Here's how you'd go about solving this problem.
Start with the ICE table for the equilibrium reaction
" "N_2O_(4(g)) rightleftharpoons 2NO_(2(g)) N2O4(g)⇌2NO2(g)
I..........n_0n0..................0
C.......-epsilon−ε..............+2epsilon+2ε
E...n_0-epsilonn0−ε..............2epsilon2ε
Express the mole fraction of the two species like this
chi_(N_2O_4) = (n_0 - epsilon)/(n_0 - epsilon + 2epsilon) = (n_0 - epsilon)/(n_0 + epsilon)χN2O4=n0−εn0−ε+2ε=n0−εn0+ε
chi_(NO_2) = (2epsilon)/(n_0 + epsilon)χNO2=2εn0+ε
Now just use the definition of K_pKp to solve for epsilonε
K_p = P_(NO_2)^(2)/P_(N_2O_4) = ((2epsilon)^(2)/(n_0 + epsilon)^(2) * P^(2))/((n_0 - epsilon)/(n_0 + epsilon) * P)Kp=P2NO2PN2O4=(2ε)2(n0+ε)2⋅P2n0−εn0+ε⋅P
K_p = ((2epsilon)^(2) * P^(cancel(2)))/((n_0 + epsilon)^(cancel(2))) * cancel(n_o + epsilon)/((n_0 - epsilon) * cancel(P))
K_p = (4epsilon^(2) * P)/((n_o - epsilon)(n_o + epsilon)) = (4epsilon^(2) * P)/(n_0^(2) - epsilon^(2))
K_p(n_0^(2) - epsilon^(2)) = 4epsilon^(2) P
4epsilon^(2)P + epsilon^(2)K_p - n_0^(2)K_p = 0
epsilon^(2)(4P + K_p) = n_0^(2) K_p | : n_0^(2)
epsilon^(2)/n_0^(2) = K_p/((4P + K_p)) |sqrt
color(green)(epsilon/n_0 = (K_p/(4P + K_p))^(1/2))