Question

the rate of effusion of co2 gas through a porous barrier is observed to be 9.31x 10^-4 mol/h. under the same conditions, the rate of effusion of N2 gas would be _______ mol/h

Answer 1

`1.167* 10^(-3)\ \text{mol/h}`

Step-by-step explanation:

`R_1` = Rate of diffusion of
`CO_2` =
`9.31* 10^(-4)\ \text{mol/h}`

`R_2` = Rate of diffusion of
`N_2`

`M_1` = Molar mass of
`CO_2` = 44.01 g/mol

`M_2` = Molar mass of
`N_2` = 28.0134 g/mol

From Graham's law we have the relation

`(R_1)/(R_2)=\sqrt{(M_2)/(M_1)}\\\Rightarrow R_2=\frac{R_1}{\sqrt{(M_2)/(M_1)}}\\\Rightarrow R_2=\frac{9.31* 10^(-4)}{\sqrt{(28.0134)/(44.01)}}\\\Rightarrow R_2=1.167* 10^(-3)\ \text{mol/h}`

The rate of effusion of the
`N_2` gas would be
`1.167* 10^(-3)\ \text{mol/h}`.