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)), (" Atm eqm. " ,1-0*1=0*9"mole",,0*1 "mole",,0*1"mole"):}`
Total no. of moles ` = 0*9 + 0*1 + 0*1 = 1 * 1 "mole " `
` p_(PCl_(5))= (0*9)/(1 *1) xx 4"atm " , p_(PCl_(3)) = (0*1)/(1*1) xx 4 " atm " , p_(Cl_(2)) = (0*1)/(1.1) xx 4 " atm " `
` :. K_(p) = (p_(PCl_(3)) xx p_(Cl_(2)))/ (p_(PCl_(5))) = ((0*4)/(1*1) xx (0*4)/(1*1))/((0*9 xx 4)/(1*1) )=0*0404 `
2nd case When `PCl_(5)` is 20% dissociated . Suppose total pressure = P atm . Then
`{: (,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Intial " ,1 "mole",,,,),("At equilibrium " ,1-0*2=0*8,,0*2,,0*2 "mole"),(,,,,," Total no. of moles"= 0*8 + 0*2 + 0*2 = 1*2 "moles"):}`
` p_(PCl_(5)) = (0*8)/(1*2) xxP atm , p_(PCl_(3)) = (0*2)/(1*2) xx P atm , p_(Cl_(2))= (0*2)/(1*2) xx P atm `
` K_(p) = ((0*2 P)/(1*2) xx(0*2)/(1*2))/(( 0*8P)/(1*2))= (0*2)/(1*2) xx (0*2)/(0*8) P = 0*0404 ("calculated above")`
which gives `P= 0* 97 ` atm