At `20^(@)C`, two balloons of equal volume and porosity are filled to a pressure of `2atm`, one with `14 kg N_(2)` and the other with `1 Kg` of `H_(2)`. The `N_(2)` balloon leaks to a pressure of `1//2 atm` in `1 hour`. How long will it take for the `H_(2)` balloon to reach a pressure of `1//2 atm`?

Explanation

At constant `V` and `T` for a gas `P prop W`
Thus, for `N_(2)`
`P_(1)=2 atm`
`P_(2)=(1)/(2)atm`
At `t=1 h`, `W_(1)=14Kg`, `W_(2)=?`
`:. (P_(1))/(P_(2))=(W_(1))/(W_(2))`
`implies (2)/((1)/(2))=(14)/(W_(2))`
`implies W_(2)=(14)/(4)kg N_(2)`
Weight of `N_(2)` diffused `=14-(14)/(4)=(42)/(4)=(21)/(2)kg`
Similarly, for `H_(2):P_(1)=2atm`, `P_(2)=(1)/(2)atm`, at `t=th`
`W_(1)=1kg W_(2)=?`
`:. (P_(1))/(P_(2))=(W_(1))/(W_(2))`
`(2)/(1//2)=(1)/(W_(2))`, `implies W_(2)=(1)/(4)kg`
Now, `(r_(N_(2))/(r_(H_(2))=sqrt((M_(H_(2))/(M_(N_(2)))))` for duffison of `N_(2)` and `H_(2)`
Now, `(w_(N_(2))/(w_(H_(2))xx(t_(H_(2))/(t_(N_(2))=sqrt((M_(H_(2))/(M_(N_(2)))))`
`(21//2)/(3//4)xx(t_(H_(2)))/(60)=sqrt(((2)/(28)))`
`t_(N_(2))=1h=60m`
`t_(H_(2))=1.145 min`