1 / 72

Chapter 12 The Behavior of gases

Chapter 12 The Behavior of gases. 12.3 The Gas Laws. All gases. Are less dense when warmer Are less dense when under less pressure Are less dense when there are less particles in a volume Obey laws, within limits, which allow us to predict their behavior under most conditions.

trung
Download Presentation

Chapter 12 The Behavior of gases

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. Chapter 12The Behavior of gases 12.3 The Gas Laws

  2. All gases • Are less dense when warmer • Are less dense when under less pressure • Are less dense when there are less particles in a volume • Obey laws, within limits, which allow us to predict their behavior under most conditions

  3. Things you will learn • Understand Boyle’s Law • Understand Charles’s Law • Understand Gay-Lussac’s Law • Understand and be able to solve problems using the combined gas laws

  4. Boyle’s LawThe pressure-volume relationship of gases

  5. Pressure and volume

  6. Pressure and volume have an inverse relationship

  7. Boyles Law essentials • The product of a pressure and volume of any two sets of conditions at a given temperature is constant: • P1 x V1 = P2 x V2 • 100 kPa @ 1 L = 50 kPa @ 2 L = 200 kPa @ .5L • The two conditions are inversely proportional • P~1/V

  8. A high altitude balloon contains 30 L of He gas at 103 kPa. What is the volume when it gets to an altitude where the pressure is 25 kPa?

  9. A high altitude balloon contains 30 L of He gas at 103 kPa. What is the volume when it gets to an altitude where the pressure is 25 kPa? • Knowns: • V1 = 30 L • P1 = 103 kPa • P2 = 25 kPa • Unknown: • V2

  10. A high altitude balloon contains 30 L of He gas at 103 kPa. What is the volume when it gets to an altitude where the pressure is 25 kPa? • Our equation is V1 x P1 = V2 x P2 • 30L x 103 kPa = v2 x 25 kPa • 30 L x 103 kPa 25 kPa • V2 = 124 L V2

  11. Charles’s LawThe temperature-volume relationship of gases

  12. Charles’s Law esentials • An increase in the temperature of a gas yields an increase in the volume of a gas • V1/T1 = V2/T2 • 1 L @ 300K = 2 L @ 600k • The two conditions are directly proportional • V~T remember all temperature measurements are in kelvins

  13. A balloon inflated in a room at temperature of 24°C has a volume of 4 L. The balloon is then heated to a temperature of 58°C. What is the new volume if the temperature remains constant?

  14. A baloon inflated in a room at temperature of 24°C has a volume of 4 L. The balloon is then heated to a temperature of 58°C. What is the new volume if the temperature remains constant? • Knowns: • V1 = 4 L • T1 = 24°C • T2 = 58°C • Unknown: • V2

  15. A baloon inflated in a room at temperature of 24°C has a volume of 4 L. The balloon is then heated to a temperature of 58°C. What is the new volume if the temperature remains constant? • Our equation is V1/T1= V2 /T2 • 4L /24°C = xL/58°C • 58C x 4L / 24°C = xL wait, we haven’t converted to Kelvins! 331K x 4L = 4.46 L 297K

  16. Gay-Lussac The temperature-volume relationship of gases

  17. Gay-Lussac’s Law • The pressure of a gas is directly proportional to the temperature of a gas (in Kelvins) if the volume remains constant • Because these relations are directly related, they obey the formula: • P1/T1 = P2/T2

  18. Sample problemThe gas in an aerosol can may be 103 kPa, meaning it won’t squirt, at 25°C, but if thrown in a fire, the pressure could be quite dangerous. How high is the pressure if the fire is 928°C ?

  19. Sample problemThe pressure in a tire is 198 kPa at the start of a trip at 27°C. At the end of the trip, it is 225 kPa. What is the internal temperature of the tire

  20. Combining the gas laws • Boyles’ Law is • Charles’s Law is • Gay-Lussac’s Law is

  21. Combining the gas laws • Boyles’ Law is P1 x V1 = P2 x V2 • Charles’s Law is V1/T1 = V2/T2 • Gay-Lussac’s Law is P1/T1 = P2/T2

  22. The combined gas law P1 x V1 T1 P2 x V2 T2 If you hold the temperature constant, you have Boyle’s Law If you hold volume constant, you have Gay-Lussac’s Law If you hold the pressure constant, you have Charles’s Law

  23. Sometimes you are not able to hold any of the variables constant • The volume of a gas filled balloon is 30L and 153 kPa. What volume will the balloon be at STP?

  24. The volume of a gas filled balloon is 30L and 153 kPa. What volume will the balloon be at STP? • Knowns: V1=30 L T1=40C T2= 273K (standard temp) P1= 153 kPa P2=101.3 kPa (standard pressure

  25. The volume of a gas filled balloon is 30L and 153 kPa. What volume will the balloon be at STP? • Knowns: V1=30 L T1=40C T2= 273K (standard temp) P1= 153 kPa P2=101.3 kPa (standard pressure Change all temps to Kelvins Isolate V2 and solve

  26. Practice problems • A gas at 155 kPa and 25C occupies a container with an initial volume of 1 L. By changing the volume, the pressure of the gas increases to 605 kPa as the temperature is raised to 125C. What is the new volume?

  27. Practice problems • A 5 L air sample at a temperature of -50C has a pressure of 107 kPa. What will be the new pressure if the temperature is raised to 102C and the volume expands to 7 L?

  28. Ideal gas law • Up to this point, we have left out another thing that can change pressure, volume and temperature

  29. We have used the combined gas law to find changes in a system when we change conditions such as temperature, pressure or volume. P1 x V1 T1 P2 x V2 T2

  30. But the amount of gas- the number of moles- can change the pressure and volume also • It makes sense that the volume and pressure in a container must be proportional to the number of moles • We can add the term for moles (n) to the combined gas laws that we just looked at : so P1V1/T1 becomes P1V1 / T1n1

  31. P1 x V1 T1 x n1 P2 x V2 T2 x n2 • This shows that the term P x V/ T x n is a constant • This will allow us to find unknowns in a system where we are not changing conditions

  32. P1V1 = T1n1 • This is called the Ideal Gas Law • We need to evaluate this in order to come up with a constant which will make the equation work • We do this at STP • What is STP

  33. We use STP as a starting point • P= 101.3 kPa (one ATM) • V= 22.4 Liters (volume of 1 mole at STP) • N= 1 mole • T= 273K (melting point of ice in K) • The constant, R, = (P x V) / (T x n) or 8.31 (L x kPa) / (K x mol)

  34. The ideal gas law becomes: • PV=nRT • Pressure x volume = # moles x temp x constant • This will allow us to figure any one of the variables in any system if we know the other three

  35. Sample problem • You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final pressure of 2 x 104kPa at 28C. How many moles of gas are in the cylinder?

  36. You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final pressure of 2 x 104kPa at 28C. How many moles of gas are in the cylinder? • Knowns: P = 2x104kPa V = 20 L T = 28C • Unknown: N = ?

  37. You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final pressure of 2 x 104kPa at 28C. How many moles of gas are in the cylinder? • Convert 28C to K 301K • Isolate n (number of moles) n = P x V R x T

  38. You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final pressure of 2 x 104kPa at 28C. How many moles of gas are in the cylinder? n = 2x104kPa x 20L 8.31 L x kPa x 301K K x mol

  39. You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final pressure of 2 x 104kPa at 28C. How many moles of gas are in the cylinder? n = 2x104kPa x 20L 8.31 L x kPa x 301K K x mol

  40. You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final pressure of 2 x 104kPa at 28C. How many moles of gas are in the cylinder? n = 1.60 x 102 mol N2

  41. When the pressure of a rigid hollow sphere containing 685 L of helium gas is held at 621K, the pressure of the gas is 1.89 x 103kPa. How many moles of helium does the sphere contain?

  42. What pressure will be exerted by .450 mol of a gas at 25C if it is contained in a .650 L vessel?

  43. A deep underground cavern contains 2.24 x 106 L of methane (CH4) at a pressure of 1.50 x 103kPa and a temperature of 42C. How many kilograms of CH4 does this deposit contain?

  44. What volume will 12 g of oxygen gas (O2) occupy at 25C and a pressure of 52.7 kPa?

  45. Ideal gas?? • The ideal gas law assumes that gas particles occupy no volume and that there are no attractions between particles. This is never true. • Gases under high pressure and low temperatures turn to liquids. Gases do occupy volume (obviously), and there are attractions (intermolecular forces) as we learned in chapter 10. • Boyle’s law implies that gases are infinitely compressible, but this is not the case

More Related