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23 November, 06:11

What is the temperature of 0.257 mol of O2 occupying 6.78 L at 0.856 atm?

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  1. 23 November, 08:02
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    From the ideal gas law, PV = n RT

    T = PV / n R

    Given that n = 0.257 mole

    V = 6.78 L

    P = 0.856 atm

    And R = gas constant = 0.0821 L. atm/K/mol

    Therefore;

    T = PV / n R

    T = 0.856 atm * 6.78 L / 0.257 mole * 0.0821 L. atm/K/mol

    = 275.1 K

    =2.1 degree C
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