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1. A room has a volume of 60 m3 and is filled with air of an average molecular mass of 29 u. What is the mass of the air in the room at a pressure of 1.0 atm and temperature of 22°C? R = 0.082 L

Answer 1

1.

Volume  of the room = 60m³

Average molecular mass = 29 u

Pressure = 1.0 atm

Temperature = 22 ° C

Gas consatant (R) = 0.082 L

By using ideal gas equation 

                         P* V = [ (mass) /(mass of 1 mole ) ]*RT

                    =>Mass = ( P * V  *mass of 1 mole) /R*T

Now substituting the given values,

                    Mass = (1 * 60 * 29) / (0.082 * 22)

                             =1740 / 1.804

                             = 964.5 u

So , the mass of air in the room is 964.5 u