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Gases Entry Task: Oct 29 th Monday Question: Calculate the number of moles, if pressure is 2 atm, volume is 500.0 ml and temperature is 300K. You have 5 minutes!!

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Gases Agenda: Discuss Ch. 10 sec 4-6 HW: Combo, Ideal, Stocih Partial pressure ws (75% is review from last year!)

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Gases BREAK OUT AP EQUATION SHEET These formulas are rarely or not at all on the AP Exam Ideal gas law Van der Waals equation Daltons Partial pressure Moles= molar mass/molarity Kelvin/Celsius Combination gas law

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Gases BREAK OUT AP EQUATION SHEET These formulas are rarely or not at all on the AP Exam Density of gas Root mean Speed of gas Kinetic energy of gas molecules and moles of gas Grahams Law Osmotic pressure and Beers Law

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Gases Chapter 10 Gases

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Gases I can… Apply the ideal gas law equation to solve stoichiometric gas calculations. Explain how partial pressure relates to gas mixtures.

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Gases 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

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Gases Ideal-Gas Equation The constant of proportionality is known as R, the gas constant.

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Gases Ideal-Gas Equation The relationship then becomes nT P V V nT P V = R or PV = nRT

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Gases 10.3 problem Tennis balls are usually filled with either air or N 2 gas to a pressure above atmospheric pressure to increase their bounce. If a tennis ball has a volume of 144 cm 3 and contains 0.33 g of N 2 gas, what is the pressure inside the ball at 24 C? P= ? V= 144 cm 3 n= convert 0.33g to moles R= 0.08206 T= 297K 0.33 g/28g = 0.01178 moles of N 2 (X)(0.144) = (0.01178)(0.08206)(297) (X)(0.144L) = (0.28723) P = 1.99 or 2.0 atm Convert cm 3 to liters 144 cm 3 = 0.144 L

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Gases Application of Ideal Gas Law Calculate the number of moles of gas contained in a 3.0-L vessel at 300 K with a pressure of 1.50 atm. P= 1.50 atm V= 3.0 L T = 300 K n = X mol R= 0.0821 PV = nRT

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Gases Application of Ideal Gas Law (1.50 atm) (3.0 L) = (X mol)(0.0821 )(300K) P= 1.50 atm V= 3.0 L T = 300 K n = X mol R= 0.0821

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Gases GET X by its self!! (1.50 atm)(3.0L) (0.0821 )(300K) X mol= (1.50 atm) (3.0 L) = (X mol)(0.0821 )(300K)

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Gases DO the MATH 4.5 24.63 = 0.18 mol (1.50)(3.0) (0.0821 mol )(300) X mol=

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Gases Application of Ideal Gas Law If the pressure exerted by a gas at 25 C in a volume of 0.44L is 3.81 atm, how many moles of gas are present? P= 3.81 atm V= 0.44 L T = 25 + 273= 298 K n = X mol R= 0.0821 PV = nRT

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Gases Application of Ideal Gas Law ( 3.81 atm) ( 0.44 L) = (X mol)(0.0821 )( 298 K) P= 3.81 atm V= 0.44 L T = 25 + 273= 298 K n = X mol R= 0.0821

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Gases GET X by its self!! ( 3.81 atm)( 0.44 L) (0.0821 )(298 K) X mol= ( 3.81 atm) ( 0.44 L) = (X mol)(0.0821 )( 298 K)

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Gases DO the MATH 1.68 24.47 = 0.069 mol X mol= ( 3.81 )( 0.44 ) (0.0821 mol )(298 )

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Gases Application of Ideal Gas Law 2.50 g of XeF 4 gas is placed into an evacuated 3.00 liter container at 80°C. What is the pressure (atm) in the container? P= X atm V= 3.00 L T = 80 + 273 = 353K n = 2.50 g XeF 4 R= 0.0821 PV = nRT

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Gases Application of Ideal Gas Law 2.50 g of XeF 4 gas is placed into an evacuated 3.00 liter container at 80°C. What is the pressure (atm) in the container? P= X atm V= 3.00 L T = 80 + 273 = 353K R= 0.0821 n = 2.50 g XeF 4 CONVERT TO MOLES PV = nRT 2.50 g XeF 4 1 mole XeF 4 207.3 g XeF 4 = 0.012 mole XeF 4

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Gases Application of Ideal Gas Law (X atm) (3.00 L) = (0.012 mol)(0.0821 )(353 K) P= X atmT = 80 + 273 = 353K V= 3.00 L R= 0.0821 n= 0.012 XeF 4

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Gases GET X by its self!! 3.00 L X atm= (X atm) (3.00 L) = (0.012 mol)(0.0821 )(353 K)

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Gases DO the MATH 0.3477756 3.00 = 0.116 atm 3.00 X atm= = (0.012)(0.0821 atm)(353)

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Gases Combined Gas Law This lesson combines pressure, volume and temperature into a single equation.

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Gases Combined Gas Law P 1 V 1 = P 2 V 2 T1T1 T2T2

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Gases Application of Combined Gas Law A toy balloon has an internal pressure of 1.05 atm and a volume of 5.0 L. If the temperature where the balloon is released is 20˚ C, what will happen to the volume when the balloon rises to an altitude where the pressure is 0.65 atm and the temperature is –15˚ C? P 1 = 1.05 atm V 1 = 5.0 L T 1 = 20 + 273 = 293K P 2 = 0.65 atm V 2 = X T 2 = -15 + 273 = 258 K

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Gases Application of Combined Gas Law P 1 = 1.05 atm V 1 = 5.0 L T 1 = 20 + 273 = 293K P 2 = 0.65 atm V 2 = X T 2 = -15 + 273 = 258 K (1.05 atm) (5.0L) 293 K 258 K = (0.65 atm) (X L)

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Gases GET X by its self!! (1.05 atm)(5.0L)(258K) 293 K 258 K = (0.65 atm) (X L) (1.05 atm) (5.0L) (293 K)(0.65 atm) = X L

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Gases DO the MATH (1.05)(5.0L)(258) (293)(0.65) = X L 1354.5 L 190.45 = 7.11 L

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Gases YOU TRY!! Combined Gas Law A closed gas system initially has pressure and temperature of 1150 torr and 692.0°C with the volume unknown. If the same closed system has values of 242 torr, 7.37 L and -48.00°C, what was the initial volume in L? P 1 = 1150 torr V 1 = X L T 1 = 692 + 273 = 965K P 2 = 242 torr V 2 = 7.37 L T 2 = -48 + 273 = 225 K

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Gases Application of Combined Gas Law (1150 torr) (X L) 965 K 225 K = (242 torr) (7.37 L) P 1 = 1150 torr V 1 = X L T 1 = 692 + 273 = 965K P 2 = 242 torr V 2 = 7.37 L T 2 = -48 + 273 = 225 K

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Gases GET X by its self!! (965 K)(242 torr)(7.37 L) (1150 torr)(225 K) X L= (1150 torr) (X L) 965 K 225 K = (242 torr) (7.37 L)

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Gases DO the MATH X L= 1721116 L 258750 = 6.65 L (965)(242)(7.37 L) (1150)(225)

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Gases Densities of Gases If we divide both sides of the ideal-gas equation by V and by RT, we get nVnV P RT =

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Gases We know that moles molecular mass = mass Densities of Gases So multiplying both sides by the molecular mass ( ) gives n = m P RT mVmV =

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Gases Densities of Gases Mass volume = density So, Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas. P RT mVmV = d =

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Gases 10.6 problem The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressure is 1.6 atm. Assuming ideal behavior, calculate the density of Titan’s atmosphere. PRTPRT mVmV = d = (1.6)(28.6) (0.08206)(95K) 45.76 7.7957 = 5.86 or 5.9 g/L

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Gases 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 =

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Gases 10.7 problem Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21 C and 740.0 torr. (1.17) (62.36)(294K) (740) 21450.59 740 = 28.98 or 29.0g dRTPdRTP =

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Gases 10.8 problem Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21 C and 740.0 torr. (1.17) (62.36)(294K) (740) 21450.59 740 = 28.98 or 29.0g dRTPdRTP =

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Gases 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 + …

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42 Flash Animation - Click to Continue Dalton’s Law of Partial Pressures

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43 In a mixture of gases the total pressure, P tot, is the sum of the partial pressures of the gases: Dalton’s law allows us to work with mixtures of gases. Dalton’s Law of Partial Pressures

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44 Dalton’s Law of Partial Pressures For a two-component system, the moles of components A and B can be represented by the mole fractions (X A and X B ).

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45 Dalton’s Law of Partial Pressures What is the mole fraction of each component in a mixture of 12.45 g of H 2, 60.67 g of N 2, and 2.38 g of NH 3 ? Find out how many moles are for each gas then add up all the moles for total mole. Divide the individual mole by the total mole to give mole fraction 12.45g/2.0 g = 6.225 mol H 2 60.67g/28.0 g = 2.166 mol N 2 2.38g/17.0 g = 0.14 mol NH 3 8.531 total mol of all gases 6.225 mol H 2 / 8.531 = 0.73 mol H2 2.166 mol H 2 / 8.531 = 0.25 mol N2 0.14 mol H 2 / 8.531 = 0.016 mol NH3

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46 Dalton’s Law of Partial Pressures Mole fraction is related to the total pressure by: On a humid day in summer, the mole fraction of gaseous H 2 O (water vapor) in the air at 25°C can be as high as 0.0287. Assuming a total pressure of 0.977 atm, what is the partial pressure (in atm) of H 2 O in the air?

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Gases 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.

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Gases 10.11 problem Ammonium nitrite, NH 4 NO 2, decomposes on heating to form N 2 gas: NH 4 NO 2 (s) N 2 (g) + 2 H 2 O(l) When a sample of NH 4 NO 2 is decomposed in the apparatus of Figure 10.15, 511 mL of N 2 gas is collected over water at 26 C and 745 torr total pressure. How many grams of NH 4 NO 2 were decomposed? Typical gas stoichiometry problem and using Ideal gas law P= 719.79 torr V= 0.511 L n= XN 2 R= 62.36 T= 299K (719.79)(0.511) = X mol N 2 (62.36)(299) 367.81 = 0.0197 mol N 2 18645.64 Now look at equation, there is a 1:1 ratio of N 2 to NH 4 NO 2 0.0197 mol NH 4 NO 2 1 mol NH 4 NO 2 64.02 g NH 4 NO 2 1.26 g NH 4 NO 2 745 torr is NOT the correct pressure- we have to subtract water vapor pressure- see Appendix B. 745-25.21 = 719.79 torr

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Gases HW: Ideal Combo P. Pressure ws

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