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GASES

GASES. Gases (Vapors). Gases expand to fill any container. Therefore, gases are highly compressible. Kinetic Molecular Theory ( of an Ideal Gas ):.

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GASES

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  1. GASES

  2. Gases (Vapors) Gases expand to fill any container. Therefore, gases are highly compressible.

  3. Kinetic Molecular Theory (of an Ideal Gas): 1. Gases are composed of molecules or atoms whose size is negligible compared to the average distance between them. (Most of the space in the gas container is empty.) 2. Gas molecules move randomly in straight lines in all directions at various speeds. 3. The forces of attraction or repulsion between gas molecules are very weak or negligible (except during collisions) 4. Collisions between gas molecules are considered elastic. 5. The average kinetic energy of a molecule is proportional to the absolute temperature.

  4. Pressure and Volume: Boyle’s Law How is the pressure applied to a gas related to its volume? Gas molecules Piston Let’s apply pressure

  5. Pressure and Volume: Boyle’s Law How is the pressure applied to a gas related to its volume? Gas molecules Gas molecules Piston Piston Volume is inversely proportional to applied pressure. Boyle’s Law: P1V1 = P2V2

  6. The Harder we Push the smaller the gas volume gets! Boyle’s Law: P1V1 = P2V2

  7. We live in “sea of air” molecules of air 3 2 Where is the pressure the greatest? 1 Why does a diver get the bends?

  8. Pressure: force per unit area of surface Units lbs per in2 (psi) mm of Hg (torr) atmospheres (atm) Pascal (Pa) 1 atm = 760 mm of Hg =760 torr = 14.70 psi = 101.325 kPa Pairs of these can be used as conversion factors.

  9. Temperature and Volume: Charles’s Law How is the volume of a gas related to its temperature? moveable mass (constant pressure) gas molecules What happens if heat is applied to the gas?

  10. Temperature and Volume: Charles’s Law How is the volume of a gas related to its temperature? moveable mass (constant pressure) gas molecules Why did the volume change? What happens to the average speed of the gas molecules? .

  11. Temperature and Volume: Charles’s Law How is the volume of a gas related to its temperature? moveable mass (constant pressure) gas molecules The volume of a gas is directly proportional to its Temperature (temperature must be in Kelvin) Charles’s Law: V1/T1 = V2/T2

  12. Combined Gas Law (Boyle and Charles): T must be in Kelvin Can be rearranged to: P1V1T2 = P2V2T1 A combined gas law problem can be recognized by having two sets of conditions. Note: if one set of parameters is unchanged that term will cancel on each side.

  13. A balloon contains helium gas with a volume of 2.60 L at 25 oC and 768 mmHg. If the balloon ascends to an altitude where the helium pressure is 590 mmHg and the temperature is 15 oC, what is the volume of the balloon? What type of problem is this? There are 2 sets of conditions.

  14. A balloon contains helium gas with a volume of 2.60 L at 25 oC and 768 mmHg. If the balloon ascends to an altitude where the helium pressure is 590 mmHg and the temperature is 15 oC, what is the volume of the balloon? P1V1T2 = P2V2T1 P1= V1= T1= 768 torr 2.60 L 25 + 273 = 298 K = (768 torr)(2.60 L)(288 K) (590 torr)(298 K) P2= V2= T2= 590 torr ? 15 + 273 = 288 K = 3.27 L

  15. Ideal Gases and the Ideal Gas Law: PV = nRT Temperature in K *gas constant 0.0821 L•atm = 62.37 L•torr mol•K mol•K moles of gas volume in L pressure in units to match *R units Note: there is only one set of conditions.

  16. Avogadro’s Law: Equal volumes of any two gases (ideal) at the same temperature and pressure contain the same number of molecules (they also occupy equal volumes). STP Pressure 1 atm (760 mm Hg) Temperature 0oC (273 K) Standard At STP one mole of ideal gas occupies 22.4 L

  17. A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC. Determine the pressure in the cylinder. What type of problem is this? Only one set of conditions

  18. A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC. Determine the pressure in the cylinder. P = nRT V PV = nRT = (3.74 mol)(62.4L•torr)(297.5K) (12.25 L) mol•K ? P = V = n = R = T = 12.25 L mol = 5667.7 torr 75.5 g = mol 3.74 20.18 g = 5670 torr 62.4 L•torr mol•K How many atmospheres is this? 24.5 + 273 = 297.5 K

  19. What is the density of carbon dioxide gas at 25 oC and 725 mmHg pressure? Density = g/L = g  L so if we can find g and L, division will work! P = V = n = R = T = 725mmHg What do we do now? 62.4 L• torr mol•K 25 + 273 = 298 K

  20. What is the density of carbon dioxide gas at 25 oC and 725 mmHg pressure? Density = g/L = g  L so if we can find g and L division will work! P = V = n = R = T = 725mmHg Two variables! Let’s pick an amount for one and calculate the other! Let’s choose 1 mol of CO2 and find the number of Liters. 62.4 L•torr mol•K 25 + 273 = 298 K

  21. What is the density of carbon dioxide gas at 25 oC and 725 mmHg pressure? Density = g/L = g  L so if we can find g and L division will work! V = nRT P P = V = n = R = T = 725mmHg = (1 mol) (62.4 L•torr) (298 K) ( mol•K ) (725 torr) 1.0 mol (44.0 g) 62.4 L•torr mol•K = 25.6 L 44.0 g ___________ = g L NOW: 1.72 25 + 273 = 298 K 25.6 L

  22. A 2.50 gram sample of a solid was vaporized in a 505 mL vessel. If the vapor pressure of the solid was 755 mmHg at 155 oC, what is the molecular weight of the solid? molecular weight ~ molar mass = g/mol = g  mol ..so if we can find grams and moles and divide.... ...we already have grams!! We’re halfway there! P = V = n = R = T = 755 torr n = PV RT 0.505 L = 755 torr | 0.505 L | mol•K_____|______ | 62.4 L•torr | 428 K 62.4 L•torr mol•K = 0.01428 mol 155 + 273 = 428 K NOW: 2.50 g = g 0.01428 mol mol 175.1

  23. So Density is g/L (g ÷ L) and molar mass is g/mol (g ÷ mol).

  24. Dalton’s Law of Partial Pressures: He H2N2 Ptotal = P1 + P2 + P3 +...

  25. Dalton’s Law of Partial Pressures: Ptotal = P1 + P2 + P3 +... Since they are considered to be ideal gases, the attractions and repulsions between molecules are ignored. ... and... PV=nRT so: PV = (n1 + n2 + n3)RT or: We also refer to mole fractions:

  26. To find the gas pressure, the pressure of the water vapor must be subtracted from the total pressure.

  27. A 250.0 mL flask contains 1.00 mg of He and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres. The total pressure is due to the partial pressures of each of these gases. so: For He: 1.00 x 10-3 g He mol _____________________ = mol He 2.50 x 10-4 4.00 g For H2: 2.00 x 10-3 g H2 mol ______________________ = mol H2 9.92 x 10-4 2.016 g

  28. A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres. so: 1.00 x 10-3 g He mol _____________________ = mol He For He: 2.50 x 10-4 4.00 g For H2: 2.00 x 10-3 g H2 mol ______________________ = mol H2 9.92 x 10-4 2.016 g And: Ptotal = (2.50 x 10-4 + 9.92 x 10-4)(RT/V) = (0.001242 mol)(0.0821 L•atm)(25 + 273)K mol•K (0.2500 L) Ptotal= 0.1216 atm

  29. A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres. so: 1.00 x 10-3 g He mol _____________________ = mol He For He: 2.50 x 10-4 4.00 g For H2: 2.00 x 10-3 g H2 mol ______________________ = mol H2 9.92 x 10-4 2.016 g Calculate the pressure due just to He (you have 37 seconds): = 0.0245 atm and Phydrogen= ? 0.1216 - 0.0245 = 0.0971 atm

  30. Magnesium is an active metal that replaces hydrogen from an acid by the following reaction: Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) How many g of Mg are needed to produce 5.0 L of H2 at a temperature of 25 oC and a pressure of 745 mmHg? Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) ? g 5.0 L Hint: find moles of H2 using PV = nRT then work as a stoichiometry problem. n = PV RT 745 mmHg 5.0 L mol•K =____________________________________ 62.4 L•mmHg 298 K n = 0.20 mol

  31. Magnesium is an active metal that replaces hydrogen from an acid by the following reaction: Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) How many g of Mg are needed to produce 5.0 L of H2 at a temperature of 25 oC and a pressure of 745 mmHg? Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) ? g 5.0 L 0.20 mol 0.20 mol H2 1 mol Mg 24.3 g Mg ____________________________________ = g Mg 4.87 1 mol H2 mol Mg

  32. Molecular Speeds: K.E. = ½ mv2 Average kinetic energy of a gas molecule:  = ½ m2 Where = the rms (root-mean-square) speed of the molecules at each temperature. From kinetic-molecular theory: At any given temperature the molecules of all gases have the same average kinetic energy. Which molecules travel faster, big or little?

  33. At room temperature, the average speed of an N2 molecule is ........ 1150 mi/hr

  34. Molecular diffusion and effusion: Diffusion: “gas molecules spreading out to fill a room are diffusing.” Its not easy since an average gas molecule at room temperature and pressure will experience about 10 billion collisions per second! It only travels about 60 nm between collisions!

  35. Effusion: “A Helium filled balloon loses He by effusion.” escaping molecule Small hole or pore

  36. Which molecules will effuse faster from this semiporous container? Graham’s Law of effusion: effusion rate is inversely proportional to the square root of its molar mass. For 2 gases:

  37. r = rate of effusion u = root mean speed (~average speed) of molecules M = molar mass Compare the rates of effusion of He and N2. He effuses 2.65 times as fast as N2.

  38. He N2 Which balloon will lose pressure sooner?

  39. N2 He Little molecules (escape more easily) Big molecules Which balloon will lose pressure sooner?

  40. Real Gases: When do gases become non-ideal? As they approach the liquid state, attractions between molecules increase and they become less ideal. Temperature: low Pressure: high van der Waal’s equation is one equation used to treat non-ideal gases. a and b are constants found in tables for each gas.

  41. Which gas would deviate the most from the ideal gas law at room temperature (25oC)? C3H8 boiling Pt. 231K PH3 boiling Pt. 188K SiH4 boiling Pt. 161K CO boiling Pt. 81K

  42. Which gas would deviate the most from the ideal gas law at room temperature (25oC)? 300K 298K C3H8(l) boiling Pt. 231 200K PH3(l) boiling Pt. 188K SiH4(l) boiling Pt. 161K 100K CO(l) boiling Pt. 81K

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