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Ratio Method of Solving Basic Gas Laws

0. Ratio Method of Solving Basic Gas Laws. 0. Boyle’s Law. Don’t hate me just because I’m beautiful. Robert Boyle. 0. Boyle’s Law. Demonstrations that illustrate Boyle’s Law Marshmallow or balloon in the bell jar (marshmallow or balloon gets big when pressure gets low). 0. Boyle’s Law.

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Ratio Method of Solving Basic Gas Laws

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  1. 0 Ratio Method of SolvingBasic Gas Laws

  2. 0 Boyle’s Law Don’t hate me just because I’m beautiful Robert Boyle

  3. 0 Boyle’s Law Demonstrations that illustrate Boyle’s Law • Marshmallow or balloon in the bell jar (marshmallow or balloon gets big when pressure gets low)

  4. 0 Boyle’s Law Volume is inversely related to Pressure If volume decreases, pressure increases If volume increases, pressure decreases (they move in opposite directions) You can see this relationship when temperature and moles of gas are held constant

  5. 0 Boyle’s Law Mathematically, PV = k Where “k” is a constant

  6. Boyle’s Law • Another way to write Boyle’s law is • P1V1 = P2V2 • Temperature is constant!! If it is not, then you can not use Boyle’s Law

  7. 0 Boyle’s Law Problem #1 A 0.030L marshmallow undergoes a drop in pressure from 1.1 atm to 0.20 atm. What will be the new volume? • Start with a “data table” to organize what you know and what you don’t know • V1 = 0.030L • V2 = ? • P1 = 1.1 atm • P2 = 0.20 atm

  8. 0 Boyle’s Law Problem #1 A 0.030L marshmallow undergoes a drop in pressure from 1.1 atm to 0.20 atm. What will be the new volume? 1.1 atm 0.20 atm • V1 = 0.030L • V2 = ? • P1 = 1.1 atm • P2 = 0.20 atm V2 = 0.030L x = 0.17 L Bigger number on top makes Volume get larger Pressure got smaller so Volume must get larger

  9. 0 Boyle’s Law Problem #2 A 0.025L marshmallow at 0.60 atm experiences a pressure increase to 0.95 atm. What will be the new volume? • Start with your data table • V1 = 0.025L • V2 = ? • P1 = 0.60 atm • P2 = 0.95 atm

  10. 0 Boyle’s Law Problem #2 A 0.025L marshmallow at 0.60 atm experiences a pressure increase to 0.95 atm. What will be the new volume? 0.60 atm 0.95 atm V2 = 0.025L x = 0.016 L • V1 = 0.025L • V2 = ? • P1 = 0.60 atm • P2 = 0.95 atm Smaller number on top Makes volume get smaller Pressure got larger, so Volume must get smaller

  11. 0 Charles’ Law Hey Ladies!!!

  12. 0 Charles’ Law Demonstrations that illustrate Charles’ Law • Balloon on flask (heat the flask then the balloon expands)

  13. Charles’ Law Volume is directly related to Temperature If temperature increases, volume increases If temperature decreases, volume decreases (they move in the same direction) You can see this relationship when pressure and moles of gas are held constant

  14. Charles’ Law V = kT Where “k” is a constant of proportionality Note that V does NOT have to equal T, but they are directly proportional to each other

  15. Charles’ Law • Another way to write this is: • (V1/ T1) = (V2/T2) • Or • V1T2 = V2T1 • Pressure must remain constant or you can not use this law!

  16. Charles’ Law Problem #1 A 2.0L sample of air is collected at 298K then cooled to 278K. What will be the new volume? • Start with a data table to organize what you know and what you don’t know • V1 = 2.0 L • V2 = ? • T1 = 298 K • T2 = 278 K

  17. Charles’ Law Problem #1 A 2.0L sample of air is collected at 298K then cooled to 278K. What will be the new volume? 278 K 298 K V2 = 2.0 L x = 1.9 L • V1 = 2.0 L • V2 = ? • T1 = 298 K • T2 = 278 K Smaller number on top makes volume get smaller Temperature decreased so volume must decrease

  18. Charles’ Law Problem #2 A 3.25L balloon at 298K changes volume to 2.50L. What temperature would cause this to happen? • Start with your data table • V1 = 3.25 L • V2 = 2.50 L • T1 = 298 K • T2 = ?

  19. Charles’ Law Problem #2 A 3.25L balloon at 298K changes volume to 2.50L. What temperature would cause this to happen? 2.50 L 3.25 L T2 = 298 K x = 229 K • V1 = 3.25 L • V2 = 2.50 L • T1 = 298 K • T2 = ? Volume got smaller so temperature must get smaller Smaller number on top makes temperature smaller

  20. Gay-Lussac Law O.K. So I have a funny name. Go ahead get it out of your system so we may continue.

  21. Gay-Lussac Law Demonstrations that illustrate Gay-Lussac’s Law Aerosol can in the campfire (As temperature increases, pressure increases until the can explodes)

  22. Gay-Lussac Law Pressure is directly related to Temperature If temperature increases, pressure increases If temperature decreases, pressure decreases (they move in the same direction) You can see this relationship when volume and moles of gas are held constant

  23. Gay-Lussac Law P = kT Where “k” is a constant of proportionality

  24. Gay-Lussac Law • Another way to write it: • (P1 / T1) = (P2 / T2) • Or • T1P2 = P1T2 • Volume stays constant!

  25. Gay-Lussac Law Problem #1 A can of hairspray contains a gaseous propellant at 1.50 atm and 298K. What is the new pressure if the can is heated to 500.K? • Start with a data table • T1 = 298K • T2 = 500.K • P1 = 1.50 atm • P2 = ?

  26. Gay-Lussac Law Problem #1 A can of hairspray contains a gaseous propellant at 1.50 atm and 298K. What is the new pressure if the can is heated to 500.K? 500 K 298 K P2 = 1.50 atm x = 2.52 atm • T1 = 298K • T2 = 500.K • P1 = 1.50 atm • P2 = ? Big number on top makes pressure get bigger Temperature increased so pressure must increase

  27. Gay-Lussac Law Problem #2 A can of spraypaint at 298K experiences an increase in pressure from 101.3 kPa to 425 kPa. What temperature would cause such a change? • Start with a data table • T1 = 298 K • T2 = ? • P1 = 101.3 kPa • P2 = 425 kPa

  28. Gay-Lussac Law Problem #2 A can of spraypaint at 298K experiences an increase in pressure from 101.3 kPa to 425 kPa. What temperature would cause such a change? 425 kPa 101.3 kPa T2 = 298 K x = 1250 K • T1 = 298 K • T2 = ? • P1 = 101.3 kPa • P2 = 425 kPa Big number on top makes Temperature get bigger Pressure increased so temperature must increase

  29. Avogadro’s Law I’m dead sexy and you know it!! Text me. You got my number.

  30. Avogadro’s Law Demonstration Blowing up a balloon As you add moles of gas, the volume increases

  31. Avogadro’s Law Volume is directly related to Moles If moles of gas increases, volume increases If moles of gas decreases, volume decreases (they move in the same direction) You can see this relationship when temperature and pressure of gas are held constant

  32. Avogadro’s Law V = kn Where “k” is a constant of proportionality

  33. Avogadro’s Law • Another way to write this: • V1/n1 = V2/n2 • Or • V1n2 = V2n1

  34. Avogadro’s Law Problem #1 A 12.2 L sample of gas contains 0.500 mol of O2 at 1.00 atm and 25.0oC. If 3.00 mol of O2 are added, what will be the new volume? • Start with a data table • V1 = 12.2 L • V2 = ? • n1 = 0.500 mol • n2 = 3.500 mol

  35. Avogadro’s Law Problem #1 A 12.2 L sample of gas contains 0.500 mol of O2 at 1.00 atm and 25.0oC. If 3.00 mol of O2 are added, what will be the new volume? 3.500 mol 0.500 mol = 85.4 L V2 = 12.2 L x • V1 = 12.2 L • V2 = ? • n1 = 0.500 mol • n2 = 3.500 mol Bigger number on top Makes volume get bigger Moles increased so volume must increase

  36. Avogadro’s Law Problem #2 A 2.25 L balloon contains 0.475 mol of helium. To increase the volume of the balloon to 6.50 L, how many moles of helium must it contain? • Start with a data table • V1 = 2.25 L • V2 = 6.50 L • n1 = 0.475 mol • n2 = ?

  37. Avogadro’s Law Problem #2 A 2.25 L balloon contains 0.475 mol of helium. To increase the volume of the balloon to 6.50 L, how many moles of helium must it contain? 6.50 L 2.25 L n2 = 0.475 mol x = 1.37 mol • V1 = 2.25 L • V2 = 6.50 L • n1 = 0.475 mol • n2 = ? Volume increased so moles must increase Bigger number on top makes Moles get bigger

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