posted by Oli 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.
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.
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?