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Gases

Gases. Chapter 12. Properties of Gases. Compressibility Mass Volume (fill container) Exert Pressure Diffuse through other gases Low Density. Kinetic Molecular Theory (KMT). All gases are matter (have mass and take up space). These particles are in constant, rapid and random motion.

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Gases

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  1. Gases Chapter 12

  2. Properties of Gases • Compressibility • Mass • Volume (fill container) • Exert Pressure • Diffuse through other gases • Low Density

  3. Kinetic Molecular Theory (KMT) • All gases are matter (have mass and take up space). • These particles are in constant, rapid and random motion. • All collisions are perfectly elastic. (no energy is lost in the collision) • The force of gas particle collisions on the walls of the container creates pressure; however, gas particles do not exert force on each other. • At a given temperature, all gas particles have the same amount of kinetic energy (temperature). • The distance between gas particles is very large.

  4. Gas Pressure Gas pressure results from the force of colliding particles on a given area. The metric unit of force is the newton (similar to the ounce in the English system). A newton of force acting on a square meter is called a pascal. Normal atmospheric pressure at sea level is 101.3 kPa.

  5. Measuring Gases • In order to describe a gas, we use four variables to make predictions about the behavior of the particles. • Amount of a gas (n) expressed in moles • Amount of space the gas takes up, volume (V), as measured in liters. *note: this is the volume of the container • Measure of kinetic energy, temperature (T), in degrees Kelvin • *note: K = C + 273 • Measure of the force exerted by the gas particles on the walls of the container, pressure (P)

  6. Kinetic Energy and Temperature Temperature is an indication of kinetic energy. T1 T2 Fraction of Particles T2 > T1 Speed of Particles

  7. The Kelvin Temperature Scale (Absolute Zero Scale) Kinetic Energy of Molecules 0 K = oC + 273 -273oC -300 -200 -100 0 100 oCelcius scale 0 73 173 273 373 Kelvin scale

  8. Pressure • Pressure is the measure of the amount of force of an object per unit area. • Newton / m2 = Pascal • Atmospheric Pressure is the amount of pressure exerted by the entire atmosphere on the surface of the Earth. • A barometer is an instrument that measures atmospheric pressure • A manometer is an instrument that measures the pressure of an enclosed gas

  9. Mercury Barometer

  10. Manometer

  11. STP • Pressure and temperature changes affect gas volume. • To compare experimental results, scientist convert their results to standard temperature and pressure (STP) • Standard Conditions: Standard Temperature = 0oC(273 K) Standard Pressure = 101.3 kPa (1atm) {760 torr}

  12. Pressure Units • Kilopascals (kPa) • Atmospheres (atm) 1 atm = 101.3 kPa • Torricellis (torr) 1 torr = 1 mm Hg 760 torr = 101.3 kPa (other units are listed on page 420)

  13. Pressure Units Atmosphere (atm) Millimeter Hg (mmHg) Pascal (Pa) Pounds per square inch (psi) Kilopascal (kPa) • 1 atm = Standard Pressure • 1 atm = 760 mmHg • 1 atm = 101,325 Pa • 1 atm = 14.7 psi • 1 atm = 101.3 kPa

  14. Conversions • Covert using Dimensional Analysis • Information you have x conversion factor (unknown/known) • Convert 665 mmHg to kPa • 665 mmHg x 101.3 kPa = 88.6 kPa 760 mmHg • Classwork: page 421 #1-4 • Homework: page 422 #1-12

  15. Boyle’s Law Pressure and Volume are inversely proportional at a given temperature and number of molecules. VP = k (a constant value)

  16. Pressure Effects on Volume V  1/P

  17. Inverse Proportionality

  18. Boyle’s Law continued • V1P1 = k1 and V2P2 = k2 since k1 = k2 (that’s why its called a constant) V1P1 = V2P2 This is the mathematical expression for Boyle’s Law

  19. A Boyle’s Law Problem A gas in a 242 cm3 container exerts a pressure of 87.6 kPa. What volume would the gas occupy at standard atmospheric pressure? Identify Variables V1 = V2 = P1 = P2 = 242 cm3 ? 87.6 kPa 101.3 kPa

  20. Two Methods of Solving • Algebraic: • V1P1 = V2P2 • V2 = V1P1 • P2 V2 = 242 cm3x 87.6 kPa 101.3 kPa V2 = 209 cm3

  21. Another Method 2. Logic and reason: The new volume (V2) depends upon the old volume (V1) multiplied by a pressure change. V2 = 242 cm3x _______ Since the pressure is increasing, the volume should decrease. Multiply by a fraction less than 1. 87.6 101.3

  22. Charles’s Law At a given pressure and number of molecules, Volume and Temperature are directly proportional. V = k T

  23. Two Temperature Scales b y = mx + b V = kT + b V = kT + 0 V = kT 0 Volume b -300 -200 -100 0 100 oCelcius scale 0 73 173 273 373 Kelvin scale

  24. Charles’s Law Continued • V1 = k1 T1 and V2 = k2 T2 since k1 = k2 V1 = V2 T1 T2 This is the mathematical expression for Charles’s Law

  25. A Charles’s Law Problem A 225 cm3 volume of gas is collected at 58oC. What volume would the gas occupy at standard temperature? Identify Variables V1 = V2 = T1 = T2 = 225 cm3 ? 58oC + 273 = 331 K 0oC + 273 = 273 K

  26. Two Methods of Solving • Algebraic: • V1 / T1 = V2 / T2 • V2 = V1T2 • T1 • V2 = 225 cm3 x 273 K • 331 K • V2 = 186 cm3

  27. The Other Method 2. Logic and reason: The new volume (V2) depends upon the old volume (V1) multiplied by a temperature change. V2 = 225 cm3x _______ Since the temperature is decreasing, the volume should decrease. Multiply by a fraction less than 1. 273 331

  28. Gay-Lussac’s Law P1 = k1 T1 and P2 = k2 T2 since k1 = k2 P1 = P2 T1 T2 This is the mathematical expression for Gay-Lussac’s Law

  29. The Combined Gas Law • In most laboratory situations, both temperature and pressure change. • The Combined Gas Law is used to calculate the affects of these changes.

  30. A Combination of Gas Laws Boyle’s Law: V  1/P Charles’s Law: V  T  V  T/ P To change a proportionality to an equality… multiply by a proportionality constant. V = kT P VP = k T Gay-Lussac’s Law: P  T V1P1 = V2P2 T1 T2

  31. V1P1=V2P2 T1 T2Solve for each variable V2 = V1P1T2 / T1P2 T2 = V2P2T1 / V1P1 P2 = V1P1T2 / T1V2

  32. Sample Problem What volume would 955 cm3 of a gas measured at 58 oC and 108.0 kPaoccupy at 76 oC and 123.0 kPa? V1 = V2 = T1 = T2 = P1 = P2 = 955 cm3 ? 331 K 349 K 108.0 kPa 123.0 kPa

  33. Method #1 V1P1 / T1 = V2P2 / T2 V2 = V1P1T2 / (T1P2) V2 = (955 cm3) (108.0 kPa) (349 K) (331 K) (123.0 kPa) V2 = 884 cm3

  34. Method #2 V2 = 955 cm3 x --------- x --------- temperature increases volume increases 108 349 331 123 pressure increases volume decreases V2 = 884 cm3

  35. Gas Law Variables P V T When T is constant P V When P is constant TV When V is constant T P

  36. Dalton’s Law of Partial Pressures • Gases are often collected by water displacement. • These gases contain water vapor. • To find the pressure of the dry gas alone, Dalton’s Law of Partial Pressures is used. • The total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each gas in the mixture.

  37. Dalton’s Law of Partial Pressures • Gases are often collected by water displacement. • These gases contain water vapor. • To find the pressure of the dry gas alone, Dalton’s Law of Partial Pressures is used. • The total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each gas in the mixture.

  38. The Mathematical Expression of Dalton’s Law PT = P1 + P2 + …. Pn Example: Gas A = 1.0 atm Gas B = 1.5 atm Gas C = 0.5 atm PT = 3.0 atm

  39. The Pressure of a “Dry” Gas When a gas is collected over water, the total pressure is the result of the dry gas plus the water vapor pressure. PT = Pgas + Pvapor To determine the pressure of the dry gas alone, subtract the water vapor pressure from the total pressure. ( P. 401 ) Pgas = PT - Pvapor

  40. Diffusion • If a bottle of ammonia were opened at the front of the room, the odor could soon be detected at the back of the room. Diffusion is the random movement of particles through another substance. Effusion is gas under pressure escaping through a small opening.

  41. Diffusion

  42. Kinetic Energy The equation for finding the kinetic energy of a particle: K.E. = ½ mv2 m = mass of the particle v = velocity (speed) of the particle

  43. Kinetic Energy K.E. = ½ mv2 #@*~!!*~##~@!&

  44. Graham’s Law • If gases A and B are at the same temperature….. ____ ____ K.E.A = K.E.B • ½ mAv2A = ½ mBv2B • mAv2A = mBv2B Collecting terms… • v2A=mB v2B mA vA=mB vB mA This is the mathematical expression for Graham’s Law

  45. The Meaning of Graham’s Law “The ratio of the velocities of two gases is equal to the square root of the inverse ratios of their masses.” mAv2A = mBv2B if: mA > mB then: vB > vA (The lightest gas always travels faster) See practice problems on page 438

  46. HCl NH3 A Graham’s Law Experiment HCl(g) + NH3(g) NH4Cl(s) vA / vB = √mB / mA (accepted) • v = d/t • vA = dA/tA • vB = dB/tB • tA = tB • vA / vB = dA / dB (experimental) % error: | exp. – accepted | accepted x 100

  47. Ideal Gases • The gas laws we have studied only apply to ideal gases. • There are no ideal gases.

  48. Properties of Ideal Gases • Ideal gases have: 1. Point mass (Particles have mass but occupy no volume) 2. No mutual attractions (No matter how close or how slow the particles are, they do not attract each other)

  49. Ideal Gas Laws Work • Real gas particles occupy negligible volume compared to the total volume occupied by the gas… (unless under very high pressures)

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