Gas Laws problem...Help

A(n) 0.65 mole quantity of O2 originally at 85◦C is cooled such that it now occupies 4.0 L at 20◦C. What is the final pressure exerted by the gas?
1. 0.92 atm
2. 3.9 atm
3. 1.1 atm
4. 0.27 atm
5. 6.2 atm
6. Not enough information is given

Using the ideal gas law: PV=nRT

P= pressure in atm (unknown)
V= volume in liters (4.0 L)
n = # of moles of O2 (0.65 mol)
R= ideal gas constant (0.08206 L*atm / mol*K)
T= Temperature in KELVIN ( They give you 20 C, but that equals 293 K)

P(4.0)= (0.65)(0.08206)(293)

P= 3.907 RT 3.9 atm

So the correct answer is 2.

Use ideal gas law: PV = NRT

(They do the whole changing temperature in order to confuse you into thinking it is a different type of problem. This is just an ideal gas law problem)

You are given everything except P, so plug in what you know and solve. They gave you 2 temperatures, and each will result in one of those answers, pick the correct one.

Hints:
Double check units to make sure they work. With wrong units, I bet the result is still one of the options.
R = 0.0821 L-atm / mol-K

Answer:

T = 20C + 273 = 293 K

P (4 L) = (0.65 mol) * (0.0821 L-atm/mol-K) * (293 K)

(4 L) * P = 15.6395 L-atm

P = 3.9 atm