Value of K$$_P$$ for the equilibrium reaction N$$_2$$O$$_4$$(g) $$\rightleftharpoons$$ 2NO$$_2$$(g) at 288 K is 47.9. The K$$_C$$ for this reaction at same temperature is (Nearest integer)
(R = 0.083 L bar K$$^{-1}$$ mol$$^{-1}$$)

For the equilibrium $$\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$$, the relationship between $$K_P$$ and $$K_C$$ is given by $$K_P = K_C(RT)^{\Delta n}$$, where $$\Delta n$$ is the change in moles of gaseous species.

Here $$\Delta n = 2 - 1 = 1$$. Substituting the values: $$47.9 = K_C \times (0.083 \times 288)^1 = K_C \times 23.904$$.

Solving for $$K_C$$: $$K_C = \frac{47.9}{23.904} = 2.004$$.

Rounding to the nearest integer, the answer is $$2$$.