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)),("Mole before dissociation",1,,0,,0),("Mole after dissociation",(1-alpha),,alpha,,alpha):}`
`K_(p)=(n_(PCl_(3))xxn_(Cl_(2)))/(n_(PCl_(5)))xx[(P)/(sumn)]^(Deltan)` .....(`1`)
At `P=4 atm` and `alpha=0.1`, By eq. (`1`),
`:. K_(p)=(0.1xx0.1)/(0.9)xx[(4)/(1.1)]^(1)=0.0404 atm`
At `alpha=0.2`,
`0.0404=(0.2xx0.2)/(0.8)xx[(P)/(1.2)]^(1)`
`:. P=0.969 atm`