Posted by **Physics/Math** on Friday, February 23, 2007 at 2:48am.

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.

Thanks!

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