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9 May, 18:33

A 13.0 L sample of helium gas has a pressure of 29.0 atm. What volume would this gas occupy at 3.50 atm? Assume ideal behavior.

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  1. 9 May, 18:57
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    From the equation;

    P1V1=P2V2

    V2=P1V1:P2

    V2=29.0*13.0:3.50

    V2=107.71L

    therefore V2=107.7L
  2. 9 May, 21:16
    0
    This uses an equation known as Boyle's law. Boyle's law states that at a constant temperature, the product of the pressure that a gas exerts and the volume it occupies is constant, meaning they are inversely proportional. This means the equation is:

    P1V1 = P2V2

    We have 3 knowns:

    P1 = 29.0 atm

    V1 = 13.0 L

    P2 = 3.50 atm

    V2 = unknown

    Plugging into our equation:

    (29.0 atm) * (13.0 L) = (3.50 atm) * (V2)

    V2 = (29.0 atm * 13.0 L) / (3.50 atm)

    V2 = 108 L = Volume occupied at 3.50 atm

    This makes sense, as when the pressure decreased, Boyle's Law predicts that the volume will increase to compensate for the lost pressure, and it did.
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