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9 October, 05:40

What is the volume of a gas at 2.00 atm and 200.0 K if its original volume was 300.0 L at 0.250 atm and 400.0 K

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  1. 9 October, 07:07
    0
    Given:

    P1 = 2.00 atm

    T1 = 200.0 K

    V2 = 300.0 L

    P2 = 0.250 atm

    T2 = 400.0 K

    Required:

    V1

    Solution:

    P1V1/T1 = P2V2/T2

    (2.00 atm) V1 / (200.0 K) = (300.0 L) (0.250 atm) / (400.0 K)

    V1 = 18.75 L
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