Combined Gas Law?

When 0.600 L of Ar at 1.20 atm and 227 degrees celsius is mixed with 0.200 L of O2 at 501 torr and 127 degrees celsius in a 400 mL flask at 27 degrees celsius, what is the presssure in the flask?

I think I am completely WRONG on this one. Please help, thanks!

Answer:
Get the number of moles of each gas:

n = PV / RT

For Ar, n = (1.20atm*0.600L) / (0.082057L-atm/mol-K)*500K or 0.01755 mol.
For O2, n = (0.659atm*0.200L) / (0.082057L-atm/mol-K)*400K or 0.0040155 mol.
Then, get the total number of moles, 0.01755 mol + 0.0040155 mol = 0.021566 mol.
Use this total number of moles to get the final pressure in the flask after mixing.

P = nRT / V
= 0.021566mol* 0.082057L- atm/mol-K*300K / 0.4L
= 1.33 atm

the eq being PV = nRT
first we find the moles of both Ar and O2
moles of Ar, n = PV/RT
n= 1.2 x 0.6/R (227 +273)
n = 0.00144/R

moles of O2, n* = 501/760 x 0.2/R (127 +273)
n* = 0.0003296/R
Thus total no of moles = n + n*
= 0.00177/R

now using PV = nRT
P = 0.00177/R x R x (27 +273) / 0.4
P = 1.3275 atm

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