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27 April, 08:59

A sample of neon occupies a volume of 461 mL at 1 atm and 273 K. What will be the volume of the neon when the pressure is reduced to 0.92 atm? (0.501 L)

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  1. 27 April, 09:59
    0
    0.501 L

    Explanation:

    To solve this problem we will use Boyle, s Law. According to this law "The volume of given amount of gas is inversely proportional to applied pressure at constant temperature".

    V∝ 1/P

    V = K/P

    VP=K

    Here the K is proportionality constant.

    so,

    P1V1 = P2V2

    P = pressure

    V = volume

    Given dа ta:

    P1 = 1 atm

    V1 = 461 mL

    P2 = 0.92 atm

    V2=? (L)

    To solve this problem we have to convert the mL into L first.

    1 L = 1000 mL

    461/1000 = 0.461 L

    Now we will put the values in the equation,

    P1V1 = P2V2

    V2 = P1V1 / P2

    V2 = 1 atm * 0.461 L / 0.92 atm

    V2 = 0.501 L
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