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25 July, 16:10

A gas that has a volume of 28 Liters, a Temperature of 318 K, and an unknown pressure initially

has its Volume increased to 34 Liters and its Temperature decreased to 308 K. If I measure the

pressure after the change to be 20 atm, what was the original pressure of the gas?

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Answers (1)
  1. 25 July, 16:20
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    Answer: 25.07 atm

    Explanation:

    Given that,

    Initial Pressure of gas (P1) = ?

    Initial Volume of gas (V1) = 28L

    Initial Temperature of gas (T1) = 318K

    Final pressure of helium = 20 atm

    Final volume of gas (V2) = 34L

    Final temperature of gas (T2) = 308K

    Since pressure, volume and temperature are given, apply the combined gas equation

    (P1V1) / T1 = (P2V2) / T2

    (P1 x 28L) / 318K = (20 atm x 34L) / 308K

    28L•P1/318K = 680 atm•L/308K

    To get P1, cross multiply

    28L•P1 x 308K = 680 atm•L x 318K

    8624L•K•P1 = 216240 atm•L•K

    Divide both sides by 8624L•K

    8624L•K•P1/8624L•K = 216240 atm•L•K/8624L•K

    P1 = 25.07 atm

    Thus, the original pressure of the gas is 25.07 atmosphere
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