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

(I) Calculate of `K_(p).`
`{:(,PCl_(5)(g),hArr,PCl_(3)(g),+,C_(2)(g)),("Initial conc": ,1,,0,,0),("Eqm. conc" : ,1-alpha,, alpha,,alpha):}`
Total number of moles in the equilibrium mixture `=1-alpha+alpha+alpha=(1+alpha) " mol ".`
Let the total pressure of equilibrium mixture =P atm
`:." Partial pressure of " PCl_(5) i.e. P_(PCl_5) = ((1-alpha))/((1+alpha))xx P" atm ".`
partial pressure of `PCl_(3) i.e. P_(PCl_3) = (alpha)/((1+alpha)) xx P " atm ".`
Partial pressure of `Cl_(2) i.e. P_(PCl_2) = (alpha)/((1+alpha)) xx P " atm "`
`K_(c) = (P_(PCl_3)xx P_(Cl_(2)))/(P_(PCl_(5)))=(((alpha)/(1+alpha)P atm)xx((alpha)/(1+alpha)P atm))/(((1-alpha)/(1+alpha) P atm))=(alpha^(2)P)/(1-alpha)atm`
`P= 4 " atm and " alpha = 10% = (10.0)/(100)=0.1`
`K_(P) =((0.1)xx(0.1)xx(4 " atm "))/(1=(0.1)^(2)) = (0.04)/(0.99) =0.04 " atm "`
(II) Calculation of P.
Calculation P under new conditions
`alpha =0.2 K_(p) =0.04 " atm "`
`K_(p) = (alpha^(2)P)/(1-alpha^(2)) or P = (K_(p)(1-alpha^(2)))/(alpha^(2))`
`P=((0.04 " atm " ) xx (1-(0.2)^(2)))/((0.2)^(2))= ((0.04 " atm ")xx(0.96))/(0.04)=0.96 " atm ".`