1. Ideal Gas Law & Gas Stoichiometry

2. Ideal Gas Law P V = n R T 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

3. PV = nRT Calculate the number of moles of a gas contained in a 3.0 L vessel at 300.0K with a pressure of 1.50 atm

4. PV = nRT n = ? V = 3.0 L T = 300.0 K P = 1.50 atm PV = nRT (1.50 atm)(3.0 L) = n (0.0821L atm / mol K)(300.0 K) n = 0.18 mol

5. 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 (mmHg) in the tank in the dentist office? Note: 1atm = 760 mm Hg

6. n = 2.86 mol V = 20.0 L T = 23 °C = 273 + 23 = 296 K P = ? PV = nRT (P)(20.0 L) = (2.86) (0.0821L atm / mol K)(296 K) P = 3.5 atm P = 2600 mm Hg Example

7. Permutations of the Ideal Gas Law PV = mRT 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

8. 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 = 300.0K M = 28.02 g/mol

9. Example PV = mRT M P(5.0) = (2.0)(0.0821)(300.0) 28.04 P = 2.8 atm

10. 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

11. 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

12. 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

13. 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

14. 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)

15. Example Calculate the volume that 2.000 kg of methane would occupy at STP. 2.000 kg x 1x10 3 g x 1 mol x 22.4 L = 1kg 16.05g 1 mol 2791 L CH 4