1. Gases

2. Kinetic Molecular Theory The kinetic molecular theory is used to explain the behavior of gases. All matter is made up of particles called atoms. The atoms are in constant motion. Collisions between the particles are elastic (no energy is gained or lost due to the collision).

3. Properties used to describe gases. The following four properties are used to describe gases: Temperature ( o C or K)-how fast the particles making up the gas are moving. Volume (L)-the amount of space the gas occupies (usually the size of the container) Amount of sample (moles)-# of particles in the sample Pressure (atm, kPa, mm Hg, torrs)

4. What is temperature? Temperature is a measure of the average kinetic energy (energy of motion) of particles in matter. As the kinetic energy of the particles increases, the temperature increases. As the kinetic energy of the particles decreases, the temperature decreases. The lowest possible temperature (the temperature at which the particles stop moving) is called absolute zero. Temperature is an intensive property and does not depend on the amount of matter present.

5. How is temperature measured? Temperature is measured with a thermometer. Almost all substances expand with an increase in temperature (exception is water). Thermometers are designed so that the substances they contain (mercury, etc.) expand and contract more than the volume of the glass tube that contains them so that the column height of the substance changes.

6. Temperature Scales Several temperature scales have been devised: - Fahrenheit ( o F): weather is measured using this scale. - Celsius ( o C): the scale used in the metric system. -Kelvin (K): based on absolute zero; used in the International System (SI).

7. Comparison of Temperature Scales Celsius degrees are larger in size than Fahrenheit degrees. Celsius degrees are the same size as kelvins. The Kelvin scale is the same as the Celsius scale (only 273 o higher). The only temperature that is the same on the Celsius and Fahrenheit scale is -40 o.

8. Converting Between Scales Celsius to Fahrenheit o F = 9/5 o C + 32 Fahrenheit to Celsius o C = ( o F – 32)5/9 Celsius to Kelvin K = o C + 273 Kelvin to Celsius o C = K - 273

9. Practice Problems 65 o C = _______K 25 o C = _______K -50 o C = ______K 300 K = _______ o C 273 K = _______ o C 150 K = _______ o C a)338 Kb) 298 Kc) 223 Kd) 27 o C e) 0 o Cf) -123 o C

10. Pressure Pressure is the force that is exerted on a surface. P = F/A Units for pressure are N/m 2 One pascal (Pa) = 1 N/m 2 Since the pascal is a very small unit, the kPa is more commonly used. 1 kPa = 1000 Pa

11. Practice Problem A car tire makes contact with the ground on a rectangular area of.12 m x.18 m. If the car’s mass is 9300 N, what pressure does it exert on ground as it rests on all four tires?

12. Gases and Pressure Gases exert pressure due to the collisions between the molecules and the walls of the container. The atmosphere exerts a pressure of 14.7 lbs/in 2. Gravity acts as the atmosphere’s container.

13. The atmosphere exerts a pressure that can support a column of mercury 760 mm tall. 1 atm of pressure = 760 mm Hg = 14.7 lbs/in2 = 101.3 kPa

14. Complete the following conversion problems: 1)Convert 2.2 atm to kPa. 2)Convert 750 torrs to atm. 3)Convert 202.5 kPa to mmHg. 4)Convert 10.5 lbs/in 2 to atm.

15. Measuring Pressure The device used to measure pressure is a manometer. A barometer is a specific type of manometer used to measure air pressure. Open ManometerClosed Manometer

16. Manometer Problems

17. The Components of the Atmosphere The atmosphere consists of the following gases: Nitrogen (N 2 ) = 78% Oxygen (O 2 ) = 21 % Argon (Ar) = 0.93% Carbon Dioxide (CO 2 ) = 0.038%

18. The Structure of the Atmosphere The atmosphere consists of five principal layers (starting closest to the earth and moving outward): 1)Troposphere (<20 km): where weather occurs, planes fly, etc. 2)Stratosphere (20-50 km) : where ozone layer is located, weather balloons fly, etc. 3)Mesosphere (50-85 km): where most meteors burn up 4)Thermosphere (85-690 km): International Space Station located here, aurora borealis, etc. 5)Exosphere (690-10,000 km): mostly consists of hydrogen and helium

19. Layers of the atmosphere

20. Gas Laws

21. Boyle’s Law At constant temperature, the pressure and volume of a gas are inversely related. (The number of gas particles remains constant). As pressure increases, volume decreases. Examples/Applications of Boyle’s Law: Opening a soft drink bottle Ears popping when an airplane takes off A bicycle pump Scuba diving

22. Predicting Pressure and Volume Using Boyle’s Law P 1 x V 1 = P 2 x V 2 P 1 : Original/Starting Pressure V 1 : Original/Starting Volume P 2 : New/Resulting Pressure V 2 : New/Resulting Volume

23. The volume of a gas at 99.0 kPa is 300.0 mL. If the pressure is increased to 188 kPa, what will be the new volume? P 1 = 99.0 kPa V 1 = 300.0 mL P 2 = 188 kPa V 2 = ? P 1 x V 1 = P 2 x V 2 (99.0) (300.0) = (188) x X = 158 mL (The pressure increased, so the volume decreased).

24. Charles’ Law At constant pressure, the Kelvin temperature and volume of a gas are directly related. (The number of gas particles remains constant). As temperature increases, volume increases. Examples/Applications of Charles’ Law: Bread rising in the oven Car tires appearing to be flat in the morning A balloon pops if left in a hot car

25. Predicting Temperature and Volume Using Charles’ Law V 1 = V 2 T 1 T 2 V 1 : Original/Starting Volume T 1 : Original/Starting Temperature (Kelvin) V 2 : New/Resulting Volume T 2 : New/Resulting Temperature (Kelvin)

26. A helium balloon in a closed car occupies a volume of 2.32 L at 40.0 o C. If the car is parked on a hot day and the temperature inside rises to 75.0 o C, what is the new volume of the balloon (assuming the pressure remains constant)? V 1 = 2.32 L T 1 = 40.0 o + 273 = 313 K T 2 = 75.0 o + 273 = 348 K V 2 = ? V 1 = V 2 T 1 T 2 2.32 = x 313 348 x = 2.58 L (The temperature increased, so the volume increased also)

27. Gay-Lussac’s Law At constant volume, the Kelvin temperature and pressure of a gas are directly related. (The number of gas particles remains constant). As temperature increases, pressure increases. Everyday Examples/Applications of Gay-Lussac’s Law: An aerosol can explodes if left in a fire. A pressure cooker http://preparatorychemistry.com/Bishop_Gay_Lussac_frames.htm

28. Predicting Temperature and Pressure Using Gay-Lussac’s Law P 1 = P 2 T 1 T 2 P 1 : Original/Starting Pressure T 1 : Original /Starting Temperature (in Kelvin) P 2 : New/Resulting Pressure T 2 : New/Resulting Temperature

29. A cylinder contain a gas which has a pressure of 125kPa at a temperature of 200 K. Find the temperature of the gas which has a pressure of 100 kPa. P 1 = 125 kPa T 1 = 200 K P 2 = 100 kPa T 2 = ? P 1 = P 2 T 1 T 2 125 = 100 200 x X = 160 K

30. Combined Gas Law The combined gas law states the relationship among pressure, temperature, and volume of a fixed amount of gas. All three variables have the same relationship to each other as they have in the other gas laws. This law can be represented by the following equation: V 1 P 1 = V 2 P 2 T 1 T 2

31. Practice Problem A sample of air in a syringe exerts a pressure of 1.02 atm at 22.0 o C. The syringe is placed in a boiling water bath at 100.0 o C. The pressure is increased to 1.23 atm by pushing the plunger in, which reduces the volume to 0.224 mL. What was the initial volume? P 1 = 1.02 atmP1V1 = P2V2 T 1 = 22.0 o C + 273 = 295 K T1 T2 V 1 =? P 2 = 1.23 atm (1.02)(x) = (1.23)(0.224) T 2 = 100.0 o C + 273= 373.0 K 295 373 V 2 = 0.224 mL x = 0.214 mL

32. Avogadro’s Principle Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. 1 mole of any gas at standard temperature (0 o C) and pressure (1 atm) occupies 22.4 L of space. (STP) 22. 4 L is known as the molar volume.

33. What is the volume of 2 moles of H 2 at STP? 44.8 L

34. Predicting Volume and Moles of Gases Using Avogadro’s Principle. V 1 = V 2 n 1 n 2 V 1 : Original/Starting Volume n 1 : Original/Starting Number of Moles V 2 : New/Resulting Volume n 2 : New/Resulting Number of Moles

35. What size container do you need to hold 0.05 moles of N 2 gas at STP? V 1 = ? n 1 = 0.05 moles V 2 = 22.4 L n 2 = 1 mole V1 = V2 n1 n2 x = 22.4.05 1 X = 1.12 L

36. Graham’s Law Diffusion is the term used to describe the movement of one material through another. Effusion is a term used to describe the escape of a gas through an opening. Example: Helium will effuse through a hole in a balloon. Graham’s law states that the rate of effusion for a gas is inversely proportional to the square root of its molar mass. In other words; the heavier the gas, the slower it’s rate of effusion.

37. Predicting Rates of Effusion Using Graham’s Law Rate of gas A = Rate of gas B Molar Mass of B Molar Mass of A

38. Which will deflate faster; a balloon filled with carbon dioxide or a balloon filled with helium? Rate of gas A = Rate of gas B Gas A is Helium (He) Gas B is Carbon Dioxide (CO 2 ) Molar mass of CO 2 = 44 g Molar mass of He = 4 g 44/4 Helium diffuses 3.31 times faster Molar Mass of B Molar Mass of A

39. Dalton’s Law of Partial Pressures The total pressure of a mixture of gases is equal to the sum of the pressures of all of the gases in the mixture. P total = P 1 + P 2 + P 3 + … A mixture of oxygen, nitrogen and hydrogen gases have a total pressure of 0.97 atm. What is the partial pressure of oxygen if the partial pressure of nitrogen is 0.70 atm and the partial pressure of hydrogen is 0.12 atm? 0.97 = x + 0.70 + 0.12 0.15 atm

40. Ideal Gas Law The ideal gas law is a mathematical equation that describes the relationship among all 4 properties. The formula works best when gases obey the assumptions of the kinetic molecular theory. PV = nRT P is pressure; V is volume in liters; T is temperature in Kelvin; n is the number of moles; and R is the gas constant R = 8.314 if pressure is in kPA; R= 0.0821 if pressure is in atm; R= 62.4 if pressure is in mm Hg or torrs.

41. Calculate the number of moles of ammonia gas contained in a 3.0 L container at 300. K with a pressure of 1.50 atm. V = 3.0 L P = 1.50 atm T = 300. K R = 0.0821 n = ? PV = nRT (1.50)(3.0) = x (0.0821)(300) X = 0.18 mol