A mixture in which the mole ratio of `H_(2)` and `O_(2)` is `2:1` is used to prepare water by the reaction.
`2H_(2(g))+O_(2(g))rarr2H_(2)O_((g))`
The total pressure in the container is `0.8 atm` at `20^(@)C` before the reaction. Determine the final pressure at `120^(@)C` after reaction assuming `80%` yield of water.

Explanation

`p_(H_(2))=(2)/(3)xx0.8=0.533atma`
`P_(O_(2))=(1)/(3)xx0.8=0.266atm`

Total pressure =0.1066+0.4264=0.5863atm
Using Gay-Lussac's law`(P_(1))/(T_(1))=(P_(2))/(T_(2))`
`=(0.5863)/(293)=(P_(2))/(393)`
`P_(2)=0.7864atm`