Question

If the rms speed of the molecules of a quantity of helium at temperature 142 K is the same as the rms speed of neon atoms at a second temperature, what is the temperature of the neon gas? Answer in Kelvins. The answer is NOT 710, 713.55 or 284.

Answer 1

The rms speed of a gas is given by

$v_{rms} = \sqrt{\frac{3 k_{B}T}{m}} = \sqrt{\frac{3 RT}{M}}$

where k_B is Boltzmann's Constant, T is temprature in Kelvin R is gas constant = 8.31 JK^-1mol^-1 and M is molecular mass,

Now For Helium M = 4

$v_{rms}(He) = \sqrt{\frac{3 \times 8.31\times 142}{4}}$

Now For Neon M = 20

$v_{rms}(Ne) = \sqrt{\frac{3 \times 8.31\times T}{20}}$

From the data given

$\sqrt{\frac{3 \times 8.31\times T}{20.1797}} = \sqrt{\frac{3 \times 8.31\times 142}{4.00}}$

$\frac{3 \times 8.31\times T}{20.1797} = \frac{3 \times 8.31\times 142}{4}$

$T = \frac{20.1797 \times 142}{ 4} = 716.379 K \approx 716.4 K$