0.1 mole of `N_(2O_(4)(g)` was sealed in a tude under one atmospheric conditions at `25^(@)C` Calculate the number of moles of `NO_(2)(g)` preesent , if the equilibrium `N_(2)O_(4)(g)hArr2NO_(2)(g)(K_(P)=0.14)` is reached after some time :

Explanation

`{:(N_2O_4 hArr , 2NO_2),(0.1,0),((0.1-alpha), 2alpha):}`
`K_P=((2alpha)^2)/((0.1-alpha))xx[P/(0.1+alpha)]'` or `K_P=(40alpha^2)/((0.1-alpha)) " " [ {:(? P prop 0.1, therefore 1 prop 0.1),("(If V and T are constant)", P prop 0.1+alpha) :}]`
or `(40alpha^2)/((0.1-alpha))=0.14, alpha =0.017` , moles of `NO_2=0.017xx2` = 0.034 mole