A gas mixture contains each of the following gases at the indicated partial pressures: N2, 315 torr; O2, 134 torr; and He, 219 torr?

What is the total pressure of the mixture? What mass of each gas is present in a 2.15 L sample of this mixture at 25.0 C?

2 Answers

  • total pressure = partial pressure due to N2 + partial pressure due to O2 + partial pressure due to He = 315 + 134 + 219 = 668 torr or 668/760 = 0.879 atm ( as 1 atm = 760 torr)

    now PV = nRT in order to calculate total mole of gas

    P = 0.879 atm

    V = 2.15 L

    n = total no.of moles of all three gases = ?

    R = 0.0821 L atm/K/mole

    T = 25 + 273 = 298 K

    0.879 X 2.15 = n X 0.0821 X 298

    1.889 = n X 24.466

    n = 1.889/24.466 = 0.077

    now partial pressure = mole fraction X total pressure

    partial pressure of N2 = mole fraction of N2 X total pressure

    315 = no.of moles of N2/total no.of moles X 668

    315 = no.of moles of N2/ 0.077 X 668

    no.of moles of N2 = 315 X 0.077 / 668 = 0.036

    molar mass of N2 = 28 g/mole

    so amount of N2 = 28 X 0036 = 1.008 g

    similalrly ...partial pressure of O2 = mole fraction of O2 X total pressure

    134 = no.of moles of O2/0.077 X 668

    no.of moles of O2 = 134 X 0.077 / 668 = 0.015

    molar mass of O2 = 32 g/mole

    so amount of O2 = 32 X 0.015 = 0.48 g

    and finally....partial pressure of He = mole fracion of He X total pressure

    219 = no.of moles of He/0.077 X 668

    no.of moles of He = 219 X 0.077/668 = 0.025

    molar mass of He = 4 g/mole

    amount of He = 0.025 X 4 = 0.1 g

  • (315 torr) + (134 torr) + (219 torr) = 668 torr total

    n = PV / RT = (668 torr) x (2.15 L) / ((62.36367 L torr/K mol) x (25.0 + 273.15) K) = 0.07724 mol total

    (0.07724 mol) x (315 torr N2 / 668 torr total) x (28.01344 g N2/mol) = 1.02 g N2

    (0.07724 mol) x (134 torr O2 / 668 torr total) x (31.99886 g O2/mol) = 0.496 g O2

    (0.07724 mol) x (219 torr He) / 668 torr total) x (4.002602 g He/mol) = 0.101 g He

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