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15 November, 13:59

A sample of helium gas initially at 37.0°C, 785 torr and 2.00 L was heated to 58.0°C while the volume expanded to 3.24 L. What is the final pressure in atm?

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  1. 15 November, 14:06
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    0.681 atm

    Explanation:

    To solve this problem, we make use of the General gas equation.

    Given:

    P1 = 785 torr

    V1 = 2L

    T1 = 37 = 37 + 273.15 = 310.15K

    P2 = ?

    V2 = 3.24L

    T2 = 58 = 58+273.15 = 331.15K

    P1V1/T1 = P2V2/T2

    Now, making P2 the subject of the formula,

    P2 = P1V1T2/T1V2

    P2 = [785 * 2 * 331.15]/[310.15 * 3.24]

    P2 = 515.715 Torr

    We convert this to atm: 1 torr = 0.00132 atm

    515.715 Torr = 515.715 * 0.00132 = 0.681 atm
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