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23 February, 14:05

Air trapped in a cylinder fitted with a piston occupies 145 mL at 1.08 atm

pressure. What is the new pressure of air when the volume is decreased to 111 L?

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  1. 23 February, 14:27
    0
    Answer: 0.0014 atm

    Explanation:

    Given that,

    Original pressure of air (P1) = 1.08 atm

    Original volume of air (T1) = 145mL

    [Convert 145mL to liters

    If 1000mL = 1l

    145mL = 145/1000 = 0.145L]

    New volume of air (V2) = 111L

    New pressure of air (P2) = ?

    Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law

    P1V1 = P2V2

    1.08 atm x 0.145L = P2 x 111L

    0.1566 atm•L = 111L•P2

    Divide both sides by 111L

    0.1566 atm•L/111L = 111L•P2/111L

    0.0014 atm = P2

    Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
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