A vessel contains `14` g (`7` moles ) of hydrogen and `96` g (`3`moles) of oxygen at `STP`. Chemical reaction is induced by passing electric spark in the vessel till one of the gases is consumed. The temperature is brought back to it's starting value `273 K` .The pressure in the vessel is-

Explanation

Step I. Calculation of volume of the container.
No. of moles of `H_(2) = (14 g)/((2 g mol^(-1))) = mol`
No. of moles of `O_(2) = (96 g)/((32 gmol^(-1))) = 3` mol
Total no. of moles in the container `= (7+3) = 10` mol
According to ideal gas equation, `PV = nRT` or `V = (nRT)/(P)`
Under S.T.P. conditions `V = ((10 mol) xx (0.083 dm^(3) "bar" K^(-1) mol^(-1)) xx (273 K))/((1 "bar"))`
`= 226.59 dm^(3)`
Volume of container (Volume of gaseous mixture enclosed) `= 226.59 dm^(3)`
Step II. Calcualation of pressure in the container
`{:(underset(2 mol)(2H_(2)(g)),+,underset(1 mol)(O_(2))(g)rarr2H_(2)O(l)):}`
1 mole of `O_(2)` react with `H_(2) = 2` mol
3 moles of `O_(2)` react with `H_(2) = 6` mol
No. of mass of `H_(2)` left in the container
`= (7-6) = 1` mol
(Neglect the volume of the water is in the liquid state)
According to ideal gas equation
`P' = (n'RT)/(V), P = (nRT)/(V)`
`:. (P')/(P) = (n')/(n) = 1/10` or `P' = (P)/(10) = ((1 "bar"))/(10) = 0.1` bar