A container encloses 6.3 mol of an ideal gas that has molar mass M1 and 0.83 mol

Question

A container encloses 6.3 mol of an ideal gas that has molar mass M1 and 0.83 mol of a second gas that has a molar mass of M2 = 7.5M1. What fraction of the total pressure on the container wall is attributable to the second gas? (The kinetic theory explanation of pressure leads to the experimentally discovered law of partial pressures for a mixture of gases that do not react chemically: The total pressure exerted by the mixture is equal to the sum of the pressures that the several gases would exert separately if each were to occupy the vessel alone.)

Explanation / Answer

The pressure P1 due to the first gas is P1 = n1RT / V

and the pressure P2 due to thE second gas is P2 = n2RT / V

Thus the total pressure on the container wall

is P = P1 + P2 = (n1 + n2)RT/V

The fraction of the total pressure due to the second

gas is then P2/P = n2RT/V ×1/(n1+n2)RT/V = n2/n1+n2 = 0.83/6.3+0.83 =0.116

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