Ammonia dissociates into ${\mathrm{N}}_2$ and ${\mathrm{H}}_2$ such that degree of dissociation $\mathrm{\alpha }$ is much less than 1 and equilibrium pressure is ${\mathrm{P}}_0$ then the value of $\mathrm{\alpha }$ is, if $K_p$ for $2{\mathrm{NH}}_3\left(g\right)\rightleftharpoons {\mathrm{N}}_2\left(g\right)+3{\mathrm{H}}_2\left(g\right)$ is $27\times {10}^{-8}{P}_0^2\mathit{ }:$