A closed container with volume of 7.20 L holds 10.9 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?

The final pressure inside the container is 107.7 atm

Explanation:

Given;

volume of the closed container =  7.20 L

mass of liquid helium in the container = 10.9 g

initial temperature of the container = 3.00 K

initial pressure of the container = 1.00 atm.

final temperature of the container = 293 K

Number of moles of He in the container = 10.9 / 4 = 2.725 moles

At 273˚K and 1 atm pressure (STP), the volume of 1 mole of a gas = 22.4 L

Volume of 2.725 moles of He at STP = 2.725 x  22.4 = 61.04 L

Volume of 2.725 moles of He at 3 K and 1 atm = (3/273) x 61.04 = 0.671 L

Thus, volume of He = 0.671 L

Volume of air in the container = 7.2 L – 0.671 L = 6.529 L

Since, the liquid helium evaporated, we only have air left in the container,

thus, final volume of the container, V₂ =  6.529 L

Initial volume of the container, V₁ = 7.20 L

Initial pressure of the in the container, P₁ = 1 atm

initial temperature of the container, T₁ = 3 K

Final temperature of the container, T₂ = 293 K

Final pressure of the container, P₂ = ?

Apply general gas law, in order to estimate P₂

Therefore, the final pressure inside the container is 107.7 atm