1. 1 © 2006 Brooks/Cole - Thomson Importance of Gases Airbags fill with N 2 gas in an accident.Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3.Gas is generated by the decomposition of sodium azide, NaN 3. 2 NaN 3 ---> 2 Na + 3 N 22 NaN 3 ---> 2 Na + 3 N 2 How much sodium azide is required to fill an air bag? How much sodium azide is required to fill an air bag?

2. 2 © 2006 Brooks/Cole - Thomson Hot Air Balloons How Do They Work?

3. 3 © 2006 Brooks/Cole - Thomson THREE STATES OF MATTER

4. 4 © 2006 Brooks/Cole - Thomson General Properties of Gases There is a lot of free space in a gas.There is a lot of free space in a gas. Gases can be expanded infinitely.Gases can be expanded infinitely. Gases occupy containers uniformly and completely.Gases occupy containers uniformly and completely. Gases diffuse and mix rapidly.Gases diffuse and mix rapidly.

5. 5 © 2006 Brooks/Cole - Thomson Properties of Gases Gas properties can be modeled using math. Model depends on V = volume of the gas (L)V = volume of the gas (L) T = temperature (K)T = temperature (K) n = amount (moles)n = amount (moles) P = pressure (atmospheres)P = pressure (atmospheres)

6. 6 © 2006 Brooks/Cole - Thomson Pressure Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643)

7. 7 © 2006 Brooks/Cole - Thomson Pressure Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up into the cylinder). P of Hg pushing down related to Hg density (13.55 g/mL)Hg density (13.55 g/mL) column heightcolumn height

8. 8 © 2006 Brooks/Cole - Thomson Pressure Column height measures P of atmosphere 1 standard atm = 760 mm Hg1 standard atm = 760 mm Hg = 29.9 inches Hg = about 34 feet of water SI unit is PASCAL, Pa, where 1 atm = 101.325 kPa

9. 9 © 2006 Brooks/Cole - Thomson IDEAL GAS LAW Brings together gas properties. Can be derived from experiment and theory. P V = n R T

10. 10 © 2006 Brooks/Cole - Thomson Boyle s Law If n and T are constant, then PV = nRT = k P = k (1/V) = nRT This means, for example, that P goes up as V goes down. They are inversely proportional. For changing pressure or volume of a confined gas, PV=PV Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

11. 11 © 2006 Brooks/Cole - Thomson Charles s Law If n and P are constant, then V = (nR/P)T = kT V and T are directly related. For changing temperature or volume of a confined gas, V/T=V/T V/T=V/T Jacques Charles (1746- 1823). Isolated boron and studied gases. Balloonist.

12. 12 © 2006 Brooks/Cole - Thomson Charless original balloon Modern long-distance balloon

13. 13 © 2006 Brooks/Cole - Thomson Charles s Law Balloons immersed in liquid N 2 (at -196 ˚C) will shrink as the air cools (and is liquefied).

14. 14 © 2006 Brooks/Cole - Thomson Avogadro s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V = n (RT/P) = kn V and n are directly related. For changing moles of a confined gas, V/n=V/n twice as many molecules

15. 15 © 2006 Brooks/Cole - Thomson KINETIC MOLECULAR THEORY (KMT) Theory used to explain gas laws. KMT assumptions are Gases consist of molecules in constant, random motion.Gases consist of molecules in constant, random motion. P arises from collisions with container walls.P arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic.No attractive or repulsive forces between molecules. Collisions elastic. Volume of molecules is negligible.Volume of molecules is negligible.

16. 16 © 2006 Brooks/Cole - Thomson Kinetic Molecular Theory Because we assume molecules are in motion, they have a kinetic energy. KE = (1/2)(mass)(speed) 2 At the same T, all gases have the same average KE. As T goes up for a gas, KE also increases and so does speed.

17. 17 © 2006 Brooks/Cole - Thomson Using KMT to Understand Gas Laws Recall that KMT assumptions are Gases consist of molecules in constant, random motion.Gases consist of molecules in constant, random motion. P arises from collisions with container walls.P arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic.No attractive or repulsive forces between molecules. Collisions elastic. Volume of molecules is negligible.Volume of molecules is negligible.

18. 18 © 2006 Brooks/Cole - Thomson Deviations from Ideal Gas Law Real molecules have volume.Real molecules have volume. There are intermolecular forces.There are intermolecular forces. –Otherwise a gas could not become a liquid.

19. 19 © 2006 Brooks/Cole - Thomson Deviations from Ideal Gas Law Account for volume of molecules and intermolecular forces with VAN DER WAALSs EQUATION. Measured V = V(ideal) Measured P intermol. forces vol. correction J. van der Waals, 1837-1923, Professor of Physics, Amsterdam. Nobel Prize 1910. nRT V - nb V 2 n 2 a P + ----- )(

20. 20 © 2006 Brooks/Cole - Thomson Combining the Gas Laws V proportional to 1/PV proportional to 1/P V prop. to TV prop. to T V prop. to nV prop. to n Therefore, V prop. to nT/PTherefore, V prop. to nT/P V = 22.4 L for 1.00 mol whenV = 22.4 L for 1.00 mol when –Standard pressure and temperature (STP) –T = 273 K –P = 1.00 atm

21. 21 © 2006 Brooks/Cole - Thomson Using PV = nRT How much N 2 is reqd to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? R = 0.082057 Latm/Kmol R = 0.082057 Latm/KmolSolution 1. Get all data into proper units V = 27,000 L V = 27,000 L T = 25 o C + 273 = 298 K T = 25 o C + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm

22. 22 © 2006 Brooks/Cole - Thomson Using PV = nRT How much N 2 is reqd to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? R = 0.082057 Latm/Kmol R = 0.082057 Latm/KmolSolution 2. Now calc. n = PV / RT n = 1.1 x 10 3 mol (or about 30 kg of gas)

23. 23 © 2006 Brooks/Cole - Thomson Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

24. 24 © 2006 Brooks/Cole - Thomson Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution Strategy: Calculate moles of H 2 O 2 and then moles of O 2 and H 2 O. Finally, calc. P from n, R, T, and V.

25. 25 © 2006 Brooks/Cole - Thomson Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution

26. 26 © 2006 Brooks/Cole - Thomson Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution P of O 2 = 0.16 atm

27. 27 © 2006 Brooks/Cole - Thomson GAS DENSITY Screen 12.5 PV = nRT d and M proportional and density (d) = m/V

28. 28 © 2006 Brooks/Cole - Thomson The Disaster of Lake Nyos Lake Nyos in Cameroon was a volcanic lake. CO 2 built up in the lake and was released explosively on August 21, 1986. 1700 people and hundreds of animals died.