Post a New Question


posted by on .

A sealed container of volume 0.10 m3 holds a sample of 3.0 x 10^24 atoms of helium gas in equilibrium. The distribution of speeds of helium atoms shows a peak at 1100 ms-1.

Caluculate the temperature and pressure of the helium gas. (helium atom 4.0 amu)

How do I calculate the mass of helium? I got 2.0 x 10^-2 kg from (sample size/avogrado's constant) x 4.0 x 10^-3 kg.
Is this correct as when I put it into

T = (v^2 x m)/2k

I get a huge temp in the 8^30 K region which cannot be right!

What am I doing wrong here?

When I get the temp I can get the pressure from the ideal gas equation.

  • Physics - ,

    You can get the temperature from the velocity distribution curve. The average kinetic energy is (1/2) m V^2 = (3/2) k T
    where V^2 is the average V^2 of the atoms.
    This tells you that the average V^2 is
    V^2 = 3 k T/m, so
    T = mV^2/(3 k)
    k is the Boltzmann constant 1.3804*10^-23 Joule/K
    m is the mass of a helium atom, which is 4.0 g/6.023*10^23 = 6.64*10^-27 kg.

    The peak of the velocity distribution is not the same as the square root of the averge V^2, but is is pretty close. You can worry about the difference later. Using sqrt(Vpeak^2) for average V^2, I get
    T = 6.64*10^-27 kg*(1100^2)m^2/s^2/3*(1.38*10^-23)J/K = 194 K

    Now that you know T and the number density n (atoms/m^3), the equation
    P = n k T
    can be used for the pressure.
    k is Boltzmann's constant. You could also use P = N R T, where R is the gas constant and N is the number of MOLES per m^3 and r is the gas constant.

    Now for a fine point. In the Maxwellian velocity distribution of atoms, which is what you have, the peak velocity Vp differs from Vrms.
    Vp = sqrt(2/3)* Vrms = 0.816 Vrms
    for an explanation.

    To get the temperature, I should have used Vp instead of Vrms. This has the effect of lowering the temperature by a factor 2/3.

  • Physics - ,

    Thanks for the help drwls.

    I have now also found the average kinectic energy of the helium atoms.

    But what is the position of the maximum in the energy distribution? Is this the peak energy that I need to find?

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question