Question

During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. Show the calculation that supports this value. The molar mass of 235UF6 = 235.043930 + 6 x 18.998403 = 349.034348 g/mol, and the molar mass of 238UF6 = 238.050788 + 6 x 18.998403 = 352.041206 g/mol.

Answer 1

Answer: The below calculations proves that the rate of diffusion of
`^(235)UF_6` is 0.4 % faster than the rate of diffusion of
`^(238)UF_6`

Step-by-step explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

`\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}`

We are given:

Molar mass of
`^(235)UF_6=349.034348g/mol`

Molar mass of
`^(238)UF_6=352.041206g/mol`

By taking their ratio, we get:

`(Rate_((^(235)UF_6)))/(Rate_((^(238)UF_6)))=\sqrt{(M_((^(238)UF_6)))/(M_((^(235)UF_6)))}`

`(Rate_((^(235)UF_6)))/(Rate_((^(238)UF_6)))=\sqrt{(352.041206)/(349.034348)}\\\\(Rate_((^(235)UF_6)))/(Rate_((^(238)UF_6)))=(1.00429816)/(1)`

From the above relation, it is clear that rate of effusion of
`^(235)UF_6` is faster than
`^(238)UF_6`

Difference in the rate of both the gases,
`Rate_((^(235)UF_6))-Rate_((^(238)UF_6))=1.00429816-1=0.00429816`

To calculate the percentage increase in the rate, we use the equation:

`\%\text{ increase}=(\Delta R)/(Rate_((^(235)UF_6)))* 100`

Putting values in above equation, we get:

`\%\text{ increase}=(0.00429816)/(1.00429816)* 100\\\\\%\text{ increase}=0.4\%`

The above calculations proves that the rate of diffusion of
`^(235)UF_6` is 0.4 % faster than the rate of diffusion of
`^(238)UF_6`