Gases © 2009, Prentice-Hall, Inc. Chapter 5: Gases (Zumdahl) John Bookstaver St. Charles Community College Cottleville, MO Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Gases © 2009, Prentice-Hall, Inc. Characteristics of Gases Unlike liquids and solids, gases –expand to fill their containers; –are highly compressible; –have extremely low densities.
Gases © 2009, Prentice-Hall, Inc. Pressure is the amount of force applied to an area. Pressure Atmospheric pressure is the weight of air per unit of area. P = FAFA
Gases © 2009, Prentice-Hall, Inc. Units of Pressure Pascals –1 Pa = 1 N/m 2 Bar –1 bar = 10 5 Pa = 100 kPa
Gases © 2009, Prentice-Hall, Inc. Units of Pressure mm Hg or torr –These units are literally the difference in the heights measured in mm ( h ) of two connected columns of mercury. Atmosphere –1.00 atm = 760 torr
Gases © 2009, Prentice-Hall, Inc. Manometer This device is used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.
Gases Sample Problem: Manometer pressure # 29, p 232 If the open-tube manometer contains a nonvolatile silicone oil (density = 1.30 g/cm 3 ) instead of mercury (density 12.6 g/cm 3 ) what are the pressures in the flask as shown in parts a and b in torr, atm, and Pa? SOLUTION: If the levels of Hg in each arm of the manometer are unequal, the difference in height in mm will be equal to the difference in pressure in mm Hg between the flask and the atmosphere. Which level is higher will tell us whether the pressure in the flask is less than or greater than atmospheric. P flask < P atm ; P flask = = 642 mm Hg= 642 torr. 642 torr x 1 atm = atm. 760 torr atm x x 10 5 Pa = 8.56 x 10 4 Pa 1 atm © 2009, Prentice-Hall, Inc.
Gases © 2009, Prentice-Hall, Inc. Standard Pressure Normal atmospheric pressure at sea level is referred to as standard pressure. It is equal to –1.00 atm –760 torr (760 mm Hg) – kPa –14.7 psi (pounds per in 2 )
Gases © 2009, Prentice-Hall, Inc. Boyle’s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.
Gases © 2009, Prentice-Hall, Inc. As P and V are inversely proportional A plot of V versus P results in a curve. Since V = k (1/P) This means a plot of V versus 1/P will be a straight line. PV = k
Gases Boyle’s Law Since PV=k P 1 V 1 = k AND P 2 V 2 = k © 2009, Prentice-Hall, Inc. This equation can be used to solve for a missing value when 3 of the values are known. THEN P 1 V 1 = P 2 V 2 Where P 1 & V 1 are the pressure and volume occurring at the same time Where P 2 & V 2 are the pressure and volume occurring at the same time, but a different time than P 1 & V 1
Gases Boyle’s Law: Practice Problem Consider a 1.53 L sample of gaseous SO 2 at a pressure of 5.6 x 10 3 Pa. If the pressure is change to 1.5 x 10 4 Pa at a constant temperature, what will be the new volume of the gas? © 2009, Prentice-Hall, Inc. V1V1 P1P1 P2P2 V2V2 (5.6 x 10 3 Pa) (1.53 L) = V x 10 4 Pa V 2 = 0.57 L
Gases © 2009, Prentice-Hall, Inc. Charles’s Law The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. A plot of V versus T will be a straight line. i.e., VTVT = k
Gases Charles’s Law Since V = k T V 1 = k AND V 2 = k T 1 T 2 © 2009, Prentice-Hall, Inc. This equation can be used to solve for a missing value when 3 of the values are known. THEN V 1 = V 2 T 1 T 2 Where V 1 & T 1 are the volume and temperature occurring at the same time Where V 2 & T 2 are the volume and temperature occurring at the same time, but a different time than V 1 & V 2 NOTE: all temperatures for gas law calcs must be in K. (K = ⁰ C)
Gases Charles’s Law Sample Problem A sample of gas at 15 C and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38 C and 1 atm? © 2009, Prentice-Hall, Inc. T1 = 288K P1P1 V1V1 T 2 = 311KP2P2 V2V2 V 2 = (2.58 L) (311K) 288K V 2 = 2.79 L
Gases © 2009, Prentice-Hall, Inc. Avogadro’s Law The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. Mathematically, this means V = kn
Gases © 2009, Prentice-Hall, Inc. Ideal-Gas Equation V 1/P (Boyle’s law) V T (Charles’s law) V n (Avogadro’s law) So far we’ve seen that Combining these, we get V V nT P
Gases © 2009, Prentice-Hall, Inc. Ideal-Gas Equation The constant of proportionality is known as R, the gas constant.
Gases © 2009, Prentice-Hall, Inc. Ideal-Gas Equation The relationship then becomes nT P V V nT P V = R or PV = nRT
Gases © 2009, Prentice-Hall, Inc. Densities of Gases If we divide both sides of the ideal-gas equation by V and by RT, we get nVnV P RT =
Gases © 2009, Prentice-Hall, Inc. We know that –moles molecular mass = mass Densities of Gases So multiplying both sides by the molecular mass ( ) gives n = m P RT mVmV =
Gases © 2009, Prentice-Hall, Inc. Densities of Gases Mass volume = density So, Note: One only to know the molecular mass, the pressure, and the temperature to calculate the density of a gas THIS EQUATION IS NOT ON THE AP FORMULA SHEET. P RT mVmV = d =
Gases Sample Problem: density of gases Uranium hexafluoride is a solid at room temperature, but it boils at 56 C. Determine the density of uranium hexafluoride at 60. C and 745 torr. © 2009, Prentice-Hall, Inc. T= 333K P = 745torr x 1 atm 760 torr = atm d= ( atm) (352 g/mol) ( L*atm/mol*K)(333K) d= 12.6 g/L
Gases © 2009, Prentice-Hall, Inc. Molecular Mass We can manipulate the density equation to enable us to find the molecular mass of a gas: Becomes P RT d = dRT P =
Gases © 2009, Prentice-Hall, Inc. Dalton’s Law of Partial Pressures The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. In other words, P total = P 1 + P 2 + P 3 + …
Gases Sample problem: partial pressure A mixture of 1.00 g H2 and 1.00 g He is placed in a 1.00 L container at 27C. Calculate the partial pressure of each gas and the total pressure. Copy solution from board. © 2009, Prentice-Hall, Inc.
Gases © 2009, Prentice-Hall, Inc. Partial Pressures When one collects a gas over water, there is water vapor mixed in with the gas. To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.
Gases © 2009, Prentice-Hall, Inc. Kinetic-Molecular Theory This is a model that aids in our understanding of what happens to gas particles as environmental conditions change.
Gases © 2009, Prentice-Hall, Inc. Main Tenets of Kinetic- Molecular Theory Gases consist of large numbers of molecules that are in continuous, random motion.
Gases © 2009, Prentice-Hall, Inc. Main Tenets of Kinetic- Molecular Theory The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained.
Gases © 2009, Prentice-Hall, Inc. Main Tenets of Kinetic- Molecular Theory Attractive and repulsive forces between gas molecules are negligible.
Gases © 2009, Prentice-Hall, Inc. Main Tenets of Kinetic- Molecular Theory Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant.
Gases © 2009, Prentice-Hall, Inc. Main Tenets of Kinetic- Molecular Theory The average kinetic energy of the molecules is proportional to the absolute temperature (Kelvin measures absolute temperature.)
Gases © 2009, Prentice-Hall, Inc. Effusion Effusion is the escape of gas molecules through a tiny hole into an evacuated space.
Gases © 2009, Prentice-Hall, Inc. Effusion The difference in the rates of effusion for helium and nitrogen, for example, explains a helium balloon would deflate faster.
Gases © 2009, Prentice-Hall, Inc. Diffusion Diffusion is the spread of one substance throughout a space or throughout a second substance.
Gases © 2009, Prentice-Hall, Inc. Graham's Law (re: rate of effusion) KE 1 KE 2 = 1/2 m 1 v 1 2 1/2 m 2 v 2 2 = = m1m1 m2m2 v22v22 v12v12 m1m1 m2m2 v 2 2 v 1 2 = v2v2 v1v1 =
Gases Graham’s Law of Effusion “the rate of effusion is equal to the inverse ratio of the masses” © 2009, Prentice-Hall, Inc.
Gases Graham’s Law of Effusion: Sample Problem #82, p 236: the rate of effusion of a particular gas was measured and found to be 24.0 mL/min. Under identical experimental conditions, the effusion rate of pure methane gas is 47.8 mL/min. What is the molar mass of the unknown gas? SOLUTION: © 2009, Prentice-Hall, Inc.
Gases © 2009, Prentice-Hall, Inc. Deviations from Ideal Behavior The assumptions made in the kinetic-molecular model (negligible volume of gas molecules themselves, no attractive forces between gas molecules, etc.) break down at high pressure and/or low temperature.
Gases © 2009, Prentice-Hall, Inc. Real Gases In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.
Gases © 2009, Prentice-Hall, Inc. Real Gases Even the same gas will show wildly different behavior under high pressure at different temperatures.
Gases © 2009, Prentice-Hall, Inc. Corrections for Nonideal Behavior The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account. The corrected ideal-gas equation is known as the van der Waals equation.
Gases © 2009, Prentice-Hall, Inc. The van der Waals Equation: predicts behavior of gases at low temps & high pressures ) (V − nb) = nRT n2aV2n2aV2 (P + Note: the greater the molecular mass and complexity, the greater the deviation from ideal.
Gases Kinetic Energy We can calculate the kinetic energy (KE) of a molecule or a mole of molecules. KE molecule = ½ mv 2 KE mole = RT © 2009, Prentice-Hall, Inc.
Gases Sample Problem: Kinetic Energy © 2009, Prentice-Hall, Inc.