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26 June, 17:01

At what temperature will 6.21 g of oxygen gas exert a pressure of 5.00 atm in a 10.0-L container?

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  1. 26 June, 18:03
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    PV = nRT

    P = 5 atm

    V = 10 L

    molecular mass of oxygen gas = 16 X 2 = 32 g/mole

    so number of moles of oxygen gas =

    6.21/32 = 0.194

    so n = 0.194

    R = 0.0821 L atm K^-1 mole^-1

    T = ? K

    5 X 10 = 0.194 X 0.0821 X T

    50 = 0.0159 X T

    T = 50/0.0159 = 3144.654 K

    in degree c = 3144.654 - 273

    = 2871.65 degree c
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