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Entry Task: Oct 29 th Monday

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

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Entry Task: Oct 29 th Monday

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  1. Entry Task: Oct 29th 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!!

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

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

  5. Chapter 10Gases

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

  7. Combining these, we get nT P V Ideal-Gas Equation • So far we’ve seen that V 1/P (Boyle’s law) VT (Charles’s law) Vn (Avogadro’s law)

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

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

  10. 10.3 problem • Tennis 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= 0.08206 T= 297K Convert cm3 to liters 144 cm3 = 0.144 L 0.33 g/28g = 0.01178 moles of N2 (X)(0.144) = (0.01178)(0.08206)(297) (X)(0.144L) = (0.28723) P = 1.99 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 = nRT T = 300 K n = X mol R= 0.0821 P= 1.50 atm V= 3.0 L

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

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

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

  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 = nRT T = 25 + 273= 298 K n = X mol R= 0.0821 P= 3.81 atm V= 0.44 L

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

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

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

  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 = nRT T = 80 + 273 = 353K n = 2.50 g XeF4 R= 0.0821 P= X atm V= 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 = nRT P= X atm V= 3.00 L T = 80 + 273 = 353K R= 0.0821 n = 2.50 g XeF4 CONVERT TO MOLES 1 mole XeF4 2.50 g XeF4 = 0.012 mole XeF4 207.3 g XeF4

  21. Application of Ideal Gas Law P= X atm T = 80 + 273 = 353K V= 3.00 L R= 0.0821 n= 0.012 XeF4 (X atm) (3.00 L) = (0.012 mol)(0.0821 )(353 K)

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

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

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

  25. Combined Gas Law P1 V1 = P2 V2 T1 T2

  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 atm V1= 5.0 L T1 = 20 + 273 = 293K P2 = 0.65 atm V2 = X T2 = -15 + 273 = 258 K

  27. Application of Combined Gas Law P1 = 1.05 atm V1= 5.0 L T1 = 20 + 273 = 293K P2 = 0.65 atm V2 = X T2 = -15 + 273 = 258 K (1.05 atm) (5.0L) = (0.65 atm) (X L) 258 K 293 K

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

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

  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 -48.00°C, what was the initial volume in L? P1 = 1150 torr V1= X L T1 = 692 + 273 = 965K P2 = 242 torr V2 = 7.37 L T2 = -48 + 273 = 225 K

  31. Application of Combined Gas Law P1 = 1150 torr V1= X L T1 = 692 + 273 = 965K P2 = 242 torr V2 = 7.37 L T2 = -48 + 273 = 225 K (1150 torr) (X L) = (242 torr) (7.37 L) 225 K 965 K

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

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

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

  35. m V P RT = Densities of Gases • We know that • moles  molecular mass = mass n  = m • So multiplying both sides by the molecular mass ( ) gives

  36. m V P RT d = = 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.

  37. m V P RT d = = 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. (1.6)(28.6) (0.08206)(95K) 45.76 7.7957 = 5.86 or 5.9 g/L

  38. P RT dRT P d =  = Molecular Mass We can manipulate the density equation to enable us to find the molecular mass of a gas: Becomes

  39. dRT P  = 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

  40. dRT P  = 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

  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 12.45 g of H2, 60.67 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 = 6.225 mol H2 8.531 total mol of all gases 60.67g/28.0 g = 2.166 mol N2 2.38g/17.0 g = 0.14 mol NH3 Divide the individual mole by the total mole to give mole fraction 6.225 mol H2/ 8.531 = 0.73 molH2 2.166 mol H2/ 8.531 = 0.25 molN2 0.14 mol H2/ 8.531 = 0.016 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 0.0287. Assuming a total pressure of 0.977 atm, what is the partial pressure (in atm) of H2O in the air?

  47. 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.

  48. 10.11 problem Ammonium 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 NH4NO2were decomposed? 745 torr is NOT the correct pressure- we have to subtract water vapor pressure- see Appendix B. 745-25.21 = 719.79 torr Typical gas stoichiometry problem and using Ideal gas law P= 719.79 torr V= 0.511 L n= XN2 R= 62.36 T= 299K (719.79)(0.511) = X mol N2 (62.36)(299) 367.81 = 0.0197 mol N2 18645.64 Now look at equation, there is a 1:1 ratio of N2 to NH4NO2 64.02 g NH4NO2 0.0197 molNH4NO2 1.26 g NH4NO2 1 molNH4NO2

  49. HW: Ideal Combo P. Pressure ws

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