A vessel has `N_(2)` gas and water vapours at a total pressure of `1 atm`. The partial pressure of water vapours is `0.3atm`. The contents of this vessel are transferred to another vessel having one-third of the capacity of original volume, completely at the same temperature the total pressure of this system in the new vessel is

Answer

`2.4` atm

Explanation

`P_(N_(2))+P_(H_(2)O_((V))=1` atm, `becauseP_(H_(2)O)=0.3` atm ltbr. `thereforeP_(N_(2))=0.7` atm
Now new pressure of `N_(2)` in another vessel of volue `V//3` at same `T` is given by:
`P_(N_(2))=2.1` atm
Since aqueous tension remains constant, and thus total pressure in new vessel.
`=P_(N_(2))+P_(H_(2)O)=2.1+0.3=2.4` atm.