With air inside, assuming for simplicity constant volume and ideal gas, its pressure variation from 50F to 212F can be computed from T in K: P_{2}/P_{1} = T_{2}/T_{1} = 373.15/283.15 = 1.32. Since P1 = 1.04 atm we have: P_{2} = 1.32*1.04 = 1.37 atm so the variation is 1.37-1.04 = 0.33 atm

Adding the vapour pressure, at 100°C the pressure inside is 1.37 + 1.00 = 2.37 atm.

This doesn't consider however the volume variation. Since some liquid water becomes vapour, the volume available for the gases increases and this reduces the air pressure, so the final result would be less than 2.37 atm.