1. I. Physical Properties Ch. 12 - Gases

2. A. Kinetic Molecular Theory b Particles in an ideal gas… have mass but no definite volume. have elastic collisions (no energy of motion is lost). are in constant, random motion colliding with the walls of the container. These collisions cause the pressure exerted by the gas. don’t attract or repel each other. have an avg. KE directly related to Kelvin temperature. If temp. goes up, KE goes up.

3. B. Real Gases b Particles in a REAL gas… have their own volume attract each other b Gas behavior is most ideal… at low pressures at high temperatures in nonpolar atoms/molecules

4. B. Characteristics of Gases 1.Gases expand to fill any container. random motion, no attraction 2.Gases are fluids (like liquids). no attraction 3.Gases have very low densities. no volume = lots of empty space

5. B. Characteristics of Gases 4. Gases can be compressed. no volume = lots of empty space 5. Gases undergo diffusion & effusion. random motion (Defns next slide)

6. B. Characteristics of Gases b Diffusion Spreading of gas molecules throughout a container until evenly distributed. b Effusion Passing of gas molecules through a tiny opening in a container

7. C. Temperature ºF ºC K -45932212 -2730100 0273373 K = ºC + 273 b Always use absolute temperature (Kelvin) when working with gases.

8. D. Pressure Which shoes create the most pressure?

9. D. Pressure b Barometer b Measures atmospheric pressure b The pressure of the atmosphere at sea level will hold a column of mercury 760 mm Hg. 1 atm Pressure 760 mm Hg Vacuum

10. D. Pressure b KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm (atmosphere) 760 mm Hg 760 torr 14.7 psi 29.94 in Hg

11. E. Example Problems 1. The column of mercury in a barometer is 745 mm high. What is the atmospheric pressure in kPa? 745 mm Hg 101.325 kPa 760 mm Hg = 99.3 kPa

12. F. STP Standard Temperature & Pressure 0°C 273 K 1 atm Or101.325 kPa Or … Or STP

13. Describe the behavior of gases in regards to Pressure, Volume, & Temperature The Gas Laws:

14. A. Boyle’s Law b The pressure and volume of a gas are inversely related at constant mass & temp P V PV = k

15. V T B. Charles’ Law b The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

16. P T C. Gay-Lussac’s Law b The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume

17. V n D. Avogadro’s Law b Equal volumes of gases contain equal numbers of moles (n) at constant temp & pressure true for any gas

18. D. Combined Gas Law P1V1n1T1P1V1n1T1 = P2V2n2T2P2V2n2T2 P 1 V 1 = P 2 V 2 What other equations? Play video!

19. GIVEN: V 1 = 473 cm 3 T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: V 1 / T 1 = V 2 / T 2 E. Gas Law Problems b A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. CHARLES’ LAW TT VV 473 cm 3 )/(309 K)=V 2 /(367 K) V 2 = 562 cm 3

20. GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 1.50 x10 3 mmHg = 200. kPa WORK: P 1 V 1 = P 2 V 2 E. Gas Law Problems b A gas occupies 100. mL at 150. kPa. Find its volume at 1.50 x10 3 mm Hg. BOYLE’S LAW PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

21. GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25°C = 298 K V2 = ?V2 = ? P 2 = 101.325 kPa T 2 = 273 K WORK: P 1 V 1 = P 2 V 2 (71.8 kPa)(7.84 cm 3 )/(298 K) =(101.325 kPa) V 2 /(273 K) V 2 = 5.09 cm 3 E. Gas Law Problems b A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. P  T  VV COMBINED GAS LAW T 1 T 2

22. GIVEN: P 1 = 2.25 atm T 1 = 20°C = 293K P 2 = ? T 2 = 45°C = 318K WORK: P 1 /T 1 = P 2 /T 2 E. Gas Law Problems b On a spring morning, 20°C, you fill your tires to a pressure of 2.25 atm. As you ride along, the tire heats up to 45°C from the friction on the road. What is the pressure in the tires now in units of psi? GAY-LUSSAC’S LAW PP TT (2.25 atm)/(293 K) = ( ?)/ (318 K) P 2 = 2.44 atm = 35.9 psi

23. F. Dalton’s Law b The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. P total = P 1 + P 2 +... When a H 2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H 2 and water vapor.

24. F. Dalton’s Law & Water Displacement P total = P gas + P water vapor Measured at several temps. (see chart)

25. GIVEN: P gas = ? P total = 742.0 mm Hg P H2O = 42.18 mm Hg WORK: P total = P gas + P H2O 742.0 mm = P gas +42.18 mm P gas = 699.8 mm Hg b A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 mm Hg. What is the partial pressure of the dry gas? Look up water-vapor pressure on the chart for 35.0°C. Sig Figs: Round to least number of decimal places. F. Dalton’s Law The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.

26. GIVEN: P H2 = ? P total = 94.4 kPa P H2O = 19.827 mmHg WORK: P total = P H2 + P H2O 94.4 kPa = P H2 + 2.64 kPa P H2 = 91.8 kPa F. Dalton’s Law b Hydrogen gas is collected over water at 22°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. Look up water-vapor pressure and convert for 22°C. Sig Figs: Round to least number of decimal places. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H 2 and water vapor.