posted by Physics/Math on .
Compute the rms speed of a nitrogen molecule at 31.0°C. At what temperature will the rms speed be half that value?
At what temperature will the rms speed be twice that value?
T = 31 C = 304 K
Mass M = 28 g/(6.02*10^23)
= 4.65*10^-23 g = 4.65*10^-26 kg
The average kinetic energy per molecule is
(1/2) M V^2 = (3/2) k T
where k = 1.38*10^-23 Joule/K is Boltzmann's constant and V^2 is the average value of (molecular velcoity)^2
(For explanation, see http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html )
Solve for V^2, making sure T is in K and M is in kg. The V^2 that you get will be in m^2/s^2.
The square root of V^2 is the rms (root mean square) velocity.