Two gas samples are mixed in a volume of 14.0 dm3 and at a temperature of 420.0

Question

Two gas samples are mixed in a volume of 14.0 dm3 and at a temperature of 420.0 K.
The amounts by weight are 3.00 g of argon gas (Ar) and 6.00 g of oxygen gas (O2). Both
gases and the mixture are considered to follow the perfect gas law. Calculate the total
pressure p of this mixture in units of atm. The atomic weights of the elements involved
are 39.95 g mol?1 for Ar and 16.00 g mol?1 for O.

Explanation / Answer

Argon: PV = nRT =>P = ((3/39.95)*8.314*420)/(14*0.001) = 18729.91239N/m^2 Oxygen: PV = nRT =>P = ((6/16)*8.314*420)/(14*0.001) = 93532.5N/m^2 Total pressure = 112262.4124N/m^2 = 1.1082173atm

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