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The Gas Laws

The Gas Laws. Part 2. Standard Temperature and Pressure. STP= A temperature of 273 k and a pressure of 1.00 atm

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The Gas Laws

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  1. The Gas Laws Part 2

  2. Standard Temperature and Pressure STP= A temperature of 273 k and a pressure of 1.00 atm Anytime a gas has a temperature close to or larger than 273 k and a pressure that is near (or lower than) 1.00 atm, the gas will behave as an ideal gas. The ideal gas laws and formulas then apply. 1.00 atm = 760.0 torr 0r 760.0 mm Hg

  3. Dalton’s Law of Partial Pressure When two or more ideal gases are mixed together, the total pressure of the mixture is equal to the sum of the pressures of each individual gas. Ptotal = P1 + P2 + P3 + P4 + …… The pressure of an ideal gas does not depend on the identity of the gas. It depends only on the quantity of that gas

  4. Eample: The Atmosphere is composed of 21% O2, 78% N2 and 1% “Other”. P total of the atmosphere = PO2 + PN2 + Pother 1.00 atm = 0.21 atm + 0.78 atm + 0.01 atm

  5. Vapor Pressure Vapor pressure: The pressure exerted by the vapor which sits on top of any liquid. Vapor pressure increases with increasing temperature

  6. What is the boiling point? Boiling point is the temperature at which the vapor pressure of a liquid is equal to normal atmospheric pressure. At 100.0 C the vapor pressure of water is 760.0 torr or 1.00 atm http://www.youtube.com/watch?v=9q5gEZGoBnk

  7. Example problem A mixture of nitrogen and oxygen gas are stored at a temperature of 25˚C and a pressure of 1.0 atm. If the partial pressure of Nitrogen in the mixture is 0.7 atm, what is the partial pressure of oxygen? Dalton’s law of partial pressure: Ptotal = PN2 + PO2 So 1.0 atm = 0.7 atm + PO2 PO2 = 0.3 atm

  8. Another way to use Dalton’s Law of Partial Pressure Dalton’s Law of Partial Pressure can be used to calculate the mole fraction X = # moles of component Total # moles in the mixture The mole fraction can then be used to calculate the partial pressure of a gas P1 = X1Ptotal

  9. Example • A mixture of gases has a pressure of 1.5 atm. The mixture is analyzed and found to contain 0.200 g of H2, 4.00 g CO2 and 1.45 g N2. What is the partial pressure of each gas in the mixture? • Dalton’s Law of Partial pressure • Find the mole fraction for each gas • Multiply the mole fraction of each gas by the total pressure to get the pressure for each gas.

  10. Find the mole fraction for each gas X = # moles of component total # moles in the mixture Need to convert each mass to moles. 0.200 g H2 X 1 mole H2 = 0.099 moles H2 2.02 g H2 4.00 g CO2 X 1 mole CO2= 0.091 moles CO2 44 g CO2 1.45 g N2 X 1 mole N2= 0.052 moles N2 28 g N2

  11. Now we can calculate mole fractions. XH2 = # moles of H2 total # moles of mixture Where do we find the total number of moles? 0.099 moles H2 + 0.091 moles CO2+ 0.052 moles N2 = 0.242 Moles XH2= # 0.099 moles H2 = 0.409 0.242 moles XCO2 = 0.091 moles CO2 = 0.376 0.242 moles XN2 = 0.052 moles N2 = 0.214 0.242 moles

  12. Now that we have the mole fractions, we can use that information to calculate the partial pressure of each gas. PH2 = XH2Ptotal PH2 = (0.409)(1.5 atm) PH2 = 0.61 atm PCO2 = (0.376)(1.5 atm) PCO2 = 0.56 atm PN2 = (0.214) (1.5 atm) PN2= 0.32 atm

  13. The Ideal Gas Law PV=nRT P= pressure V=Volume N=number of moles R=Ideal gas constant T=Temperature (in K) R=0.0821 L·atm mole·K

  14. Example What is the volume of 12.1 grams of nitrogen gas at STP? PV=nRT What do we know? P= 1.00 atm V= ? n= need to calculate from our mass R= constant…0.0821 L· atm Mole · K T=273 K

  15. Calculate n (number of moles) 12.1 g N2 X 1 mole N2 = .432 moles N2 28 g N2 Now we can solve for V PV=nRT Or V=nRT P V= (.432 moles N2) (0.0821 L· atm) (273.0K) = 9.68 L Mole · K 1.00 atm

  16. Stoichiometry with Gases A chemist performs the following reaction: CS2 (s) + 3 O2 (g) -------------- CO2 (g) + 2 SO2 (g) If the chemist starts with 110.0 grams of CS2 and an excess of O2, what volume of sulfur dioxide will be produced at a temperature of 341˚C and a pressure of 2.1 atm? Will eventually use the Ideal Gas Equation PV=nRT What do we know? P=2.1 atm V=? n= need to figure out using grams of CS2 R=constant constant…0.0821 L· atm Mole · K T= 341 ˚C….need to convert to K

  17. Let’s convert our temperature K = ˚C + 273.15 K= 341 ˚C + 273.15 K= 614 K Now we can convert grams of CS2 to moles of SO2 using stoichiometry 110.0 g CS2 X 1 mole CS2 = 1.44 moles CS2 76.2 g CS2 1.44 moles CS2 X 2 moles SO2= 2.88 moles SO2 1 mole CS2 Using the Ideal Gas Law V=nRT= (2.88 moles)(0.0821 L· atm) (614K) = 69 L SO2 P Mole · K_____ 2.1 atm

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