calculate the mass of 500cm3 of oxygen at 25°c under 90k Pa pressure
Answers
molar mass of O2= 32 grams per mole.
At STP, 1 mole of any gas will have a volume of 22.4 litres.
STP conditions are as follows: 273 K temperature and 1.013*10^5 Pa.
=>At 273 K temperature and 1.013 * 10^2 KPa Pressure, 32 grams of O2 has volume of 22.4 litres or 22,400 cm3.
mass of 500 cm3 O2 at 90 KPa Pressure & 25°C temperature =?
First, we have to convert 500 cm3 of O2 at 90 KPa and 25°C to 101.3 KPa and 0°C.
V1= 500 cm3 V2= ?
P1= 90 KPa P2= 101.3 KPa
T1 = 25 C => 298 K T2= 273 K
P1V1/T1=P2V2/T2 ( By gas equation)
90*500/298 = 101.3*V2/273
45,000/298 =101.3 * V2/273
V2= 45,000*273/101.3*298
V2 = 406.96 cm3 (approximately)
Hence, the sample of Oxygen gas has a volume of 406.96 cm3 at STP .
As, 1 mole of Oxygen gas has a volume of 22,400 cm3 at STP.
So, 406.96 cm3 of O2 at STP will be 406.96/22,400 moles.
=>0.018 moles (approximately)
=> 0.018 X 32 grams of Oxygen gas. => 0.576 grams
So, 0.576 grams of Oxygen gas will have a volume of 500 cm3 at 25°C temperature and 90 KPa pressure