1 / 53

Chapter 7 Gases

Chapter 7 Gases. 7.1 Properties of Gases 7.2 Gas Pressure. 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

bonifacy-ania
Download Presentation

Chapter 7 Gases

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 7 Gases 7.1 Properties of Gases 7.2 Gas Pressure

  2. 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

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

  4. 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

  5. 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

  6. 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

  7. Chapter 7 Gases 7.3Pressure and Volume(Boyle’s Law)

  8. 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 8

  9. 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)

  10. 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

  11. 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

  12. 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

  13. Guide to Calculations with Gas Laws

  14. 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

  15. 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

  16. Chapter 7 Gases 7.4 Temperature and Volume (Charles’s Law)

  17. 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

  18. 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

  19. 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.

  20. 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

  21. Chapter 7 Gases 7.5 Temperature and Pressure (Gay-Lussac’s Law)

  22. 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

  23. Chapter 7 Gases 7.6The Combined Gas Law

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

  25. 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

  26. 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)? Step 1 Set up data table: Conditions 1 Conditions 2 P1 = 0.800 atm P2 = 3.20 atm V1 = 0.180 L (180 mL) V2 = 90.0 mL T1 = 29 °C + 273 = 302 K T2 = ?

  27. 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

  28. Chapter 7 Gases 7.7Volume and Moles(Avogadro’s Law)

  29. 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

  30. 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

  31. 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

  32. 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)

  33. 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

  34. 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

  35. Guide to Using Molar Volume

  36. 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.

  37. 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

  38. Chapter 7 Gases 7.8The Ideal Gas Law

  39. Ideal Gas Law • The relationship between the four properties (P, V, n, and T) of gases can be written equal to a constant R. PV = R nT • Rearranging this expression gives the expression called the ideal gas law. PV = nRT

  40. Universal Gas Constant, R The universal gas constant, R, • can be calculated using the molar volume at STP • when calculated at STP, uses a temperature of 273 K, a pressure of 1.00 atm, a quantity of 1.00 mole of a gas, and a molar volume of 22.4 L. PV R = PV = (1.00 atm)(22.4 L) nT (1.00 mole)(273K) n T = 0.0821 Latm moleK

  41. Summary of Units for Ideal Gas Constants

  42. Guide to Using the Ideal Gas Law

  43. Chapter 7 Gases 7.9Partial Pressure (Dalton’s Law)

  44. Partial Pressure The partial pressure of a gas • is the pressure of each gas in a mixture • is the pressure that gas would exert if it were by itself in the container

  45. Dalton’s Law of Partial Pressures Dalton’s Law of Partial Pressures indicates that • pressure depends on the total number of gas particles, not on the types of particles • the total pressure exerted by gases in a mixture is the sum of the partial pressures of those gases PT = P1 + P2 + P3 + .....

  46. Dalton’s Law of Partial Pressures (continued)

  47. Total Pressure • For example, at STP, one mole of a pure gas in a volume of 22.4 L will exert the same pressure as one mole of a gas mixture in 22.4 L. V = 22.4 L Gas mixtures 1.0 mole N2 0.4 mole O2 0.6 mole He 1.0 mole 0.5 mole O2 0.3 mole He 0.2 mole Ar 1.0 mole 1.0 atm 1.0 atm 1.0 atm

  48. Scuba Diving • When a scuba diver is below the ocean surface, the increased pressure causes more N2(g)to dissolve in the blood. • If a diver rises too fast, the dissolved N2 gas can form bubbles in the blood, a dangerous and painful condition called “the bends.” • For deep descents, helium, which does not dissolve in the blood, is added to O2.

  49. Guide to Solving for Partial Pressure

  50. Gases We Breathe The air we breathe • is a gas mixture • contains mostly N2 and O2 and small amounts of other gases

More Related