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The Gaseous Phase

The Gaseous Phase

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The Gaseous Phase

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  1. The Gaseous Phase The three phases of matter, solids, liquids and gases, have different characteristics. A gas expands to fill any container it occupies A liquid has a fixed volume but takes the shape of the volume of the container it occupies A solid has both fixed volume and shape. These characteristics originate from the nature of the interactions between the atoms or molecules

  2. On a macroscopic scale, gases are distinguished from solids and liquids by their much smaller values of density. On the microscopic scale, the smaller values of density arise due to the much lower NUMBER DENSITY (number of molecules per cm3 of the sample) compared with liquids and solids.

  3. Understanding the behavior of gases and how reactions occur in the gas phase is of practical importance CH4(g) + O2(g) --> CO2(g) + H2O(g) - combustion of fuels N2(g) + H2(g) --> NH3(g) - production of ammonia for fertilizers 2NO(g) + O2 (g) -> 2NO2 (g) - responsible for acid rain

  4. Properties of Gases - A gas will fill the volume of the container which contains it. - The volume of the gas equals the volume of its container - Gases are highly compressible; when pressure is applied to a gas, its volume readily decreases - Gases form homogenous mixtures with each other regardless of their identity or relative proportions These properties arise because the individual atoms/ molecules are relatively far apart

  5. Three properties of gases that are used to describe gases are pressure (P), volume (V) and temperature (T). The volume of a gas is defined by the volume of the container. Typical units for volume of gases is the liter, L.

  6. Pressure = Force Area SI Units for pressure Force is newton, N (=kg m/s2) Area - m2 Pressure - N/m2 or pascal (Pa) PRESSURE The force exerted by a gas on a unit area of the walls of its container is the pressure exerted by the gas.

  7. Atmospheric Pressure Because of gravity, the earth’s atmosphere exerts a downward force and consequently a pressure on the earth’s surface. Atmospheric pressure: pressure exerted by the atmosphere around us A column of air 1m2 in cross section extending through the atmosphere has a mass or roughly 10,000 kg.

  8. The pressure exerted by this air column ~ 1 x 105 N 1 m2 The acceleration produced by earth’s gravity is 9.8 m/s2 force = mass x acceleration Force exerted by this air column is ~ 1 x 105 N ~ 1 x 105 Pa More precisely, 1.01325 x 105 Pa = 1 atmosphere (atm)

  9. A barometer operates on the principle that the height of a liquid in a closed tube depends on the atmospheric pressure Pressure = g h d g ~ 9.8 m/s h is the height of the liquid in the sealed tube d is the density of the liquid

  10. a) What is the height of a mercury column in a barometer at atmospheric pressure? b) What is the height of a water column in a barometer at atmospheric pressure? Explains why mercury is used in barometers and not water

  11. Units of pressure 1 atm = 760 mm Hg = 760 torr = 1.01325 x 105 Pa There are other units of pressure (lbs/in2, bar) but we will typically deal with atm, mm of Hg or torr and Pa.

  12. The Gas Laws Through experimental observations, relationships have been established between the pressure (P), temperature (T) and volume (V) and number of moles (n) of gases. These relationships are called the GAS LAWS. Having defined P, V, T, and n for a gas, this information defines the physical condition or state of a gas. The relationships between P, V, T and n that will be discussed hold for IDEAL gases (or for low pressures; “ideal” conditions)

  13. Relationship between Pressure and Volume: Boyle’s Law Boyle noted from the experiments he performed that at a fixed temperature and for a fixed amount of gas, as pressure on a gas increases, the volume occupied by a gas decreases.

  14. At a fixed temperature and for a fixed amount of gas P V = constant P a 1 V P = constant Volume Boyle’s Law The product of pressure and volume of a sample of gas is a constant, at constant temperature and for a fixed amount of gas.

  15. Plot illustrating P-V relationship

  16. Since, PV = constant P1V1 = P2V2 The conditions of 1.00 atm pressure and 0oC are called standard temperature and pressure (STP). Under STP conditions, the volume occupied by the gas in the J-tube is 22.4 L.

  17. Temperature-Volume Relation - Charles Law The volume of a fixed quantity of gas at constant pressure increases linearly with temperature. V = V0 + a V0 t V0 is the volume of the gas at 0oC t is the temperature in oC a is the coefficient of thermal expansion

  18. Volume V = V0 + a V0 t y = mx + b

  19. Volume From a plot of V vs t we can determine V0 from the y-intercept. From the slope = a V0, a can be determined

  20. a = 1 (oC-1) 273.15 Since gases expand by the same relative amount when heated between the same two temperatures (at low pressure) implies that a is the same for all ideal gases. For gases, at low pressure For liquids and solids a varies from substance to substance

  21. V V t = -1 V0 V0 a t = 273.15 oC [ -1 ] Re-writing the expression connecting V and t: Gas thermometer: By measuring the volume of a gas at 0oC and measuring the volume change as temperature changes, the temperature can be calculated

  22. t V = V0 [ 1 + ] 273.15 oC At t = -273.15 oC => volume of gas is zero temperatures < -273.15oC => negative volume which is physically impossible. Hence 273.15oC is the lowest temperature that can be physically attained and is the fundamental limit on temperature. Absolute temperature - Kelvin Scale

  23. t V = V0 [ 1 + ] 273.15 oC V0 T V = 273.15 This temperature is called ABSOLUTE ZERO and is defined to be the zero point on the Kelvin scale (K) T (Kelvin) = 273.15 + t (Celsius) If we substitute the above expression in and solve for V

  24. V0 273.15 at a fixed pressure and for a fixed amount of gas V1 T1 = V2 T2 Hence, V a T Charles’ Law In other words, on an absolute temperature scale, at a constant pressure and for a fixed amount if gas, the volume of the gas is proportional to the temperature is a constant Hence, Note: T is temperature in Kelvin

  25. Volume is affected not just by pressure and temperature, but also by the amount of gas. Avogadro’s hypothesis - Equal volumes of gases at the same temperature and pressure contain the same number of molecules. Quantity-Volume relation - Avogadro’s Law

  26. Avogadro’s law: the volume of a gas maintained at constant pressure and temperature is directly proportional to the number of moles of gas. V = constant x n Hence, doubling the moles of gas will cause the volume to double (as long as T and P remain constant)

  27. n T V a P Boyle’s law: V a P-1 (constant n, T) Charles’ law: V a T (constant n, P) Avogadro’s law: V a n (constant P, T) The Ideal-Gas Equation Putting the three laws together:

  28. V = R n T P P V = n R T IDEAL GAS EQUATION An ideal gas is a gas whose pressure, volume and temperature behavior is completely described by this equation. R is called the universalgas constant since it is the same for all gases. Note: The ideal gas equation is just that - ideal. The equation is valid for the most gases at low pressures. Deviations from “ideal” behavior are observed as pressure increases.

  29. Values of R Units Numerical value L-atm/(mol-K) 0.08206 cal/(mol-K) 1.987 J/(mol-K) 8.314 m3-Pa/(mol-K) 8.314 L-torr/(mol-K) 62.36 The value and units of R depends on the units of P, V, n and T Temperature, T, MUST ALWAYS BE IN KELVIN n is expressed in MOLES P is often in atm and V in liters, but other units can be used.

  30. Given: Volume of CO2 = 250 mL = 0.250 L Pressure of CO2 = 1.3 atm temperature of CO2 = 31oC Example: Calcium carbonate, CaCO3(s), decomposes upon heating to give CO2(g) and CaO(s). A sample of CaCO3 is decomposed, and the CO2 collected in a 250. mL flask. After the decomposition is complete, the gas has a pressure of 1.3 atm at a temperature of 31oC. How many moles of CO2 were generated? First convert temperature to K T = 31 + 273 = 304 K

  31. P V n = R T (1.3 atm)(0.250 L) n = (0.08206 L-atm/(mol-K)) (304K) Based on the units given for P and V, use appropriate value for R R = 0.08206 L-atm/(mol K) To calculate n: n = 0.013 mol CO2

  32. Pi n R P Pf = = constant Tf Ti V T Pi Tf Pf Pf = 3.6 atm = = Ti Since, the can is sealed, both V and n stay fixed. P (atm) t (oC) T(K) Initial 1.5 25 298 Final ? 450 723 Problem The gas pressure in a closed aerosol can is 1.5 atm at 25oC. Assuming that the gas inside obeys the ideal-gas equation, what would the pressure be if the can was heated to 450oC?

  33. The number of moles of a compound = mass of gas sample (m) Molar Mass (M) m n = M Molar Mass and Gas Density The ideal gas law, P V = n R T can be used to determine the molar mass of gaseous compounds.

  34. Substituting this in the ideal gas equation P M m m d = P V = P V = R T R T R T M M Solving for M, the molar mass m R T M = P V The ideal gas equation can be also be used to determine the density of the gas

  35. Gas Stoichiometry If the conditions of pressure and temperature are known, then the ideal gas law can be used to convert between chemical amounts i.e. moles, and gas volume. Hence in dealing with chemical reactions involving gases, we can deal with volumes of gases instead of moles of gases, being that volume is usually an easier quantity to measure.

  36. Problem Dinitrogen monoxide, N2O, better known as nitrous oxide or laughing gas, is shipped in steel cylinders as a liquid at pressures of 10 MPa. It is produced as a gas in aluminium trays by the decomposition of ammonium nitrate at 200oC. NH4NO3(s) --> N2O(g) + 2H2O(g) What volume of N2O(g) at 1.00 atm would be produced from 100.0g of NH4NO3(s) after separating out the H2O and cooling the N2O gas to 273K. Assume a 100% yield in the production of N2O(g) . Assuming a 100% yield => all the NH4NO3(s) is converted to N2O(g) Moles of NH4NO3(s) decomposed = 100.0 g/80.04g/mol = 1.249 mol

  37. Hence, 1.249 mol of N2O(g) formed. To calculate the volume of N2O(g) produced, use the ideal gas equation. V = n R T/ P V =[ (1.249 mol) (0.08206 L-atm/(mol-K)) (273)]/(1.00 atm) = 28.0 L N2O(g)

  38. Can we use the ideal gas equation to determine the properties of gases in a mixture? Dalton observed that the total pressure of a mixture of gases equals the sum of the pressures that each would exert if each were present alone. Mixtures of Gases The partial pressure of a gas in a mixture of gases is defined as the pressure it exerts if it were present alone in the container.

  39. Dalton’s law states that the total pressure is the sum of the partial pressures of each gas in the mixture. For example, consider a mixture of two gases A and B in a closed container Assuming that the pressure is low enough, A and B obey the ideal gas equation. The fact that A and B behave as ideal gases implies that A and B do not interact with each other.

  40. The pressure exerted by A, PA is then and that exerted by B is: PA = nA RT PB = nB RT V V

  41. nA RT nB RT = + V V = (nA + nB) RT V = ntotal RT V From Dalton’s law: Ptotal = PA + PB Where ntotal = nA + nB is the total number of moles

  42. Mole Fraction The quantity is called the MOLE FRACTION of A, XA nA ntotal What is the fraction of the number of moles of A in the mixture? To find this out, we need to divide the number of moles of A by the total number of moles of gases in the mixture Note: mole fraction is unitless since it is a fraction of two quantities with the same unit. Also, sum of mole fractions of the components in a mixture =1

  43. For the component A in the mixture, we can write its pressure as The total pressure is Ptotal = ntotal RT V PA = nA RT PA nA V = = XA ntotal Ptotal Dividing PA/Ptotal Hence, PA = XA Ptotal

  44. a) PO2 = XO2 Ptotal = (0.180) (745 torr) = 134 torr b) PV = n R T (134 torr) (1 atm) (120 L) = n (0.08206 L-atm/molK)(295 K) (760 torr) n = 0.872 mol A study of the effect of certain gases on plant growth requires a synthetic atmosphere composed of 1.5 mol percent of CO2, 18.0 mol percent of O2 and 80.5 mol percent Ar. a) Calculate the partial pressure of O2 in the mixture if the total pressure of the atmosphere is to be 745 torr? b) If this atmosphere is to be held in a 120-L space at 295 K, how many moles of O2 are needed?

  45. Kinetic Theory of Gases The ideal gas equation describes how gases behave. In the 19th century, scientists applied Newton’s laws of motion to develop a model to explain the behavior of gases. This model, called the kinetic theory of gases, assumes that the atoms or molecules in a gas behave like billiard balls. In the gas phase, atoms and molecules behave like hard spheres and do not interact with each other.

  46. Assumptions of Kinetic Theory of Gases 1) A gas consists of a large number of particles that are so small compared to the average distance separating them, that their own size can be considered negligible. 2) The particles of an ideal gas behave totally independent, neither attracting nor repelling each other. 3) Gas particles are in constant, rapid, straight-line motion, incessantly colliding with each other and with the walls of the container. All collisions between particles are elastic. 4) A collection of gas particles can be characterized by its average kinetic energy, which is proportional to the temperature on the absolute scale.

  47. Gas particles are constantly colliding with each other and the walls of the container. It is the collisions between the gas particles and the walls of the container that define the pressure of the gas. Every time a gas particle collides with the wall of the container, the gas particle imparts its momentum to the wall momentum = mass x velocity

  48. P a (m x u) x [ N x u] V The pressure exerted by the gas is proportional to the momentum of the particle and the number of collisions per unit time, the collision frequency. Pressure a (momentum of particle) x (rate of collisions with the wall) The rate of collision is proportional to the number of particles per unit volume (N/V) and the speed of the particle (u).

  49. P aN m u2 V Replace u2 with the mean-square speed, u2 P V a N m u2 P V a N m u2 The speed, u, is the average speed of the particles, since not all the particles move with the same speed.

  50. 1 P V = N m u2 3 1 n R T = N m u2 3 Particles are moving in a 3-D space: Comparing this equation with the ideal gas equation P V = n R T This equation relates the speed of the gas particles with the temperature of the gas