At some temperature and under a pressure of `4` atm, `PCl_(5)` is `10%` dissociated. Calculated the pressure at which `PCl_(5)` will be `20%` dissociated temperature remaining same.

Explanation

`{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),(""moles" of start",1,,0,,0),(""moles" at equilibrium",1-alpha,,alpha,,alpha):}`
`:.` Total "moles" at equilibrium `=1-alpha+alpha+alpha`
`=1+alpha`
Let the pressure of `PCl_(5)=p "atm"`
`:. p_(PCl_(3))=alpha/(1+alpha)xxP, p_(Cl_(2))=alpha/(1+alpha)xxP`
`p_(PCl_(5))=(1-alpha)/(1+alpha)xxP`
Substituting the values in the relation
`K_(p)=([p_(Cl_(3))][p_(Cl_(2))])/([p_(PCl_(5))])=([alpha/(1+alpha)]P[alpha/(1+alpha)]P)/([(1-alpha)/(1+alpha)]P)=(alpha^(2)P)/((1-alpha^(2)))`
When `P=4` "atm", `alpha=10//100=0.1`
`:. K_(p)=(0.1xx0.1xx4)/(1-(0.1)^(2))=0.04/0.99=4/99=0.04` "atm"
Calculation of pressure when `alpha=0.2`
Here `K_(p)=0.04` "atm" (since temperature is same)
`:.` Substituting the value in the relation,
`K_(p)=(a^(2)P')/((1-a^(2)))`
`0.04=(0.2xx0.2xxP')/(1-(0.2)^(2)) :. P'=(0.04xx0.96)/0.04=0.96 "atm"`