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Gases and Gas Laws Chemistry– Unit 11: Chapter 14

Gases and Gas Laws Chemistry– Unit 11: Chapter 14. Kinetic Molecular Theory (THIS IS IMPORTANT!!). Particles in an ideal gas… have no volume . have elastic collisions. are in constant , random , straight-line motion. don’t attract or repel each other.

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Gases and Gas Laws Chemistry– Unit 11: Chapter 14

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  1. Gases and Gas LawsChemistry– Unit 11: Chapter 14

  2. Kinetic Molecular Theory (THIS IS IMPORTANT!!) • Particles in an ideal gas… • have no volume. • have elastic collisions. • are in constant, random, straight-line motion. • don’t attract or repel each other. • have an avg. KE directly related to Kelvin temperature.

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

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

  5. Characteristics of Gases • Gases can be compressed. • no volume = lots of empty space • Gases undergo diffusion & effusion. • random motion

  6. 3 = (KE) RT avg 2 The Meaning of Temperature • Kelvin temperature is an index of the random motions of gas particles (higherT means greater motion.)

  7. 1 = 2 KE mv 2 Kinetic Energy of Gas Particles At the same conditions of temperature, all gases have the same average kinetic energy. m = mass v =velocity

  8. K = ºC + 273 Temperature • Always use absolute temperature (Kelvin) when working with gases. ºF -459 32 212 ºC -273 0 100 K 0 273 373

  9. 9 + o C 32 5 5 o ( F - 32) 9 ºF = ºC = Absolute Zero: _____________________________________ Absolute Zero = ____K = _____ºC Temperature at which motion stops. 0 - 273

  10. 5 = (355 - 32) K =179 oC + 273 = 452 K 9 K =- 40 oC + 273 = 233 K Example Conversions:   (a) Convert 355 ºF to ºC and K (b) Convert -40 ºC to K and ºF oC = 179oC 9 oF = -40 oF o 40 C + 32 = - 5

  11. Pressure Which shoes create the most pressure?

  12. Measuring Pressure The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17th century. The device was called a “barometer”. • Baro=weight • Meter =measure

  13. Aneroid Barometer Mercury Barometer Pressure • Barometer • measures atmospheric pressure

  14. Pressure • KEY UNITS AT SEA LEVEL 101.3 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi ** All of these amounts are equal to each other (conversion factors), just in different units!

  15. Pressure Conversions A. What is 475 mm Hg expressed in atm? 1 atm 760 mm Hg B. The pressure of a tire is measured as 29.4 psi.What is this pressure in mm Hg? 760 mm Hg 14.7 psi 475 mm Hg x = 0.625 atm 29.4 psi x = 1.52 x 103 mm Hg

  16. Pressure Conversions A. What is 2 atm expressed in torr? B. The pressure of a tire is measured as 32.0 psi.What is this pressure in kPa?

  17. IQ 1 • List the properties of an ideal gas. • What is this theory called? • How is an ideal gas different from a real gas?

  18. IQ #3 • What are 2 ways to increase the pressure of a gas (Think about the formula for pressure)? • What effect does temperature have on kinetic energy? • What do you expect would happen to the volume of a balloon if the gas inside it made more collisions with it?

  19. P V Boyle’s Law Robert Boyle (1627-1691). Son of Earl of Cork, Ireland. PV = k

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

  21. Boyle’s Law This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down. P1V1 = P2 V2 As the volume of the air trapped in the bicycle pump is reduced, its pressure goes up, and air is forced into the tire.

  22. Gas Law Problems • A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa P V WORK: P1V1= P2V2 V2= P1V1 P2 (150.kPa)(100.mL)=(200.kPa)V2 V2 = 75.0 mL

  23. V T Charles’ Law

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

  25. Charles’s Law V and T are directly proportional. V1 V2 = T1 T2 • If one temperature goes up, the volume goes up! Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.

  26. Charles’s original balloon Modern long-distance balloon

  27. Gas Law Problems • A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW GIVEN: V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K T V WORK: V1 = V2 T1 T2 V2 = V1T2 T1 (473 cm3 x 367 K)/309 K =V2 V2 = 562 cm3

  28. P T Gay-Lussac’s Law

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

  30. Gay-Lussac’s Law P and T are directly proportional. P1 P2 = T1 T2 • If one temperature goes up, the pressure goes up! Joseph Louis Gay-Lussac (1778-1850)

  31. Gas Law Problems • A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW GIVEN: P1 = 765 torr T1 = 23°C = 296K P2 = 560.0 torr T2 = ? P T WORK: P1 = P2 T1 = T2 T2 = T1P2 P1 T2 = (560.0 torr)(296K)/765 torr T2 = 217 K = -56°C

  32. Combined Gas Law • The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P1 V1 P2 V2 = T1 T2

  33. P1 V1 P2 V2 T1 T2 Combined Gas Law If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law! = Boyle’s Law Charles’ Law Gay-Lussac’s Law

  34. Gas Law Problems • A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. COMBINED GAS LAW P T V GIVEN: V1=7.84 cm3 P1=71.8 kPa T1=25°C = 298 K V2=? P2=101.325 kPa T2=273 K WORK: P1V1T2 = P2V2T1 (71.8 kPa)(7.84 cm3)(273 K) =(101.325 kPa)V2 (298 K) V2 = 5.09 cm3

  35. Combined Gas Law Problem #2 A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? Set up Data Table P1 = 0.800 atm V1 = 180 mL T1 = 302 K P2 = 3.20 atm V2= 90 mL T2 = ??

  36. Calculation P1 = 0.800 atm V1 = 180 mL T1 = 302 K P2 = 3.20 atm V2= 90 mL T2 = ?? P1 V1 P2 V2 = P1 V1T2= P2 V2 T1 T1 T2 T2= P2 V2 T1 P1 V1 T2 = 3.20 atm x 90.0 mL x 302 K 0.800 atm x 180.0 mL T2 = 604 K - 273 = 331 °C = 604 K

  37. Learning Check (Group) A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

  38. IQ 2 1. Covert -56 °C to Kelvin. 2. How many torr in 37.8 psi? 3. Covert 0 °C into °F.

  39. One More Practice Problem (Group) A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

  40. Try This One on Your Own! A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

  41. Standard Temperature & Pressure 0°C273 K 1 atm101.3 kPa -OR- STP STP allows us to compare amounts of gases between different pressures and temperatures

  42. 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. twice as many molecules

  43. Avagadro’s Hypothesis and the Kinetic Molecular Theory The gases in this experiment are all measured at the same T and V. P proportional to n

  44. IDEAL GAS LAW P V = n R T Brings together gas properties. Can be derived from experiment and theory. BE SURE YOU KNOW THIS EQUATION!

  45. Using PV = nRT P = Pressure V = Volume T = Temperature n = number of moles R is a constant, called theIdeal Gas Constant R = 0.0821 L • atm Mol • K

  46. Sample Problem #1 How much N2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to 745 mm Hg at 25 oC? Solution 1. Get all data into proper units V = 27,000 L T = 25 oC + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm And we always know R, 0.0821 L atm / mol K

  47. Using PV = nRT How much N2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC? Solution 2. Now plug in those values and solve for the unknown. PV = nRT RT RT n = 1.1 x 103 mol (or about 30 kg of gas)

  48. Learning Check Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office?

  49. Learning Check #2 A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

  50. IQ #3 • What are “STP” conditions? • What is the volume of an ideal gas? • A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder? 273 K/ 1 atm An ideal gas has no volume!! 6.4 g

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