value, and it's good for approximation.

    by definition, rms quantity is

    <[itex]x^{2}[/itex]>=[itex]σ^{2}[/itex]+[itex]<x>^{2}[/itex]
     
  9. Dec 3, 2013 #8

    jtbell

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    Staff: Mentor

    It's because we want the average pressure over a long period of time (many round-trips between the walls of the container), not the pressure during (only) the particle-wall collisions.

    During a particle-wall collision, the force is (relatively) large, but during the flight between walls, the force is zero.
     
    Last edited: Dec 3, 2013
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