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27 December, 06:00

A 8.65-L container holds a mixture of two gases at 11 °C. The partial pressures of gas A and gas B, respectively, are 0.205 atm and 0.658 atm. If 0.200 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

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  1. 27 December, 06:10
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    The total pressure = 1.402 atm

    calculation

    Total pressure = partial pressure of gas A + partial pressure of gas B + partial pressure of third gas

    partial pressure of gas A = 0.205 atm

    Partial pressure of gas B = 0.658 atm

    partial pressure for third gas is calculated using ideal gas equation

    that is PV=nRT where,

    p (pressure) = ? atm

    V (volume) = 8.65 L

    n (moles) = 0.200 moles

    R (gas constant) = 0.0821 L. atm/mol. k

    T (temperature) = 11°c into kelvin = 11+273 = 284 k

    make p the subject of the formula by diving both side by V

    p = nRT/v

    p = [ (0.200 moles x 0.0821 L. atm/mol. K x 284 K) / 8.65L) ] = 0.539 atm

    Total pressure is therefore = 0.205 atm + 0.658 atm + 0.539 atm

    =1.402 atm
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