Question

If gas A (particle mass 46 g/mol) effuses with an average speed of 515 m/s, find the rate of effusion of gas B (particle mass = 92g/mol)

Answer 1

The rate of effusion of gas B is 728.32 m/s

Step-by-step explanation:

Given:

Mass of A, m₁ = 46 g/mol

Rate of effusion of A, R₁ = 515 m/s

Mass of B, m₂ = 92 g/mol

Rate of effusion of B, R₂ = ?

We know:

`(R_1)/(R_2) = \sqrt{(M_2)/(M_1) }`

Substituting the value we get:

`(515)/(R_2) = \sqrt{(92)/(46) } \\\\(515)/(R_2) = √(2) \\\\R_2 = (515)/(√(2) ) \\\\R_2 = 728.32 m/s`

Therefore, the rate of effusion of gas B is 728.32 m/s