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A 334-mL cylinder for use in chemistry lectures contains 5.225 g of helium at 23 C. How many grams of helium must be releasedto reduce the pressure to 75 atm assuming ideal gas behavior?

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Dre

  1. find grams at 75
    PV=nRT solve for n, then convert to grams.

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    bobpursley

  2. He is 4 grams/mol
    so we have 5.225/4 = 1.31 mols in 0.334 Liter at 273 + 23 = 296 K

    P V = n R T

    R = 0.082 L atm deg K / mol
    V = 0.334 L
    n = 1.31 mol
    T = 296

    so
    P = 1.31 * 0.082 * 296 / 0.334
    = 95.2 atm
    new P = 75
    T R V the same
    so P is proportional to n
    75/95.2 = new n/ 1.31
    new n = 1.03 mols
    1.03 * 4 g/mol = 4.13 grams remain
    5.225 - 4.13 = 1.1 grams released

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    Damon

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