A container is divided into two chambers by a partition. The volume of first chamber is 4.5 litre and second chamber is 5.5 litre. The first chamber contain 3.0 moles of gas at pressure 2.0 atm and second chamber contain 4.0 moles of gas at pressure 3.0 atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is $$x \times 10^{-1}$$ atm. Value of $$x$$ (nearest integer) is ______

Initially, the first chamber has volume $$V_1 = 4.5$$ L at pressure $$P_1 = 2.0$$ atm with $$n_1 = 3.0$$ moles, and the second chamber has volume $$V_2 = 5.5$$ L at pressure $$P_2 = 3.0$$ atm with $$n_2 = 4.0$$ moles. When the partition is removed, the gases mix and reach a common equilibrium pressure in the total volume $$V = V_1 + V_2 = 4.5 + 5.5 = 10$$ L.

Since the temperature remains constant throughout the process, we can use the ideal gas law. For each chamber individually, $$P_1V_1 = n_1RT$$ and $$P_2V_2 = n_2RT$$. After mixing, the total number of moles is $$n = n_1 + n_2 = 3.0 + 4.0 = 7.0$$ moles, and the total volume is $$V = 10$$ L.

At equilibrium, $$PV = nRT = (n_1 + n_2)RT$$. Since $$n_1RT = P_1V_1$$ and $$n_2RT = P_2V_2$$, we get:

$$PV = P_1V_1 + P_2V_2$$

$$P = \frac{P_1V_1 + P_2V_2}{V} = \frac{2.0 \times 4.5 + 3.0 \times 5.5}{10} = \frac{9.0 + 16.5}{10} = \frac{25.5}{10} = 2.55$$ atm

Expressing this as $$x \times 10^{-1}$$ atm: $$2.55 = 25.5 \times 10^{-1}$$. Rounding to the nearest integer, $$x \approx \mathbf{26}$$.