The reaction will be given as:

PCl_{5} ⇌ PCl_{3} Cl_{2}

**Initial moles:**

PCl_{5} = 0.10

PCl_{3} = 0

Cl_{2} = 0

**Equivalent moles:**

PCl_{5} = 0.10 – α

PCl_{3} = α

Cl_{2} = α

Total number of moles = 0.10 – α + α + α = 0.10 + α

Now, we know that pV = nRT

And it is given that V = 8.0 litre, p = 1 atm, T = 540 K and R = 0.082

Therefore,

p= nRT / V

1 = (0.10 + α) X 0.082 X 540 / 8.0

(0.10 + α) = 8 X 1 / 0.082 X 540 = 0.180

Therefore, α = 0.180 – 0.10 = 0.8

Now

K_{p} = p (PCl_{3}) X p (Cl_{2}) / p (PCl_{5})

p (PCl_{3}) = α p / 0.1 + α

p (Cl_{2}) = α p / 0.1 + α

p (PCl_{5}) = (0.1 – α) p / 0.1 + α

Therefore,

K_{p} = (.008) ^{2} X 1 / (0.1 + 0.08) X (0.1 – 0.08)

= 0.0064 / 0.18 X 0.02

K_{p} = 1.78 atm

Relation between K_{p} and K_{c} is given as:

K_{p} = K_{c} (RT) ^{∆n}

Therefore,

K_{c} = K_{p} / (RT) ^{∆n}

∆n = 1 + 1 – 1 = 1

Hence,

K_{c} = 1.78 / (0.082 X 540) ^{1}

= 0.042 mol litre^{-1}

**Hence the value of K _{p} = 1.78 atm and K_{c} = 0.042 mol litre^{-1}**