To find the equilibrium pressure of the system after the partition is removed, we can apply the principle of conservation of moles (or the ideal gas law in combined volumes).

Initially, we have:

Chamber 1: $ P_1 = 1 \, \text{atm}, \, V_1 = 2 \, \text{L} $

Chamber 2: $ P_2 = 2 \, \text{atm}, \, V_2 = 3 \, \text{L} $

The total pressure after the partition is removed and the gases mix can be calculated using:

$ P_{\text{final}} = \frac{P_1 V_1 + P_2 V_2}{V_1 + V_2} $

Substituting the given values:

$ P_{\text{final}} = \frac{1 \times 2 + 2 \times 3}{2 + 3} = \frac{2 + 6}{5} = \frac{8}{5} $

This gives us:

$ P_{\text{final}} = 1.6 \, \text{atm} $

Therefore, the equilibrium pressure when the gases mix is $ 1.6 \, \text{atm} $.