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O.A # 21

O.A # 21. Pressure. O.A # 22. Barometer. O.A # 23. V 1. P 1. A gas sample occupies 35.0 mL at 255 mmHg. What is the new pressure if the gas occupies 51.0 mL?. V 2. P 2 ?. V 2. V 2. P 2 = 255 mmHg x 35.0 mL 51.0 mL. P 2 = 175 mmHg. O.A # 24.

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O.A # 21

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  1. O.A # 21 Pressure

  2. O.A # 22 Barometer

  3. O.A # 23 V1 P1 A gas sample occupies 35.0 mL at 255 mmHg. What is the new pressure if the gas occupies 51.0 mL? V2 P2 ? V2 V2 P2 =255 mmHg x 35.0 mL 51.0 mL P2 = 175 mmHg

  4. O.A # 24 Find pressure of O2 at 101.3 Kpa if: PNO2=79.10 Kpa, PCO2= 0.040 Kpa, and Pothers= 0.94 Kpa Dalton’s Law Ptotal = PO2+ PNO2+ PCO2+ Pothers PO2=101.3 – (79.10 + 0.040 + 0.94) PO2= 21.2 Kpa

  5. O.A # 25 Gay Lussac’s Law

  6. O.A # 26 V1 P1 A gas sample occupies 89.6 mL at 214 torr. What is the new pressure if the gas occupies 34.2 mL? V2 P2 ? V2 V2 P2 =214 torr. x 89.6mL 34.2 mL P2 = 561 torr Boyle’s Law

  7. O.A # 27 Combined Gas Law

  8. O.A # 28 P1 V1 A gas has a volume of 7.84 mL at 71.8 Kpa and 25.0°C. What would be the volume at STP? V2? T1 T2, P2 =25.0°C + 273 = 298 Kelvin T2 = 0°C + 273 = 273 Kelvin P2 = 101.325 Kpa V2 = P1 x V1 x T2 (T1 x P2) V2 = 71.8 Kpa x 7.84 mL x 273 Kelvin ( 298 Kelvin x 101.325 Kpa ) V2 = 5.09 mL Combined Gas Law

  9. O.A # 29 P=? n Calculate the pressure of 1.82 moles of a gas in a 5143 mL container at 69 °C V T Ideal gas law PV = nRT 69 °C + 273 = 342 kelvin 5143 mL x 1 L____ = 5.143 L 1000 mL P= 1.82 x 0.0821 x 342 5.143 P= 9.9 atm

  10. O.A # 30 A 18.6 Liter flask contains 23.2 g of an unknown gas at STP. Find molecular mass of the gas. a) Find moles (Use Molar volume at STP) 18.6 L x 1 mole = 0.830 moles L 22.4 b) Find molecular mass g/moles 23.2 g 0.830 moles = 28.0 g/moles c) Identify the gas Nitrogen= 14.0 x 2 = 28.0 g

  11. O.A # A helium-filled balloon at sea level has a volume of 2.1 L at 0.998 atm and 36 °C. If it is released and it rises to an elevation at which the pressure is 0.900 atm and the temperature is 28°C, what will be the new volume of the balloon? V1 = 2.1L P1 = 0.998 atm T1 = 36 °C V2 = ? P2 = 0.900 atm T2 = 28 °C + 273 = 309 Kelvin + 273 = 301 Kelvin V2 = P1 V1 T2 (T1 P2) V2 = 0.998 atm x 2.1 L x 301 Kelvin (309 Kelvin x 0.900 atm ) V2 = 2.3 L

  12. Unit 5(chapter 13.1 & 14) GAS LAWS

  13. Pressure Force applied on some given area • Gas Pressure: Pressure created by molecules crashing into walls and each other • Increase collisions = increase pressure Atmospheric Pressure: Pressure applied in all directions because molecules move in all directions

  14. Page #11 Converting Units of Pressure STP: Standard Temperature and Pressure Standard Temperature: 0°Celsius Standard Pressure: ……101325 Pa = ……101.325 KPa= ……1.00 atm = …… 760 mmHg = …… 760 torr =

  15. Conversions using standard pressure values: (Show work) 1. 87 atm = _______ Kpa 2. 42.5 mmHg = __________ atm 3. 35.41 Pa = ______________atm ……101325 Pa = ……101.325 KPa= ……1.00 atm = …… 760 mmHg = …… 760 torr = 8800 87 atm x 101.325 KPa 1.00 atm 0.0559 42.5 mmHg x 1.00 atm___ 760 mmHg 0.0003495 35.41 Pa x 1.00 atm___ 101325 Pa

  16. 0.332 0.829 6.30 x102 8.5 x 102 6.4 x 103 d) 33.6 kPa = __________ atm e) 84.0 kPa = __________ atm __________ torr f) 8.4 atm = __________ kPa __________ mmHg

  17. Page #12 Dalton’s Law of Partial Pressure • Ptotal represents the total pressure of a mixture of gases. • P1, P2, and so on represent the partial pressures of each gas in the mixture.

  18. Practice Problems on pg.392 (#4-6) 4) Hydrogen + Helium = Total pressure ? + 439 mmHg = 600 mmHg 600 mmHg - 439 mmHg = 161 mmHg Hydrogen = 161 mmHg 5) total pressure = 5.00 KPa + 4.56 Kpa + 3.02Kpa + 1.20Kpa total pressure = 13.78 Kpa 6) Total pressure = CO2 + gas1 + gas2 30.4 Kpa = CO2 + 16.5 Kpa + 3.7 Kpa CO2 = 30.4 Kpa – (16.5 Kpa + 3.7 Kpa) CO2 = 10.2 Kpa

  19. Practice Problems page 415 # 65-67 65) 0.01 atm 66) 0.99 atm 67) 1.1 atm

  20. Page # 13 Boyle’s Law the pressure and volume of a gas at constant temperature are inversely proportional Equation for Boyle’s Law

  21. Problems on pg. 422 (#1-5) 1) The volume of a gas at 99.0 Kpa is 300.0 mL. If the pressure is increased to 188 Kpa, what will be the new volume? P2=188 KPa V2 = ? P2 P2 P1=99.0 KPa V1= 300.0 mL V2= 99.0 KPa x 300.0 mL 188 KPa V2= 158 mL

  22. 2) The pressure of a sample of helium in a 1.00-Liter container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00-Liter container? P2 ? V2 V2 P2 =0.988 atm x 1.00 L 2.00 L P2 = 0.494 atm

  23. Problems on pg. 422 (#3-5) 3) V1 = 145.7 mL P1 = 1.08 atm V2 = ? P2 = 1.43 atm V2 = 1.08 atm x 145.7 mL 1.43 atm V2 = 1.10 x 102mL 4) V1 = 0.0500 L P1 = ? V2 = 4.00 L P2 = 0.980 atm P1 = 0.980 atm x 4.00 L 0.0500 L P1 = 78.4 atm

  24. 5) V1 = 0.220 L P1 = 0.860 atm V2 = ? P2 = 29.2 Kpa 29.2 Kpa x 1.00 atm 101.325 Kpa V2 = 0.220 L x 0.860 atm 0.288 atm V2 = 0.657L

  25. Page # 14Charles’ Law Volume of a gas is directly proportional to its Kelvin temperature when pressure is constant. Equation for Charles’ Law Gay-Lussac Law: K = °C + 273 New temperature scale: Kelvin Must use Kelvin temperature for ALL gas laws

  26. V2 T2 1. A sample of Helium occupies 473 cm³ at 36 °C. What will be the volume when the initial temperatureis 94 °C? V1 =? T1 T1: 94 °C + 273 = 367 kelvin T2: 36 °C + 273 = 309 kelvin V1 = V2 T1 T2 V1 = 473 cm³ x 367 Kelvin 309 Kelvin V1 = 560 cm³ = 5.6 x 102 cm³

  27. 2) A gas at 97°C occupies a volume of 1.5 L. At what temperature will the volume decrease at 1.0 L? T1 = 97 °C V1 = 1.5 L T2 = ? V2 = 1.0 L T1 = 97 °C + 273 = 370 Kelvin T2= V2 T1V1 T2 = V2 T1 V1 T2 = 1.0 L x 370 Kelvin 1.5 L T2 = 250 Kelvin

  28. Practice Problems on pg. 425 (#6-8) 6) A gas at 89°C occupies a volume of 0.67 L. At what Celsius temperature will the volume increase at 1.12 L T1 = 89 °C V1 = 0.67 L T2 = ? V2 = 1.12 L T1 = 89 °C + 273 = 362 Kelvin T2= V2 T1V1 T2 = V2 T1 V1 T2 = 1.12 L x 362 Kelvin 0.67 L T2 = 330 °C T2= 605 kelvin - 273

  29. 7) The Celsius temperature of a 3.00-L sample of gas is lowered from 80.0 °C to 30.0 °C. What will be the resulting volume of this gas? V1 = 3.00 L T1 = 80.0 °C V2 = ? T2 = 30.0 °C T1 = 80.0 °C + 273 = 353 Kelvin T2 = 30.0 °C + 273 = 303 Kelvin V2 = V1 T2 T1 V2 = 3.00 L x 303 Kelvin 353 Kelvin V2= 2.58 L

  30. 8) V1 = 0.620 L T1 = 25°C V2 = ? T2 = 0.00 °C T1 = 25°C + 273 = 298 Kelvin T2 = 0.00 °C + 273 = 273 Kelvin V2 = V1 T2 T1 V2 = 0.620 x 273 298 V2 = 0.57 L

  31. Page # 15 The Combined Gas Law States the relationship among pressure, volume and temperature of a fixed amount of gas Equation for the Combined Gas Law

  32. At 0.00 °C and 1.00 atm pressure, a sample of a gas occupies 30.0 mL. If the temperature is increased to 30.0 °C and the entire gas sample is transferred to a 20.0 mL container, what will be the gas pressure inside the container? P1 = V1 = T1 = P2 = V2 = T2 = 1.00 atm 30.0 mL 0.00 °C = 273 Kelvin = 303 Kelvin ? 20.0 mL 30.0 °C P2 = P1 V1 T2 ( T1 V2 ) P2 = 1.00 atm x 30.0 mL X 303 Kelvin = ( 273 Kelvin X 20.0 mL) P2 = 1.66 atm

  33. Page # 16The Ideal Gas Law The pressure, volume, temperature, and number of moles of gas can be related by using the ideal gas law Ideal Gas Law Equation PV = nRT • P: pressure, atm • V: volume, Liters • n: number of moles • T: temperature, Kelvin • R: ideal gas law constant R= 0.0821 Lx atm Kelvin.moles

  34. 1) Determine the Celsius temperature of 2.49 moles of gas contained in a 1.00 L container at a pressure of 143 Kpa. T= ?°C n = 2.49 moles V= 1.00 L P = 143 kpa x 1.00 atm 101.325 kpa = 1.41 atm PV = nRT T = 1.41 x 1.00 (2.49 x 0.0821) nR - 273 = T = 6.90 K -266 °C

  35. 2) If the pressure exerted by a gas at 25 °C in a volume of 0.044 L is 3.81 atm, how many moles of gas are present? T= 25 °C V= 0.044 L P = 3.81 atm = 298 Kelvin PV = nRT n = 3.81 x 0.044 (0.0821 x 298) RT n = 0.0069 moles = 6.9 x10-3 moles

  36. The equal volumes of gases at the same temperature and pressure contain equal number of particles Page # 17 Avogadro’s Principle: Molar Volume: Is the volume that one mole occupies at 0.00 °C and 1.00 atm pressure. These conditions are known as STP 1 mole of any gas at STP will occupy… 22.4 L Conversion factor: 22.4 L = 1 mole

  37. Practice Problems 1) Determine the volume of a container that holds 2.4 moles of gas at STP. Conversion factor: 22.4 L = 1 mol 2.4 mol x 22.4 L = 1 mol 54 L 2) A 15.3 Liter flask contains 21.86 g of an unknown gas at STP. Find molecular mass of the gas. a) Find moles (Use Molar volume at STP) 15.3 L x 1 mole = 0.683 moles L 22.4 b) Find molecular mass g/moles 21.86 g 0.683 moles = 32.0 g/moles O2

  38. Graham’s Law of Effusion Page # 18 • Measures how fast a gas moves through a medium. • The lighter gas moves faster • A is the lighter gas Rate A= Rate B molar mass B molar mass A

  39. Rate A= Rate B molar mass B molar mass A Practice Problems • Calculate the ratio of effusion rates for hydrogen, H2 • and Neon, Ne Molar mass: H2 = Ne= 2x1.0 = 2.0 g 20.2 g = 3.18 Hydrogen moves 3.18 times faster than Neon 20.2 g 2.0 g 2. Calculate the ratio of effusion rates for COand CO2 Molar mass: CO = CO2= 44.0 g 28.0 g Rate CO = Rate CO2 = 1.2, CO is 1.25 times faster than CO2 44.0 g 28.0 g

  40. Practice Problems on pg. 433 (#27-33) 27) How many moles of nitrogen gas will be contained in a 2.00 L flask at STP? 2.00 L x 1 mol = 22.4 L 0.0893 mol 28) If a balloon will rise off the ground when it contains 0.0226 mol of helium in a volume of 0.460 L, how many moles of helium are needed to make the balloon rise when its volume is 0.865 L? Assume that temperature and pressure stay constant. SET UP PROBLEM AS RATIO 0. 865 L x 0.0226 mol = 0.460 L ? mol 0. 865 L = 0.0425 mol = 4.25 x10-2mol

  41. 29) How many grams of carbon dioxide gas are in a 1.0 L balloon at STP? 1.0 L x 1 mol = 22.4 L 0.045 mol x 44.0 g = 1 mol 2.0 g 30) What volume in milliliters will 0.00922 g H2 gas occupy at STP? 0.00922 g x 1 mol = 2.0 g x 22.4 L = 1 mol 0.00461 mol 0.103 L 0.103 L x 1000 mL 1 L = 103 mL

  42. 31) 0.416 g x 1 mol = 83.8 g 0.00496 mol x 22.4 L= 1 mol 0.111 L • Set up problem as mole ratio 0.860 g – 0.205 g = 0.655 g V = 0.655 g 19.2 L 0.860 g x 0.655 g V = 14.6 L 33) 4.5 Kg x 1000 g = 1 Kg 4500 g x 1 mole 28.0 g = 161 moles 161 moles x 22.4 L 1 mole = 3600 L = 3.6 x 103 L

  43. Practice Problems pg. 437 (#44-45) 44) P=? n= 0.108 moles T=20°C =293Kelvin V=0.505L P = 0.108 x 0.0821 x 293 0.505 P = 5.14 atm 45) n= 0.0470 moles V=1.20 L P = 0.988 atm T =? T = 0.988 x 1.20 (0.0470 x 0.0821) T = 307 Kelvin

  44. O.A # A sample of Oxygen occupies 305 cm³ at 15 °C. What will be the volume when temperatureis increased to 48 °C? V1 T1 V2 =? T2 T1: 15 °C + 273 = 288 kelvin T2: 48 °C + 273 = 321 kelvin V2 = V1 T2 T1 V2 = 305 x 321 288 V2 = 340 cm³ Charles’ Law

  45. O.A # P=? n Calculate the pressure of 1.82 moles of a gas in a 5143 mL container at 69.5 °C V T Ideal gas law PV = nRT 69.5 °C + 273 = 342.5 kelvin 5143 mL x 1 L____ = 5.143 L 1000 mL P= 1.82 x 0.0821 x 342.5 5.143 P= 9.95 atm

  46. O.A # A 35.7 Liter flask contains 88.7 g of an unknown gas at STP. Find molecular mass of the gas. a) Find moles (Use Molar volume at STP) 35.7 L x 1 mole = 1.59 moles L 22.4 b) Find molecular mass g/moles 88.7 g 1.59 moles = 55.8 g/moles

  47. Calculate the volume occupied by 7.40 g of NH3 at STP O.A # 98 V = ? g - moles 7.40 g x 1 mole = 17.0 g 0.435 moles 0.435 moles x 22.4 L 1 mole V = 9.74 L

  48. Quick Write I. Label the following as being representative of inverse or direct relationship, indicate the gas law, and the equation • Pressure versus volume of a gas • Volume versus temperature of a gas • Pressure versus temperature of a gas Inverse, Boyle’s direct, Charles’ direct, Lussac’s II. Write about the following two questions: If two variables have inverse relationship, what happens to the value of one as the value of the other increases? If two variables have a direct relationship, what happens to the value of one as the value of the other increases?

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