For the equilibrium reaction $$2*NO_2 \rightleftharpoons N_2O_4$$, the equilibrium constant is defined as $$K_p=\frac{P_{N_2O_4}}{(P_{NO_2})^2}$$. If initially $$P_{NO_2}=0.50\,atm$$ and $$P_{N_2O_4}=0.10\,atm$$ (so that $$K_p=0.10/(0.50)^2=0.40$$), what will be the new value of $$K_p$$ if the system is compressed so that the total pressure doubles while the mole ratio remains constant?