CSK_Marco 發問於 科學及數學化學 · 1 十年前

[AL CHEM] Pressure 問題

A containing vessel holds a gaseous mixture of nitrogen and butane. The pressure in the vessel at 126.9 oC is 3.0atm. At 0 oC the butane completely condenses and the pressure drops to 1.0 atm. Calculate the mole fraction of nitrogen in the original gaseous mixture.

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  • 1 十年前
    最愛解答

    PV=nRT

    1atm=101325pa

    假設 nitrogen mol=n1 butane mol=n2

    3x101325 x V= (n1 +n2) R (126.9+273)---1

    after completely condenses

    PV=nRT

    1x101325 x V=(n1) R (0+273)---2

    2 divide by 1

    1/3=(n1)/(n1+n2) x (0.683)

    (n1/n1+n2)=0.488

    fraction of nitrogen=nitrogen mol/total mol=(n1/n1+n2)=0.488

    2010-09-08 19:34:45 補充:

    In closed system PV=nRT (V is vessel volume)

    In open system Pv=nRT (V is gas volume)

    資料來源: my
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