Liquid helium at 4.2 K has a density of 0.147 g/mL. Suppose that a 1.00-L metal bottle that contains air at 118K and 1.0 atm pressure is sealed off. If we inject 120.0 mL of liquid helium and allow the entire system to warm to room temperature (25 °C), what is the pressure inside the bottle?

Question

Add the partial pressures of the air and that of the vaporized helium.
120 mL of LHe has a mass of 0.147 g/ml x 120 mL = 17.64 g. That is = 4.406 moles. From that and the volume and temperature, you can calculate the partial pressure of He, using
P = nRT/V
R = 8.205*10^-2 atm*L/mole K
P(He)= 107.7 atm
The air partial pressure is much less. It starts out at 1.0 atm at 118K and rises to 1.0 x (298/118) = 2.5 atm at the higher temperature of 298 K.
Add the two partical pressures for the total pressure.
I don't guarantee my numbers, so check them yourself. The method should work.

thanks