Under similar conditions of temperature and pressure, the times it takes for the effusion of the same volume of `H_(2),N_(2) and O_(2)` gases through the same porous wall are `t_(1),t_(2) and t_(3)`, respectively. Arrange `t_(1),t_(2)` and `t_(3)` in order of their increasing values.

Explanation

Let the volume of effused gas be V. therefore, the rates of effusion of the three gases will be as follows:
For `H_(2):(V)/(t_(1)) prop(1)/(sqrt(M_(H_(2))))` . . . [1] For `N_(2):(V)/(t_(2))prop(1)/(sqrt(M_(N_(2))))` . . [2]
For `O_(2):(V)/(t_(3)) prop (1)/(sqrt(M_(O_(2))))` . . [3]
from [1] and [2] we get, `t_(2)//t_(1)=sqrt(M_(N_(2))//M_(H_(2)))` and from [2] and [3] we get, `t_(3)//t_(2)=sqrt(M_(O_(2))//M_(N_(2)))`.
Since, `M_(H_(2)) lt M_(N_(2)) lt M_(O_(2)), t_(2) gt t_(1) and t_(3) gt t_(2)`. thus,
`t_(1),t_(2)` and `t_(3)` will follow the order: `t_(1) lt t_(2) lt t_(3)`.