A sample of O2 with an initial temperature of 50.0 oC and a volume of 105 L is cooled to -25 oC. The new pressure is 105.4 kPa and the new v

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A sample of O2 with an initial temperature of 50.0 oC and a volume of 105 L is cooled to -25 oC. The new pressure is 105.4 kPa and the new volume is 55.0 L. What was the initial pressure of the sample? Show your work, using the G.U.E.S.S. method.

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  1. Answer:

    71.92 kPa

    Explanation:

    Using the combined gas law equation;

    P1V1/T1 = P2V2/T2

    Where;

    P1 = initial pressure (kPa)

    P2 = final pressure (kPa)

    V1 = initial volume (L)

    V2 = final volume (L)

    T1 = initial temperature (K)

    T2 = final temperature (K)

    According to the information provided in this question;

    T1 = 50°C = 50 + 273 = 323K

    V1 = 105L

    T2 = -25°C = -25 + 273 = 248K

    P2 = 105.4 kPa

    P1 = ?

    V2 = 55.0 L

    Using P1V1/T1 = P2V2/T2

    P1 × 105/323 = 105.4 × 55/248

    105P1/323 = 5797/248

    0.325P1 = 23.375

    P1 = 23.375 ÷ 0.325

    P1 = 71.92 kPa

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