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Gases. Chapter 14. Sections. The Gas Laws The Combined Gas Laws and Avogadro’s Principle Ideal Gas Law Gas Stoichiometry. The Gas Laws. Chapter 14.1. The Gas Laws. Boyles Law Charles Law Gay-Lussac’s Law Each law relates two variables to the behavior of gasses. Pressure Temperature I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
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1. Gases Chapter 14

2. Sections • The Gas Laws • The Combined Gas Laws and Avogadro’s Principle • Ideal Gas Law • Gas Stoichiometry

3. The Gas Laws Chapter 14.1

4. The Gas Laws • Boyles Law • Charles Law • Gay-Lussac’s Law • Each law relates two variables to the behavior of gasses. • Pressure • Temperature • Volume • amount

5. Kinetic Theory Review • K-M theory suggests gas particles behave differently than liquids and solids. • K-M assumes the following are true about gasses: • Gas particles do not attract or repel each other • Gas particles are much smaller than the distances between them • Gas particles are in constant, random motion • No kinetic energy is lost when gas particles collide with each other or with the walls of the container • All gasses have the same average kinetic energy at a given temperature

6. The Nature of Gasses • Actual gasses don’t obey all the assumptions made by K-M theory • But their behavior approximates the theory • Notice that all assumptions of K-M theory are based on the four factors mentioned about the gas laws: • Number of particles present • Temperature • Pressure • Volume of gas

7. Boyle’s Law • Robert Boyle (1627-1691), Irish, studied relationship between pressure and volume. • An inverse relationship • Boyle’s Law: • The volume of a given amount of gas at constant temperature varies inversely with the pressure • So at any two different times, for the same gas, the product of pressure and volume will be the same • P1V1 = P2V2

8. Boyle’s Law • http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/gaslaw/boyles_law_graph.html

9. Boyle’s Law Problem • The volume of a gas at 99.0 kPa is 300.0 mL. If the pressure is increased to 188 kPa, what is the new volume? • P1V1 = P2V2 • 99.0kPa x 300.0mL = 188kPa x ? 99.0kPa x 300.0mL = ? 188kPa 158.0mL

10. Charles’s Law • Jacque Charles (1746-1823) French • Studied relationship between volume and temperature • As temperature increases , so does volume of gas • MUST BE IN KELVIN • Tk = Tc + 273 • K-M Theory says as temperature increases, particles move faster, so must be further away. • Relationship between temperature and volume is a straight line with positive slope V1 T1 V2 T2 =

11. Charles’s Law • Law: The volume of a gas is directly proportional to the temperature, expressed in kelvin, at a constant pressure

12. Charles’s Law Problem • A gas at 89°C occupies a volume of 0.67L. At what Celsius temperature will the volume increase to 1.12L? V1 T1 V2 T2 = 1.12L ? 0.67L 89°C + 273 1.12L ? x0.67L 362 °K = = 1.12L x 362°K 0.67L ? = = 605.1 K 605.1°K – 273K = 332°C

13. Gay-Lussac’s Law • Joseph Gay-Lussac studied relationship between temperature and pressure of a constained gas at fixed volume: • Found direct proportion exists between temperature (KELVIN) and pressure • Mathematically: P1 T1 P2 T2 =

14. Gay-Lussac’s Law Problem • A gas in a sealed container has a pressure of 125kPa at a temperature of 30°C. If the pressure of the container is increased to 201kPa, what is the new temperature? • Convert temp to K: 30+273 = 303K P1 T1 P2 T2 125kPa 303K 201kPa T2 = = 201kPa x 303K 125kPa T2 = = 487K 487K – 273 = 214°C

15. Combined Gas Law and Avogadro’s Principle Chapter 14, Section 2

16. Combined Gas Law • Combine Boyle’s Law, Charles’s Law, and Gay-Lussac’s Law into one: • P1V1/T1 = P2V2/T2 • Let’s you work out problems involving more variables • Use known variables under one set of conditions to find a value of a missing variable.

17. Avogadro’s Principle • 1811, Avogadro suggested that equal volumes of gasses at the same temperature and pressure contain the same number of particles: • K-M theory says the same thing • Particles are so far apart that size doesn’t matter • 1 mole of particles is 6.02 x 1023 particles • 1 mole of gas particles is 22.4 L at 0.00° C and 1.00 atm pressure • Standard Temperature and Pressure (STP)

18. Example Problems • Determine the volume of a container that holds 2.4 mole of gas at STP: • 2.4 mol x 22.4L/1mol = 53.76 L = 54L • What volume will 1.02 mol of carbon monoxide gas occupy at STP? • 1.02 mol x 22.4L/1 mol = 22.8 L • A balloon will rise off the ground when it contains 0.0226 mol of He in a volume of 0.460L. How many moles of He are needed to make the balloon rise when volume is 0.885L, assuming temperature and pressure are constant? • 0.0226 mol x 0.865L/0.460L = 0.0425 mol

19. Ideal Gas Law Chapter 14, Section 3

20. Ideal Gas Law • P1V1/T1 = P2V2/T2 can be interpreted as saying that PV/T is a constant, k • k is based on the amount of gas present, n • Experiments have shown that k = nR, where n is number of moles and R is a constant. • Ideal Gas Law is therefore: • PV = nRT • Ideal Gas Constant (R), is: • 0.0821 L*atm/mol-K (most often used) • 8.314 L*kPa/mol K • 62.4 L*mm Hg/mol K

21. Real vs. Ideal • Ideal Gas Assumptions • Particles take up no space • No intermolecular forces • Follows ideal gas law under all temperatures and pressures • Real Gas • All gas particles take up some space • All gas particles are subject to intermolecular forces • Most gasses act like ideal gasses at many/most temperatures and pressures • Deviations occur at extremely high pressures and extremely low temperatures because intermolecular forces start to show effects.

22. Applying Ideal Gas Law • If you know any three variables, you can solve for the fourth: • You can solve for n (number of moles) • Combined gas law, you cannot. • You can use ideal gas law allows you to solve for molar mass and density, if mass is known • n (number of moles) = m (mass) / M (molar mass) • Sustitute m/M for n • PV = mRT/M

23. Solving for Density • Density is m (mass) / V (volume) • Substitute D (Density) for m/V • M = mRT/PV = DRT/P • Or D = MP/RT

24. Example Problem • Calculate the number of moles of gas contained in a 3.0L vessel at 3.00x102K with a pressure of 1.50 atm • Find: moles • Known: V = 3.0L • Known: T = 3.00x102 • Known: P = 1.50 atm • Known: R = 0.0821 L*Atm/mol*K • PV=nRT • Solve for n • N = PV/RT

25. Example Problem • n = PV/RT – substitute in known information: • n = (1.50atm)(3.0L) • (0.0821 L*atm/mol*K)(3.00 x 102 K) n = 0.18 mole Evaluation: 1 mole of gas occupies 22.4L at STP. The volume is much less than 1 mole. Temperature and pressures are not dramatically different than STP. Slightly higher pressure means more gas. Slightly higher temperature means a little less gas. Units of the answer is moles – every unit cancels except 1/ 1/mol.

26. Example Problem – Using Molar Mass • What is the molar mass of a pure gas that has a density of 1.40g/L at STP? • Find: molar mass (g/mole) • Known: D = 1.40g/L • Known: T=0.00°C • Known: P = 1.00atm • Known: R = 0.0821 L*atm/mol*K • Step1: Convert T to Kelvin • Tk = Tc+273 • Tk = 273

27. Molar Mass Problem • Use density form of ideal gas law • M = DRT/P • Substitute known values: M = (1.40 g/L)(0.0821 L*atm/mol*K)(273K) 1 atm M = 31.4 g/mol

28. Gas Stoichiometry Chapter 14, Section 4

29. Gas Stoichiometry – Volume only • Example: • 2C4H10(g) + 13 O2(g)→ 8 CO2(g) + 10 H2O(g) • Remember: Avogadro’s principle states that 1 mole of gas is 22.4 L • For volume to volume calculations, you only need to know mole ratios • 2 L of butane reacts, it involves • 13 L of O2 (2L x 13 O2/2 C4H10) • 8 L of CO2 (2L x 8CO2/2C4H10) • 10 L of H2O

30. Gas Stoichiometry – Volume & Mass • Ammonia is synthesized fro hydrogen and nitrogen • N2(g) + 3 H2(g)→ 2 NH3(g) • If 5.00 L of nitrogen reacts completely by this reaction at 3.0 atm and 298K, how many grams of ammonia are produced? • Analyze: you are given L, pressure, temperature. You are asked to find grams of NH3. • Known: • VN = 5.00L, P = 3.00 atm, T = 298K

31. Gas Stoichiometry (Cont) • Determine the mole ratio needed • 1 volume N2 / 2 volumes NH3 • Determine volume of ammonia produced: 1 volume N2 2 volumes NH3 = 10L NH3 5.00L N2 x 3. Rearrange Gas Law to solve for n (moles): PV RT PV = nRT n = 4. Substitute values and solve for n

32. Gas Stoichiometry (Cont). 4. Substitute values and solve for n (3.00 atm)(10.00L) (0.0821 L*atm/mol*k)(298K) = 1.23 mol NH3 n = 5. Find the mass (M) of NH3 by finding the molecular mass and converting the moles in grams Molecular mass = 1N(14.01amu) + 3 H(1.01amu) = 17.04amu 1.23 mol NH3 x 17.04 g/mol = 21.0 g NH3 Done