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Chapter 11 Gases

Chapter 11 Gases. 2006, Prentice Hall. CHAPTER OUTLINE. PROPERTIES OF GASES. Gases occupy much greater space than the same amount of liquid or solid. This is because the gas particles are spaced apart from one another and are therefore compressible.

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Chapter 11 Gases

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  1. Chapter 11 Gases 2006, Prentice Hall

  2. CHAPTER OUTLINE

  3. PROPERTIESOF GASES • Gases occupy much greater space than the same amount of liquid or solid. • This is because the gas particles are spaced apart from one another and are therefore compressible. • Solid or liquid particles are spaced much closer and cannot be compressed. • Gases are the least dense and most mobile of the three phases of matter. • Particles of matter in the gas phase are spaced far apart from one another and move rapidly and collide with each other often.

  4. PROPERTIESOF GASES • Gases are characterized by four properties. • These are: 1. Pressure (P) 2. Volume (V) 3. Temperature (T) 4. Amount (n)

  5. KINETIC-MOLECULARTHEORY • The KMT consists of several postulates: • Scientists use the kinetic-molecular theory (KMT) to describe the behavior of gases. 1. Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities. 2. Gas particles have little attraction for one another. Therefore attractive forces between gas molecules can be ignored. 3. The distance between the particles is large compared to their size. Therefore the volume occupied by gas molecules is small compared to the volume of the gas.

  6. KINETIC-MOLECULARTHEORY 4. Gas particles move in straight lines and collide with each other and the container frequently. The force of collision of the gas particles with the walls of the container causes pressure. 5. The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin). Animation

  7. Kinetic Molecular Theory Tro's Introductory Chemistry, Chapter

  8. PRESSURE &ITS MEASUREMENT • Pressure is the result of collision of gas particles with the sides of the container. Pressure is defined as the force per unit area. • Pressure is measured in units of atmosphere (atm) or mmHg or torr. The SI unit of pressure is pascal (Pa) or kilopascal (kPa). 1 atm = 760 mmHg = 101.325 kPa 1 mmHg = 1 torr

  9. Common Units of Pressure Tro's Introductory Chemistry, Chapter

  10. PRESSURE &ITS MEASUREMENT • Atmospheric pressure can be measured with the use of a barometer. • Mercury is used in a barometer due to its high density. • At sea level, the mercury stands at 760 mm above its base.

  11. Atmospheric Pressure & Altitude • the higher up in the atmosphere you go, the lower the atmospheric pressure is around you • at the surface the atmospheric pressure is 14.7 psi, but at 10,000 ft is is only 10.0 psi • rapid changes in atmospheric pressure may cause your ears to “pop” due to an imbalance in pressure on either side of your ear drum Tro's Introductory Chemistry, Chapter

  12. Pressure Imbalance in Ear If there is a difference in pressure across the eardrum membrane, the membrane will be pushed out – what we commonly call a “popped eardrum.” Tro's Introductory Chemistry, Chapter

  13. 1 760 Example 1: The atmospheric pressure at Walnut, CA is 740. mmHg. Calculate this pressure in torr and atm. 1 mmHg = 1 torr 740. mmHg = 740. torr 740. mmHg = 0.974 atm

  14. 760 1 Example 2: The barometer at a location reads 1.12 atm. Calculate the pressure in mmHg and torr. 1.12 atm = 851 mmHg 1 mmHg = 1 torr 851 mmHg = 851 torr

  15. PRESSURE & MOLES OF A GAS • The pressure of a gas is directly proportional to the number of particles (moles) present. The greater the moles of the gas, the greater the pressure

  16. BOYLE’S LAW • At constant temperature, the volume of a fixed amount of gas is inversely proportional to its pressure. • The relationship of pressure and volume in gases is called Boyle’s Law. As pressure increases, the volume of the gas decreases Pressure and volume of a gas are inversely proportional

  17. BOYLE’S LAW • Boyle’s Law can be mathematically expressed as P1 x V1 = P2 x V2 P and V under initial condition P and V under final condition

  18. Example 1: A sample of gas has a volume of 12-L and a pressure of 4500 mmHg. What is the volume of the gas when the pressure is reduced to 750 mmHg? Pressure decreases Volume increases P1= 4500 mmHg P2= 750 mmHg V1 = 12 L V2 = ??? P1 x V1 = P2 x V2 V2 = 72 L

  19. Example 2: A sample of hydrogen gas occupies 4.0 L at 650 mmHg. What volume would it occupy at 2.0 atm? Pressure increases Volume decreases P2 = 2.0 atm = 1520 mmHg P1= 650 mmHg P2= 2.0 atm V1 = 4.0 L V2 = ??? P1 x V1 = P2 x V2 Pressures must be in the same unit V2 = 1.7 L

  20. CHARLES’S LAW • At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature. • The relationship of temperature and volume in gases is called Charles’ Law. Temperature and volume of a gas are directly proportional As temperature increases, the volume of the gas increases

  21. T must be in units of K CHARLES’S LAW • Charles’ Law can be mathematically expressed as V and T under initial condition V and T under final condition

  22. Example 1: A 2.0-L sample of a gas is cooled from 298K to 278 K, at constant pressure. What is the new volume of the gas? Temperature decreases Volume decreases V1 = 2.0 L V2 = ??? T1 = 298 K T2 = 278 K V2 = 1.9 L

  23. Example 2: If 20.0 L of oxygen is cooled from 100ºC to 0ºC, what is the new volume? Temperatures must be in K Volume decreases Temperature decreases T1= 100ºC + 273 = 373 K V1 = 20.0 L V2 = ??? T1 = 100ºC T2 = 0ºC T2= 0ºC + 273 = 273 K V2 = 14.6 L

  24. GAY-LUSSAC’SLAW • The relationship of temperature and pressure in gases is called Gay-Lussac’s Law. • At constant volume, the pressure of a fixed amount of gas is directly proportional to its absolute temperature. Temperature and pressure of a gas are directly proportional As temperature increases, the pressure of the gas increases

  25. T must be in units of K GAY-LUSSAC’SLAW • Gay-Lussac’s Law can be mathematically expressed as P and T under final condition P and T under initial condition

  26. Example 1: An aerosol spray can has a pressure of 4.0 atm at 25C. What pressure will the can have if it is placed in a fire and reaches temperature of 400C ? Temperatures must be in K P1 = 4.0 atm P2 = ??? T1 = 25ºC T2 = 400ºC T1= 25ºC + 273 = 298 K T2= 400ºC + 273 = 673 K

  27. Example 1: An aerosol spray can has a pressure of 4.0 atm at 25C. What pressure will the can have if it is placed in a fire and reaches temperature of 400C ? Temperature increases Pressure increases P1 = 4.0 atm P2 = ??? T1 = 298 K T2 = 673 K P2 = 9.0 atm

  28. Example 2: A cylinder of gas with a volume of 15.0-L and a pressure of 965 mmHg is stored at a temperature of 55C. To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg? Pressure decreases Temperature decreases P1 = 965 mmHg P2 = 850 mmHg T1 = 55ºC T2 = ??? T1= 55ºC + 273 = 328 K T2 = 289 K T2 = 16ºC

  29. VAPOR PRESSURE& BOILING POINT • In an open container, liquid molecules at the surface that possess sufficient energy, can break away from the surface and become gas particles or vapor. • In a closed container, these gas particles can accumulate and create pressure called vapor pressure. • Vapor pressure is defined as the pressure above a liquid at a given temperature. Vapor pressure varies with each liquid and increases with temperature.

  30. VAPOR PRESSURE& BOILING POINT • Listed below is the vapor pressure of water at various temperatures.

  31. VAPOR PRESSURE& BOILING POINT • A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure). • For example, at sea level, water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature.

  32. VAPOR PRESSURE& BOILING POINT • At higher altitudes, where atmospheric pressure is lower, water reaches boiling point at temperatures lower than 100C. • For example, in Denver, where atmospheric pressure is 630 mmHg, water boils at 95C, since its vapor pressure is 630 mmHg at this temperature.

  33. COMBINEDGAS LAW • This law is useful for studying the effect of changes in two variables. • All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law. Initial condition Final condition

  34. Boyle’s Law Charles’s Law Gay-Lussac’s Law COMBINEDGAS LAW • The individual gas laws studied previously are embodied in the combined gas law.

  35. Example 1: A 25.0-mL sample of gas has a pressure of 4.00 atm at a temperature of 10C. What is the volume of the gas at a pressure of 1.00 atm and a temperature of 18C? Both pressure and temp. change P1= 4.00 atm V1= 25.0 mL T1= 283 K P2= 1.00 atm V2= ??? T2 = 291 K Must use combined gas law V2 = 103 mL

  36. AVOGADRO’SLAW • At constant temperature and pressure, the volume of a fixed amount of gas is directly proportional to the number of moles. • The relationship of moles and volume in gases is called Avogadro’s Law. As number of moles increases, the volume of the gas increases Number of moles and volume of a gas are directly proportional

  37. Avogadro’s Law Tro's Introductory Chemistry, Chapter

  38. AVOGADRO’SLAW • As a result of Avogadro’s Law, equal volumes of different gases at the same temp. and pressure contain equal number of moles (molecules). 2 Tanks of gas of equal volume at the same T & P contain the same number of molecules

  39. AVOGADRO’SLAW • Avogadro’s Law also allows chemists to relate volumes and moles of a gas in a chemical reaction. This relationship is only valid for gaseous substances For example: 2 H2 (g) + 1 O2 (g)  2 H2O(g) 2 molecules 1 molecule 2 molecules 2 moles 1 mole 2 moles 2 Liters 1 Liter 2 Liters

  40. Example 1: A sample of helium gas with a mass of 18.0 g occupies 1.6 Liters at a particular temperature and pressure. What mass of oxygen would occupy1.6 L at the same temperature and pressure? Same Temp. & Pressure mol of He = 4.50 mol mol of O2 = mol of He = 4.50 mol mass of O2 = 144 g

  41. 2 3 Example 2: How many Liters of NH3 can be produced from reaction of 1.8 L of H2 with excess N2, as shown below? N2 (g) + 3 H2 (g)  2 NH3 (g) 1.8 L H2 = 1.2 L NH3

  42. STP & MOLAR VOLUME • To better understand the factors that affect gas behavior, a set of standard conditions have been chosen for use, and are referred to as Standard Temperature and Pressure (STP). STP = 0ºC (273 K) 1 atm (760 mmHg)

  43. Molar Volume at STP STP & MOLAR VOLUME • At STP conditions, one mole of any gas is observed to occupy a volume of 22.4 L. V = 22.4 L

  44. 1 22.4 Example 1: If 2.00 L of a gas at STP has a mass of 3.23 g, what is the molar mass of the gas? mol of gas = = 0.0893 mol Molar volume at STP Molar mass = = 36.2 g/mol

  45. Example 2: A sample of gas has a volume of 2.50 L at 730 mmHg and 20C. What is the volume of this gas at STP? P1= 730 mmHg V1= 2.50 L T1= 293 K P2= 760 mmHg V2= ??? T2 = 273 K V2 = 2.24 L

  46. Universal gas constant IDEAL GAS LAW • Combining all the laws that describe the behavior of gases, one can obtain a useful relationship that relates the volume of a gas to the temperature, pressure and number of moles. R = 0.0821 L atm/mol K

  47. IDEAL GAS LAW • This relationship is called the Ideal Gas Law, and commonly written as: P V = n R T Temp. in K Pressure in atm Number of moles Volume in Liters

  48. Example 1: A sample of H2 gas has a volume of 8.56 L at a temp. of 0C and pressure of 1.5 atm. Calculate the moles of gas present. PV = nRT P= 1.5 atm V= 8.56 L T= 273 K n = ??? R = 0.0821 = 0.57 mol

  49. Example 2: What volume does 40.0 g of N2 gas occupy at 10C and 750 mmHg? Moles must be calculated from mass mol of N2 = = 1.43 mol P=750 mmHg V= ??? T= 283 K n = not given m = 40.0 g R = 0.0821 Pressure must be converted to atm P = 750 mmHg P = 0.987 atm

  50. Example 2: What volume does 40.0 g of N2 gas occupy at 10C and 750 mmHg? P=0.987 atm V= ??? T= 283 K n = 1.43 R = 0.0821 V = 33.7 L

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