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21 October, 23:12

A balloon containing 0.500 mol at 0.00°c and 65.0kpa pressure is expanded by adding more argon. How many moles of argon are added to bring the sample to a final volume of 60.0 L at 30.0°c and 45.0kpa

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  1. 22 October, 00:01
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    n = 0.572 moles of Argon added to reach a volumen of 60.0 L

    Explanation:

    PV = RTn

    ∴ n1 = 0.500 mol

    ∴ T1 = 0.0°C = 273.15 K

    ∴ P1 = 65.0 KPa * (0.009869 atm / KPa) = 0.6415 atm

    ∴ R = 0.082 atm. L/K. mol

    ⇒ V1 = RT1n1 / P1 = 17.46 L

    n2 = P2V2 / RT2

    ∴ P2 = 45.0 KPa = 0.444 atm

    ∴ V2 = 60.0 L

    ∴ T2 = 30.0 °C = 303.15 K

    ⇒ n2 = ((0.444 atm) * (60.0 L)) / ((0.082 atm. L/K. mol) * (303.15 K))

    ⇒ n2 = 1.072 mol

    moles of Argon added:

    n = n2 - n1 = 1.072 - 0.500 = 0.572 mol added
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