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

Question

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?

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Orla Orla 3 years 2021-09-05T07:30:48+00:00 1 Answers 0 views 0

Answers ( )

  1. Euphemia
    0
    2021-09-05T07:32:19+00:00
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    Answer:

    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

    \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \\\\P_2 = \frac{P_1V_1T_2}{T_1V_2} \\\\P_2 = \frac{1*7.2*293}{3* 6.529}\\\\P_2 = 107.7 \ atm

    Therefore, the final pressure inside the container is 107.7 atm

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