A closed container with volume of 4.60 L holds 10.7 g of liquid helium at 3.00 K and enough air to fill the re

A closed container with volume of 4.60 L holds 10.7 g of liquid helium at 3.00 K and enough air to fill the rest of its volume at a pressure of 1.00 atm. The helium then evaporates and the container warms to room temperature (293 K). What is the final pressure inside the container?

Do I solve for it at 3K then at 293K? Or use the fact that:

Pi*Vi/Ti = Pf*Vf/Tf
then the V cancel because they don't change
Pi/Ti=Pf/Tf
so 1atm/3K = Pf/293K
then solve for Pfinal?
I get the wrong answer, so are my units wrong?

The part we're missing here is the temperature of the air at the begining of the process. I have to assume it's at 293°K, because if it's the same temp as the He, it will be a liquid......
In that case, the initial volume of the air is ≈ 4.6 L.

10.7 g of LHe = 10.7/4 = 2.675 GMW = 2.675 GMV @ 273°K = 59.92 L @ 273 and 59.92*[293/273] =

64.31 L @ 293°K
Since there is already ≈ 4.6 L of air inside, the total gas volume @ std pressure = 64.31+4.6 = 68.91 L

Ergo, with 68.91 L confined to a space of 4.60 L, the final pressure is Patm*[68.91/4.60] = 14.98*Patm

Moles of He = 10.7 ÷ 4 = 2.675
At 273˚K and 1 atm pressure, the volume of 1 mole of a gas = 22.4 liters
Volume of 2.675 moles of He at STP = 2.675 * 22.4 = 59.92 liters
Volume of 2.675 moles of He at 293˚K = 293/273 * 59.92 = 64.3 liters

Let’s determine the volume of 2.675 moles of He at 3˚K and 1 atm pressure.
59.92/273 = V/3
Volume of He = 0.658 liters
Initial volume of air in the container = 4.6 – 0.658 = 3.942 Liters

Final pressure = initial pressure * (59.92 m+ 3.942) ÷ 4.6
Final pressure = 1 * (59.92 m+ 3.942) ÷ 4.6 = 13.88 atm