Ch 5 Ideal and real gases.pdf

Chapter 5: Gases Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior writte...

Chapter 5: Gases Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. The Three States of Matter Figure 5.1 ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer 5.1 Distinguishing gases from liquids and solids 1) Gas volume changes significantly with pressure. Example: piston air pumps for inflating tires. – Solids and liquids can hardly be compressed. 2) Gas volume changes significantly with temperature. Examples: rocket lifting, popping corn. – Gases expand when heated and shrink when cooled. – The volume change is 50 to 100 times greater for gases than for liquids and solids. 3) Gases flow much more freely than liquids. Examples: gas transporting through pipes, gas leaking. 4) Gases have much lower densities than liquids (x1000). 5) Gases mix (form a solution) in any proportions. – Gases experience no restrictions in mixing with each other. ©McGraw-Hill Education. 5.2 Gas Pressure and its Measurement Gas molecules or atoms move with high speed at room temperature. Average speed is 600 m/s. Molecules collide frequently with the container wall and exert a pressure: force 𝐹 Pressure = 𝑜𝑟 𝑝 = area 𝐴 [𝐹] 𝑁 𝑘𝑔 𝑚 𝑝 = = 1 2 = 1 𝑃𝑎 𝑃𝑎𝑠𝑐𝑎𝑙 , 𝑤ℎ𝑒𝑟𝑒 1 𝑁 = 1 2 [𝐴] 𝑚 𝑠 Gas pressure is uniform everywhere on the wall as the gas motion is random (all directions are equally probable or isotropic). Atmospheric pressure is the air pressure at the ground level. For uniformity, ground level is defined as the sea level (water level in the oceans). ©McGraw-Hill Education. Effect of Atmospheric Pressure on a Familiar Object Figure 5.2. Vacuum is pulled in a thin Aluminum container with a pump. The atmospheric pressure immediately crushes the can (deflation). The value of the atmospheric pressure is about 14.7 lb/in2. ©McGraw-Hill Education. © McGraw-Hill Education/Charles Winters/Timeframe Photography, Inc. A Mercury Barometer A barometer is the oldest pressure gauge in history on which the pressure unit was defined in 1643 by the Italian physicist Evangelista Torricelli. It is a 1 m long glass tube of constant diameter, sealed at one end, calibrated in mm, and filled with liquid Mercury to the brim. The filled tube is flipped upside down to place the open end of the tube in a Mercury reservoir. When the tube is inverted, mercury flows out of the tube creating vacuum at the sealed end. Mercury stops flowing out when 𝑝𝐻𝑔 = 𝑝𝑎𝑖𝑟. Fig. 5.3 ©McGraw-Hill Education. A Mercury Barometer – cont’d Mercury level drops at the sealed end until the hydrostatic pressure of the Hg column matches the external atmospheric pressure. At the sea level and 0°C, the Hg column height is 760. mm (0.760 m). The oldest pressure unit is mmHg or Torr: 1 𝑡𝑜𝑟𝑟 = 1 𝑚𝑚𝐻𝑔. Atmospheric pressure decreases with altitude for 2 reasons: 1) Gravity makes the atmosphere most dense at the ground level. Density of air decreases with altitude. 2) The air column is tallest at the ground level. ©McGraw-Hill Education. Two Types of Manometer Figure 5.4 A manometer is a barometer without the mercury reservoir. It measures the pressure of a gas either relative to vacuum (closed end) or relative to the atmospheric pressure (open end). ©McGraw-Hill Education. Common Units of Pressure Hydrostatic pressure of a mercury column: 𝑝𝐻𝑔 = 𝑑(𝐻𝑔, 0℃) × 𝑔 × ∆ℎ 𝑑 𝐻𝑔, 0℃ = 13 595.1 𝑚 𝑘𝑔 𝑚3 𝑔𝑠𝑡𝑑 = 9.80665 2 𝑠 ∆ℎ𝐻𝑔 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙 = 0.760 000 𝑚 𝑘𝑔 𝑚 𝑝𝑎𝑖𝑟 0℃, 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙 = 13 595.1 3 × 9.80665 2 × 0.760 000 𝑚 = 𝑚 𝑠 𝑘𝑔 𝑁 = 101 325 = 101 325 = 101 325 𝑃𝑎 = 1 𝑎𝑡𝑚 = 760 𝑚𝑚𝐻𝑔 2 2 𝑚𝑠 𝑚 Unit Normal Atmospheric Pressure at Sea Level and 0°C pascal (Pa); kilopascal (kPa) 1.01325×105 Pa; 101.325 kPa atmosphere (atm) 1 atm (exact) millimeters of mercury (mmHg) 760 mmHg (exact) torr 760 torr (exact) pounds per square inch (lb/in2 or psi) 14.7 lb/in2 bar (1 x 105 Pa) 1.01325 bar ©McGraw-Hill Education. *These are exact quantities; in calculations, we use as many significant figures as necessary. Common Units of Pressure – cont’d Pascal (SI unit): 𝑁 𝑚 1 𝑃𝑎 = 1 2 = 1 𝑘𝑔 2 𝑚 𝑠 Atmosphere (non-SI): 1 𝑎𝑡𝑚 = 101 325 𝑃𝑎 torr (non-SI) or mmHg: 1 101 325 𝑃𝑎 1 𝑡𝑜𝑟𝑟 = 𝑎𝑡𝑚 = = 133.322 𝑃𝑎 760 760 bar (non-SI, used in Chemistry and Engineering): 1 𝑏𝑎𝑟 = 1 × 105 𝑃𝑎 = 0.986923 𝑎𝑡𝑚 𝑂𝑅 1 𝑎𝑡𝑚 = 1.01325 𝑏𝑎𝑟 psi (pounds-per-square-inch), USA: 𝑙𝑏 1 1 𝑝𝑠𝑖 = 1 2 = 𝑎𝑡𝑚 = 6886 𝑃𝑎 𝑖𝑛 14.715 ©McGraw-Hill Education. Sample Problem 5.1 – Problem and Plan Converting Units of Pressure PROBLEM: A geochemist heats a limestone (CaCO3) sample and collects the CO2 released in an evacuated flask attached to a closed-end manometer. After the system comes to room temperature, Δh = 291.4 mm Hg. Calculate the CO2 pressure in torr, atm, and kilopascals. PLAN: Construct conversion factors to find the other units of pressure. ©McGraw-Hill Education. Sample Problem 5.1 –Solution SOLUTION: 1torr 291.4 mmHg x = 291.4 torr 1 mmHg 1 atm 291.4 torr x = 0.3834 atm 760 torr 101.325 kPa 0.3834 atm x = 38.85 kPa 1 atm ©McGraw-Hill Education. 5.3 The Gas Laws The gas laws describe the physical behavior of gases in terms of 4 interdependent variables: – pressure (p) – temperature (T) – volume (V) – amount (number of moles, n) An ideal gas is a gas that exhibits linear relationships among these variables. No ideal gas actually exists, but most simple gases behave nearly ideally at high temperatures and low pressures. ©McGraw-Hill Education. Boyle’s Law in Images Figure 5.5 ©McGraw-Hill Education. Boyle(1662)-Mariotte(1676) Law Boyle’s Law: “At constant temperature, the volume occupied by a fixed amount of gas is inversely proportional to the external pressure.” 1 V∝ or PV = constant P At fixed temperature and number of moles of gas: – P decreases as V increases – P increases as V decreases The value of the constant is the slope of the linear function: 1 𝑉=𝑓 𝑃 𝑚 𝑜𝑟 𝑉 = 𝑤ℎ𝑒𝑟𝑒 𝑚 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑃 The value of the constant changes with the temperature and the amount (mol) of gas in the experiment but is independent of the nature (chemical identity) of the ideal gas. ©McGraw-Hill Education. Charles(1787)-Gay-Lussac(1802) Law in Images Figure 5.6 ©McGraw-Hill Education. Charles’s Law “At constant pressure, the volume occupied by a fixed amount of gas is directly proportional to its absolute (Kelvin) temperature.” V V ∝ T or = constant T At fixed pressure and number of moles: – V decreases as T decreases – V increases as T increases The value of the constant is the slope of the linear function: 𝑉=𝑓 𝑇 𝑜𝑟 𝑉 = 𝑚𝑇 𝑤ℎ𝑒𝑟𝑒 𝑚 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 The value of the constant changes with the pressure and the amount (mol) of gas in the experiment but is independent of the nature (chemical identity) of the ideal gas. ©McGraw-Hill Education. Amonton-Charles Law “At constant volume, the pressure exerted by a fixed amount of gas is directly proportional to its absolute (Kelvin) temperature.” P P ∝ T or = constant T At fixed volume and number of moles: – P decreases as T decreases – P increases as T increases The value of the constant is the slope of the linear function: 𝑃=𝑓 𝑇 𝑜𝑟 𝑃 = 𝑚𝑇 𝑤ℎ𝑒𝑟𝑒 𝑚 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 The value of the constant changes with the volume and the amount (mol) of gas in the experiment but is independent of the nature (chemical identity) of the ideal gas. ©McGraw-Hill Education. Combined Gas Law By combining the 3 gas laws we get: PV = constant (at fixed amount of gas, n) T The value of the constant is the slope of the linear function: 𝑃𝑉 = 𝑓 𝑇 𝑜𝑟 𝑃𝑉 = 𝑚𝑇 𝑤ℎ𝑒𝑟𝑒 𝑚 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 The value of the constant changes with the amount (mol) of gas in the experiment but is independent of the nature (chemical identity) of the ideal gas. ©McGraw-Hill Education. Avogadro’s Law “At fixed temperature and pressure, the volume occupied by a gas is directly proportional to the amount (moles) of gas.” Avogadro’s Law: at fixed temperature and pressure, equal volumes of any ideal gas contain equal numbers of particles (or moles). 𝑉 𝑉 ∝ 𝑛 𝑝 𝑎𝑛𝑑 𝑇 𝑓𝑖𝑥𝑒𝑑 𝑜𝑟 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑛 ©McGraw-Hill Education. Familiar Application of the Gas Laws Figure 5.8 The lungs and the breathing. ©McGraw-Hill Education. Ideal Gas Behavior at Standard Conditions STP or standard temperature and pressure specifies a pressure of 1 atm (760 torr) and a temperature of 0°C (273.15 K). The standard molar volume is the volume of 1 mol of an ideal gas at STP. Standard molar volume V(1 mol) = 22.4141 L or 22.4 L ©McGraw-Hill Education. Standard Molar Volume Figure 5.9 ©McGraw-Hill Education. Volumes of Some Familiar Objects 1 gallon = 3.79 L 1 bottle = 2.00 L 1 basketball = 7.50 L Figure 5.10 ©McGraw-Hill Education. © McGraw-Hill Education/Charles Winters/Timeframe Photography, Inc. The Ideal Gas Law PV=nRT R is the universal gas constant; the numerical value of R depends on the units used. PV 1 atm x 22.4141 L atm∙L R= = =0.082058 nT 1 mol x 273.15K mol∙K The ideal gas law can also be expressed by the combined equation: P1V1 P2V2 = n1T1 n2T2 R in SI units: 𝐽 𝑅 = 8.31447 𝑚𝑜𝑙 𝐾 ©McGraw-Hill Education. Individual Gas Laws as Special Cases Figure 5.11 ©McGraw-Hill Education. Sample Problem 5.2: Problem and Plan Applying the Volume-Pressure Relationship PROBLEM: A 3.50-L sample of neon gas has a pressure of 924 torr. What pressure (in atm) is required to compress the gas into a tank with a volume of 1350 cm3 at constant temperature? PLAN: We must find the final pressure (P2) in atm, given the initial volume (V1), initial pressure(P1), and final volume (V2). The temperature and amount of gas are fixed. We must use consistent units of volume, so we convert the unit of V2 from cm3 to mL and then to L. We then convert the unit of P1 from torr to atm, rearrange the ideal gas law to the appropriate form, and solve for P2. © McGraw Hill Sample Problem 5.2: Plan © McGraw Hill Sample Problem 5.2: Solution SOLUTION: 1 atm P1 ( atm ) = 924 torr  = 1.22 atm 760 torr V1 ( L ) = 1350 cm3  1 mL 1L  = 1.35 L 3 1 cm 1000 mL P1V1 P2 V2 = or P1V1 = P2 V2 n 1 T1 n 2 T2 V1 3.50 L P2 = P1  = 1.22 atm  = 3.16 atm V2 1.35 L © McGraw Hill Sample Problem 5.2A – Problem and Plan Applying the Volume-Pressure Relationship PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm3 at 851 torr. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)? PLAN: We must find the final volume (V2) in liters, given the initial volume (V1), initial pressure(P1), and final pressure (P2). The temperature and amount of gas are fixed. We must use consistent units of pressure, so we convert the unit of P1 from torr to atm. We then convert the unit of V1 from cm3 to mL and then to L, rearrange the ideal gas law to the appropriate form, and solve for V2. ©McGraw-Hill Education. Sample Problem 5.2A –Plan ©McGraw-Hill Education. Sample Problem 5.2A - Solution SOLUTION: 1 atm P1 atm =851 torr× =1.12 atm 760 torr 1 mL 1L 3 V1 L =24.8 cm × × =0.0248 L 3 1 cm 1000 mL P1V1 PV = 2 2 or P1V1=P2V2 n1T1 n2T2 1.12 atm P1 V2=V1× =0.0248 L× =0.0105 L 2.64 atm P2 ©McGraw-Hill Education. Sample Problem 5.3 – Problem and Plan Applying the Volume-Temperature and Pressure-Temperature Relationships PROBLEM: A balloon is filled with 1.95 L of air at 25°C and then placed in a car in the sun. What is the volume of the balloon when the temperature in the car reaches 90°C? PLAN: We know the initial volume (V1) and the initial (T1) and final temperatures (T2) of the gas; we must find the final volume (V2). The pressure of the gas is fixed since the balloon is subjected to atmospheric pressure and n is fixed since air cannot escape or enter the balloon. We convert both T values to degrees Kelvin, rearrange the ideal gas law, and solve for V2. ©McGraw-Hill Education. Sample Problem 5.3 - Solution SOLUTION: V1 = 1.95 L T1 = 25ºC (convert to K) V2 = 1.95 L T2 = 90ºC (convert to K) P and n remain constant T1 (K) = 25°C + 273.15 = 298 K T2 (K) = 90°C + 273.15 = 363 K P1V1 PV V V = 2 2 or 1 = 2 n1T1 n2T2 T1 T2 363 K T2 V2=V1× =1.95L× =2.38L 298 K T1 ©McGraw-Hill Education. Sample Problem 5.4 – Problem and Plan Applying the Volume-Amount Relationship PROBLEM: A scale model of a blimp rises when it is filled with helium to a volume of 55.0 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. How many more grams of He must be added to make it rise? Assume constant T and P. PLAN: The initial amount of helium (n1) is given, as well as the initial volume (V1) and the volume needed to make it rise (V2). We need to calculate n2, and hence the mass of He to be added. ©McGraw-Hill Education. Sample Problem 5.4 - Solution SOLUTION: n1 = 1.10 mol n2 = unknown V1 = 26.2 dm3 V2 = 55.0 dm3 T and P are constant P1V1 P2V2 V1 V2 = or = n1T1 n2T2 n1 n2 V2 55.0 dm3 n2 =n1 × = 1.10 mol x = 2.31 mol He 3 V1 26.2 dm Additional amount of He needed = 2.31 mol – 1.10 mol = 1.21 mol He 4.003 g He 1.21 mol He x = 4.84 g He 1 mol He ©McGraw-Hill Education. Sample Problem 5.5 – Problem and Plan Applying the Volume-Pressure-Temperature Relationship PROBLEM: A helium-filled balloon has a volume of 15.8 L at a pressure of 0.980 atm and 22°C. What is its volume at the summit of Mt. Hood, Oregon’s highest mountain, where the atmospheric pressure is 532 mmHg and the temperature is 0°C? PLAN: We know the initial volume (V1), pressure (P1), and temperature (T1) of the gas; we also know the final pressure (P2) and temperature (T2) and we must find the final volume (V2). Since the amount of helium in the balloon does not change, n is fixed. We convert both T values to degrees Kelvin, the final pressure to atm, rearrange the generalized ideal gas equation, and solve for V2. ©McGraw-Hill Education. Sample Problem 5.5 - Solution Solution: V1 = 15.8 L T1 = 22°C (convert to K) P1 = 0.980 atm V2 = unknown T2 = 0°C (convert to K) P2 = 523 mmHg (convert to atm) n remains constant T1 (K) = 22°C + 273.15 = 295 K T2 (K) = 0°C + 273.15 = 273 K 1 atm P2 (atm) = 532 mmHg x = 0.700 𝑎𝑡𝑚 760 mmHg P1V1 PV PV PV = 2 2 or 1 1 = 2 2 n1T1 n2T2 T1 T2 V2 = V1 x ©McGraw-Hill Education. P1T2 (0.980 atm) (273 K) = 15.8 L x = 𝟐𝟎. 𝟓 𝐋 P2T1 (0.700 atm) (295 K) Sample Problem 5.6 – Problem, Plan and Solution Solving for an Unknown Gas Variable at Fixed Conditions PROBLEM: A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2 at 21°C. PLAN: We are given V, T, and mass, which can be converted to moles (n). Use the ideal gas law to find P. SOLUTION: V = 438 L, T = 21°C = 294 K, m = 0.885 kg O2 P=? 103 g 1 mol O2 n = 0.885 kg O2 x x = 27.7 mol O2 1 kg 32.00 g O2 atm∙L nRT 27.7 mol x 0.0821mol∙K ×294 K P= = =1.53 atm V 438 L ©McGraw-Hill Education. Sample Problem 5.7 - Problem Using Gas Laws to Determine a Balanced Equation PROBLEM: The piston-cylinder is depicted before and after a gaseous reaction that is carried out at constant pressure. The temperature is 150 K before the reaction and 300 K after the reaction. (Assume the cylinder is insulated.) Which of the following balanced equations describes the reaction? (1) A2 (g) + B2 (g) → 2AB (g) (2) 2AB (g) + B2 (g) → 2AB2 (g) (3) A (g) + B2 (g) → AB2 (g) (4) 2AB2 (g) → A2 (g) + 2B2 (g) ©McGraw-Hill Education. Sample Problem 5.7 – Plan and Solution PLAN: We are told that P is constant for this system, and the depiction shows that V does not change either. Since T changes, the volume could not remain the same unless the amount of gas in the system also changes. SOLUTION: n2 T1 150 K 1 n1T1 = n2T2 or = = = n1 T2 300 K 2 Since T doubles, the total number of moles of gas must halve – i.e., the moles of product must be half the moles of reactant. This relationship is shown by equation (3). A (g) + B2 (g) → AB2 (g) ©McGraw-Hill Education. The Ideal Gas Law and Gas Density The density of a gas is – directly proportional to its molar mass and pressure – inversely proportional to its temperature. m density =d= V m PV = RT M OR and m n(mol) = M 𝑚 PM= 𝑅𝑇 = 𝑑𝑅𝑇 𝑉 m PM =d= V RT ©McGraw-Hill Education. Sample Problem 5.8 – Problem and Plan Calculating Gas Density PROBLEM: To apply a green chemistry approach, a chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene. Find the density (in g/L) of CO2 and the number of molecules per liter (a) at STP (0°C and 1 atm) and (b) at room conditions (20.°C and 1.00 atm). PLAN: We must find the density (d) and the number of molecules of CO2, given two sets of P and T data. We find ​, convert T to kelvins, and calculate d. Then we convert the mass per liter to molecules per liter with Avogadro’s number. ©McGraw-Hill Education. Sample Problem 5.8 – Solution SOLUTION: (a) At 0 °C and 1 atm: T = 0°C + 273.15 = 273K p = 1 atm M (CO2) = 44.01 g/mol d= M x P 44.01 g/mol x 1.00 atm RT = 0.0821 atm∙L x 273 K mol∙K =1.96 g/L 1.96 g CO2 1 mol CO2 6.022 x 1023 × × 1L 44.01 g CO2 1 mol = 2.68 × 10 ©McGraw-Hill Education. 22 1 𝐿 Sample Problem 5.8 – Solution, Cont’d SOLUTION: (b) At 20. °C and 1.00 atm: T = 20.°C + 273.15 = 293K p = 1.00 atm M (CO2) = 44.01 g/mol d= M x p 44.01 g/mol x 1.00 atm RT = 0.0821 atm∙L x 293 K mol∙K =1.83 g/L 1.83 g CO2 1 mol CO2 6.022 x 1023 × × 1L 44.01 g CO2 1 mol = 2.50 x 𝟏𝟎𝟐𝟐 /L ©McGraw-Hill Education. Molar Mass from the Ideal Gas Law m PV n= = M RT mRT M= PV ©McGraw-Hill Education. Sample Problem 5.9 – Problem and Plan Finding the Molar Mass of a Volatile Liquid PROBLEM: An organic chemist isolates a colorless liquid from a petroleum sample. She places the liquid in a preweighed flask and puts the flask in boiling water, causing the liquid to vaporize and fill the flask with gas. She closes the flask and reweighs it. She obtains the following data: Volume (V) of flask = 213 mL T = 100.0°C p = 754 torr mass of flask + gas = 78.416 g mass of flask = 77.834 g Calculate the molar mass of the liquid. PLAN: The variables V, T and P are given. We find the mass of the gas by subtracting the mass of the flask from the mass of the flask with the gas in it, and use this information to calculate M. ©McGraw-Hill Education. Sample Problem 5.9 - Solution SOLUTION: m of gas = (78.416 g - 77.834 g) = 0.582 g 1L V = 213 mL x 3 = 0.213 L 10 mL T = 100.0°C + 273.15 = 373.2 K 1 atm P = 754 torr x = 0.992 atm 760 torr atm∙L x 373.2 K 0.582 g x0.0821 mRT mol∙K M= = = 84.4 g/mol PV 0.213 L x0.992 atm ©McGraw-Hill Education. Mixtures of Gases Gases mix homogeneously in any proportions. The pressure exerted by each gas in a mixture is called its partial pressure. Partial pressure of a gas in a mixture is the pressure that the pure gas would exert if it were by itself in the same container, in the same amount and at the same temperature. Dalton’s Law of partial pressures states that “the total pressure in a mixture of unreacting gases is the sum of the partial pressures of the component gases:” 𝑝𝑡𝑜𝑡𝑎𝑙 = 𝑝1 + 𝑝2 + ⋯ + 𝑝𝑖 ©McGraw-Hill Education. Mixtures of Gases – cont’d 𝑛1 𝑅𝑇 𝑛2 𝑅𝑇 𝐼𝑓 𝑃1 = 𝑎𝑛𝑑 𝑃2 = 𝑉 𝑉 𝑛1 𝑅𝑇 𝑛2 𝑅𝑇 (𝑛1 + 𝑛2 )𝑅𝑇 𝑛𝑡𝑜𝑡𝑎𝑙 𝑅𝑇 𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃1 + 𝑃2 = + = = 𝑉 𝑉 𝑉 𝑉 Mole fractions, Xi: 𝑛1 𝑛2 𝑋1 = 𝑎𝑛𝑑 𝑋2 = 𝑛1 + 𝑛2 𝑛1 + 𝑛2 The partial pressure of a gas A is proportional to its mole fraction XA : nA PA = ΧA x Ptotal XA= ntotal Demonstration: 𝑛𝐴 𝑅𝑇 𝑃𝐴 𝑛𝐴 𝑉 = = = 𝑋𝐴 𝑃𝑡𝑜𝑡𝑎𝑙 𝑛𝑡𝑜𝑡𝑎𝑙 𝑅𝑇 𝑛𝑡𝑜𝑡𝑎𝑙 𝑉 ©McGraw-Hill Education. Sample Problem 5.10 – Problem and Plan Applying Dalton’s Law of Partial Pressures PROBLEM: In a study of O2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mole % N2, 17 mole % 16O2, and 4.0 mole % 18O. (The isotope 18O will be measured to 2 determine the O2 uptake.) The total pressure of the mixture is 0.75 atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18O2 in the mixture. PLAN: Find X18O2 and P18O2 from Ptotal and mol % 18O2. Dividing the mole % by 100 gives the mole fraction, X18O2. Then we multiply X18O2 by Ptotal to find P18O2. ©McGraw-Hill Education. Sample Problem 5.10 - Solution SOLUTION: X18O2 = 4.0 mol %18O2 = 0.040 100 p18O2 = X18O2× 0.75atm = 0.030 atm ©McGraw-Hill Education. Determining Pressure of a Water-insoluble Gaseous Product Figure 5.12 A water-insoluble gas collected above water in a belljar contains water vapors. ©McGraw-Hill Education. Vapor Pressure of Water at Different Temperatures T (°C) PH2O (torr) T (°C) PH2O (torr) T (°C) PH2O (torr) T (°C) PH2O (torr) 0 4.6 20 17.5 40 55.3 75 289.1 5 6.5 22 19.8 45 71.9 80 355.1 10 9.2 24 22.4 50 92.5 85 433.6 12 10.5 26 25.2 55 118.0 90 525.8 14 12.0 28 28.3 60 149.4 95 633.9 16 13.6 30 31.8 65 187.5 100 760.0 18 15.5 35 42.2 70 233.7 ©McGraw-Hill Education. Sample Problem 5.11 – Problem and Plan Calculating the Amount of Gas Collected over Water PROBLEM: Acetylene (C2H2) is produced in the laboratory when calcium carbide (CaC2) reacts with water: CaC2 (s) + 2H2O (l) → C2H2 (g) + Ca(OH)2 (aq) A collected sample of acetylene has a total gas pressure of 738 torr and a volume of 523 mL. At the temperature of the gas (23°C), the vapor pressure of water is 21 torr. How many grams of acetylene are collected? PLAN: The difference in pressures will give P for the C2H2. The number of moles (n) is calculated from the ideal gas law and converted to mass using the molar mass. ©McGraw-Hill Education. Sample Problem 5.11 - Solution SOLUTION: PC2H2 = 738−21 torr = 717 torr T = 23°C + 273.15 K = 296 K 1 atm P = 717 torr x = 0.943 atm 760 torr 1L V = 523 mL x 3 = 0.523 L 10 mL T = 23°C + 273.15 K = 296 K PV 0.943 atm ×0.523 L nC2H2 = = = 0.0203 mol atm∙L RT 0.0821 ×296 K mol∙K 26.04 g C2H2 0.0203 mol x = 0.529 g C2H2 1 mol C2H2 ©McGraw-Hill Education. The Ideal Gas Law and Stoichiometry 𝑷𝑽 𝒏(𝒎𝒐𝒍) = 𝑹𝑻 ©McGraw-Hill Education. Sample Problem 5.12 – Problem and Plan Using Gas Variables to Find Amounts of Reactants and Products I PROBLEM: Solid lithium hydroxide is used to "scrub" CO2 from the air in spacecraft and submarines; it reacts with the CO2 to produce lithium carbonate and water. What volume of CO2 at 23°C and 716 torr can be removed by reaction with 395 g of lithium hydroxide? PLAN: Write a balanced equation. Next, we convert the given mass (395 g) of lithium hydroxide, LiOH, to amount (mol) and use the molar ratio to find amount (mol) of CO2 that reacts (stoichiometry portion). Then, we use the ideal gas law to convert moles of CO2 to liters (gas law portion). ©McGraw-Hill Education. Sample Problem 5.12: Solution SOLUTION: 2LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l) 1 mol CO 2 1 mol LiOH 395 g LiOH   = 8.25 mol CO 2 23.95 g LiOH 2 mol LiOH 1 atm P = 716 torr  = 0.942 atm 760 torr T = 23˚C + 273.15 K = 296 K nRT V= = P © McGraw Hill atm  L  296 K mol  K = 213 L CO 2 0.942 atm 8.25 mol CO 2  0.0821 Sample Problem 5.13 – Problem Using Gas Variables to Find Amounts of Reactants and Products II PROBLEM: The alkali metals [Group 1A(1)]react with the halogens [Group 7A(17)] to form ionic metal halides. What mass of potassium chloride forms when 5.25 L of chlorine gas at 0.950 atm and 293 K reacts with 17.0 g of potassium? ©McGraw-Hill Education. © McGraw-Hill Education/Stephen Frisch, photographer Sample Problem 5.13 – Plan and Solution PLAN: The amounts of two reactants are given, so this is a limiting-reactant problem. We use the ideal gas law to find the amount (n)of gaseous reactant from the known V, P, and T. We first write the balanced equation and then use it to find the limiting reactant and the amount and mass of product. SOLUTION: Cl2 (g) + 2K (s) → 2KCl (s) PV 0.950 atm × 5.25 L 𝑛𝐶𝑙2 = = = 0.207 mol Cl2 atm∙L RT 0.0821 x293 K mol∙K ©McGraw-Hill Education. Sample Problem 5.13 – Solution For Cl2: 0.207 mol Cl2 x For K: 17.0 g K x 2 mol KCl = 0.414 mol KCl 1 mol Cl2 1 mol K 2 mol KCl x = 0.435 KCl 39.10 g K 2 mol K Cl2 is the limiting reactant because the given amount produces less KCl. 74.55 g KCl 0.414 mol KCl x = 30.9 g KCl 1 mol KCl ©McGraw-Hill Education. Kinetic-Molecular Theory Postulate 1: Gas particles are tiny with large spaces between them. The volume of each particle is so small compared to the total volume of the gas that it is assumed to be zero. Postulate 2: Gas particles are in constant, random, straightline motion except when they collide with each other or with the container walls. Postulate 3: Collisions are elastic, meaning that colliding particles exchange energy but do not lose any energy due to friction. Their total kinetic energy is constant. Between collisions the particles do not influence each other by attractive or repulsive forces. ©McGraw-Hill Education. Kinetic Energy and Gas Behavior 1 1 2 𝐸𝐾 = mass × (speed) = 𝑚𝑢2 2 2 At a given T, all gases in a sample have the same average kinetic energy: 3𝑅 1 𝐸ത𝐾 = 𝑇 = 𝑚𝑢2 , 𝑤ℎ𝑒𝑟𝑒 𝑢2 ≠ 𝑢ത 2 2 𝑁𝐴 2 Kinetic energy depends on both the mass and the speed of a particle. At the same T, a heavier gas particle moves more slowly than a lighter one. ©McGraw-Hill Education. Molecular Speeds at Three Temperatures Figure 5.14 ©McGraw-Hill Education. Molar Mass and Molecular Speed Figure 5.20 ©McGraw-Hill Education. Root-Mean-Square Speed Root-mean-square speed (rms speed, urms): the speed of a molecule having the average kinetic energy. urms = u2 = 3RT = m × NA 3RT M The velocity of a gas particle increases with increasing temperature and decreases with increasing molar mass. © McGraw Hill Graham’s Law of Effusion Effusion is the process by which a gas escapes through a small hole in its container into an evacuated space. Graham’s law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. – A lighter gas moves more quickly and therefore has a higher rate of effusion than a heavier gas at the same T. Rate of effusion ∝ ©McGraw-Hill Education. 1 ℳ Process of Effusion Figure 5.21 ©McGraw-Hill Education. Sample Problem 5.14 – Problem and Plan Applying Graham’s Law of Effusion PROBLEM: A mixture of helium (He) and methane (CH4) is placed in an effusion apparatus. (a) Calculate the ratio of the effusion rates of the two gases. (b) If it takes 7.55 min for a given volume of CH4 to effuse from the apparatus, how long will it take for the same volume of He to effuse? PLAN: (a) The effusion rate is inversely proportional to , so we find the molar mass of each substance from the formula and take its square root. The inverse of the ratio of the square roots is the ratio of the effusion rates. (b) Once we know the ratio of effusion rates, we can apply that ratio to the time. ©McGraw-Hill Education. Sample Problem 5.14 – Solution SOLUTION: M of CH4 = 16.04 g/mol (a) 1 effusion rateHe MHe = = 1 effusion rateCH4 MCH4 (b) ©McGraw-Hill Education. M of He = 4.003 g/mol MCH4 = MCH4 7.55 min = 3.77 min 2.002 16.04 = 2.002 4.001 The Origin of Pressure Figure 5.15 ©McGraw-Hill Education. Molecular View of Boyle’s Law Figure 5.16 ©McGraw-Hill Education. Molecular View of Dalton’s Law Figure 5.17 ©McGraw-Hill Education. Molecular View of Charles’s Law Figure 5.18 ©McGraw-Hill Education. Molecular View of Avogadro’s Law Figure 5.19 ©McGraw-Hill Education. Diffusion of Gases Figure 5.22 ©McGraw-Hill Education. Variations in P and T with altitude on Earth Figure B5.1 ©McGraw-Hill Education. Composition of Air at Sea Level Component Mole Fraction Nitrogen (N2) Oxygen (O2) Argon (Ar) Carbon dioxide (CO2) Neon (Ne) Helium (He) Methane (CH4) Krypton (Kr) Hydrogen (H2) Dinitrogen monoxide (N2O) Carbon monoxide (CO) Xenon (Xe) Ozone (O3) Ammonia (NH3) Nitrogen dioxide (NO2) Nitrogen monoxide (NO) Sulfur dioxide (SO2) Hydrogen sulfide (H2S) 0.78084 0.20946 0.00934 0.00040 1.818×10−5 5.24×10−6 2×10−6 1.14×10−6 5×10−7 5×10−7 1×10−7 8×10−8 2×10−8 6×10−9 6×10−9 6×10−10 2×10−10 2×10−10 ©McGraw-Hill Education. Real Gases: Deviations from Ideal Behavior The kinetic-molecular model describes the behavior of ideal gases. Real gases deviate from this behavior. Real gases have real volume. – Gas particles are not points of mass, but have volumes determined by the sizes of their atoms and the bonds between them. Real gases do experience attractive and repulsive forces between their particles. Real gases deviate most from ideal behavior at low temperature and high pressure. ©McGraw-Hill Education. Molar Volume of Some Common Gases at STP Gas Molar Volume (L/mol) Boiling Point (°C) He 22.435 −268.9 H2 22.432 −252.8 Ne 22.422 −246.1 Ideal gas 22.414 Ar 22.397 −185.9 N2 22.396 −195.8 O2 22.390 −183.0 CO 22.388 −191.5 Cl2 22.184 −34.0 NH3 22.079 −33.4 ©McGraw-Hill Education. — Deviations From Ideal Behavior With Increasing External Pressure Figure 5.23 ©McGraw-Hill Education. Effect of Interparticle Attractions on Measured Gas Pressure Figure 5.24 ©McGraw-Hill Education. Effect of Particle Volume on Measured Gas Volume Figure 5.25 ©McGraw-Hill Education. Van der Waals Equation The van der Waals equation adjusts the ideal gas law to take into account – the real volume of the gas particles and – the effect of interparticle attractions. Van der Waals equation for n moles of a real gas 𝑛2 𝑎 𝑃 + 2 𝑉 − 𝑛𝑏 = 𝑛𝑅𝑇 𝑉 The constant a relates to factors that influence the attraction between particles. The constant b relates to particle volume. ©McGraw-Hill Education. Van der Waals Constants for Some Common Gases Gas a​​ (atm  L2/mol2) ​b (L/mol) He 0.034 0.0237 Ne 0.211 0.0171 Ar 1.35 0.0322 Kr 2.32 0.0398 Xe 4.19 0.0511 H2 0.244 0.0266 N2 1.39 0.0391 O2 1.36 0.0318 Cl2 6.49 0.0562 CH4 2.25 0.0428 CO 1.45 0.0395 CO2 3.59 0.0427 NH3 4.17 0.0371 H2O 5.46 0.0305 ©McGraw-Hill Education.