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1 As P (h) increases V decreases Apparatus for Studying the Relationship Between Pressure and Volume of a Gas

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2 P 1/V P x V = constant P 1 x V 1 = P 2 x V 2 Boyle’s Law Constant temperature Constant amount of gas

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3 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 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg P x V = constant

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4 As T increasesV increases Variation in Gas Volume with Temperature at Constant Pressure

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5 Variation of Gas Volume with Temperature at Constant Pressure V TV T V = constant x T V 1 /T 1 = V 2 /T 2 T (K) = t ( 0 C) Charles’ & Gay-Lussac’s Law Temperature must be in Kelvin

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6 A sample of carbon monoxide gas occupies 3.20 L at C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V L x K 3.20 L = = 192 K V 1 /T 1 = V 2 /T 2 T 1 = 125 ( 0 C) (K) = K

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7 Avogadro’s Law V number of moles (n) V = constant x n V 1 / n 1 = V 2 / n 2 Constant temperature Constant pressure

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8 Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO

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9 Summary of Gas Laws Boyle’s Law

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10 Charles Law

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11 Avogadro’s Law

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12 Ideal Gas Equation Charles’ law: V T (at constant n and P) Avogadro’s law: V n (at constant P and T) Boyle’s law: P (at constant n and T) 1 V V V nT P V = constant x = R nT P P R is the gas constant PV = nRT

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13 The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)( K) R = L atm / (mol K) Experiments show that at STP, 1 mole of an ideal gas occupies L.

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14 What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = K P = 1 atm n = 49.8 g x 1 mol HCl g HCl = 1.37 mol V = 1 atm 1.37 mol x x K Latm molK V = 30.7 L

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15 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 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm

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Sample Problem 5.2 Applying the Volume-Pressure Relationship PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm 3 at 1.12 atm. 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: SOLUTION: V 1 in cm 3 V 1 in mL V 1 in L V 2 in L unit conversion gas law calculation P 1 = 1.12 atmP 2 = 2.64 atm V 1 = 24.8 cm 3 V 2 = unknown P and T are constant 24.8 cm 3 1 mL 1 cm 3 L 10 3 mL = L P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = P 1 V 1 = P 2 V 2 P1V1P1V1 P2P2 V 2 = = L 1.12 atm 2.46 atm = L 1cm 3 =1mL 10 3 mL=1L xP 1 /P 2

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Sample Problem 5.3Applying the Pressure-Temperature Relationship PROBLEM: A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x10 3 torr. It is filled with methane at 23 0 C and atm and placed in boiling water at exactly C. Will the safety valve open? PLAN:SOLUTION: P 1 (atm)T 1 and T 2 ( 0 C) P 1 (torr)T 1 and T 2 (K) P 1 = 0.991atmP 2 = unknown T 1 = 23 0 CT 2 = C P 2 (torr) 1atm=760torr x T 2 /T 1 K= 0 C P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = P1P1 T1T1 P2P2 T2T2 = atm 1 atm 760 torr = 753 torr P 2 =P1P1 T2T2 T1T1 = 753 torr 373K 296K = 949 torr

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Sample Problem 5.4 Applying the Volume-Amount Relationship PROBLEM: A scale model of a blimp rises when it is filled with helium to a volume of 55 dm 3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm 3. How many more grams of He must be added to make it rise? Assume constant T and P. PLAN: SOLUTION: We are given initial n 1 and V 1 as well as the final V 2. We have to find n 2 and convert it from moles to grams. n 1 (mol) of He n 2 (mol) of He mol to be added g to be added x V 2 /V 1 x M subtract n 1 n 1 = 1.10 moln 2 = unknown V 1 = 26.2 dm 3 V 2 = 55.0 dm 3 P and T are constant P1V1P1V1 n1T1n1T1 P2V2P2V2 n2T2n2T2 = V1V1 n1n1 V2V2 n2n2 = n 2 = n 1 V2V2 V1V1 n 2 = 1.10 mol 55.0 dm dm 3 = 2.31 mol g He mol He = 9.24 g He

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Sample Problem 5.5Solving for an Unknown Gas Variable at Fixed Conditions PROBLEM: A steel tank has a volume of 438 L and is filled with kg of O 2. Calculate the pressure of O 2 at 21 0 C. PLAN: SOLUTION: V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P. V = 438 LT = 21 0 C (convert to K) n = kg (convert to mol)P = unknown 21 0 C = K 0.885kg 10 3 g kg mol O g O 2 = 27.7 mol O 2 P = nRT V = 24.7 mol K atm*L mol*K x x 438 L = 1.53 atm

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Sample Problem 5.6 Using Gas Laws to Determine a Balanced Equation PROBLEM: The piston-cylinders below depict a gaseous reaction carried out at constant pressure. Before the reaction, the temperature is 150K; when it is complete, the temperature is 300K. PLAN: SOLUTION: We know P, T, and V, initial and final, from the pictures. Note that the volume doesn’t change even though the temperature is doubled. With a doubling of T then, the number of moles of gas must have been halved in order to maintain the volume. Which of the following balanced equations describes the reaction? (1) A 2 + B 2 2AB(2) 2AB + B 2 2AB 2 (4) 2AB 2 A 2 + 2B 2 (3) A + B 2 AB 2 Looking at the relationships, the equation that shows a decrease in the number of moles of gas from 2 to 1 is equation (3). New figures go here.

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