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11. Gases

11. Gases. 11.1 Properties of Gases 11.2 The Kinetic Molecular Theory of Gases Molecular Speed Diffusion and Effusion 11.3 Gas Pressure Definition and Units of Pressure Calculation of Pressure Measurement of Pressure 11.4 The Gas Laws

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11. Gases

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  1. 11. Gases 11.1 Properties of Gases 11.2 The Kinetic Molecular Theory of Gases Molecular Speed Diffusion and Effusion 11.3 Gas Pressure Definition and Units of Pressure Calculation of Pressure Measurement of Pressure 11.4 The Gas Laws Boyle’s Law: The Pressure-Volume Relationship Charles’s and Gay-Lussac’s Law: The Temperature-Volume Relationship Avogadro’s Law: The Amount-Volume Relationship The Gas Laws and Kinetic Molecular Theory The Combined Gas Law: The Pressure-Temperature-Amount-Volume Relationship 11.5 The Ideal Gas Equation Applications of the Ideal Gas Equation

  2. 11. Gases 11.6 Real Gases Factors That Cause Deviation from Ideal Behavior The van der Waals Equation van der Waals Constants 11.7 Gas Mixtures Dalton’s Law of Partial Pressures Mole Fractions 11.6 Reactions with Gaseous Reactants and Products Calculating the Required Volume of a Gaseous Reactant Determining the Amount of Reactant Consumed Using Change in Pressure Using Partial Pressures to Solve Problems

  3. A theory is a model explaining chemical behavior. Examples – Atomic Theory (Classical Mechanics & Quantum Mechanics) Standard model (QM applied to fundamental particles) Relativity (gravity) The Big Bang Theory (Inflation) String Theory The Theory of Evolution It must account for the facts (laws = statement of reproducible observation). Dalton’s atomic theory explained Law of conservation of mass … etc. If observation goes against theory, either the theory or the observation must be incorrect. Theories are fluid (not dogmatic) – they can change with new data or be discarded A theory with a visual model is easier to understand A theory has to begin with assumptions (premise)

  4. Kinetic Molecular Theory of Gases The following assumptions hold for any ‘ideal gas’. 1. Gas particles are hard spheres that are so far apart that the volume of the molecule is negligible. Therefore, V = volume of container. 2. Gas particles are in constant random motion. All collisions are ‘elastic’ meaning that no energy is lost as the molecules collide with the container walls or each other. 3. Gas particles do not attract or repel other gas particles. 4. Temperature (T) is a measure of the average kinetic energy (motion) of gas particles. <KE> = 3/2kt = ½m<v2>. (k = Boltmann’s constant) At absolute Zero (0K) all motion ceases.

  5. Gas Model

  6. Gas Variables P = Pressure (atm) = F/A, due to collisions with walls V = Volume (Liters) = Volume of container n = # moles of gas ‘particles’ T = Temperature (Kelvin) – causes motion of particles Water freezes at: 273K, 0°C, 32°F Water boils at: 373K, 100°C, 212°F Body T: 98.6˚F = 37˚C = 310K Room T: 68˚F = 20˚C = 293K C = (F – 32)  5/9 & K = C + 273.15 ˚F = 9/5 ˚C + 32(DT: 1.0K = 1.0˚C ~ 1.8˚F)

  7. Kinetic Molecular Theory (KMT) Which two of the variables changed when I blew up the balloon? a. P & V b. P & n c. P & T d. V & n Balloon – deflated V ~ 0 n ~ 0 Balloon – inflating V↑ = volume inside of balloon n↑ = # moles of gas molecules inside balloon P – Pressure pushing in = pressure pushing out (at equilibrium) T – Temperature of room (not exact as air leaving lungs is a bit warmer) Balloon – inflated V = cst = volume inside of balloon n = cst = # moles of gas molecules inside balloon Pin = Patm + Pelastic T = Temperature of room

  8. Units of Pressure SI: Pa = N m-2 = kg m-1 s-2 Pa = Paschal; N = Newton (force); Area (A) = m2 (101,325 Pa = 1 atm) 1.00 atm P = …… 101,325 Pa 760 torr (mm of Hg) 29.9 in (of Hg) 14.7 psi (lbs/in2)

  9. Hg Pressure = F/A = mg (V = areah) ….. P = mgh = dgh area V 760 mm Air Pressure Air Pressure P = 1 atm when h = 760 mm 850 mm What will happen to Hg in sealed tube? a. Stays at top of tube b. Drops part-way down tube c. Completely drains into Hg bath Hg = 13.6 g ml-1 H2O = 1.0 g ml-1 Hg(l)

  10. P = (density) (g) (h) For mercury (Hg) at 1 atm Pressure 101,325 Pa (kg m-1s-2) = 1.36 x 104 (kg m-3) 9.8 (m s-2) h (m) h = 0.760 m or 760 mm For water at 1 atm Pressure 101,325 Pa (kg m-1s-2) = 1.0 x 103 (kg m-3) 9.8 (m s-2) h (m) h = 10.34 m or 33.9 ft (a water barometer would occupy ~ 3 stories) Volume Units 1 L = 1000 ml = 1 dm3 1 ml = 1 cm3 1 m3 = 1000 L = 1000 dm3 = 1 x 106 ml

  11. P = (density) (g) (h) For mercury (Hg) at 1 atm Pressure 101,325 Pa (kg m s-2) = 1.36 x 104 (kg m-3) 9.8 (m s-2) h (m) h = 0.760 m or 760 mm For water at 1 atm Pressure 101,325 Pa (kg m s-2) = 1.0 x 103 (kg m-3) 9.8 (m s-2) h (m) h = 10.34 m or 33.9 ft (a water barometer would occupy ~ 3 stories) Volume Units 1 L = 1000 ml = 1 dm3 1 ml = 1 cm3 1 m3 = 1000 L = 1000 dm3 = 1 x 106 ml

  12. Kinetic Molecular Theory of Gases The following assumptions hold for any ‘ideal gas’. P V T n ? ↓ same same P will a) ↑ b) ↓ c) same

  13. Boyle’s Law P & V P (atm) As V↑ …. P↓ V (L) P  1/V P = cst  1/V PV = cst P1V1= P2V2 P (atm) If you decrease the volume then gas particles will collide with the walls more often and P will increase. 1/V (L-1)

  14. Kinetic Molecular Theory of Gases The following assumptions hold for any ‘ideal gas’. P V T n same ? ↑ same V will a) ↑ b) ↓ c) same

  15. V T (°C) Charles’ Law … V & T As T↑ …. V ↑ V  T V = cst  T V/T = cst V1/T1= V2/T2 Xint = Absolute 0 = -273.15 °C If you increase T then the particles will collide more frequently with the walls and push them out (↑) until the internal and external pressures are the same.

  16. Kinetic Molecular Theory of Gases The following assumptions hold for any ‘ideal gas’. P V T n same ? Same ↑ V will a) ↑ b) ↓ c) same Balloon – inflating V↑ = volume inside of balloon n↑ = # moles of gas molecules inside balloon P – Pressure pushing in = pressure pushing out (at equilibrium = same) T = Temperature of room (same)

  17. P = Pressure (atm) = F/A, due to collisions with walls V = Volume (Liters) = Volume of container n = # moles of gas ‘particles’ T = Temperature (Kelvin) – causes motion of particles Which two of the variables above changed when I blew up the beach ball? a. P & V b. P & n c. P & T d. V & n • Which of the following would cause the pressure of the beach ball to increase? • a. Decrease V at constant n & T • Increase V at constant n & T. • Decrease T at constant V & n.

  18. Boyle’s Law: V  1/P V  nT/P or PV = cst  nT Charles’ Law: V  T Avogadro’s Law: V  n cst = R = ideal gas constant = NAk At a constant T and P equal amounts of any gas will occupy the same volume. The Ideal Gas Equation PV = nRT R = 0.08206 L atm mol-1 K-1 or …. 8.314 J mol-1 K-1 82.06 cm3atm mol-1 K-1 STP (Standard Temperature & Pressure) T = 0ºC (273K) and P = 1.00 atm(1 bar or 1.00 x 105 Pa = 0.987 atm). At 1.00 atm P and 273 K, 1.00 mol of any ideal gas occupies 22.4L V = nRT/P = 1.0 • 0.08206 • 273 = 22.4 = mols •L atm mol-1 K-1 •K 1.00atm

  19. R = 0.08206 L atm mol-1 K-1 or 8.314 J mol-1K-1 or 82.06 cm3atm mol-1 K-1 Two ways to use IG Equation: PV = nRT • Given any 3 of 4 variables, find the 4th What is the pressure of 3.5 mols of Cl2 gas at 34°C and 22.0 L? • 2.0 atm • 3.0 atm • 4.0 atm • 8.0 atm P = nRT/V = 3.5 • 0.08206 • 307 ÷ 22.0 = 4.0 atm. Would the answer change if the gas was N2 instead of Cl2? a) yes b) no Although Cl2 is heavier than N2, and thus strikes the wall with more force it is also slower and thus strikes the wall less often. These two variables cancel to make P the same regardless of the FW of the gas.

  20. Show Charles’, Boyle’s, & Avogadro’s Laws from here P1V1= P2V2 n1T1 n2T2 Boyle’s Law (PV): P1V1= P2V2 Boyle’s Law (PV): Two ways to use IG Equation: PV = nRT Charles’ Law (VT): V1/T1= V2/T2 Charles’ Law (VT): 2. changes of multiple variables from one set of conditions to another. PV = nRT or …. R = PV/nT or … What if T also dropped to 13°C? Note that only n is now constant. a. < 3.32 L b. > 3.32L c. = 3.32L A balloon is filled with He gas has a volume of 2.68L at 23°C and 1.04 atm. It is released up into the atmosphere where P drops to 0.839 atm. How will the volume change if T and n remain constant? a. decrease b. increase c. Stay the same Avogadro’s Law (Vn): V1/n1= V2/n2 Avogadro’s Law (Vn): V2= V1 • _P1_ P2 V2= V1 • _P1_ • _T2_ P2 T1

  21. PV = nRT One can determine the molecular weight (M)of an unknown gaseous material if you measure the volume that a given mass of the gas occupies at a known T and P M = m/n m = mass (g) 0.620 grams of an unknown gas occupies 0.35 L at 298K and 1.0 atm. What is the MW (or simply M) of the gas? Which of the following is useful to solve this? a. V1/n1 = V2/n2 b. n = PV/RT c. P1V1 = P2V2d. all of these n = PV/RT = 1.0 • 0.35 ÷ (0.08206 • 298) = 0.014 mols M = 0.620 g ÷ 0.014 mols = 44.0 g mol-1 (possibly CO2)

  22. Problem – When coal is burned the sulfur in the coal reacts as follows…. S(s) + O2(g) → SO2(g) If 2.54 kg of S react with excess oxygen what volume of SO2(g) is Produced at 30.5°C and 1.12 atm? • Find moles of known (S) = 2540 g • 1 mole S = 79.1 moles S • 32.1 g S 2. Find moles of unknown (SO2) = 79.1 moles 3. Find V of SO2 using PV = nRT or V = nRT/P = 79.1 • 0.08206 • 303.5 1.12 V of SO2= 1760 L

  23. Dalton’s Law of Partial Pressures Applies to gas mixtures – e.g O2 + N2 Since the KMT proposes that all collisions are elastic and there are no interactions between different gases, each gas behaves as if it were the only gas in the container. Partial Pressure of N2 = PN2 = nN2RT/V Partial Pressure of O2= PO2 = nO2RT/V P = PO2 + PN2(or P = Si Pi) The total Pressure of a gas mixture is equal to The sum of the partial pressures of each gas in the mixture. Total Pressure P = nRT/V = (nO2 + nN2)RT/V

  24. Diffusion The mixing of gases as the result of random motion and frequent collisions.

  25. What is final P, PCO, and PO2? 300 K 2.00 L CO at 0.500 atm 1.00 L O2 at 0.500 atm n = PV/RT PCO = ? a. 0.5 atm b. 1.0 atm c. 0.25 atm d. 0.33 atm V = 3L T = 300K nCO = nO2 = n = nCO + nO2 = 0.041 0.020 0.061 V = 3L T = 300K PCO = 0.333 atm PO2 = 0.167 atm P = 0.5 atm = PCO + PO2 P2= P1 • _V1_ V2

  26. Effusion The escape of gas molecules from a container to a region of vacuum.

  27. Gas facts: Equal volumes contain equal numbers of any gas. Equal numbers exert the same pressure at constant V and T. Pressure is force exerted per unit area. Smaller molecules exert less pressure per collision but move faster and collide with the wall much more frequently. Effusion of gases If a container has a hole in it – how fast will the gas escape? Which type of balloon loses its gas faster? a. Helium balloon b. air balloon c. they both lose gas at same rate Speed (root mean square) of gas: urms = (3RT/M)½ where M – molecular weight of gas The rate of gas effusion is inversely proportional to the square root of M. rHe/rN2 = (MN2/MHe)½ r1/r2 = (M2/M1)½ How much faster will the He balloon lose it gas? (assume M = 29 g/mol for air) a. 7.3 times b. 3.8 times c. 1.3 times d. 2.7 times

  28. Dalton’s Law of Partial Pressure: P = Si Pi V = 3L T = 300K PCO = 0.333 atm PO2 = 0.167 atm P = 0.5 atm = PCO + PO2 V = 3L T = 300K nCO = nO2 = n = nCO + nO2 = 0.041 0.020 0.061 Mole fraction: like % only 0 → 1 rather than 0 → 100 cO2= nO2/n = nO2/(nO2 + nCO) = 0.33 (or ci = ni/ntot) PO2/Ptotal = nO2RT• __V__ = nO2/ntotal = cO2 V nCORT cO2 = PO2/Ptotal = nO2/ntotal …. So PO2 = cO2• Ptot Dalton’s law can also be stated ….. Pi= ciPtot

  29. Dalton’s Law of Partial Pressure: P = SiPi or Pi = ciPtot The atmosphere contains ….. 79% N2 If rel humidity = 0% 20% O2 4% H2O if 100 % humidity 1% Ar 0.04% CO2 What is cO2 in the atmosphere? a. 1.0 b. 0.5 c. 0.25 d 0.20 What is PO2 in the atmosphere? (in atm) 0.20 atm What is PO2 in the atmosphere? (in torr) 1 atm = 760 torr a. 760 torr b. 20 torr c. 150 torr d. 0.20 torr What is PO2 in the atmosphere? (in torr) 1 atm = 760 torr a. 760 torr b. 20 torrc. 150 torrd. 0.20 torr

  30. Vapor Pressure Water

  31. Gas Mixtures Calculate the mass of O2 produced by the decomposition of KClO3 when 821 mL of O2 is collected over water at 30.0°C and 1.015 atm. Solution: Step 1:Use Table 11.5 to determine the vapor pressure of water at 30.0°C.

  32. What is PCO2 in the atmosphere? cCO2= 370 ppm = 370/1 x 106 = 0.00037 PCO2 = cCO2 Ptot = 0.00037  760 = 0.28 torr What was PCO2 in 1800? cCO2 = 0.00028 PCO2 = 0.21 torr 33%↑ Projection = 2x by 2050 if we freeze C emission at present Levels.

  33. Deviation from Ideal Behavior • Gas particles do occupy V – significant as V • (or P at cst T, n) 2. Gas particles attract each other when close, some more than others. - significant as P • Collisions are not elastic; • although total energy is conserved. This is why not all • like molecules travel at the same speed. At low T molecules • tend to stick together more (condensation) At low T - high P - low V: gases deviate from ideal behavior -

  34. Real Gases The van der Waals equation is useful for gases that do not behave ideally. Experimentally measured pressure Container volume corrected pressure term corrected volume term

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