We can use ideal gas law here assuming both gases act as ideal gasses.
First we should find the moles of each gas before mixing.
`PV = nRT`
`n = (PV)/(RT)`
Assume the temperature is T in Kelvins.
For gas A;
`n_A = (1xx2.5)/(0.08206T) = 30.466/T`
For gas B;
`n_B = (0.75xx0.5)/(0.08206T) = 4.569/T`
`A+3B rarr AB_3`
To complete the reaction we need 3times B than A. But as you can see above it doen't have that much of B in the flask to react with all the A.
So what will happen is all the B will be reacted.
Amount of B reacted `= 4.569/T`
Amount of A reacted `= 1/3xx(4.569/T) = 1.523/T`
So the rest A in flask `= 30.466/T-1.523/T = 28.477/T`
Amount of `AB_3` formed = reacted amount of A
Amount of `AB_3` formed `= 1.523/T`
So in the final mixture we have excess A and produced AB3.
Total moles in the mixture `= (1.523/T+28.477/T)`
Volume of flask `= (2.5+0.5) = 3L`
Using PV = nRT
`P = (nRT)/V`
`P = (1.523/T+28.477/T)xx0.08206xxT/3 = 0.8206atm`
So the total pressure of the system after joining the flasks is 0.8206atm.