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)

Respuesta :

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

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:

[tex]\frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1} }[/tex]

Substituting the value we get:

[tex]\frac{515}{R_2} = \sqrt{\frac{92}{46} } \\\\\frac{515}{R_2} = \sqrt{2} \\\\R_2 = \frac{515}{\sqrt{2} } \\\\R_2 = 728.32 m/s[/tex]

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