2. NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

3. BEHAVIOR OF GASES Gases have weight Gases take up space Gases exert pressure Gases fill their containers Gases doing all of these things!

4. Kinetic Theory of Gases The basic assumptions of the kinetic molecular theory are:  Gases are mostly empty space  The molecules in a gas are separate, very small and very far apart

5. Kinetic Theory of Gases The basic assumptions of the kinetic molecular theory are:  Gas molecules are in constant, chaotic motion  Collisions between gas molecules are elastic (there is no energy gain or loss)

6. Kinetic Theory of Gases The basic assumptions of the kinetic molecular theory are:  The average kinetic energy of gas molecules is directly proportional to the absolute temperature  Gas pressure is caused by collisions of molecules with the walls of the container

7. Measurements of Gases: To describe a gas, its volume, amount, temperature, and pressure are measured. Volume: measured in L, mL, cm 3 (1 mL = 1 cm 3 ) Amount: measured in moles (mol), grams (g) Temperature: measured in KELVIN (K)  K = ºC + 273 Pressure: measured in mm Hg, torr, atm, etc.  P = F / A (force per unit area)

8. Moderate Force (about 100 lbs) Small Area (0.0625 in 2 ) Enormous Pressure (1600 psi) P = F / A

9. Bed of Nails Large Surface Area (lots of nails) Moderate Force Small Pressure P = F / A

10. Units of Pressure:  Units of Pressure:  1 atm = 760 mm Hg  1 atm = 760 torr  1 atm = 1.013 x 10 5 Pa  1 atm = 101.3 kPa

11. A.5 - BOYLE’S LAW: As P , V  and vice versa…. INVERSE RELATIONSHIP P 1 V 1 = P 2 V 2 For a given number of molecules of gas at a constant temperature, the volume of the gas varies inversely with the pressure.

12. Boyle’s Law in action… Recall: in the lab, when you added more mass (force), and therefore more pressure, to the syringe the smaller the volume of gas inside the syringe became. Example: if the volume of the gas in the syringe were changed to half of its original volume by pushing on the plunger, the pressure of the gas sample would be DOUBLED.

13. Boyle’s Law in action… Example: If the gas volume in the syringe were reduced to ¼ of its original volume, the gas pressure would be 4 TIMES larger.

14. Boyle’s Law and Kinetic Molecular Theory: How does kinetic molecular theory explain Boyle’s Law?  gas molecules are in constant, random motion;  gas pressure is the result of molecules colliding with the walls of the container;  as the volume of a container becomes smaller, the collisions over a particular area of container wall increase…the gas pressure INCREASES!

16. Pressure-Volume Calculations: Example: Consider the syringe. Initially, the gas occupies a volume of 8 mL and exerts a pressure of 1 atm. What would the pressure of the gas become if its volume were increased to 10 mL?

17. “Reason and Ratio” method: First, reason or predict: If the volume INCREASES from 8 mL to 10 mL, the pressure must DECREASE by a proportional amount. To determine the proportional amount, determine the volume ratio (2 possibilities): 8 mLOR10 mL 10 mL8 mL

18. “Reason and Ratio” method: Which ratio is correct? Your “reasoning” indicates the pressure should DECREASE, so use the ratio that is less than 1:8 mL / 10 mL Now, multiply the original pressure by this volume ratio: 1 atmx(8 mL / 10 mL) = 0.8 atm

19. Equation for Boyle’s Law: P 1 V 1 =P 2 V 2 where: P 1 = initial pressure V 1 = initial volume P 2 = final pressure V 2 = final volume

20. P1V1=P2V2P1V1=P2V2 Using the same syringe example, just “plug in” the values: P 1 V 1 = P 2 V 2 (1 atm) (8 mL)=(P 2 ) (10 mL)

21. P1V1=P2V2P1V1=P2V2 (1 atm) (8 mL)=P 2 (10 mL) P 2 = 0.8 atm

22. Example: A sample of gas occupies 12 L under a pressure of 1.2 atm. What would its volume be if the pressure were increased to 3.6 atm? (assume temp is constant)  P 1 V 1 = P 2 V 2  (1.2 atm)(12 L) = (3.6 atm)V 2  V 2 = 4.0 L

23. Example: A sample of gas occupies 28 L under a pressure of 200 kPa. If the volume is decreased to 17 L, what be the new pressure? (assume temp is constant)  P 1 V 1 = P 2 V 2  (200 kPa)(28 L) = (P 2 )(17 L)  P 2 = 329 kPa

24. A.7: Temperature – Volume Relationships What happens to matter when it is heated? It EXPANDS. What happens to matter when it is cooled? It CONTRACTS. Gas samples expand and shrink to a much greater extent than either solids or liquids.

25. Charles’ Law: Jacques Charles (1746-1828) The volume of a given number of molecules is directly proportional to the Kelvin temperature. As T , V  (when P and amt. of gas are constant) and vice versa…. DIRECT RELATIONSHIP Equation:

28. Temperature – Volume Relationship: So, doubling the Kelvin temperature of a gas doubles its volume; reducing the Kelvin temperature by one half causes the gas volume to decrease by one half… WHY KELVIN?  the Kelvin scale never reaches “zero” or has negative values

30. Converting Kelvin: To convert from Celsius to Kelvin: add 273. Example: What is 110 ºC in Kelvin? 110 ºC+273=383 K

31. Converting Kelvin: To convert from Kelvin to Celsius: subtract 273. Example: 555 K in Celsius? 555 K-273=282 ºC

32. Example: A sample of nitrogen gas occupies 117 mL at 100.°C. At what temperature would it occupy 234 mL if the pressure does not change? (express answer in K and °C)  V 1 = 117 mL;T 1 = 100 + 273 = 373 K  V 2 = 234 mL;T 2 = ???  V 1 / T 1 = V 2 / T 2  (117 mL) / (373 K) = (234 mL) / T 2  T 2 = 746 K(now subtract 273 to get ºC)  T 2 = 473 ºC

33. Example: A sample of oxygen gas occupies 65 mL at 28.8°C. If the temperature is raised to 72.2°C, what will the new volume of the gas?  V 1 = 65 mL;T 1 = 28.8 + 273 = 301.8 K  V 2 = ??? mL;T 2 = 72.2 + 273 = 345.2 K  V 1 / T 1 = V 2 / T 2  (65 mL) / (301.8 K) = (V 2 ) / 345.2 K  V 2 = (65 mL) (345.2 K) / (301.8 K)  V 2 = 74.3 mL

34. A.8: Temperature – Pressure Relationships Picture a closed, rigid container of gas (such as a scuba tank) – the volume is CONSTANT. So, what would happen to the kinetic energy of the gas molecules in the container if you were to heat it up? How would this affect pressure?

35. Temperature – Pressure Relationships: Raising the Kelvin temperature of the gas will cause an INCREASE in the gas pressure. WHY? With increasing temperature, the K.E. of the gas particles increases – they move faster! They collide more often and with more energy with the walls of the container…

37. Temperature – Pressure Relationships: The pressure increases! So, as temperature INCREASES, pressure INCREASES, and; As temperature DECREASES, pressure DECREASES. DIRECT RELATIONSHIP! Equation: P 1 =P 2 T 1 T 2

38. Example: A sample of oxygen gas is in a rigid steel container. The pressure inside the container is 2 atm and the temperature is 45ºC. If the temperature is cooled to 32ºC, what will be the new pressure inside the container?  P 1 = 2 atm;T 1 = 45 + 273 = 318 K  P 2 = ??? mL;T 2 = 32 + 273 = 305 K  P 1 / T 1 = P 2 / T 2  (2 atm) / (318 K) = (P 2 ) / (305 K)  V 2 = (2 atm) (305 K) / (318 K)  V 2 = 1.9 atm