Helium gas with T1 = 500K and P1 = 0.02MPa in a rigid container with volume V = 1 cm^3.
Then Helium goes through a process where atoms with kinetic energies greater than kB*T1, where kB is Boltzmann constant, are instantaneously removed from the container.
Atoms remaining in the container attain a M-B velocity distribution with a final Temperature T2.
Calculate T2 and the final pressure P2.
(1) PV = N * kB * T
M-B speed distribution:
(2) f(v) = 4 * pi * v^2 * (m / (2*pi*kB * T))^(3/2) * exp( - m * v^2 / (2 * kB * T))
The Attempt at a Solution
First, I find the number of molecules using (1) and the total number of atoms in state 1 is 2.987e18.
Then find the limiting velocity by setting E = kB* T = 1/2 * m * v^2 to solve for v,
integrate from 0 to v ( f(v) dv ) to find the fraction of the atoms up to this speed v = 0.463.
The remaining atoms will be this fraction multiply by the number of atoms = 1.34e18 atoms remaining.
Then I am lost on solving for T2 with this remaining atoms that attain M-B velocity distribution.
Any help will be appreciated. Thanks!