`N_(2)O_(4)` dissociates as
`N_(2)O_(4)(g)hArr2NO_(2)(g)`
At `40^(@)C` and one atmosphere `%` decomposition of `N_(2)O_(4)` is `50.3%`. At what pressure and same temperature, the equilibrium mixture has the ratio of `N_(2)O_(4): NO_(2)` as `1:8`?

Explanation

`{:(,N_(2)O_(4),hArr,2NO_(2)),("moles before equilibrium",1,,0),("moles at equilibrium",1-alpha,,2alpha),("where" alpha "is degree of dissociation =0.503",,,):}`
`K_(p)=(.^(n)NO_(2))^(2)/((.^(n)N_(2)O_(4)))xx(P/(Sigman))^(Deltan)=((2alpha)^(2))/((1-alpha))[P/((1+alpha))]`
`K_(p)=(4alpha^(2)P)/(1-alpha^(2))=(4xx(0.503)^(2)xx1)/(1-(0.503)^(2))=1.355`
`{:("Now,",,N_(2)O_(4),hArr,2NO_(2),,"(ar pressure P)"),(,,1,,1,,),(,,(1-x),,2x,,):}`
Given `(N_(2)O_(4))/(NO_(2))=(1-x)/(2x)=1/8, ( :. x=8//10=0.8)`
Using again at pressure `P`
`:. K_(p)=(4x^(2)xxP)/(1-x^(2))`
`1.355=(4xx(0.8)^(2)xxP)/(1-(0.8)^(2)) :. P=0.19 "atm"`