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Chapter 5: the Gaseous state

Chapter 5: the Gaseous state. Vanessa Prasad- Permaul Valencia Community College CHM 1045. GAS LAWS: PRESSURE AND MEASUREMENT. Elements that exist as gases at 25 0 C and 1 atmosphere. GAS LAWS: PRESSURE AND MEASUREMENT. Physical Characteristics of Gases.

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Chapter 5: the Gaseous state

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  1. Chapter 5: the Gaseous state Vanessa Prasad-Permaul Valencia Community College CHM 1045

  2. GAS LAWS: PRESSURE AND MEASUREMENT Elements that exist as gases at 250C and 1 atmosphere

  3. GAS LAWS: PRESSURE AND MEASUREMENT Physical Characteristics of Gases • Gases assume the volume and shape of their containers. • Gases are the most compressible state of matter. • Gases will mix evenly and completely when confined to the same container. • Gases have much lower densities than liquids and solids.

  4. GAS LAWS: PRESSURE AND MEASUREMENT a) Gas is a large collection of particles moving at random throughout a volume b) Collisions of randomly moving particles with the walls of the container exert a force per unit area that we perceive as gas pressure

  5. Force Area GAS LAWS: PRESSURE AND MEASUREMENT HOW IS PRESSURE DEFINED? The force the gas exerts on a given area of the container in which it is contained. The SI unit for pressure is the Pascal, Pa. Pressure = • If you’ve ever inflated a tire, you’ve probably made a pressure measurement in pounds (force) per square inch (area).

  6. Barometer GAS LAWS: PRESSURE AND MEASUREMENT Units of Pressure Height is proportional to the barometric pressure 1 pascal (Pa) = 1 kg/m·s2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa 1 bar = 105 Pa 1 atm = 14.69 lb/in2 760mm • Hg is used instead of H2O : • more dense • better visibility • accuracy mercury

  7. GAS LAWS: PRESSURE AND MEASUREMENT EXERCISE 5.1 A GAS CONTAINER HAD A MEASURED PRESSURE OF 57kPa. CALCULATE THE PRESSURE IN UNITS OF ATM AND mmHg. • First, convert to atm (57 kPa = 57 x 103 Pa). 57 x 103 Pa x 1 atm = 0.562 = 0.56 atm 1.01325 x 105 Pa • Next, convert to mmHg. 57 x 103 Pa x 760 mmHg = 427.5 = 4.3 x 102 mmHg 1.01325 x 105 Pa

  8. GAS LAWS: EMPIRICAL GAS LAWS EMPIRICAL GAS LAWS You can predict the behavior of gases based on the following properties: • Volume • Amount (moles) • Pressure • Temperature * If two of these physical properties are held constant, it is possible to show a simple relationship between the other two…

  9. GAS LAWS: EMPIRICAL GAS LAWS Boyle’s experiment: A manometer to study the relationship between pressure (P) & Volume (V) of a gas Hg Hg As P (h) increases V decreases

  10. GAS LAWS: EMPIRICAL GAS LAWS BOYLE’S LAW the volume of a sample of gas at a given temperature varies inversely with the applied pressure So…. For a given amount of gas (n) & @ constant temperature (T) : If pressure (P) increases, the volume (V) of the gas decreases P1 * V1 = P2 * V2

  11. Boyle’s Law • Pressure–Volume Law (Boyle’s Law):

  12. GAS LAWS: EMPIRICAL GAS LAWS Boyle’s Law P a 1/V P * V = constant P1 * V1 = P2 * V2 Constant temperature Any given amount of gas

  13. GAS LAWS: EMPIRICAL GAS LAWS EXERCISE 5.2 A volume of carbon dioxide gas equivalent to 20.0 L was collected @ 23oC and 1.00atm pressure. What would be the volume of gas at constant temperature and 0.830atm? P1 * V1 = P2 * V2 Application of Boyle’s law gives V2 = V1 x P1 = 20.0 L x 1.00atm = 24.096 = 24.1 L P2 0.830atm

  14. GAS LAWS: EMPIRICAL GAS LAWS Charles’ Law: relating volume and temperature As T increases V increases

  15. Charles’ Law • Temperature–Volume Law (Charles’ Law):

  16. GAS LAWS: EMPIRICAL GAS LAWS Charles’ Law Variation of gas volume with temperature V a T For a given amount of gas at constant pressure Temperature must be in Kelvin V / T = constant T (K) = t (0C) + 273.15

  17. GAS LAWS: EMPIRICAL GAS LAWS CHARLES’ LAW the volume occupied by any sample of gas at a constant pressure is directly proportional to the absolute temperature So…. For a given amount of gas (n) & @ constant pressure (P) : If temperature (T) increases, the volume (V) of the gas increases V1V2 T1 =T2

  18. GAS LAWS: EMPIRICAL GAS LAWS EXERCISE 5.3 A chemical reaction is expected to produce 4.3dm3 of oxygen at 19oC and 101kPa. What will the volume be at constant pressure and 25oC? First, convert the temperatures to the Kelvin scale. Ti = (19 + 273) = 292 K Tf = (25 + 273) = 298 K Following is the data table. Vi = 4.38 dm3Pi = 101 kPaTi = 292 K Vf = ? Pf = 101 kPaTf = 298 K Apply Charles’s law to obtain Vf = Vi x Tf = 4.38 dm3 x 298K = 4.470 = 4.47 dm3 Ti 292K

  19. GAS LAWS: EMPIRICAL GAS LAWS COMBINED GAS LAW Relating Volume, Temperature and Pressure Taking Boyle’s Law and Charles’ Law: The volume occupied by a given amount of gas is proportional to the absolute temperature divided by the pressure: V = constant x T or PV = constant (for a given amount of gas) P T P1V1 = P2V2 T1 T2

  20. Combined Gas Law We can combine Boyle’s and charles’ law to come up with the combined gas law Use Kelvins for temp, any pressure, any volume

  21. GAS LAWS: EMPIRICAL GAS LAWS EXERCISE 5.4 A balloon contains 5.41dm3 of helium at 24oC and 10.5kPa. Suppose the gas in the balloon is heated to 35oC and the pressure is now 102.8kPa, what is the volume of the gas? First, convert the temperatures to kelvins. Ti = (24 + 273) = 297 K Tf = (35 + 273) = 308 K Following is the data table. Vi = 5.41 dm3Pi = 101.5 kPaTi = 297 K Vf = ? Pf = 102.8 kPaTf = 308 K Apply both Boyle’s law and Charles’s law combined to get Vf = Vi x Pi x Tf =5.41 dm3 x 101.5kPa x 308K = 5.539 = 5.54 dm3 PfTi 102.8kPa 297K

  22. GAS LAWS: EMPIRICAL GAS LAWS AVOGADRO’S LAWrelating volume and amount Va number of moles (n) V = constant x n equal volumes of any two gases at the same temperature & pressure contain the same number of molecules V1/n1 = V2/n2 Constant temperature Constant pressure

  23. Avogadro’s Law • The Volume–Amount Law (Avogadro’s Law):

  24. Avogadro’s Law • The Volume–Amount Law (Avogadro’s Law): • At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present. • Use any volume and moles • V1 = V2 • n1 n2

  25. Experiments show that at STP, 1 mole of an ideal gas occupies 22.4 L. GAS LAWS: EMPIRICAL GAS LAWS Avogadro’s Number = one mole of any gas contains the same number of molecules 6.023 x 1023 Must occupy the same volume at a given temperature and pressure The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

  26. Boyle’s law: V a (at constant n and T) V a nT nT nT P P P V = constant x = R 1 P GAS LAWS: THE IDEAL GAS LAW IDEAL GAS LAW Charles’ law: VaT (at constant n and P) Avogadro’s law: V an (at constant P and T) R is the molar gasconstant PV = nRT

  27. R = (1 atm)(22.414L) PV = nT (1 mol)(273.15 K) GAS LAWS: THE IDEAL GAS LAW IDEAL GAS EQUATION PV = nRT R = 0.082057 L • atm / (mol • K) R = 8.3145 J / (mol • K) R = 1.9872 cal / (mol • K) *The units of pressure times volume are the units of energy joules (J) or calories (cal)

  28. The Ideal Gas Law • Ideal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro. • 1 mole of an ideal gas occupies 22.414 L at STP • STP conditions are 273.15 K and 1 atm pressure • The gas constant R = 0.08206 L·atm·K–1·mol–1 • P has to be in atm • V has to be in L • T has to be in K

  29. GAS LAWS: THE IDEAL GAS LAW EXERCISE 5.5 Show that the moles of gas are proportional to the pressure for constant volume and temperature Use the ideal gas law, PV = nRT, and solve for n: n = PV RT n = V x P RT Note that everything in parentheses is constant. Therefore, you can write n = constant x P Or, expressing this as a proportion, nP

  30. GAS LAWS: THE IDEAL GAS LAW EXERCISE 5.6 What is the pressure in a 50.0L gas cylinder that contains 3.03kg of oxygen at 23oC? T = 23oC + 273K = 296K V = 50.0L R = 0.08206 L . atm/K . mol n = 3.03kg x 1000g x 1mol = 94.688 mol O2 1 kg 31.998g P = ? PV= nRT P = nRT = 0.0347mol x 0.08206 L . atm x 296K V K . mol = 46.0 atm 50.0L

  31. The Ideal Gas Law • Density and Molar Mass Calculations: • You can calculate the density or molar mass (MM) of a gas. The density of a gas is usually very low under atmospheric conditions.

  32. GAS LAWS: THE IDEAL GAS LAW EXERCISE 5.7 Calculate the density of helium (g/L) at 21oC and 752mmHg. The density of air under these conditions is 1.188g/L. What is the difference in mass between 1 liter of air and 1 liter of helium? 32

  33. GAS LAWS: THE IDEAL GAS LAW Using the ideal gas law, solve for n, the moles of helium. n = PV = 0.98947 atm x 1L = RT 0.08206 L . atm/ K . mol x 294K 0.4101 mol Now convert mol He to grams. 0.04101 mol He x 4.00g He = 0.16404g He = 0.164g/L 1.00 mol He 1 L Therefore, the density of He at 21°C and 752 mmHg is 0.164 g/L. The difference in mass between one liter of air and one liter of He: mass air − mass He = 1.188 g − 0.16404 g = 1.02396 = 1.024 g difference

  34. GAS LAWS: THE IDEAL GAS LAW EXERCISE 5.8 A sample of a gaseous substance at 25oC and 0.862 atm has a density of 2.26 g/L. What is the molecular mass of the substance? Variable Value P 0.862 atm V 1 L (exact number) T (25 + 273) = 298 K n ? From the ideal gas law, PV = nRT, you obtain n = PV = 0.862 atm x 1 L = 0.03525 mol RT 0.08206 L . atm/K . Mol x 298K 34 34

  35. GAS LAWS: THE IDEAL GAS LAW Dividing the mass of the gas by moles gives you the mass per mole (the molar mass). Molar mass = grams of gas = 2.26g = 64.114 g/mol moles of gas 0.03525mol Therefore, the molecular mass is 64.1 amu.

  36. GAS LAWS: GAS MIXTURES Dalton’s Law of Partial Pressures V and T are constant P1 P2 Ptotal= P1 + P2

  37. Dalton’s Law of Partial Pressures • For a two-component system, the moles of components A and B can be represented by the mole fractions (XA and XB). Mole fraction is related to the total pressure by:

  38. GAS LAWS: GAS MIXTURES EXERCISE 5.10 A 10.0L flask contains 1.031g O2 and 0.572g CO2 at 18oC. What are the partial pressures of each gas? What is the total pressure? What is the mole fraction of oxygen in this mixture? Each gas obeys the ideal gas law. 1.031 g O2 x 1 mol = 0.0322188 mol O2 32.00g P = nRT = 0.0322mol x 0.0821L.atm/K.mol x 291K = 0.076936 atm V 10.0L 0.572 g CO2 x 1 mol = 0.012997 mol CO2 44.01g P = nRT = 0.0130mol x 0.0821L.atm/K.mol x 291K = 0.031036 atm V 10.0L

  39. GAS LAWS: GAS MIXTURES The total pressure is equal to the sum of the partial pressures: PT = PO2 + PCO2 = 0.076936atm + 0.031036atm = 0.10797 = 0.1080 atm The mole fraction of oxygen in the mixture is Mole fraction O2 = PO2 = 0.076936atm = 0.7122 = 0.712 PT 0.1080atm 0.712 x 100% = 71.2 mole % of O2 in this gas mixture

  40. Kinetic Molecular Theory • This theory presents physical properties of gases in terms of the motion of individual molecules. • Average Kinetic Energy  Kelvin Temperature • Gas molecules are points separated by a great distance • Particle volume is negligible compared to gas volume • Gas molecules are in constant random motion • Gas collisions are perfectly elastic • Gas molecules experience no attraction or repulsion

  41. Kinetic Molecular Theory

  42. Average Kinetic Energy (KE) is given by: U = average speed of a gas particle R = 8.314 J/K mol m = mass in kg MM = molar mass, in kg/mol NA = 6.022 x 1023

  43. The Root–Mean–Square Speed: is a measure of the average molecular speed. Taking square root of both sides gives the equation

  44. Example 17: Calculate the root–mean–square speeds of helium atoms and nitrogen molecules in m/s at 25°C.

  45. Kinetic Molecular Theory • Maxwell speed distribution curves.

  46. Graham’s Law • Diffusion is the mixing of different gases by random molecular motion and collision.

  47. Graham’s Law • Effusion is when gas molecules escape without collision, through a tiny hole into a vacuum.

  48. Graham’s Law • Graham’s Law: Rate of effusion is proportional to its rms speed, urms. • For two gases at same temperature and pressure:

  49. Behavior of Real Gases • At higher pressures, particles are much closer together and attractive forces become more important than at lower pressures.

  50. Behavior of Real Gases • The volume taken up by gas particles is actually less important at lower pressures than at higher pressure. As a result, the volume at high pressure will be greater than the ideal value.

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