`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

Case I `N_(2)O_(4)(g)hArr2NO_(2)(g)`
`{:("At Eq",,(1-x),,2x,),(,,"total moles" =1+x,,):}`
`p_(N_(2)O_(4))=((1-x))/((1+x))xxP, p_(NO_(2))=(2x)/((1+x))xxP`
`K_(p)=(((2x)/(1+x).P)^(2))/(((1-x)/(1+x).P))=(4x^(2)P)/((1-x^(2)))`
The `%` dissociation of `N_(2)O_(4)=50.3%` (given) Hence, degree of dissociation `=50.3/100=0.503` and `P=1`
`:. K_(P)=(4xx(0.503)^(2)xx1)/([1-(0.503)^(2)])`
`rArr K_(p)=1.3548` atm
Case II
`N_(2)O_(4)hArr2NO_(2)`
`{:((1-x),(2x)):}`
Given, `((1-x))/(2x)=1/8`
`x=0.8`
Let the new pressure be `P` atm.
`K_(p)=(4x^(2)P)/((1-x^(2)))=(4xx0.8xx0.8xxP)/((1+0.8)(1-0.8))=1.3548`
`P=0.19 "atm"`