1 / 61

GASES Chemistry I – Chapter 11

GASES Chemistry I – Chapter 11. Importance of Gases. Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3 . 2 NaN 3 ---> 2 Na + 3 N 2. THREE STATES OF MATTER. General Properties of Gases. There is a lot of “free” space in a gas.

taniel
Download Presentation

GASES Chemistry I – Chapter 11

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. GASESChemistry I – Chapter 11

  2. Importance of Gases • Airbags fill with N2 gas in an accident. • Gas is generated by the decomposition of sodium azide, NaN3. • 2 NaN3 ---> 2 Na + 3 N2

  3. THREE STATES OF MATTER

  4. General Properties of Gases • There is a lot of “free” space in a gas. • Gases can be expanded infinitely. (they will fill whatever “container” they are in.) • Gases fill containers uniformly and completely. • Gases diffuse and mix rapidly.

  5. Pressure Pressure is the amount of force per unit area. • Gas Pressure is the pressure caused by particles of gas striking an object. • Air pressure is the pressure of the column of atmosphere above you ,pressing down on you.

  6. Atmospheric Pressure • Measured with a BAROMETER(developed by Torricelli in 1643) • Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). (Just like a straw in a soft drink) • Why is Hg so good for use in barometer?

  7. Atm. Pressure About 34 feet of water = 1 atm pressure! Column height measures Pressure of atmosphere • 1 standard atmosphere (atm) * = 760 mm Hg (or torr) * = 29.92 inches Hg * = 14.7 pounds/in2 (psi)* = 101.3 kPa (SI unit is PASCAL)

  8. Pressure Conversions • Factor Label Methodis how we will make conversions. Sample Problem, p 379 of text One standard atmosphere (1 atm) is equivalent to 760 mm of Hg. What is that height expressed in inches? (Hint: 1.00 in = 25.4 mm) Analysis: Identify • Your “given”: _____ • Your unknown (what you’re solving for):_____ • the relationship (formula, etc) between the given & the unknown: ___ Write an equation: • Your given is on the left, including the units. • The conversion factor is written next, but you must decide how to write it: • Write an “=“ • Write a blank for your answer, and the units your answer will be written in.

  9. Pressure Conversions A. What is 475 mm Hg expressed in atm? 1 atm 760 mm Hg B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 760 mm Hg 14.7 psi 475 mm Hg x = 0.625 atm 29.4 psi x = 1.52 x 103 mm Hg

  10. Pressure Conversions A. What is 2 atm expressed in torr? B. The pressure of a tire is measured as 32.0 psi. What is this pressure in kPa?

  11. Properties of Gases Gas properties can be modeled using math. Model depends on— • V = volume of the gas (L) • T = temperature (K) • ALL temperatures in the entire chapter MUST be in Kelvin!!! No Exceptions! • n = amount (moles) • P = pressure (atmospheres)

  12. Chapter 14 • The Behavior of Gases • (new page of your permanent Chem notebook)

  13. Properties of Gases • Compressibility-a measure of how much the volume of matter decreases under pressure. • Gases are compressible b/c of the space btwn particles in a gas • At room temp, distance btwn particles in an enclosed gas ~ 10x the diameter of the particles

  14. Factors to Describe Gas Pressure • Pressure in kPa • Volume in L • Temperature in K • Number of moles, n

  15. Number of Moles • Pressure is caused by the particles striking the walls of the container. • If the gas is in a rigid container, the volume is constant • If you triple the number of gas particles (n), you triple the amount of pressure. (see p 415 of text)

  16. Volume • Reducing the volume in a container will raise the pressure (inverse relationship.) • See p 416 If you press on a piston, no gas escapes, but the volume decreases. • The same # of particles are striking a smaller area, so pressure increases.

  17. Temperature • As a gas is heated, the temperature increases • If temperature increases, particle speed increases • IF the container is rigid, the pressure will increase. • If the container is not rigid, the volume will increase & pressure will remain the same.

  18. Section 14.1 Assessmentp 417 1. Why is a gas easy to compress? 2. List 3 factors that can affect gas pressure. 3. Why does a collision with an air bag cause less damage than a collision with a steering wheel? 4. How does a decrease in temp affect the pressure of a contained gas? 5. If the temp is constant, what change in volume would cause the pressure of an enclosed gas to be reduced to ¼ of its original value? 6. Assuming the gas in a container remains at a constant temp, how could you increase the gas pressure in a container a hundredfold?

  19. Properties of Gases, cont. • We can study the relationship between 2 variables if we keep the others the same.

  20. Boyle’s Law • 2 variables we will study: P & V • All other variables kept the same (T, n, etc.) • Ex: a rising balloon

  21. Boyle’s Law P α 1/V This means Pressure and Volume are INVERSELY PROPORTIONAL For example, P goes up as V goes down. P1V1 = P2 V2 Robert Boyle (1627-1691). Son of Early of Cork, Ireland.

  22. Boyle’s Law and Kinetic Molecular Theory P proportional to 1/V

  23. 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, and air is forced into the tire.

  24. Sample Problem 14.1 (p 419) • A balloon contains 30.0 L of helium gas at 103kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? Givens P1= 103 kPa P2 = 25.0kPa V1 = 30.0L Unknown V2

  25. Charles’s Law • 2 variables we will study: V & T • All other variables kept the same (P, n, etc.)

  26. Charles’s Law If n and P are constant, then V α T V and T are directly proportional. V1 V2 = T1 T2 • If one temperature goes up, the volume goes up! Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.

  27. Sample Problem 14.2 Using Charles Law (p 421) • A balloon inflated in a room at 24˚C has a volume of 4.00L. The balloon is then heated to a temperature of 48˚C. What is the new volume if the pressure remains constant? Givens: T1 = 24˚C + 273 = 297K T2 = 48 + 273 =321K V1 = 4.00 L Unknown = V2 4.00L = V2 297K 321K 4.00L*321K = V2 297K

  28. CW/HWPractice Problems, pp 419-423 # 7-12

  29. Charles’s original balloon Modern long-distance balloon

  30. Charles’s Law

  31. Gay-Lussac’s Law If n and V are constant, then P α T P and T are directly proportional. P1 P2 = T1 T2 • If one temperature goes up, the pressure goes up! Joseph Louis Gay-Lussac (1778-1850)

  32. Gas Pressure, Temperature, and Kinetic Molecular Theory P proportional to T

  33. Combined Gas Law • Since all 3 gas laws are related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P1 V1 = P2 V2 T1 T2 No, it’s not related to R2D2

  34. Combined Gas Law If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law! = = P1 V1 P2 Boyle’s Law Charles’ Law Gay-Lussac’s Law V2 T1 T2

  35. Combined Gas Law Problem A sample of helium gas has a volume of 180 mL, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? Set up Data Table P1 = 0.800 atm V1 = 180 mL T1 = 302 K P2 = 3.20 atm V2= 90 mL T2 = ??

  36. Calculation P1 = 0.800 atm V1 = 180 mL T1 = 302K P2 = 3.20 atm V2= 90 mL T2 = ?? P1 V1 P2 V2 = P1 V1T2 = P2 V2 T1 T1T2 T2 = P2 V2 T1 P1 V1 T2 = 3.20 atm x 90.0 mL x 302 K 0.800 atm x 180.0 mL T2 = 604 K - 273 = 331 °C = 604 K

  37. Learning Check A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

  38. One More Practice Problem A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

  39. And now, we pause for this commercial message from STP OK, so it’s really not THIS kind of STP… STP in chemistry stands for Standard Temperature and Pressure Standard Pressure = 1 atm (or an equivalent) Standard Temperature = 0 deg C (273 K) STP allows us to compare amounts of gases between different pressures and temperatures

  40. Try This One A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

  41. 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.

  42. Avogadro’s Hypothesis and Kinetic Molecular Theory The gases in this experiment are all measured at the same T and V. P proportional to n

  43. IDEAL GAS LAW P V = n R T Brings together gas properties. BE SURE YOU KNOW THIS EQUATION!

  44. Using PV = nRT P = Pressure n = number of moles V = Volume T = Temperature R is a constant, called the Ideal Gas Constant= 8.31 L*kPa K*mol We must convert the units to match R.

  45. Using PV = nRT • Ex: p 439, Q#55 • 1.24 moles of gas at 35 C and 96.2 kPa pressure. What is the volume the gas occupies? • V=? • n= 1.24 mol • T = 35 + 273 = 308K • P = 96.2kPa • R=8.31L*kPa/K*mol • (96.2kPa)V = (1.24 mol ) (8.21L*kPa/K*mol) (308K) • V=33L

  46. Learning Check Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office?

  47. Learning Check A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

  48. Gases in the Air The % of gases in air Partial pressure (STP) 78.08% N2 593.4 mm Hg 20.95% O2 159.2 mm Hg 0.94% Ar 7.1 mm Hg 0.03% CO2 0.2 mm Hg PAIR = PN2 + PO2 + PAr + PCO2 = 760 mm Hg (Total Pressure = 760mm Hg)

  49. Dalton’s Law of Partial Pressures • 2 H2O2 (l) ---> 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 = PH2O + PO2 = 0.48 atm Dalton’s Law: total P is sum of PARTIAL pressures.

  50. Dalton’s Law John Dalton 1766-1844

More Related