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Homework Help: Ideal gas/kinetic theory

  1. Apr 28, 2004 #1
    I have this sheet to do but I think its really hard! I've tried most of the questions but I need someone to tell me if what I've done is right or wrong and what is wrong if it is wrong because I'm not sure on a lot of it!

    1. A cylinder contains 2.3 x 10^4 cm3 of an ideal gas at a pressure of 1.8 x 10^7 Pa and a temperature of 27C.

    Calculate the amount, in mol, of gas in the cylinder

    The main thing here I'm not too sure about is the conversion of the volume into m3. Will it be 2.3 x 10^-2m3? Because this gave me an end answer of 166 moles which seems like quite a lot..

    2. Some neon-20 gas, assumed to be ideal, has a volume of 160 cm3 at a pressure of 1.2 x 10^5Pa and a temperature of 23C.

    a) the amount of substance in mol

    again not really too sure about converting to m3 but this is what ive done:
    160 cm3 = 160 x 10^-6m3

    pV = nRT
    1.2 x 10^5 x 1.6 x 10^-4 = n x 8.31 x 296
    n = 7.81 x 10^-3

    b) the number of atoms of neon present

    Ermm don't know if what is right but I multiplied my answer from a) by Avogadros constant (6.02 x 10^23) to get 4.7 x 10^21.

    c) the mass of a neon atom

    I used this moles formula here
    moles = mass
    mass = 7.81 x 10^-3 x 10 = 7.81 x 10^-2

    d) the rms speed of the atoms

    Well I guess this depends on whether my previous answers are correct but this is what I did!

    pV = 1/3Nm(mean square speed)
    1.2 x 10^5 x 160 x 10^-6 = 1/3 x 7.81 x 10^-2 x (mean square speed)
    mean square speed = 73.8
    root mean square speed = 8.59 m/s

    3. Last question now!

    Calculate the factor by which the rms speed of the molecules of an ideal gas will change when temperature of the gas changes from 20C to 120C.

    I'm just pretty much stuck with this question!

    Thanks very much for any help!!
  2. jcsd
  3. Apr 28, 2004 #2
    Question 1 to question 2b seem ok to me.
    This is correct.

    Molar mass of Neon is 20, not 10

    pV = 1/3Nm(mean square speed)
    1.2 x 10^5 x 160 x 10^-6 = 1/3 x 7.81 x 10^-2 x mass x (mean square speed)
    but not
    1.2 x 10^5 x 160 x 10^-6 = 1/3 x 7.81 x 10^-2 x (mean square speed)

    Use the following formula
    rms speed = sqrt (3RT/Molar Mass)

    I think you should know how to derive the formula above from pV = 1/3Nm(mean square speed)
  4. Sep 3, 2004 #3
    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?
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