# Ideal Gas Law Gas Stoichiometry Ideal Gas Law

• Slides: 15

Ideal Gas Law & Gas Stoichiometry

Ideal Gas Law PV=n. RT • • • P = Pressure (atm) V = Volume (L) T = Temperature (K) n = number of moles R is a constant, called the Ideal Gas Constant Instead of learning a different value for R for all the possible unit combinations, we can just memorize one value and convert the units to match R. • R = 0. 0821 L atm / mol K

PV = n. RT • Calculate the number of moles of a gas contained in a 3. 0 L vessel at 300. 0 K with a pressure of 1. 50 atm

PV = n. RT • • • n=? V = 3. 0 L T = 300. 0 K P = 1. 50 atm PV = n. RT • (1. 50 atm)(3. 0 L) = n (0. 0821 L atm / mol K)(300. 0 K) • n = 0. 18 mol

Example Dinitrogen monoxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If 2. 86 mol of gas occupies a 20. 0 L tank at 23°C, what is the pressure (mm. Hg) in the tank in the dentist office? Note: 1 atm = 760 mm Hg

Example • • n = 2. 86 mol V = 20. 0 L T = 23 °C = 273 + 23 = 296 K P=? PV = n. RT • (P)(20. 0 L) = (2. 86) (0. 0821 L atm / mol K)(296 K) • P = 3. 5 atm • P = 2600 mm Hg

Permutations of the Ideal Gas Law • • • PV = m. RT M P = Pressure (atm) V = volume (L) m = mass of the gas (g) R = 0. 0821 L atm / mol K T = Temperature (K) M = molecular mass

Example • What is the pressure 2. 0 g of nitrogen gas in a 5. 0 L container at 300. 0 K? • P=? • m = 2. 0 g • V = 5. 0 L • T = 300. 0 K • M = 28. 02 g/mol

Example PV = m. RT M P(5. 0) = (2. 0)(0. 0821)(300. 0) 28. 04 P = 2. 8 atm

Permutations of the Ideal Gas Law • • • P = DRT M P = pressure (atm) D = density (g/L) R = 0. 0821 L atm / mol K T = temperature (K) M = molecular mass

Example • What is the molar mass of a gas that has a density of 1. 40 g/L at STP? – NOTE – STP is standard temperature and pressure – At STP temperature is 273 K and pressure is 1. 00 atm

Example • • • P = 1. 00 atm D = 1. 40 g/L R = 0. 0821 L atm / mol K T = 273 K M=? P = DRT M 1. 00 = (1. 40)(0. 0821)(273) M M = 31. 4 g/mol

Avogadro’s Principle • Avogadro’s Principle – equal volumes of gases at equal temperature and pressure contain the same number of particles • Molar volume – the volume of gas that 1 mole of a substance occupies at STP • At STP 1 mol of a gas = 22. 4 L • New conversion factor at STP ONLY! 1 mol 22. 4 L

Example • Calculate the volume 0. 881 mol of a gas will occupy at STP. • 0. 881 mol x 22. 4 L = 19. 7 L 1 mol (You could also have worked this out with the ideal gas law equation)

Example • Calculate the volume that 2. 000 kg of methane would occupy at STP. • 2. 000 kg x 1 x 10 3 g x 1 mol x 22. 4 L = 1 kg 16. 05 g 1 mol • 2791 L CH 4