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4 July, 08:40

What volume is occupied by 26.6 g of argon gas at a pressure of 1.29 atm and a temperature of 355 K?

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  1. 4 July, 08:58
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    V = 15 L

    Explanation:

    PV = nRT

    P = 1.29 atm, T = 355 K mass argon 40g/mol

    n = mass / molecular mass; n = 26.6 g/40 g/mol; n = 0.665 mol

    V = (0,665 mol x 0.082 x 355 K) / 1.29 atm = 15 L
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