How many grams of oxygen gas are in the tire?

Question

pressure is 40.0lbs/in2

temperature 22.5 celcius 

volume 10.5 gallons

Answer 1

Convert lbs/in² to atmospheres

--  40.0 lbs/in² x 1 atm/14.6959 psi = 2.72 atm = Pressure

 

Convert gallons to liters

--  10.5 gallons x 3.785 liters/gallon = 39.74 liters = Volume

 

Convert temperature to degrees K

--  22.5ºC + 273 = 295.5 K

 

Now assuming that the oxygen in the tire behaves as an ideal gas, use the ideal gas law to solve for moles of O₂ present.

--  PV = nRT and n = moles = PV/RT = (2.27 atm)(39.74 L)/(0.0821 Latm/Kmol)(295.5 K) = 3.72 moles O₂

 

Finally, convert moles O₂ to grams of O₂

--  3.72 moles O₂ x 32 g/mole = 119 g O₂