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.

• 235UF₆ molar mass = 349.034348 g/mol

• 238UF₆ molar mass = 352.041206 g/mol

Answer 1

Using Graham's law of effusion, we can calculate that 235UF₆ diffuses approximately 0.4% faster than 238UF₆ by taking the square root of the inverse ratio of their molar masses and converting that ratio to a percentage.

Step-by-step explanation:

To demonstrate why 235UF₆ diffuses 0.4% faster than 238UF₆, we can use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The calculation is as follows:

Let r1 and r2 be the rates of effusion for 235UF₆ and 238UF₆, respectively, and M1 and M2 their molar masses.

According to Graham's law:

r1/r2 = √(M2/M1)

Substitute the given molar masses:

r1/r2 = √(352.041206 g/mol / 349.034348 g/mol)

r1/r2 = √(1.008594)

r1/r2 = 1.004292

To find how much faster 235UF₆ diffuses, we subtract 1 from the ratio and then multiply by 100 to get the percentage:

((1.004292 - 1) × 100) = 0.4292%

Round this value to one decimal place gives us approximately 0.4%. Hence, 235UF₆ diffuses approximately 0.4% faster than 238UF₆.