Question

It is desired to inflate a baggie with a volume of 937 milliliters by filling it with oxygen gas at a pressure of 0.557 atm and a temperature of 310 K. How many grams of O₂ gas are needed?  g

Answer 1

Solution:

According to gas equation,

PV = n R T

Given,

P = 0.557 atm

V = 937 mL = 0.937 L

T = 310 K

R = 0.0821 L atm K-1 mol-1

n = number of moles

Thus,

n = P V / R T

n = (0.557 atm x 0.937 L) / (0.0821 L atm K-1 mol-1 x 310 K)

n = 0.0205 mol

Hence,

Mass of oxygen gas = number of mol x molar mass of O2

Mass = 0.0205 mol x 32 g /mol = 0.656 g

Therefore,

Mass needed of O2 = 0.656 g