A `100 dm^(3)` flask contains `10 mol` each of `N_(2)` and `H_(2)` at `700 K`. After equilibrium was reached, partial pressure of `H_(2)` was `1 atm`. At this point, `5 L` of `H_(2)O(l)` was injected and gas mixture was cooled to `298 K`. Find out the gas pressure.

Explanation

Now, at equilibrium
Moles of `H_(2)=(P_(H_(2))V)/(RT)=(1xx100)/(0.821xx700)=1.74`
`:. 10-3x=1.74 implies x=2.75`
`implies n_(N_(2))=10-x=7.25`, `n_(H_(2))=1.74`, and `n_(NH_(3))=2x=5.5`
After `H_(2)O(l)` is injected and gas mixture is colled to `298 K`, `NH_(3)(g)` will go to aqueous state.
`NH_(3)(g)+H_(2)O(l)` `NH_(4)OH(l)`
Volume of the flask available `=100-5=95L`
Therefore, Total pressure `=(n_(total)RT)/(V)`
`=((1.74xx7.25)xx0.0821xx298)/(95)`
`=2.31 atm`