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21 August, 00:14

A simple of helium measuring 6L was kept at a pressure is of 1.5 atm, if pressure is doubled what would be its new volume?

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  1. 21 August, 01:31
    0
    3L

    Explanation:

    The following data were obtained from the question:

    V1 (original volume) = 6L

    P1 (original pressure) = 1.5 atm

    P2 (new pressure) = 2 x P1 (since the new pressure is doubled)

    P2 = 2 x 1.5

    P2 = 3 atm

    V2 (new volume) = ?

    Using the Boyle's law equation P1V1 = P2V2, the new volume can be obtain as follow:

    P1V1 = P2V2

    1.5 x 6 = 3 x V2

    Divide both side by 3

    V2 = (1.5 x 6) / 3

    V2 = 3L

    From our calculations, the new volume will be half the original volume and this will be 3L
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