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GASES

GASES. DEFINITIONS. Gas: a compound or element whose molecules or atoms are far apart and have much freedom of movement; can undergo diffusion Diffusion: the spontaneous, even mixing of a gas with other gases as gases move from areas of high concentration to low concentration

brynne-kaufman
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GASES

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  1. GASES

  2. DEFINITIONS • Gas: a compound or element whose molecules or atoms are far apart and have much freedom of movement; can undergo diffusion • Diffusion: the spontaneous, even mixing of a gas with other gases as gases move from areas of high concentration to low concentration • Miscibility:gases are infinitely miscible (“mix-able”) and form soln's w/other gases quickly and easily.

  3. PROPERTIES OF GASES • highly compressible • can expand and condense • have indefinite shape (fill their containers) • low density (about 1/1000th the density of its liquid or solid) • quick diffusion

  4. THE KINETIC MOLECULAR THEORY OF GASES • … states that gas molec's are very far apart and are in constant, random motion • kinetic energy = "energy of motion“ = KE = ½ mv2 • Note: m = mass and v = velocity

  5. THE KINETIC MOLECULAR THEORY OF GASES • gas molec's have very low intermolecular F's of attraction & high K.E. • the moving gas molec's undergo frequent collisions w/each other and w/the walls of their container; this exerts pressure or force of gaseous molec's that increases w/increasing T • at any given T, the ave K.E. of all gas molec's is the same. (so a gas with a smaller mass must be moving faster than a particle with a larger mass)

  6. LARGE MOLECULES vs. SMALL MOLECULES • at any given T, the ave K.E. of all gas molec's is the same (per Avogadro) • Therefore, KEa = KEb • If gas a = H2 & gas b = O2, KEH2 = KEO2 • or (mv2/2)H2 = (mv2/2) O2 • But H2 weighs 2.02 g/mol and O2 weighs 32 g/mol, so that could only be true if what was true about gas v?

  7. GAS PRESSURE • Pressure is force (F) per unit of area; any exertion of F applied to a definite area can be expressed as: P = F/A • In a closed container, gas exerts a measurable pressure on the walls of the container • P increases as T increases

  8. Pressure is due to all of the molecules in the column of air above your head all the way up to the edge of the atmosphere. WHY DOES AIR PRESSURE EXIST?

  9. BOYLE'S LAW • Robert Boyle, 300 years ago, noted: "if the T remains constant, the P exerted by a gas varies inversely with its V” • ie: A gas is collected in a 242 cm3 container. The P of the gas in the container is measured at 87.6 kPa. What is the V of this gas at standard P (ASSUME T REMAINS CONSTANT!)

  10. EFFECTS OF AIR PRESSURE What you’re looking at is a tank car that has been crushed by air pressure. How might that have happened? www.nsdl.arm.gov/ images/tanker.jpg

  11. UNITS FOR PRESSURE • In the metric system, we measure forces in Newtons (N): equal to 1 kg m/sec2 ; but that’s a unit for force, not pressure since pressure is dependent on area. • a)pascal (Pa): a pressure of 1 N/m2 • (kilopascals, (kPa) will be used • b)mm Hg (torr) = Torricelli's column of mercury rose 760mm at sea level • c)atmospheres (atm) = average air pressure at sea level at 0oC • d)pounds per square inch (psi) = (or inches, as in weather) Non-metric

  12. STANDARD AIR PRESSURE • Defined = average air pressure at sea level • at 0oC • 1 atm = 101.3 kPa = 760 mm Hg = 760 torr(memorize these) • Concept: Hg is poured into a tube. Outside air pressure will make the Hg in the tube rise & fall. (This would happen with any liquid but Hg doesn’t evaporate as would water or alcohol & it’s dense)

  13. CHARLES’ LAW • "at constant P, as T increases, V increases" (or: T and V are directly proportional) • ie: a 225 cm3 V of gas is collected at 58.0oC. What V would this sample occupy at standard T (assume P = constant) *T must be in K!

  14. A WORD ABOUT TEMP • Most of the substances we work with in this unit are gases, even below 0oC • Our main gas formula is • V1P1 / T1 = V2P2 / T2 • If we put in a negative value (below 0oC) for T in our formula, a calculation might give us the impression we have negative volume and there’s no such thing

  15. CONVERSIONS • C = 5/9 (F-32) • F = (9/5 C) + 32 • K = C + 273 • C = K - 273

  16. COMBINED GAS LAW • this is the one to memorize! • ie: 10.0 cm3 of a gas measured at 75.6 kPa and 60.0oC is to be corrected to correspond to the V it would occupy at STP.

  17. EXAMPLE: • ie: the V of a gas measured at 75.6 kPa and 60.0oC is to be corrected to correspond to the V it would occupy at STP. The measured V is 10.0 cm3 10.0 cm3 75.6 kPa 101.3 kPa 60.0 C +273K = 333K 273 K Now, in the “2” column, (the “after”), write the standard conditions.

  18. VACUUM PUMP DEMOS

  19. MISCELLANEOUS DEMOS

  20. MISCELLANEOUS DEMOS

  21. MANOMETERS • Manometers – devices which measure gas pressure • Two types of manometers: • Open and Closed (see handout)

  22. TEMPERATURE AND GASES • temperature is a measure of K.E. (not a measure of heat!) • Kelvin scale - based on "absolute zero" = which is -273.15 oC and is the T at which all molecular motion stops • absolute zero is a THEORETICAL POINT ONLY! • K = oC + 273 (K= "kelvins", not degrees kelvin) • the K scale is used since it will never have negative values

  23. FORMULAS • C = 5/9 (F - 32) • F = (9/5 C) + 32 • K = C + 273 • C = K - 273 28 C = _____K 520 K = _____C 10 C = _____ K

  24. ENERGY FLOW • HEAT is a form of energy • Energy always flows from a hotter object to a colder one • Heat measured in joules

  25. Avogadro’s Principle: • “At equal T's and P's, equal V's of gases contain the same # of molecules" • "n" relates to the number of moles of gas at constant T and P, if V1=V2, then n1 = n2

  26. IDEAL GAS • an imaginary gas that serves as a model only and does not factor in real-life conditions, such as V and attractive F's • STP = "standard temp and pressure" • T = 273 K (which is 0oC) • P = 1 atm = 101.3 kPa = 760 mm Hg = 760 torr

  27. Ideal Gas Equation • P V = n R T • comes from Boyle's Law, Charles's Law, & molar volume • if we substitute in all standard conditions, we can solve for "R"

  28. Solving for R • P V = n R T • Rearrange to solve for R • R = PV / nT Memorize These R = (1 atm)(22.4 L) / (1 mol)(273 K) = 0.0821 ____ L. atm/mol K R = (101.3 kPa)(22.4 L) / (1 mol)(273 K) = ___ L. kPa/mol K 8.31 R = (760 torr)(22.4 L) / (1 mol)(273 K) = 62.4 ___ L. torr/mol K

  29. Repeat: Memorize all of those values for R! • this crazy unit doesn't really exist as a measurable entity but is used for cancellation purposes

  30. Molecular Mass • what are the units for MM? How can you use PV = nRT? • can be found from lab measurements in this way: • MM = m R T • P V • ie: What is the molecular mass of a gas if 150.0 cm3 of it has a mass of 0.922 g at 99.0oC and 107.0 kPa?

  31. Molecular Mass Problem • ie: What is the molecular mass of a gas if 150.0 cm3 of it has a mass of 0.922 g at 99.0oC and 107.0 kPa? • MM = m R T • P V

  32. Molar Volume @STP • 1 mol of any gas occupies the same V as 1 mol of any other gas • molar volume = the V of 1 mole of a gas at STP = 22.4 L

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