# The two isotopes of uranium, ""^238 U and ""^235 U, can be separated by effusion of the corresponding UF_6 gases. What is the ratio (in the form of a decimal) of the RMS speed of ""^238 UF_6 to that of ""^235 UF_6 at constant temperature?

Jul 31, 2017

I got about $0.9957$.

I know it doesn't give you a temperature to work with, but it doesn't really matter because you're at constant temperature.

The root-mean-square (RMS) speed of a freely-moving gas following Maxwell-Boltzmann statistics (in the absence of an external field) is given by:

${v}_{R M S} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 {k}_{B} T}{m}}$,

where $R$ is the universal gas constant, ${k}_{B}$ is the Boltzmann constant, $M$ is the molar mass in $\text{kg/mol}$, and $m$ is the mass in $\text{kg}$.

Use whichever version you want, and you would get the same result. I will use the second version. Thus:

$\left({v}_{R M S} \left({\text{^(238) UF_6))/(v_(RMS)(""^(235) UF_6)) = sqrt((3k_BT)/m_(""^238 UF_6))/sqrt((3k_BT)/m_(}}^{235} U {F}_{6}\right)\right)$

$= \sqrt{{m}_{{\text{^235 UF_6)/m_(}}^{238} U {F}_{6}}}$

You can also see that this follows Graham's law of effusion, i.e. the ratio of the gas speeds is equal to the ratio of their rates of effusion.

Since the masses are in a ratio, we can use arbitrary units for them, rather than $\text{kg}$, if we wanted. Remember to use the isotopic masses, i.e.

m_(""^238 U) = "238.05078826 amu"

m_(""^235 U) = "235.0439299 amu"

m_(""^19 F) = "18.99840322 amu"

Therefore, we get:

color(blue)((v_(RMS)(""^(238) UF_6))/(v_(RMS)(""^(235) UF_6))) = sqrt(("235.0439299 amu" + 6 xx "18.99840322 amu")/("238.05078826 amu" + 6 xx "18.99840322 amu"))

$= \textcolor{b l u e}{0.9957}$

So, the separation process is apparently very precise... ${\text{^238 "UF}}_{6} \left(g\right)$ is only 0.43% slower than ${\text{^235 "UF}}_{6} \left(g\right)$.