Presentation on theme: "Entry Task: Oct 29th Monday"— Presentation transcript:
1 Entry Task: Oct 29th Monday Question: Calculate the number of moles, if pressure is 2 atm, volume is ml and temperature is 300K. You have 5 minutes!!
2 Agenda: Discuss Ch. 10 sec 4-6 HW: Combo, Ideal, Stocih Partial pressure ws (75% is review from last year!)
3 BREAK OUT AP EQUATION SHEET Ideal gas lawVan der Waals equationDaltons Partial pressureMoles= molar mass/molarityKelvin/CelsiusCombination gas lawThese formulas are rarely or not at all on the AP Exam
4 BREAK OUT AP EQUATION SHEET Density of gasRoot mean Speed of gasKinetic energy of gas molecules and moles of gasGrahams LawOsmotic pressure and Beers LawThese formulas are rarely or not at all on the AP Exam
6 I can…Apply the ideal gas law equation to solve stoichiometric gas calculations.Explain how partial pressure relates to gas mixtures.
7 Ideal-Gas Equation V nT P So far we’ve seen that V 1/P (Boyle’s law)V T (Charles’s law)V n (Avogadro’s law)Combining these, we getV nTP
8 Ideal-Gas EquationThe constant of proportionality is known as R, the gas constant.
9 Ideal-Gas Equation PV = nRT nT P V nT P V = R or The relationship then becomesorPV = nRT
10 10.3 problemTennis balls are usually filled with either air or N2 gas to a pressure above atmospheric pressure to increase their bounce. If a tennis ball has a volume of 144 cm3 and contains 0.33 g of N2 gas, what is the pressure inside the ball at 24 C?P= ? V= 144 cm3 n= convert 0.33g to moles R= T= 297KConvert cm3 to liters cm3 = L0.33 g/28g = moles of N2(X)(0.144) = ( )( )(297)(X)(0.144L) = ( )P = or 2.0 atm
11 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.PV = nRTT = 300 Kn = X molR=P= 1.50 atmV= 3.0 L
12 Application of Ideal Gas Law T = 300 Kn = X molR=P= 1.50 atmV= 3.0 L(1.50 atm) (3.0 L)= (X mol)( )(300K)
13 (0.0821 )(300K) GET X by its self!! (1.50 atm)(3.0L) X mol= = (X mol)( )(300K)(1.50 atm)(3.0L)X mol=( )(300K)
14 DO the MATH(1.50)(3.0)X mol=(0.0821mol)(300)4.5= 0.18 mol24.63
15 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?PV = nRTT = = 298 Kn = X molR=P= 3.81 atmV= 0.44 L
16 Application of Ideal Gas Law T = = 298 Kn = X molR=P= 3.81 atmV= 0.44 L(3.81 atm) (0.44 L)= (X mol)( )(298 K)
17 (0.0821 )(298 K) GET X by its self!! (3.81 atm)(0.44 L) X mol= = (X mol)( )(298 K)(3.81 atm)(0.44 L)X mol=( )(298 K)
18 (0.0821 mol)(298 ) = 0.069 mol 24.47 DO the MATH (3.81)(0.44 ) X mol= 1.68= mol24.47
19 Application of Ideal Gas Law 2.50 g of XeF4 gas is placed into an evacuated 3.00 liter container at 80°C. What is the pressure (atm) in the container?PV = nRTT = = 353Kn = 2.50 g XeF4R=P= X atmV= 3.00 L
20 Application of Ideal Gas Law 2.50 g of XeF4 gas is placed into an evacuated 3.00 liter container at 80°C. What is the pressure (atm) in the container?PV = nRTP= X atmV= 3.00 LT = = 353KR=n = 2.50 g XeF4CONVERT TO MOLES1 mole XeF42.50 g XeF4= mole XeF4207.3 g XeF4
21 Application of Ideal Gas Law P= X atm T = = 353KV= 3.00 L R=n= XeF4(X atm) (3.00 L)= (0.012 mol)( )(353 K)
22 3.00 L GET X by its self!! X atm= (X atm) (3.00 L) = (0.012 mol)( )(353 K)= (0.012 mol)( )(353 K)X atm=3.00 L
23 DO the MATH= (0.012)( atm)(353)X atm=3.00= atm3.00
24 Combined Gas LawThis lesson combines pressure, volume and temperature into a single equation.
26 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?P1 = 1.05 atmV1= 5.0 LT1 = = 293KP2 = 0.65 atmV2 = XT2 = = 258 K
27 Application of Combined Gas Law P1 = 1.05 atmV1= 5.0 LT1 = = 293KP2 = 0.65 atmV2 = XT2 = = 258 K(1.05 atm) (5.0L)= (0.65 atm) (X L)293 K258 K
28 258 K 293 K (293 K)(0.65 atm) GET X by its self!! (1.05 atm) (5.0L) = (0.65 atm) (X L)293 K258 K(1.05 atm)(5.0L)(258K)= X L(293 K)(0.65 atm)
29 (293)(0.65) = 7.11 L 190.45 DO the MATH (1.05)(5.0L)(258) = X L
30 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 °C, what was the initial volume in L?P1 = 1150 torrV1= X LT1 = = 965KP2 = 242 torrV2 = LT2 = = 225 K
31 Application of Combined Gas Law P1 = 1150 torrV1= X LT1 = = 965KP2 = 242 torrV2 = LT2 = = 225 K(1150 torr) (X L)= (242 torr) (7.37 L)965 K225 K
32 225 K 965 K (1150 torr)(225 K) GET X by its self!! (1150 torr) (X L) = (242 torr) (7.37 L)965 K225 K(965 K)(242 torr)(7.37 L)X L=(1150 torr)(225 K)
33 (1150)(225) = 6.65 L 258750 DO the MATH (965)(242)(7.37 L) X L=
34 Densities of Gases n P V = RT If we divide both sides of the ideal-gas equation by V and by RT, we getnVPRT=
35 Densities of Gases n = m P RT m V = We know that moles molecular mass = massn = mSo multiplying both sides by the molecular mass ( ) givesPRTmV=
36 Densities of Gases P RT m V = d = Mass volume = densitySo,PRTmV=d =Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas.
37 10.6 problem P RT m V = d = = 5.86 or 5.9 g/L 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.PRTmV=d =(1.6)(28.6)( )(95K)45.767.7957= 5.86 or 5.9 g/L
38 Molecular Mass P d = RT dRT P = We can manipulate the density equation to enable us to find the molecular mass of a gas:PRTd =BecomesdRTP =
39 10.7 problemCalculate the average molar mass of dry air if it has a density of 1.17 g/L at 21 C and torr.dRTP =(1.17) (62.36)(294K)(740)740= or 29.0g
40 10.8 problemCalculate the average molar mass of dry air if it has a density of 1.17 g/L at 21 C and torr.dRTP =(1.17) (62.36)(294K)(740)740= or 29.0g
41 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,Ptotal = P1 + P2 + P3 + …
42 Dalton’s Law of Partial Pressures Flash Animation - Click to Continue
43 Dalton’s Law of Partial Pressures In a mixture of gases the total pressure, Ptot, is the sum of the partial pressures of the gases:Dalton’s law allows us to work with mixtures of gases.
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 (XA and XB).
45 Dalton’s Law of Partial Pressures What is the mole fraction of each component in a mixture of g of H2, g of N2, and 2.38 g of NH3?Find out how many moles are for each gas then add up all the moles for total mole.12.45g/2.0 g = mol H2total mol of all gases60.67g/28.0 g = mol N22.38g/17.0 g = 0.14 mol NH3Divide the individual mole by the total mole to give mole fraction6.225 mol H2/ = 0.73 molH22.166 mol H2/ = 0.25 molN20.14 mol H2/ = molNH3
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 H2O (water vapor) in the air at 25°C can be as high as Assuming a total pressure of atm, what is the partial pressure (in atm) of H2O in the air?
47 Partial PressuresWhen 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.
48 10.11 problemAmmonium nitrite, NH4NO2, decomposes on heating to form N2 gas: NH4NO2(s) N2(g) + 2 H2O(l) When a sample of NH4NO2 is decomposed in the apparatus of Figure 10.15, 511 mL of N2 gas is collected over water at 26 C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?745 torr is NOT the correct pressure- we have to subtract water vapor pressure- see Appendix B = torrTypical gas stoichiometry problem and using Ideal gas lawP= torr V= L n= XN2 R= T= 299K(719.79)(0.511) = X mol N2(62.36)(299)= mol N2Now look at equation, there is a 1:1 ratio of N2 to NH4NO264.02 g NH4NO2mol NH4NO21.26 g NH4NO21 mol NH4NO2
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