Question

Uranium has two isotopes of masses 235 units and 238 units. If both of them are present in Uranium hexafluoride gas, which has a larger average speed? If the atomic mass of fluorine is 19 units, estimate the percentage difference in speeds at any temperature. How is the above concept of the difference in speeds used in the enrichment of uranium needed for the nuclear fission?

Answer 1

Hint: The average speed of a gas molecule at a particular temperature is dependent on the mass of the molecule. The mass of the molecule will be approximately the sum of the individual atoms that make up the molecule.

Complete step by step answer:
The average velocity of a gas is given by the formula, ${{v}_{avg}}=\sqrt{\dfrac{8RT}{\pi M}}$
Where,

R is the Universal Gas Constant
T is the temperature of the gas.
M is the mass per mole or the molar mass.

So, in our problem, the Uranium hexafluoride gas is at a constant temperature. So, T is the same for both the uranium isotopes gas molecules.

So the mass of the Uranium 235 hexafluoride gas molecule is,

${{M}_{1}}={{M}_{235}}+6\times {{M}_{F}}$
${{M}_{1}}=235units+6\times 19units$
$\therefore {{M}_{1}}=349units$

So the mass of the Uranium 238 hexafluoride gas molecule is,

${{M}_{2}}={{M}_{238}}+6\times {{M}_{F}}$
${{M}_{2}}=238units+6\times 19units$
$\therefore {{M}_{2}}=352units$


Let ${{v}_{1}}$ be the velocity of the Uranium 235 hexafluoride gas molecule. So, it can be written as,

${{v}_{1}}=\sqrt{\dfrac{8RT}{\pi {{M}_{1}}}}$ … equation (1)

Let ${{v}_{2}}$ be the velocity of the Uranium 238 hexafluoride gas molecule. So, it can be written as,

${{v}_{2}}=\sqrt{\dfrac{8RT}{\pi {{M}_{2}}}}$ …. equation (2)

Taking the ratio of equation (2) to equation (1), we get,

$\dfrac{{{v}_{2}}}{{{v}_{1}}}=\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}}$
$\dfrac{{{v}_{2}}}{{{v}_{1}}}=\sqrt{\dfrac{349}{352}}$

$\therefore \dfrac{{{v}_{2}}}{{{v}_{1}}}=0.99573$ … equation (3)

The percentage change in the speed can be written as,
$\text{Percentage Change}=\dfrac{\Delta v}{{{v}_{1}}}\times 100=\dfrac{{{v}_{1}}-{{v}_{2}}}{{{v}_{1}}}\times 100=\left( 1-\dfrac{{{v}_{2}}}{{{v}_{1}}} \right)\times 100$

From equation (3) we know the value of $\dfrac{{{v}_{2}}}{{{v}_{1}}}$, substituting the value in the equation above, we get,

$\text{Percentage Change}=\left( 1-0.9957 \right)\times 100$
$\therefore \text{Percentage Change}=0.427%$

So the difference in the average speeds of the two uranium molecules is about 0.427%.

This small velocity difference can be utilized to separate the Uranium 235 needed for nuclear fission from the Uranium 238. Gaseous diffusion enrichment is the method used to separate the two isotopes. It is based on the small mass difference between 235 $\text{U}{{\text{F}}_{\text{6}}}$ and 238 $\text{U}{{\text{F}}_{6}}$, which means that the 235 $\text{U}{{\text{F}}_{\text{6}}}$ travels slightly more quickly at a given temperature. $\text{U}{{\text{F}}_{\text{6}}}$ gas is inputted into a diffusion chamber, which contains a porous membrane and pressure is applied across the membrane to push the gas through it. The higher speed of the 235 $\text{U}{{\text{F}}_{\text{6}}}$ molecules means that they would hit the membrane more frequently than its isotope counterpart and are therefore more likely to pass through one of the pores. This way the radioactive abundant Uranium 235 is separated from the Uranium 238.

Note: Apart from the average velocity of the gas molecules, there are two other types of velocities called the
Most Probable Velocity of gas molecules given by,
${{v}_{m}}=\sqrt{\dfrac{2RT}{M}}$

Another one is the Root Mean Square Velocity of gas molecules, which is given by,
${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$

The physical behaviour of gases is explained by the kinetic molecular theory of gases.
The number of collisions that gas particles make with the walls of their container and the force at which they collide determine the magnitude of the gas pressure.