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Gases

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Gases

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  1. Gases The Gas Laws Labs #18 Molar Mass of a Volatile Liquid #19 Calcium Carbonate Analysis: Molar Volume of Carbon Dioxide

  2. Gases have general characteristics • Expansion: Expand indefinitely to fill the space available to them • Indefinite shape: Fill all parts of container evenly, so have no definite shape of their own • Compressibility: Most compressible of states of matter • Mixing: Two or more gases will mix evenly and completely when confined to same container • Low density: Have much lower densities than liquids and solids (typically about 1/1000 those of liquids/solids) • Pressure: Exert pressure on their surroundings

  3. Vapors • Gas phase at temperature where same substance can also exist in liquid or solid state • Below critical temperature of substance (vapor can be condensed by increasing pressure without reducing temperature)

  4. Pressure: force per unit area (P= f/a) • Force generated by collisions of gas particle w/container walls • Related to velocity of gas particles • Total force = sum of forces of all collisions each second per unit area • Force = m x acceleration • Pressure dependent on • Gas particle velocity • Collision frequency • Collision frequency dependent on • Gas particle velocity • Distance to container walls. • Changing temperature changes collision force, as well as collision frequency • Collision frequency changed by altering size of container • Force of collisions is not affected

  5. (force of 1 newton exerted on one square meter of area) (29.92” Hg) (psi)

  6. Manometers • Used to measure pressure of enclosed gas • U-tube partially filled with liquid, typically Hg • One end connected to container of gas being measured • Other end sealed with vacuum existing above liquid, or open to atmosphere

  7. Closed-end manometer • Pressure is just difference between two levels (in mm of Hg)-indicates pressure of system attached to apparatus • Gas connected to one arm • Space above Hg in other arm is vacuum • Liquid in tube falls to height (directly proportional to pressure exerted by gas in the tube) • Since pressure of gas causes liquid levels to be different in height, it is this difference (h) that is measure of gas pressure in container • Pgas = Ph • If left to atmosphere, it measures atmospheric pressure-barometer

  8. Open-end manometer • Used to measure pressure of gas in container • Difference in Hg levels indicates pressure difference in gas pressure and atmospheric pressure • Atmospheric pressure pushes mercury in one direction • Gas in container pushes it in the other direction • Two ends connected to gases at different pressures • Closed end to gas in bulb (gas filled) • Open end to atmosphere • If pressure of gas higher than atmospheric, Hg level lower in arm connected to gas (Pgas = Pbarometric + h) • If pressure of gas lower than atmospheric, Hg level higher in arm connected to gas (Pgas = Pbarometric - h) • If levels are equal, gas at atmospheric pressure

  9. If fluid other than mercury is used: Difference in heights of liquid levels inversely proportional to density of liquid and represents the pressure Greater density of liquid, smaller difference in height High density of mercury (13.6 g/mL) allows relatively small monometers to be built Readings must be corrected for relative densities of fluid used and of mercury (mm Hg = mm fluid) density fluid density Hg

  10. A sample of CH4 is confined in a water manometer. The temperature of the system is 30.0 °C and the atmospheric pressure is 98.70 kPa. What is the pressure of the methane gas, if the height of the water in the manometer is 30.0 mm higher on the confined gas side of the manometer than on the open to the atmosphere side. (Density of Hg is 13.534 g/mL).

  11. 1) Convert 30.0 mm of H2O to equivalent mm of mercury: • (30.0 mm) (1.00 g/mL) = (x) (13.534 g/mL) x = 2.21664 mm (I will carry some guard digits.) • 2) Convert mmHg to kPa: • 2.21664 mmHg x (101.325 kPa/760.0 mmHg) = 0.29553 kPa • 3) Determine pressure of enclosed wet CH4: • At point A in the above graphic, we know this: Patmo. press. = Pwet CH4 + Pthe 30.0 mm water column • 98.70 kPa = x + 0.29553 kPa • x = 98.4045 kPa • 4) Determine pressure of dry CH4: • From Dalton's Law, we know this: Pwet CH4 = Pdry CH4 + Pwater vapor • (Water's vapor pressure at 30.0 °C is 31.8 mmHg. Convert it to kPa.) • 98.4045 kPa = x + 4.23965 kPa • x = 94.1648 kPa • Based on provided data, use three significant figures; so 94.2 kPa.

  12. Boyle’s Law X axis independent variable Y axis dependent variable Pressure-Volume Relationship Temperature/# molecules (n) constant

  13. Inversely proportional (one goes up, other goes down) • V 1/P or • P = k/T where K is constant • PV = constant • P1V1 = P2V2 • Gas that strictly obeysBoyle’s Law is ideal gas (holds precisely for gases at very low temperatures)

  14. Pressure applied vs. volume measured • Shows inverse proportion • If pressure doubled, volume decreased by ½

  15. Pressure vs. inverse of volume • Plot of V (P) against 1/P (1/V) gives straight line • Graph of equation P = k1 x 1/V • Same relationship holds whether gas is being expanded or contracted

  16. A gas which has a pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant pressure? • P1 = 1.3 atm • V1 = 27 L • P2 = 3.9 atm • V2 = ? • P1V1 = P2V2 • 1.3 atm (27 L) = 3.9 atm (X) = 9.0 L

  17. Charles’ Law Temperature-Volume Relationship Pressure/n constant

  18. Volume of gas directly proportional to temperature (T  V), and extrapolates to zero at zero Kelvin (convert Celsius to Kelvin)

  19. Used to determine absolute zero • From extrapolated line, determine T at which ideal gas would have zero volume • Since ideal gases have infinitely small atoms, only contribution to volume of gas is pressure exerted by moving atoms bumping against walls of container • If no volume, then no kinetic energy left • Absolute zero is T at which all KE has been removed • Does not mean all energy has been removed, merely all KE http://www.absorblearning.com/media/attachment.action?quick=10o&att=2629

  20. If T↓, V↓ and vs. • V = kT • P is constant • k = proportionality constant • V/T = constant

  21. A gas at 30oC and 1.00 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60.0oC and 1.00 atm? • V1 = 0.842 atm • T1 = 30oC = 303 K • V2 = ? • T2 = 60.0oC = 333 K • V1/T1 = V2/T2 • 0.842/303 K = X/333 K = 0.925 L

  22. Gay Lussac’s Law Pressure-Temperature Relationship Volume/n constant

  23. When two gases react, do so in volume ratios always expressed as small whole numbers When H burns in O, volume of H consumed is always exactly 2x volume of O Direct relationship between pressure and temperature (P T) P/T = constant Pi/Ti = Pf/Tf

  24. Avogadro’s Law(at low pressures) Volume-Amount Relationship Temperature/Pressure constant

  25. Equal volumes of gases, measured at same temperature and pressure, contain equal numbers of molecules Avogadro's law predicts directly proportional relation between # moles of gas and its volume Helped establish formulas of simple molecules when distinction between atoms and molecules was not clearly understood, particularly existence of diatomic molecules Once shown that equal volumes of hydrogen and oxygen do not combine in manner depicted in (1), became clear that these elements exist as diatomic molecules and that formula of water must be H2O rather than HO as previously thought

  26. V of gas proportional to # of moles present (V n) • At constant T/P, V of container must increase as moles of gas increase • V/n = constant • Vi/ni = Vf/nf

  27. A 5.20 L sample at 18.0oC and 2.00 atm pressure contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? • V1 = 5.20 L • n1 = 0.436 mol • V2 = X • n2 = 0.436 + 1.27 = 1.70 mol • V1/n1 = V2/n2 • 5.20 L/0.436 mol = X/1.70 mol = 20.3 L

  28. Generalization applicable to most gases, at pressures up to about 10 atm, and at temperatures above 0°C. Ideal gas’s behavior agrees with that predicted by ideal gas law. Ideal Gas Law(Holds closely at P < 1 atm)

  29. Formulated from combination of Boyle’s law, Charles’s law, Gay-Lussac’s law, and Avogadro’s principle • Combined • If proportionality constant called R • Rearrange to form ideal gas equation

  30. Experimentally observed relationship between these properties is called the ideal gas law: PV = nRT Pressure (P) in atm Volume (V) in L Absolute temperature (T) in K (Charles’ Law) Amount (number of moles, n) R (universal ideal gas constant) 0.08206 liter ∙ atm/mole ∙ K or 0.08206 L·atm·K–1·mol–1 8.31 liter ∙ kPa/mole ∙ K 8.31 J/mole ∙ K 8.31 V ∙ C/mole ∙ K 8.31 x 10-7 g ∙ cm2/sec2 ∙ mole ∙ K 6.24 x 104 L ∙ mm Hg/mol ∙ K 1.99 cal/mol ∙ K

  31. You can use ideal gas law equation for all problems Given 3 of 4 variables and calculate 4th PiVi= niRTi PfVf nfRTf Cancel all constants and R, make appropriate substitutions from given data to perform calculation

  32. A sample containing 0.614 moles of a gas at 12.0oC occupies a volume of 12.9 L. What pressure does the gas exert? • PV = nRT • P (12.9 L) = (0.614 mol)(0.08206 L atm/K mol)(285 K) = 1.11 atm

  33. Homework: Read 5.1-5.3, pp. 189-202 Q pp. 232-234, #29, 31-34, 44

  34. Standard Temperature and Pressure (STP) • STP = 0°C (273K) and 1.00 atm pressure (760 mm Hg) • One mole of gas at STP will occupy 22.42 L • Allows you to compare gases at STP to each other

  35. What volume will 1.18 mole of O2 occupy at STP? • PV = nRT • (1atm)(X) = (1.18 mol)(0.08206 L atm/K mol)(273 K) = 26.4 L • Alternate way: • At STP, 1 mole occupies 22.4 L • Vi/ni = Vf/nf • 22.4 L/1 mol = X/1.18 mol = 26.4 L

  36. A sample containing 15.0 g of dry ice, CO2(s), is put into a balloon and allowed to sublime according to the following equation: CO2(s)  CO2(g) How big will the balloon be (what is the volume of the balloon) at 22oC and 1.04 atm after all of the dry ice has sublimed? • 15.0 g CO2 1 mol CO2 = 0.341 mol CO2 44.0 g CO2 • PV = nRT • (1.04 atm)(X) = (0.341 mol)(0.08206 L atm/K mol)(295K) = 7.94 L

  37. 0.500 L of H2(g) are reacted with 0.600 L of O2(g) according to the equation 2H2(g) + O2(g)  2H2O(g). What volume will the H2O occupy at 1.00 atm and 350oC? • Find limiting reactant: • 0.500 L H2 1 mol H2 = 0.0223 mol H2/2 22.4 L • 0.600 L O2 1 mol O2 = 0.0268 mol O2/1 22.4 L • PV = nRT • (1atm)(X) = (0.0223 mol)(0.08206 L atm/K mol)(623 K) = 1.14 L

  38. Molar Mass • Gives density which can be determined if P, T and molar mass are known • Density directly proportional to molar mass • Density increases as gas pressure increases • Density decreases as temperature increases

  39. A gas at 34.0oC and 1.75 atm has a density of 3.40 g/L. Calculate the molar mass (M.M.) of the gas. • M.M. = dRT P • M.M. = (3.40 g/L)(0.08206 L atm/K mol)(307K) (1.75 atm) • M.M. = 48.9 g/mol

  40. Example • What are the expected densities of argon, neon and air at STP? • PV = nRT • PV = g (RT) molar mass • Density = g/V = (Molar mass)P/RT • density Ar = (39.95 g/mol)(1.00 atm)/(0.0821 L-atm/mol-K)(273 K) = 1.78 g/L • density Ne = 0.900 g/L • density air = 1.28 g/L • molar mass = (0.80)(28 gN2/mol) + (0/20)(32 g O2/mol) = 28.8 g/mol

  41. Molar Volume (V/n) • V = RT n P • (0.0821 L-atm/mol-K)(273K) = 22.4 L/mol 1.00 atm

  42. Homework: Read 5.4, pp. 203-206 Q pp. 234-235, #52, 56, 58, 60

  43. Dalton’s Partial Pressure

  44. When two gases are mixed together, gas particles tend to act independently of each other • Each component gas of mixture of gases uniformly fills containing vessel • Each component exerts same pressure as it would if it occupied that volume alone • Total pressure of mixture is sum of individual pressures, called partial pressures, of each component • Pt = PA + PB + PC + …

  45. Since each component obeys ideal gas law (PV=nRT) and has same T and V, it follows that partial pressure of each gas in container is directly proportional to # moles of gas present Pi/Pt = ni/nt Pi = ni/nt x Pt Pi = XiPt Xi = mole fraction of gas component i

  46. A volume of 2.0 L of He at 46oC and 1.2 atm pressure was added to a vessel that contained 4.5 L of N2 at STP. What is the total pressure and partial pressure of each gas at STP after the He is added? • Find # moles of He at original conditions-will lead us to finding partial pressure of He at STP. • PV = nRT • (1.2 atm)(2.0 L) = n(0.08206 L atm/k mol)(319 K) • n = 0.0917 mol He • When gases are combined under STP, partial pressure of He will change while N2 will remain the same since it is already at STP. • PHeV = nRT • (X)(4.5 L) = (0.091 mol)(0.08206 L atm/K mol)(273 K) • P = 0.457 atm • Total pressure = 1.00 + 0.46 = 1.46 atm