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23 January, 13:08

A 75.0-liter canister contains 15.82 moles of argon at a pressure of 546.8 kilopascals. What is the temperature of the canister?

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  1. 23 January, 13:22
    0
    312 K

    Explanation:

    1) dа ta:

    a) V = 75.0 liter

    b) n = 15.82 mol

    c) p = 546.8 kPa

    d) T = ?

    2) Formula:

    Ideal gas equation: pV = nRT

    Where:

    n = number of moles V = volume p = absolute pressure T = absolute temperature R = Universal Gas constat: 8.314 kPa - liter / K-mol

    3) Solution:

    a) Solve the equation for T:

    T = pV / (nR)

    b) Substitute and compute:

    T = 546.8 kPa * 75.0 l iter / (15.82 mol * 8.314 kPa-liter/K-mol) = 312 K

    (since the volume is expressed with 3 significant figures, the answer must show also 3 significant figures)
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