Under similar conditions of temperature and pressures, if the time taken for effusion of the same volume of `H_(2),N_(2)` and `CO_(2)` gas through the same porour wall are `t_(1),t_(2) and t_(3)` respectively, then arrange `t_(1),t_(2) and t_(3)` in their increasing order.

Explanation

At constant temperature and pressure, if V volume of `H_(2)` gas effuses in time `t_(1)`, then according to Graham's law,
`(V)/(t_(1)) prop (1)/(sqrt(M_(H_(2))))` . . [1]
For `N_(2)` gas and `CO_(2)` gas the equations are respectively,
`(V)/(t_(2)) prop(1)/(sqrt(M_(N_(2)))) ` . . .[2]
`(V)/(t_(3))prop(1)/(sqrt(M_(CO_(2))))` . . . [3]
[`because` Volume of gas effused in each case in the same]
from equation [1] & [2] from equation [2] & [3] we get,
`(t_(2))/(t_(1))=sqrt((M_(N_(2)))/(M_(H_(2)))),(t_(3))/(t_(2))=sqrt((M_(CO_(2)))/(M_(N_(2))))`
Since, `M_(H_(2)) lt M_(N_(2)) lt M_(CO_(2))`, therefore, `t_(2) gt t_(1) and t_(3) gt t_(2)`
Hence, `t_(1),t_(2)` and `t_(3)` will be in order: `t_(1) lt t_(2) lt t_(3)`.