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

`{:("",PCI_5,hArr,PCI_3,+,CI_2),("Initial mole",1,"",0,"",0),("Mole after dissociation",(1-alpha),"",alpha,"",alpha):}`
`alpha=0.1` at 4 atmospheric pressure.
`K_p=(n_(PCI_5)xxn_(CI_2))/(n_PCI_5)xx[E/sumn]=(alpha.alpha)/((1-alpha))[P/(1+alpha)]^1`
`=(Palpha^2)/(1-alpha^2)=(4xx(0.1)^2)/(1-(0.1)^2)=0.040 atm`
When `alpha` is desired at 0.2 `K_p` remains constant
`K_p=(Palpha^2)/((1-alpha^2))rArr0.040 =(Pxx(0.2)^2)/(1-(0.2)^2)`
`therefore P= 0.96 atm`