Compute the rms speed of a nitrogen molecule at 31.0°C. At what temperature will the rms speed be half that value?

Question

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!

Answer 1

First we need to calculate V^2:

(1/2) M V^2 = (3/2) k T

V^2 = (3 k T) / M

V^2 = (3 * 1.38*10^-23 Joule/K * 304 K) / 4.65*10^-26 kg

V^2 ≈ 2845139.78 m^2/s^2

Now we can find the rms speed by taking the square root of V^2:

Vrms = √(2845139.78 m^2/s^2) ≈ 1686.75 m/s

To find the temperature at which the rms speed is half of this value, we can use this equation:

V^2 = (3 k T) / M

Or we can use the relationship between T and Vrms:

V^2 = (3 k T1) / M = (1/4) * (3 k T2) / M

T2 = (1/4) * T1

T2 = (1/4) * 304 K ≈ 76 K

To find the temperature at which the rms speed is twice the original value, we can use the same equation:

V^2 = (3 k T) / M

Or we can use the relationship between T and Vrms:

V^2 = (3 k T1) / M = (4) * (3 k T3) / M

T3 = (4) * T1

T3 = (4) * 304 K ≈ 1216 K

So the rms speed will be half the original value at 76 K, and twice the original value at 1216 K.