It is desired to inflate a baggie with a volume of 927 mm by filling it with argon gas at a pressure of 0.749 atm and a temperature of 297k how many grams of ar gas are needed? in g?

Hey here!:

Molar mass Argon = 39.948 g/mol

volume in liter = 927 mL / 1000 => 0.927 L

p = 0.749 atm

T = 297 K

R = 0.0821

Therefore:

Mass of argon=   ( p *  V * molar mass Ar ) / ( R * T )

Mass of argon = ( 0.749 * 0.927 * 39.948  ) / ( 0.0821 * 297 )

Mass of argon = 27.7368 / 24.3837

Mass of argon = 1.137 grams

Hope that helps!

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