1. The Gas Laws

2. As P (h) increases V decreases Apparatus for Studying the Relationship Between Pressure and Volume of a Gas

3. P x V = constant P 1 x V 1 = P 2 x V 2 Boyle ’ s Law Constant temperature Constant amount of gas

4. Example 1 A helium balloon was compressed from 4.0L to 2.5L at a constant temperature. If the pressure of the gas in the 4.0L balloon is 210 kPA, what will the pressure be at 2.5L? Given: V 1 = 4.0LV 2 = 2.5L P 1 = 210 kPaP 2 = ? P 1 V 1 = P 2 V 2 (210 kPa) × (4.0L) = P 2 × (2.5L) P 2 = 336 kPa ≈ 340 kPa

5. Example 2 A sample of neon gas occupies 0.200L at 0.860 atm. What will be its volume at 29.2 kPa pressure? Given: V 1 = 0.200L V 2 = ? P 1 = 0.860 atm P 2 = 29.2 kPa P 1 V 1 = P 2 V 2 (87.1 kPa) × (0.200L) = (29.2 kPa) × V 2 V 2 = 0.597L **Units must match for each variable (doesn’t matter which one is converted) 0.860 atm 101.3 kPa 1 atm = 87.1 kPa

6. As T increasesV increases Variation in Gas Volume with Temperature at Constant Pressure

7. K = 0 C + 273 **Temperature must be in Kelvin Charles ’ s Law Constant pressure Constant amount of gas

8. Example 1 A gas ample at 40.0°C occupies a volume of 2.32L. If the temperature is raised to 75.0°C, what will the volume be, assuming the pressure remains constant? Given: T 1 = 40.0°C = 313KT 2 = 75.0°C = 348K V 1 = 2.32LV 2 = ? V 1 V 2 T 1 T 2 = 2.32L V 2 313 348 = V 2 = 2.58L

9. Example 2 A gas ample at 55.0°C occupies a volume of 3.50L. At what new temperature in kelvin will the volume increase to 8.00L? Given: T 1 = 55.0°C = 328KT 2 = ? V 1 = 3.50LV 2 = 8.00L V 1 V 2 T 1 T 2 = 3.50L 8.00 328 T 2 = T 2 = 750K

10. Gay - Lussac ’ s Law K = 0 C + 273 **Temperature must be in Kelvin Constant volume Constant amount of gas

11. Example 1 The pressure of a gas in a tank is 3.20 atm at 22.0°C. If the temperature rises to 60.0°C, what will be the gas pressure in the tank? Given: P 1 = 3.20 atmP 2 = ? T 1 = 22.0°C = 295KT 2 = 60.0°C =333K P 1 P 2 T 1 T 2 = 3.20 atm P 2 295K 333K = P 2 = 3.61 atm

12. Example 2 A rigid container has a gas at constant volume at 665 torr pressure when the temperature is 22.0C. What will the pressure be if the temperature is raised to 44.6C? Given: P 1 = 665 torrP 2 = ? T 1 = 22.0°C = 295KT 2 = 44.6°C =317.6K P 1 P 2 T 1 T 2 = 665 torr P 2 295K 317.6K = P 2 = 716 torr

13. Avogadro ’ s Law Constant temperature Constant pressure 1 mol of gas at STP = 22.4L

14. Example 1 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what will be the new volume? Assume constant temperature and pressure. Given: V 1 = 5.00 LV 2 = ? n 1 = 0.965 moln 2 = 1.80 mol V 1 V 2 n 1 n 2 = 5.00 L V 2 0.965 mol 1.80 mol = V 2 = 9.33 L

15. Example 2 Calculate the volume that 0.881 mol of gas at STP will occupy. 0.881 mol22.4 L 1 mol = 19.7 L

16. Example 3 How many grams of carbon dioxide gas are in a 0.75 L balloon at STP? 0.75 L CO 2 1 mol CO 2 44.01 g CO 2 22.4 L CO 2 1 mol CO 2 = 1.5 g CO 2

17. Example 4 What volume of oxygen gas is needed for the complete combustion of 4.00 L of propane gas (C 3 H 8 )? Assume constant temperature and pressure. = 20.0 L 4.00 L C 3 H 8 5 L O 2 1 L C 3 H 8

18. Gas Law Summary

19. Gay-Lussac’s Law Charles’s Law

21. Combined Gas Law Combine all 4 to make one master equation:

22. Example 1 A gas at 110 kPa and 30.0 ˚ C fills a flexible container with an initial volume of 2.00L. If the temperature is raised to 80.0 ˚ C and the pressure increased to 440 kPa, what is the new volume? Given: P 1 = 110 kPa P 2 = 440 kPa T 1 = 30.0 ˚ C =303K T 2 = 80.0 ˚ C = 353K V 1 = 2.00L V 2 = ? 110 (2.00)440 (V 2 ) 303353 110 (2.00)440 (V 2 ) 303353 = V 2 = 0.583L ≈ 0.58L

23. Example 2 An unopened bottle of soda contains 46.0 mL of gas confined at a pressure of 1.30 atm and temperature of 5.00 ˚ C. If the bottle is dropped into a lake and sinks to a depth at which the pressure and temperature changes to 1.52 atm and 2.90 ˚ C, what will be the volume of gas in the bottle? Given: V 1 = 46.0 mL V 2 = ? P 1 = 1.30 atm P 2 = 1.52 atm T 1 = 5.00 ˚ C = 278K T 2 = 2.90 ˚ C = 275.9K 1.30 (46.0)1.52 (V 2 ) 278275.9 1.30 (46.0)1.52 (V 2 ) 278275.9 = V 2 = 39.0mL

24. Ideal Gas law R is the universal gas constant PV = nRT R = PV nT = (1 atm)(22.4L) (1 mol)(273K)

25. Example 1 Calculate the number of moles of gas contained in a 3.00L vessel at 298K with a pressure of 1.50 atm. V = 3.00L T = 298K P = 1.50atm n = ? R = 0.0821 L atm mol K PV=nRT (1.50)(3.00) = n (0.0821) (298) n= 0.184 mol

26. Example 2 What will the pressure (in kPa) be when there are 0.400 mol of gas in a 5.00L container at 17.0˚C? V = 5.00L T = 17.0 ˚ C = 290K P = ? kPa n = 0.400 mol R = 8.314 L kPa mol K PV=nRT P(5.00) = (0.400)(8.314)(290) P= 193 kPa

27. Stoichiometry

28. Example 1 If 5.00L of nitrogen reacts completely with excess hydrogen at a constant pressure and temperature of 3.00 atm and 298K, how many grams of ammonia are produced. 0.613 mol N 2 2 mol NH 3 17.04 g NH 3 1 mol N 2 1 mol NH 3 = 120.9 g NH 3

29. Example 2 How many grams of calcium carbonate will be needed to form 6.75L of carbon dioxide at a pressure of 2.00 atm and 298K? CaCO 3 (s)  CO 2 (g) + CaO(s) 0.552 mol CO 2 1 mol CaCO 3 100.09 g CaCO 3 1 mol CO 2 1 mol CaCO 3 = 55.2 g CaCO 3

30. Example 3 How many liters of chlorine will be needed to make 95.0 grams of C 2 H 2 Cl 4 at 3.50 atm and 225K? Cl 2 (g) + C 2 H 4 (g)  C 2 H 2 Cl 4 (l) 95.0 g C 2 H 2 Cl 4 1 mol C 2 H 2 Cl 4 2 mol Cl 2 167.84 g C 2 H 2 Cl 4 1 mol C 2 H 2 Cl 4 = 1.13mol Cl 2

31. 31 Density (d) Calculations n V = P RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L d = m V PMPM RT n = m M m M V = P RT so

32. Example 1 What is the molar mass of a pure gas that has a density of 1.40g?L at STP?

33. Example 2 Calculate the density a gas will have at STP if its molar mass is 39.9g/mol.