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12 August, 01:10

A gas that has a volume of 28 liters, a temperature of 45 °C, and an unknown pressure has its volume increased

to 34 liters and its temperature decreased to 35 °C. If I measure the pressure after the change to be 2.0 atm,

what was the original pressure of the gas?

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Answers (1)
  1. 12 August, 01:34
    0
    2.5 atm

    Explanation:

    V1 = 28L

    T1 = 45°C = (45 + 273.15) K = 318.15K

    V2 = 34L

    T2 = 35°C = (35 + 273.15) K = 308.15k

    P2 = 2.0atm

    P1 = ?

    From general gas equation,

    (P1 * V1) / T1 = (P2 * V2) / T2

    P2 * V2 * T1 = P1 * V1 * T2

    P1 = (P2 * V2 * T1) / (V1 * T2)

    P1 = (2.0 * 34 * 318.15) / (28 * 308.15)

    P1 = 21634.5 / 8628.2

    P1 = 2.5 atm

    The initial pressure of the gas is 2.5atm
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