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Chapter 10 Gases 1 Gases What gases are important for each of the following: O2, CO2 and/or He? A. B. C. D. Gases What gases are important for each of the following: O2, CO2 and/or He? A. CO2 B. O2/CO2 C. O2 D. He Section 13.1 Pressure Pressure • • • force Pr essure = area SI units = Newton/meter2 = 1 Pascal (Pa) 1 standard atmosphere = 101,325 Pa 1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr 1 atm = 760 torr 1 atm = 101, 325 Pa Return to TOC Section 13.1 Pressure Return to TOC 6 Section 13.1 Pressure Pressure Conversions: An Example The pressure of a gas is measured as 2.5 atm. Represent this pressure in both torr and pascals. How do we get there? 760 torr 3 = 1.9 10 torr 2.5 atm 1 atm 101,325 Pa 5 2.5 atm = 2.5 10 Pa 1 atm Return to TOC Section 13.1 Pressure Exercise The vapor pressure over a beaker of hot water is measured as 656 torr. What is this pressure in atmospheres? a) b) c) d) 1.16 atm 0.863 atm 0.756 atm 0.500 atm 656 torr 1 atm 0.863 atm 760 atm Return to TOC Four Physical Quantities for Gases Phys. Qty. Symbol SI unit Other common units pressure P Pascal (Pa) atm, mm Hg, torr, psi volume V m3 dm3, L, mL, cm3 temp. T K °C, °F moles n mol Section 13.2 Pressure and Volume: Boyle’s Law Boyle’s Law (1st ) • • In constant temperature (T), and number of moles of gas (n). PV = k (k is a constant for a given sample of air at a specific temperature) P1 V1 = P2 V2 Return to TOC Section 13.2 Pressure and Volume: Boyle’s Law Return to TOC A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P x V = constant P1 x V1 = P2 x V2 P2 = P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 V2 726 mmHg x 946 mL = = 4460 mmHg 154 mL Section 13.2 Pressure and Volume: Boyle’s Law Exercise A sample of helium gas occupies 12.4 L at 23°C and 0.956 atm. What volume will it occupy at 1.20 atm assuming that the temperature stays constant? 9.88 L P1V1 = P2V2 (0.956 atm) (12.4 L) = (1.20 atm) (V2) V2 = 9.88 L Return to TOC Section 13.3 Volume and Temperature: Charles’s Law Charles’s Law (2nd) • • • • (constant Pressure and number of moles). V=bT (b is a proportionality constant) K = °C + 273 0 K is called absolute zero. V1 V2 = T1 T2 Return to TOC Variation in Gas Volume with Temperature at Constant Pressure As T increases V increases A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1 /T1 = V2 /T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? T1 = 125 (0C) + 273.15 (K) = 398.15 K T2 = V2 x T1 V1 = 1.54 L x 398.15 K 3.20 L = 192 K Section 13.3 Volume and Temperature: Charles’s Law Exercise Suppose a balloon containing 1.30 L of air at 24.7°C is placed into a beaker containing liquid nitrogen at –78.5°C. What will the volume of the sample of air become (at constant pressure)? 0.849 L V1 V2 T1 T2 (V2 ) (1.30 L) (24.7 + 273 K) ( 78.5 + 273 K) V2 = 0.849 L Return to TOC Section 13.4 Volume and Moles: Avogadro’s Law Avogadro’s Law (3rd) • • (constant T and P). V = an (a is a proportionality constant) n1 n2 = V1 V2 Return to TOC Section 13.4 Volume and Moles: Avogadro’s Law The Relationship Between Volume and Moles Return to TOC Section 13.4 Volume and Moles: Avogadro’s Law Exercise If 2.45 mol of argon gas occupies a volume of 89.0 L, what volume will 2.10 mol of argon occupy under the same conditions of temperature and pressure? 76.3 L V1 V = 2 n1 n2 V2 89.0 L = 2.45 mol 2.10 mol V2 = 76.3 L Return to TOC If n and V are constant, then P α T P and T are directly proportional. P1 P2 T1 T2 If one temperature goes up, the pressure goes up! Joseph Louis GayLussac (1778-1850) Gay Lussac’s Law Summary The pressure and temperature of a gas are directly related, provided that the volume remains constant. P1 P2 T1 T2 P1 P2 T1 T2 3.00 atm 298 K V2 325 K 3.27 L P1 P2 T1 T2 1.8 atm 1.9 atm 273 K T2 288 K or 15.2 oC Gay-Lussac’s Law: Pressure and Temperature We can simplify this relationship by the formula: P1 P2 = T1 T2 Where, P1, P2 = pressure in any unit (atm, kPa, or mmHg), BUT they must match! T1, T2 = temperature is always in Kelvin! (Recall, just add 273 + °C) Gay-Lussac’s Law: Example A gas has a pressure of 103kPa at 25°C. What will the pressure be when the temperature reaches 928°C? P1= 103kPa P1 P2 = T1= 25°C +273= 298K T T 1 2 P2= ? T2= 928°C+273= 1201K (P2) (103kPa) = (1201K) (298K) P2 = 415kPa Section 13.5 The Ideal Gas Law • We can bring all of these laws together into one comprehensive law: V = bT (constant P and n) V = an (constant T and P) V = k (constant T and n) P PV = nRT (where R = 0.08206 L·atm/mol·K, the universal gas constant) Return to TOC The Combined Gas Law A gas occupies 3.78L at 529mmHg and 17.2°C. At what pressure would the volume of the gas be 4.54L if the temperature is increased to 34.8°C? P1 V1 P2 V2 = P1= 529mmHg T T 1 2 V1= 3.78L T1= 17.2°C + 273= 290.2K (P2)(4.54L) (529mmHg)(3.78L) P2= ? = (307.8K) (290.2K) V2= 4.54L P2 = 467mmHg T2= 34.8°C + 273= 307.8K Using PV = nRT How much N2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC? R = 0.082057 L•atm/K•mol Solution 2. Now calc. n = PV / RT 4 (0.98 atm)(2.7 x 10 L) n = (0.0821 L• atm/K • mol)(298 K) n = 1.1 x 103 mol (or about 30 kg of gas) 29 30 Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Solution Strategy: Calculate moles of H2O2 and then moles of O2 and H2O. Finally, calc. P from n, R, T, and V. 31 Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Solution 1 mol 1.1 g H2 O2 • 0.032 mol 34.0 g 1 mol O2 0.032 mol H2O2 • = 0.016 mol O2 2 mol H2 O 2 Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Solution P of O2 = nRT/V (0.016 mol)(0.0821 L• atm/K • mol)(298 K) = 2.50 L P of O2 = 0.16 atm 32 What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atm PV = nRT nRT V= P 1 mol HCl n = 49.8 g x = 1.37 mol 36.45 g HCl 1.37 mol x 0.0821 V= V = 30.7 L L•atm mol•K 1 atm x 273.15 K Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR = P = constant T V P1 P2 = T1 T2 P1 = 1.20 atm T1 = 291 K P2 = ? T2 = 358 K T2 = 1.20 atm x 358 K = 1.48 atm P2 = P1 x 291 K T1 DALTON’S LAW (of Partial Pressures) Partial Pressures: The gas in each tank on the left exerts a pressure Ptotal = Pgas1 + Pgas2 + Pgas3 … Dalton’s Law states: The total pressure of a mixture of gases is equal to the sum of the partial pressures of the gases in the mixture. Example Problem 1 An automobile tire contains a mixture of nitrogen, oxygen, and carbon dioxide with partial pressures of 25 psi, 7 psi, and 3 psi respectively. What is the total pressure inside the tire? Example Problem 1 An automobile tire contains a mixture of nitrogen, oxygen, and carbon dioxide with partial pressures of 25 psi, 7 psi, and 3 psi respectively. What is the total pressure inside the tire? Ptotal = 25 psi + 7 psi + 3 psi = 35 psi Example Problem 2 A football has a mixture of nitrogen and oxygen gases. The pressure inside the football is 760. mmHg. The partial pressure of nitrogen (PN2) is 600. mmHg. What is the partial pressure of oxygen (PO2)? Example Problem 2 A football has a mixture of nitrogen and oxygen gases. The pressure inside the football is 760. mmHg. The partial pressure of nitrogen (PN2) is 600. mmHg. What is the partial pressure of oxygen (PO2)? Ptot = PN2 + PO2 760mmHg - 600mmHg = PO2 160mmHg = PO2 Example 3 A gas mixture containing oxygen, nitrogen, and carbon dioxide has PO2 = 20.1 kPa, PN2 = 18.3 kPa, and PCO2 = 34.4 kPa. What is Ptotal? Ptotal = P1 + P2 + P3 Ptotal = PO2 + PN2 + PCO2 Ptotal = 20.1 kPa + 18.3 kPa + 34.4 kPa Ptotal = 72.8 kPa Example 4 A gas mixture containing oxygen, nitrogen, and argon has a total pressure of 50.2 kPa. If PO2 = 20.1 kPa and PN2 = 18.3 kPa what is PAr? Ptotal = P1 + P2 + P3 Ptotal = PO2 + PN2 + PAr PAr = Ptotal - PO2 - PN2 PAr = 50.2 kPa - 20.1 kPa - 18.3 kPa Ptotal = 11.8 kPa Mole fraction n1 nTotal Mole fraction = Pressure fraction n1 n1 n2 n3 .... E.g. A 1.00L sample of dry air at 25°C and 786mmHg contains 0.925g of N2, plus other gases. (a) What is the partial pressure of N2 in the air sample? (b) What is the mole fraction and mole percent of N2 in the air? 0.925gN2 x 1mol N2 = 0.330 molN2 28.0g N2 PN2 = nN2RT/V = 0.330molx0.0821 Latm/(K mol) x 298K 1.00L = 0.807 atm (=613mmHg) (b) Mole fraction of N2 = PN2 = 613mmHg = 0.780 P 786mmHg Air contains 78.0 mole percent of N2 A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = Xi PT PT = 1.37 atm 0.116 Xpropane = 8.24 + 0.421 + 0.116 = 0.0132 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm GAS DENSITY Screen 12.5 PV = nRT n P = V RT m P = M• V RT where M = molar mass m PM d = = V RT d and M proportional 47 A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0C. What is the molar mass of the gas? dRT M= P M= g 2.21 L M = 54.5 g/mol 4.65 g m = = 2.21 d= 2.10 L V x 0.0821 L•atm mol•K 1 atm x 300.15 K g L Kinetic Energy of Gas Particles At the same conditions of temperature, all gases have the same average kinetic energy. 1 2 m = mass KE mv v = velocity 2 At the same temperature, small molecules move FASTER than large molecules The Meaning of Temperature 3 ( KE ) avg RT 2 Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.) Kinetic Molecular Theory Maxwell’s equation u2 3RT M root mean square speed u where is the speed and M is the molar mass. • speed INCREASES with T • speed DECREASES with M 51 Diffusion Diffusion describes the mixing of gases. The rate of diffusion is the rate of gas mixing. Diffusion is the result of random movement of gas molecules The rate of diffusion increases with temperature Small molecules diffuse faster than large molecules Effusion Effusion: describes the passage of gas into an evacuated chamber. GAS DIFFUSION AND EFFUSION •Diffusion - The rate at which two gases mix. •Effusion - The rate at which a gas escapes through a pinhole into a vacuum. 54 GAS DIFFUSION AND EFFUSION Graham’s law governs effusion and diffusion of gas molecules. Rate for A Rate for B M of B M of A Rate of effusion is inversely proportional to its sq. root molar mass. Thomas Graham, 1805-1869. Professor in Glasgow and London. 55 Graham’s Law of Effusion Distance traveled by gas 1 M2 Distance traveled by gas 2 M1 M1 = Molar Mass of gas 1 M2 = Molar Mass of gas 2 Graham’s Law of Effusion The rate of effusion of a gas is inversely proportional to the square root of its molar mass.” Rate A Rate B MB MA Rate = Rate of effusion M=Molar Mass of gas Example: What is the relative rate of effusion of H2 vs. O2? Rate H 2 Rate O2 M 32 32 16 4 M 2 2 O2 H2 At what temperature does 16.3 g of nitrogen gas have a pressure of 1.25atm in a 25.0 L tank?? Calculate the molecular weight of a gas if 35.44 g of the gas stored in a 7.50 L tank that exerts a pressure of 60.0 atm at a constant temperature of 35.5°C. An unknown gas has a density of 3.167 g/L at STP. What is the identity of the gas? (Ar, O2, Cl2, HF, H2O)? Problem example: O2 generated in the decomposition of KClO3 is collected over water. The volume of the gas collected at 24 oC and at an atmospheric pressure of 762 torr is 128 ml. Calculate the number of moles of O2 obtained. The vapor pressure of H2O at 24 oC is 22.4 torr. 61 First step: Calculate the partial pressure of O2. PTotal Pgas PH O (Dalton’s law) 2 62 Pgas PTotal PH O 2 = 762 torr – 22.4 torr = 739.6 torr (extra sig. fig) 739.6 torr 1 atm 760 torr = 0.973 atm 63 From the ideal gas equation PV = nRT, PV n= RT (0.973 atm) (0.128 l) = (0.08206 l atm mol-1K-1)(297 K) = 0.00511 mols 64