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14 December, 19:46

An expandable container of oxygen has a volume of 30.0mL at a pressure of 36.7psi. If the pressure of the oxygen is reduced to 25.0psi what would the new volume of the contain be

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Answers (2)
  1. 14 December, 21:04
    0
    44.04 mL

    Explanation:

    From Boyle's law,

    PV = P'V' ... Equation 1

    Where P = initial pressure, V = initial volume, P' = Final pressure, V' = final volume.

    make V' the subject of the equation

    V' = PV/P' ... Equation 2

    Given: P = 36.7 psi, V = = 30 mL = (30/1000) L = 0.03 L, P' = 25 psi.

    Substitute into equation 2

    V' = 36.7 (0.03) / 25

    V' = 1.101/25

    V' = 0.04404 L

    V' = 44.04 mL.

    Hence the new volume = 44.04 mL
  2. 14 December, 23:35
    0
    Answer:44.04mL

    Explanation:Parameters given

    V1 = 30.0mL

    P1 = 36.7psi

    P2 = 25.0psi

    V2 = ?

    From Boyle's gas law, which states that "the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature"

    This means that,

    the pressure of a gas tends to increase as the volume of the container decreases, and also the pressure of a gas tends to decrease as the volume of the container increases.

    Mathematically, Boyle's can be represented as shown below

    P = k/V

    Where P = Pressure, V = Volume and k is constant

    Therefore,

    PV = k

    P1V1 = P2V2 = PnVn

    Using the formula

    P1V1 = P2V2

    V2 = P1V1/P2

    V2 = (36.7psi * 30.0mL) / 25.0psi

    V2 = 1101.0/25.0

    V2 = 44.04mL
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