The equation of the gaseuos decomposition raction of #PCl_5# with ICE table

#PCl_5(g)" "rightleftharpoons" "PCl_3(g)" "+" "Cl_2(g)#

#I" "1" "mol" "0" "mol" "0" "mol#

#C" "alpha" "mol" "alpha" "mol" "alpha" "mol#

#E" "1-alpha" "mol" "alpha" "mol" "alpha" "mol#

#color(red)("where "alpha" degree of dissociation"=10%=0.1#

*Total no. of moles in the reaction mixture at equilibrium is given by*

#n=1-alpha+alpha+alpha=1+alpha#

*Mole fractions of the components at equilibrium*

#chi_(PCl_5)=(1-alpha)/(1+alpha)#

#chi_(PCl_3)=alpha/(1+alpha)#

#chi_(Cl_2)=alpha/(1+alpha)#

*If atequilibrium the total pressure of the reaction mixture is P then the partial pressures of the components in the mixture will be*

#p_(PCl_5)=((1-alpha)P)/(1+alpha)#

#p_(PCl_3)=(alphaP)/(1+alpha)#

#p_(Cl_2)=(alphaP)/(1+alpha)#

#"The equilibrium constant in respect of preesure"#

#K_p=(p_(PCl_3)xxp_(Cl_2))/p_(PCl_5)#

Substituting respective values

#K_p= ((alphaP)/(1+alpha))^2 / ((( 1- alpha)P)/(1+alpha)) #

#=(alpha^2P)/(1-alpha^2)#

Now it is given #alpha=0.1 and P=5" atm"#

So

#K_p=((0.1)^2*5)/(1-(0.1)^2)~~0.05atm#