2.69 g of a sample of `PCl_(5)` was placed in a 1 litre flask and completely vaporised to a temperature of `250^(@)C`. The pressure observed at this temperature was 1 atm. The possibility exists that some of `PCl_(5)` may have dissociated according to the equation.
`PCl_(5)(g) hArr PCl_(3)(g) + Cl_(2)(g)`.
What are partial pressure of `PCl_(5), PCl_(3)` and `Cl_(2)` under these experimental conditions ?

Explanation

Let us first calculate the pressure supposing `PCl_(5)` does not underho dissociation.
pV = nRT
`p xx 1 = (2.69)/(208) xx 0.082 xx 523`
[mol. wt. of `PCl_(5)` = 208, R = 0.082 lit. atm/K/mole, T - (273 + 250)K]
p = 0.553 atm.
But `PCl_(5)` undergoes dissociation in the following way,
`{:("a (say)",0,,0),(PCl_(5) hArr,PCl_(3) ,+,Cl_(2)):} (alpha -= "deg. of dissociation")`
Moles at eqb. : `a(1-alpha)`
`therefore` total no. of moles `= a(1-alpha) + a alpha + a alpha = a(1+alpha)`
`because` pressure of a gas is proportional to no. of moles
`therefore ("moles before diss.")/("moles after diss.") = (a)/(a(1+alpha)) = (0.553)/(1)`
or `alpha = 0.81`.
Partial pressure of `PCl_(5) = ("moles of " PCl_(5) " at eqb.")/("total moles") xx "total pressure"`
`= (a(1-alpha))/(a(1+alpha)) xx 1 = (1-0.81)/(1+0.81) = 0.104` atm.
Partial pressure of `PCl_(5)` = partial pressure of `Cl_(2)`
`= ("moles of " PCl_(3) " or " Cl_(2))/("total mole") xx "total pressure"`
`= (a alpha)/(a(1+alpha)) xx 1 = (0.81)/(1.81) = 0.447` atm.