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4 November, 22:26

19. A sample of neon occupies a volume of 461 mL at STP.

What will be the

volume of the neon when the pressure is reduced to 93.3 kPa?

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Answers (1)
  1. 4 November, 22:45
    0
    500.65mL

    Explanation:

    The following information were obtained from the question:

    V1 (initial volume) = 461 mL

    P1 (initial pressure) = stp = 101325Pa

    P2 (final pressure) = 93.3 kPa

    Recall: 1KPa = 1000Pa

    Therefore, 93.3 kPa = 93.3x1000 = 93300Pa

    V2 (final volume) = ?

    Using the Boyle's law equation P1V1 = P2V2, the final volume of the gas can be obtained as follow:

    P1V1 = P2V2

    461 x 101325 = 93300 x V2

    Divide both side by 93300

    V2 = (461 x 101325) / 93300

    V2 = 500.65mL

    Therefore, the volume of Neon at 93.3 kPa is 500.65mL
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