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Name the following molecular shapes:

Name the following molecular shapes:. THREE STATES OF MATTER. General Properties of Gases. There is a lot of “free” space in a gas. Gases can be expanded infinitely. Gases occupy containers uniformly and completely. Gases diffuse and mix rapidly. How do gases cause pressure?

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Name the following molecular shapes:

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  1. Name the following molecular shapes:

  2. THREE STATES OF MATTER

  3. General Properties of Gases • There is a lot of “free” space in a gas. • Gases can be expanded infinitely. • Gases occupy containers uniformly and completely. • Gases diffuse and mix rapidly.

  4. How do gases cause pressure? • What affects gas pressure?

  5. Properties of Gases Gas properties can be modeled using math. Model depends on— • V = volume of the gas (L) • T = temperature (K) • n = amount (moles) • P = pressure (atmospheres)

  6. Temperature • Always use Kelvin • Kelvin was introduced in 1848 by Lord Kelvin • Proposed a temperature scale that made -273.15 the zero point Why -273.15? It was found that all gases when pressure was held constant, and the temperature lowered they reached zero volume at a common temperature…-273.15 • To convert Celsius to Kelvin: Celsius + 273.15 = Kelvin

  7. Pressure Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). P of Hg pushing down related to • Hg density • column height

  8. Pressure Column height measures P of atmosphere • 1 standard atm= 760 mm Hg = 760 torr = SI unit is PASCAL, Pa, where 1 atm = 101.325 kPa

  9. Standard Conditions • Why would you need to establish a standard? • Standard P = • 1 atm • Standard T = • 273 K

  10. IDEAL GAS LAW P V = n R T Brings together gas properties. Can be derived from experiment and theory.

  11. Boyle’s Law If n and T are constant, then THE VOLUME OF A GAS IS INVERSELY PROPORTIONAL TO THE APPLIED PRESSURE: V  1/P ( means “is proportional to”) This means, for example, that P goes up as V goes down. (P1) (V1) = (P2) (V2) Robert Boyle (1627-1691). Son of Early of Cork, Ireland.

  12. Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up.

  13. Problem A 15.0 Liter balloon contains a fixed number of moles of gas (n). If the temperature (T) stays constant, and the pressure is INCREASED from 1.0 atm to 3.0 atm, what will the new volume be? P1V1 = P2V2

  14. Problem 1 A sample of gaseous nitrogen in a 65.0L automobile air bag has a pressure of 745 mm Hg. If this sample is transferred to a 25.0L bag with the same temperature as before, what is the pressure of the gas in the new bag? (Make table of what you know and what you need)

  15. Charles’s Law If n and P are constant, then VOLUME IS DIRECTLY PROPORTIONAL TO THE KELVIN TEMPERATURE V and T are directly related. Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.

  16. Problem 2 What is the new temperature of a sample of gas if it has a volume of 300. mL at 20.0°C and the volume decreases to 250. mL?

  17. Combine Boyle’s and Charles’s Law Volume is inversely proportional to pressure Volume is directly proportional to temperature What are you holding constant?

  18. Problem 3 Helium-filled balloons are used to carry scientific investigations high into the atmosphere. Suppose that a balloon is launched when the temperature is 22.5°C and the barometric pressure is 754 mm Hg. If the balloon’s volume is 4.19 x 103 L, what will the volume be at a height of 20 miles, where the pressure is 0.10 atm and the temperature is -33.0°C?

  19. Problem 4 If a sample of neon has a volume of 250. mL at 740. mm Hg and 27.0°C, what would be the volume of the gas at STP?

  20. Gay - Lussac’s Law • Pressure and Temperature are directly proportional Problem 5 What is the new temperature of a sample of gas if it has a pressure of 300. torr at 20.0°C and the pressure increases to 550. torr at constant n and V

  21. twice as many molecules Avogadro’s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V and n are directly related.

  22. Ideal Gas Law Putting all these individual laws together:P V  n TReplace “” with “= ” and a constant,k:P V = k n T

  23. Using PV = nRT How much N2 is req’d to fill an air bag with a volume of 65 L to P = 829 mm Hg at 25 oC? R = 0.082057 L•atm/K•mol Solution 1. Get all data into proper units V = 65 L T = 25 oC + 273 = 298 K P = 829 mm Hg (1 atm/760 mm Hg) = 1.09 atm

  24. Using PV = nRT How much N2 is req’d to fill an air bag with a volume of 65 L to P = 829 mm Hg at 25 oC? R = 0.082057 L•atm/K•mol Solution 2. Now calculate n = PV / RT n = (1.09atm)(65L)/ (0.0821 L•atm/K•mol)( 298K) n = 2.89 mol (or about 81 kg of gas)

  25. Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

  26. Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Solution Strategy: Calculate moles of H2O2 and then moles of O2 and H2O. Finally, calc. P from n, R, T, and V.

  27. Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Solution

  28. Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Solution P of O2 = 0.16 atm

  29. Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) What is P of H2O? Could calculate as above. But recall Avogadro’s hypothesis. V n at same T and P P n at same T and V There are 2 times as many moles of H2O as moles of O2. P is proportional to n. Therefore, P of H2O is twice that of O2. P of H2O = 0.32 atm

  30. GAS DENSITY PV = nRT

  31. USING GAS DENSITY The density of air at 15 oC and 1.00 atm is 1.23 g/L. What is the molar mass of air? 1. Calc. moles of air. V = 1.00 L P = 1.00 atm T = 288 K n = PV/RT = 0.0423 mol 2. Calc. molar mass mass/mol = 1.23 g/0.0423 mol = 29.1 g/mol

  32. Molar Mass PV = nRT • But you can substitute n, n = g/molar mass • Rearrange: molar mass = gRT / PV

  33. Problem 7 The density of an unknown gas is 1.23g/L at STP. Calculate its molar mass.

  34. Problem 8 • Suppose you have done an experiment to determine the empirical formula of a compound now used to replace CFCs in air conditioners. • Your results give a formula of CHF2. Now you need the molar mass of the compound to find the molecular formula. • You therefore do another experiment and find that a 0.100g sample of the compound exerts a pressure of 70.5 mm Hg in a 256mL container at 22.3°C. • What is the molar mass of the compound? • What is its formula?

  35. Convert to correct units Use ideal gas law to solve for moles Molar mass = mass / moles Experimental molar mass / mass of 1 mol CHF2 Strategy:

  36. Molar Mass = 102 g/mol Exp. molar mass / mass of 1 mol CHF2 = (102 g/mol) / (51 g/mol) = 2 So the molecular formula is C2H2F4

  37. Problem 9 As an extra credit project you are asked to design an air bag for one of the new hybrid cars, Ion. You know that the bag should be filled with gas with a pressure higher than atmospheric pressure, say 829 mm Hg, at a temperature of 22.0 °C. The bag has a volume of 45.5L. What quantity of sodium azide, NaN3, should be used to generate the required quantity of gas? The chemical reaction is: 2 NaN3 (s)  2 Na (s) + 3 N2 (g) Strategy: Use PV=nRT to find moles

  38. Use PV=nRT to find moles N2 Multiply moles of N2 by stoichiometric factor Moles of NaN3 times Molar mass Mass of NaN3 What quantity of sodium azide, NaN3, should be used to generate the required quantity of gas?The chemical reaction is:2 NaN3 (s)  2 Na (s) + 3 N2 (g) Strategy:

  39. Dalton’s Law John Dalton 1766-1844

  40. The total pressure is the sum of the partial pressures • Let’s say you have a container with 3 gases: A, B, C. To find the partial pressure of each gas: PAV= nART, PBV= nBRT, PCV= nCRT Rearrange…. So P = nRT/V for each of the eqns. Ptotal = PA + PB + PC = nART/V + nBRT/V + nCRT/V Isn’t this sooooo long? Let’s simplify!!

  41. Simplify!! • Ptotal = PA + PB + PC = nART/V + nBRT/V + nCRT/V • You can simplify this by pulling out the common factor RT/V: Ptotal = (nA+ nB + nC)RT / V = ntotal (RT / V)

  42. MOLE FRACTIONS, χ • Useful when you need to find how much of each gas you have within a mixture. χA = nA / ntotal Mole fraction of each gas adds up to

  43. Combine to get another… =) Ptotal= ntotal (RT / V) & χA = nA / ntotal PA = χA Ptotal

  44. Suppose you mix 15.0 g of halothane vapor with 23.5 g of oxygen gas. If the total pressure of the mixture is 855 mm Hg, what is the partial pressure of each gas? • Convert mass to moles for each gas • Use the molar mass of each molecule • Find mole fraction of each gas: • χA = nA / ntotal Double check: Do your mass fractions add up to one? • Find the partial pressure of each gas: • PA = χA Ptotal

  45. Answer: • Halothane: 80.2 mm Hg • Oxygen: 775 mm Hg Double check your work: Do the partial pressures add up to the total pressure? (855 mm Hg)

  46. Dalton’s Law of Partial Pressures • 2 H2O2(liq) ---> 2 H2O(g) + O2(g) • 0.32 atm 0.16 atm What is the total pressure in the flask? Ptotal in gas mixture = PA + PB + ... Therefore, Ptotal = P(H2O) + P(O2) = 0.48 atm Dalton’s Law: total P is sum of PARTIAL pressures.

  47. At this point you’ve seen a bundle of symbols and equations. Take some time to summarize today – including: • What are all the variables used to describe gases? • What units are the variables measured in? • What different equations did I learn how to use? • When would I use these equations? • How do I convert from one pressure unit to another?

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