PCℓ5(g)  at an initial pressure of 1 atm is placed in a closed vessel containing chlorine at a pressure of 0.2 atm. If PCℓ5  dissociates to the extent of 20%, then the equillibrium constant Kp value is

PCℓ5(g)⇌PCℓ3(g)+Cℓ2(g) A) Partial pressure of PCℓ5=1−20100=0.8 atm B) Partial pressure of PCℓ3=20100=0.2   atm C) Partial pressure of Cℓ2=0.2+0.2 =0.4   atm D) KP  for dissociation of PCℓ5=PPCℓ3×PCℓ2PPCℓ5=0.2×0.40.8=0.1  atm