Gases Chapter 11 CHM 130 GCC

11.1 Properties of Gases 1.Gases have no shape: they take the shape of their container 2.Gases can expand & compress: V increases and atoms get farther apart, V decreases and atoms get closer together 3.Gases have low densities: air is about g /mL 4.Gases mix completely with other gases: air is a mix of many gases like N 2 (78%), O 2 (21%), CO 2, and noble gases 5.Gases exert pressure: they hit the sides of their container (exert force)

11.2 Pressure Pressure increases if –Number of collisions increase (more hits) –Energy of collisions increase (harder hits) There is no pressure in a vacuum Atmospheric pressure –Force of air above us hitting us –What happens if you climb Mt. Everest? As you climb, there is less air above you so pressure lessens, the air up high is less dense

1.00 atm = 760 torr = 760 mm Hg= 14.7 psi Barometer The pressure of the air at sea level pushes the mercury liquid up 760 mm high.

P Conversions A diver’s tank is at 4400 psi. Convert to torr, atm, and mmHg psi ( 1 atm / 14.7 psi) = 3.0 x 10 2 atm 4400 psi( 760 torr / 14.7 psi) = 230,000 torr 230,000 torr = 230,000 mmHg

11.3 What affects Pressure? 1.If V increases, P ______ 2.If T increases, P ______ 3.If # gas atoms increases, P _____ Direct or Indirect Relationship? 1.P and V 2.P and T 3.P and # atoms Decreases Increases Indirect Direct

Gas Laws 11.4 Boyle’s Law: 11.5 Charles’s Law: 11.6 Gay-Lussac’s Law: P 1 V 1 = P 2 V 2 V 1 = V 2 T 1 T 2 P 1 = P 2 T 1 T Combined Gas Law: P 1 V 1 = P 2 V 2 T 1 T 2 STP = standard temperature and pressure = 1 atm and 0°C

Videos if time Sulfur Hexafluoride gas How Density affects your voice Boyle’s law and lungs Boyle’s law and diving

Gas Law Calculations T must be in Kelvin ( o C = K) V same units on both sides to cancel P same units on both sides to cancel Identify all variables as before (1) or after the change (2) If P, V, or T stays the same or is not mentioned, it cancels out Plug in and solve, keep track of units and sig figs

A sample of krypton gas occupies 3.91 L at 105 o C. Find the temperature in o C of the gas if the volume is changed to 6.05 L. A sample of carbon dioxide gas occupies 2.25 L at 758 torr. Find the final volume if the pressure is decreased to 698 torr. P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 / P 2 V 2 = (758 torr)(2.25L) / (698 torr) V 2 = 2.44 L V 1 /T 1 = V 2 /T 2 T 2 = V 2 T 1 / V 1 T 2 = (6.05 L)(378 K) / 3.91 L T 2 = 585 K and = 312 o C

A steel container filled with nitrous oxide at 15.0 atm is cooled from 25.5°C to –40.8°C. Calculate the final pressure. A nitrogen gas sample occupies 50.5 mL at –82.8°C and 1250 torr. What is the volume at STP? P 1 /T 1 = P 2 /T 2 P 2 = P 1 T 2 / T 1 P 2 = (15.0 atm)(232.2 K) / (298.5 K) P 2 = 11.7 atm P 1 V 1 /T 1 = P 2 V 2 /T 2 V 2 = P 1 V 1 T 2 / T 1 P 2 V 2 = V 2 = 119 mL

Dalton’s Law & Respiration (for fun) Total P = sum of partial P P tot = P 1 + P 2 + P 3 etc. O 2 is 21% of air by volume, so 21% of air P is due to O 2 which is about 160 torr. When blood exposed to air in lungs, the oxygen diffuses into the blood until the partial pressure in the liquid equals that of the gas in air Gases move from a region of higher partial pressure to lower partial pressure Partial pressure of O 2 in pulmonary blood only about 40mmHg, compared to 104mmHg in alveoli. Diffusion occurs rapidly. Partial P defined Diffusion CO 2 transport &playnext=0 &playnext=0

Self Test Page 320 Try 1-7, 12-13, Answers in Appendix J