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17 August, 19:09

Two identical 1.0 L containers are filled with equimolar amounts of helium gas and argon gas, respectively. The containers are both held at 25 degrees C and a small identically sized pinhole is poked in each. After some time has passed, how will the pressure in both containers compare

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  1. 17 August, 21:15
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    The pressure of the container with Argon gas will be approximately 3 times that of the container containing Helium gas as Helium effuses 3 times more than Argon gas

    Explanation:

    Argon gas mass 39.948 g. mol - 1

    Helium gas mass

    Molar mass of Helium gas is 4.003 g/mol.

    (Rate of effusion of Helium) / (Rate of effusion of Argon) = Sqr (39.948) / Sqr (4.003) = 6.3 to 2 that is Helium effuses 3 times more than Argon gas

    Effusion is the passage of gaseous molecules through a small hole like a balloon pinhole into a vacuum. The ratios of the rates of diffusion and the rate of effusion are the same.

    Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles:

    rate of effusion∝1/√M

    rate of effusion∝1/M

    That is if two gases are at the same pressure and temperature, the ratio of their rates of effusion is varies inversely to the ratio of the square roots of their particle masses:

    (rate of effusion of A) /

    (rate of effusion of B)

    =√MB/√MA

    rate of effusion of A

    The balloon filled with helium deflates more than the balloon filled with Argon because the lighter, smaller, helium particles effuse through the pinhole in the balloon faster than the heavier molecules of Argon
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