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5 November, 18:28

If 1.00 mol of argon is placed in a 0.500-L container at 19.0 ∘C, what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation) ? For argon, a=1.345 (L2⋅atm) / mol2 and b=0.03219L/mol.

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  1. 5 November, 18:40
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    41.083 atm is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation.

    Explanation:

    Data given for argon gas:

    number of moles = 1 mole

    volume = 0.5 L

    Temperature = 19 degrees or 292.15 K

    a = 1.345 (L2⋅atm) / mol2

    b = 0.03219L/mol.

    R = 0.0821

    The real pressure equation given by Van der Waals equation:

    P = (RT : Vm-b) - a : Vm^2

    Putting the values in the equation:

    P = (0.0821 x 292.15) : (0.5 - 0.03219) - 1.345: (0.5) ^2

    = 23.98:0.4678 - 1.345 : 0.25

    = 51.26 - 5.38

    = 45.88 atm is the real pressure.

    The pressure from the ideal gas law

    PV = nRT

    P = (1 x 0.0821 x 292.15) : 0.5

    = 4.797 atm

    the difference between the ideal pressure and real pressure is

    Pressure by vander waal equation - Pressure by ideal gas law

    45.88 - 4.797

    = 41.083 atm. is the difference between the two.
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