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What do you know about GASES

What do you know about GASES

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What do you know about GASES

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  1. What do you know about GASES? Make a list of 5 things you know about gases: 1 - 5

  2. Unit 6.1 - Gases 6.1.1 Properties of Gases 6.1.2 Gas Pressure

  3. Kinetic Theory of Gases A gas consists of small particles that • move rapidly in straight lines • have essentially no attractive (or repulsive) forces • are very far apart • have very small volumes compared to the volumes of the containers they occupy • have kinetic energies that increase with an increase in temperature

  4. Properties of Gases • Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n).

  5. Gas Pressure Gas pressure • is the force acting on a specific area Pressure (P) = force area • has units of atm, mmHg, torr, lb/in.2 and kilopascals(kPa). 1 atm = 760 mmHg (exact) 1 atm = 760 torr 1 atm = 14.7 lb/in.2 1 atm = 101.325 kPa

  6. Atmospheric Pressure Atmospheric pressure • is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth

  7. Atmospheric Pressure (continued) Atmospheric pressure • is about 1 atmosphere at sea level • depends on the altitude and the weather • is lower at high altitudes where the density of air is less • is higher on a rainy day than on a sunny day

  8. Unit 6.1 - Gases 6.1.3 Pressure and Volume(Boyle’s Law)

  9. Boyle’s Law Boyle’s law states that • the pressure of a gas is inversely related to its volume when T and n are constant • if the pressure (P) increases, then the volume (V) decreases 9

  10. PV Constant in Boyle’s Law In Boyle’s law • The product P x V is constant as long as T and n do not change. P1V1 = 8.0 atm x 2.0 L = 16 atm L P2V2 = 4.0 atm x 4.0 L = 16 atm L P3V3 = 2.0 atm x 8.0 L = 16 atm L • Boyle’s law can be stated as P1V1 = P2V2 (T, n constant)

  11. Solving for a Gas Law Factor The equation for Boyle’s law can be rearranged to solve for any factor. P1V1 = P2V2 Boyle’s Law To solve for V2 , divide both sides by P2. P1V1 = P2V2 P2 P2 V1 x P1 = V2 P2

  12. Boyle’s Law and Breathing: Inhalation During inhalation, • the lungs expand • the pressure in the lungs decreases • air flows towards the lower pressure in the lungs

  13. Boyle’s Law and Breathing: Exhalation During exhalation, • lung volume decreases • pressure within the lungs increases • air flows from the higher pressure in the lungs to the outside

  14. Guide to Calculations with Gas Laws

  15. Unit Abbreviations for Gas Law Problems • Abbreviations: • atm - atmosphere • mmHg - millimeters of mercury • torr - another name for mmHg • Pa - Pascal (kPa = kilo Pascal) • K - Kelvin • °C - degrees Celsius

  16. Conversion Values for Gas Law Problems • Conversions: • K = °C + 273 • 1 cm3 (cubic centimeter) = 1 mL (milliliter) • 1 dm3 (cubic decimeter) = 1 L (liter) = 1000 mL • Standard Conditions: (same as on slide #4) • 0.00 °C = 273 K • 1.00 atm = 760.0 mmHg = 101.325 kPa = 101,325 Pa

  17. Calculation with Boyle’s Law Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of an 8.0 L sample of Freon gas after its pressure is changed from 550 mmHg to 2200 mmHg at constant T? STEP 1Set up a data table: Conditions 1 Conditions 2 Know Predict P1 = 550 mmHg P2 = 2200 mmHg P increases V1 = 8.0 L V2 = ? V decreases

  18. Calculation with Boyle’s Law (continued) STEP 2Solve Boyle’s law for V2. When pressure increases, volume decreases. P1V1 = P2V2 V2 = V1 xP1 P2 STEP 3 Set up problem V2 = 8.0 L x 550 mmHg = 2.0 L 2200 mmHg pressure ratio decreases volume

  19. Problems with Boyle’s Law • 1. A gas occupies 12.3 liters at a pressure of 40.0 mmHg. What is the volume when the pressure is increased to 60.0 mmHg? • 2. If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm? • 3. A gas occupies 1.56 L at 1.00 atm. What will be the volume of this gas if the pressure becomes 3.00 atm? • 4. A gas occupies 11.2 liters at 0.860 atm. What is the pressure if the volume becomes 15.0 L? • 5. 500.0 mL of a gas is collected at 745.0 mmHg. What will the volume be at standard pressure?

  20. Check your work 1) 8.2 L 2) 1.44L 3) 0.520 L 4) 0.642 atm 5) 490 mL

  21. Unit 6.1 - Gases 6.1.4 Temperature and Volume (Charles’s Law)

  22. Charles’s Law In Charles’s law, • the Kelvin temperature of a gas is directly related to the volume • P and n are constant • when the temperature of a gas increases, its volume increases

  23. STP represents Standard Temperature & Pressure STP conditions are: • 0 0C for temperature and 1 atmfor pressure • (but note that most of the time you will need to convert these to K or kPa values, that would be 273K and 101.3 kPa)

  24. Charles’s Law: V and T • For two conditions, Charles’s law is written V1 = V2 (P and n constant) T1T2 • Rearranging Charles’s law to solve for V2 gives T2 x V1 = V2 x T2 T1T2 V2 = V1x T2 T1

  25. Calculations Using Charles’s Law A balloon has a volume of 785 mL at 21 °C. If the temperature drops to 0 °C, what is the new volume of the balloon (P constant)? STEP 1 Set up data table: Conditions 1 Conditions 2 Know Predict V1 = 785 mLV2 = ? V decreases T1 = 21 °C T2 = 0 °C = 294 K = 273 K T decreases Be sure to use the Kelvin (K) temperature in gas calculations.

  26. Calculations Using Charles’s Law (continued) STEP 2Solve Charles’s law for V2: V1 = V2 T1T2 V2 = V1 x T2 T1 Temperature factor decreases T STEP 3 Set up calculation with data: V2 = 785 mL x 273 K = 729 mL 294 K

  27. Unit 6.1 - Gases 6.1.5 Temperature and Pressure (Gay-Lussac’s Law)

  28. Gay-Lussac’s Law: P and T In Gay-Lussac’s law, • the pressure exerted by a gas is directly related to the Kelvin temperature • V and n are constant P1 = P2 T1T2

  29. Calculation with Gay-Lussac’s Law A gas has a pressure at 2.0 atm at 18 °C. What is the new pressure when the temperature is 62 °C? (V and n constant) STEP 1 Set up a data table: Conditions 1 Conditions 2 Know Predict P1 = 2.0 atm P2 = ? P increases T1 = 18 °C + 273 T2 = 62 °C + 273 T increases = 291 K = 335 K

  30. Calculation with Gay-Lussac’s Law (continued) STEP 2Solve Gay-Lussac’s Law for P2: P1 = P2 T1T2 P2 = P1 x T2 T1 STEP 3 Substitute values to solve for unknown: P2 = 2.0 atm x 335 K = 2.3 atm 291 K Temperature ratio increases pressure

  31. Problems using Charles’s Law • 1. Calculate the decrease in temperature when 2.00 L at 20.0 °C is compressed to 1.00 L. • 2. 600.0 mL of air is at 20.0 °C. What is the volume at 60.0 °C? • 3. A gas occupies 900.0 mL at a temperature of 27.0 °C. What is the volume at 132.0 °C? • 4. What change in volume results if 60.0 mL of gas is cooled from 33.0 °C to 5.00 °C?

  32. Check your work • 147 K is new temperature • 682 mL • 1220 mL or 1.22 X 103 • 54.5 mL

  33. Problems using & Gay-Lussac’s Law • 5. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C. • 6. A gas has a pressure of 0.370 atm at 50.0 °C. What is the pressure at standard temperature? • 7. A gas has a pressure of 699.0 mmHg at 40.0 °C. What is the temperature at standard pressure? • 8. If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant what final pressure would result if the original pressure was 750.0 mmHg?

  34. Check your work • 1.03 atm • 0.313 atm • 340 K • 634.2 mm Hg

  35. Start here: Unit 6.1 - Gases 6.1.6 The Combined Gas Law (use this & leave off any variable that is a constant or not even talked about in the problem, it is understood to be a constant.)

  36. Summary of Gas Laws The gas laws can be summarized as follows:

  37. Combined Gas Law • The combined gas law uses Boyle’s Law, Charles’s Law, and Gay-Lussac’s Law (n is constant). P1 V1 = P2V2 T1T2

  38. Combined Gas Law Calculation A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29 °C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)?

  39. Combined Gas Law Calculation (continued) STEP 2 Solve for T2 P1V1 = P2V2 T1T2 T2 = T1 xP2 x V2 P1 V1 STEP 3 Substitute values to solve for unknown. T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180. mL T2 = 604 K  273 = 331 °C

  40. Combined Gas Law Problems • 1. A gas has a volume of 800.0 mL at –23.00 °C and 300.0 torr. What would the volume of the gas be at 227.0 °C and 600.0 torr of pressure? • 2. 500.0 liters of a gas are prepared at 700.0 mmHg and 200.0 °C. The gas is placed into a tank under high pressure. When the tank cools to 20.0 °C, the pressure of the gas is 30.0 atm. What is the volume of the gas?

  41. 3. What is the final volume of a 400.0 mL gas sample that is subjected to a temperature change from 22.0 °C to 30.0 °C and a pressure change from 760.0 mmHg to 360.0 mmHg? • Check your work • 1) 800 mL • 2) 9.51 L

  42. Unit 6.1 - Gases 6.1.7 Volume and Moles(Avogadro’s Law)

  43. Avogadro's Law: Volume and Moles Avogadro’s law states that • the volume of a gas is directly related to the number of moles (n) of gas • T and P are constant V1 = V2 n1n2

  44. STP The volumes of gases can be compared at STP (Standard Temperature and Pressure) when they have • the same temperature Standard temperature (T) = 0 °C or 273 K • the same pressure Standard pressure (P)= 1 atm (760 mmHg)

  45. Molar Volume The molar volumeof a gas • is measured at STP (standard temperature and pressure) • is 22.4 L for 1 mole of any gas

  46. Molar Volume as a Conversion Factor The molar volume at STP • has about the same volume as 3 basketballs • can be used to form 2 conversion factors: 22.4 L and 1 mole 1 mole 22.4 L

  47. Guide to Using Molar Volume

  48. Using Molar Volume What is the volume occupied by 2.75 moles of N2 gas at STP? STEP 1Given: 2.75 moles of N2 Need: Liters of N2 STEP 2 Write a plan: Use the molar volume to convert moles to liters.

  49. Using Molar Volume (continued) STEP 3 Write equalities and conversion factors: 1 mole of gas = 22.4 L 1 mole gas and 22.4 L 22.4 L 1 mole gas STEP 4 Substitute data and solve: 2.75 moles N2 x 22.4 L = 61.6 L of N2 1 mole N2

  50. Unit 6.1 - Gases 6.1.8 The Ideal Gas Law

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