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Gas Laws Day 3

Gas Laws Day 3. Gas Law Foldable. Fold the left and right to the middle. . Dalton’s Law of Partial Pressure. The pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. P Total = P 1 + P 2 + P 3 …. Example.

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Gas Laws Day 3

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  1. Gas LawsDay 3

  2. Gas Law Foldable Fold the left and right to the middle.
  3. Dalton’s Law of Partial Pressure The pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. PTotal = P1 + P2 + P3….
  4. Example A balloon is filled with air (O2, CO2, & N2) at a pressure of 1.3 atm. If PO2 = 0.4 atm and PCO2 = 0.3 atm, what is the partial pressure of the nitrogen gas?
  5. PTotal = P1 + P2 + P3…. Ptotal = PO2 + PCO2 + PN2 1.3 atm = 0.4 atm + 0.3 atm + PN2 PN2 = 0.6 atm
  6. Boyle’s Law: Pressure vs. Volume At a constant temperature, the volume of a fixed mass of gas varies inversely with the pressure P1V1 = P2V2 *use if temperature is constant
  7. Inverse Indirect
  8. Pressure and Volume with constant Temperature Changing the volume of the container changes the amount of space between the particles. The less space, the more the particles collide with each other and the walls.
  9. Uses of Boyle’s Law Testing materials stability and ability to maintain their shape under force. Compressing gases for use in cooking cylinders, SCUBA tanks, and shaving cream. Used to describe density relationships between gases.
  10. If a gas expands from a volume of 5 L to 25 L at an initial pressure of 3.5 atm, what will be its new pressure? P1 = 3.5 atm V1= 5 L P2 = ? V2 = 25 L Example
  11. Charles’ Law: Volume vs Temperature At constant pressure, the volume of a fixed mass of gas varies directly with the Kelvin temperature V1 = V2 T1T2 *use if pressure is constant
  12. Direct
  13. Charles’ Law Graph Temp decreases Vol decreases.
  14. Temperature and Volume with constant Pressure Changing the temperature but requiring the pressure to stay the same causes the volume to increase.
  15. Uses of Charles’ Law Describing the properties of gases, liquids, and solids at extremely low temperatures. Hot air ballooning Used to describe density relationships of gases.
  16. If 22.4 L of oxygen is heated from 23˚C to 50 ˚C, what is its new volume? V1 = 22.4 L T1= 23 + 273 = 296 K V2 = ? T2 = 50 + 273 = 323 K Example
  17. Absolute Zero Temperature at which all molecular motion stops. It is defined by 0 K or -273C. Scientists used Charles Law to extrapolate the temperature of absolute zero.
  18. Avogadro’s Law: Volume vs. Moles At a constant temperature & pressure, the volume of a gas varies directly with the moles V1= V2 n1 n2 *use if temperature and pressure are constant
  19. Avogadro’s Law As the number of moles increases, the volume expands to make room for the additional gas
  20. Example If 4.65 L of CO2 increases from 0.8 moles to 3.75 moles, what is the new volume of the gas? V1 = 4.65 L n1 = 0.8 moles V2 = ? L n2= 3.75 moles V1= V2 n1n2
  21. V1= V2 n1 n2 V2 = V1n2 n1 V2 = (4.65 L)(3.75 mol) = 21.79 L (0.8 mol) (20 L)
  22. The Combined Gas Law P1 V1= P2 V2 n1T1 n2T2
  23. The Combined Gas Law Expresses a relationship between pressure, volume, and temperature (and moles) of a fixed amount of gas. It takes all three gas laws: (Boyle’s, Charles’s, & Avogadro’s) and combines them form one usable equation.
  24. Example 1 A gas is cooled from 45˚C to 20˚C. The pressure changes from 103 kPa to 101.3 kPa as the volumes settles to 16.0 L. What was the initial volume? P1 = 103 kPa V1 = ? T1 = 45oC + 273 = 318 K P2 = 101.3 kPa V2 = 16.0 L T2 = 20oC + 273 = 293 K
  25. P1 V1= P2 V2 n1T1 n2T2 (103 kPa)(V1)= (101.3 kPa)(16.0 L) (n1) (318 K)(n2) (293 K) V1= 17.1 L
  26. Example 2 (on back of foldable) A 1.5 mole sample of methane was originally 0.5 L at 25˚C at 1.1 atm. If we decreased the volume of the container to 0.25 L, increased the pressure to 2.0 atm and added 2.5 moles, what would be the new temperature in ˚C? P1 = 1.1 atm V1 = 0.5 L T1 = 25 + 273= 298 K n1 = 1.5 moles P2 = 2.0 atm V2 = 0.25 L T2 = ? n2 = 1.5+2.5 = 4 moles
  27. *** easier if you solve for T2 first THEN plug in the #’s *** P1 V1n2T2 = P2 V2n1T1 P1 V1n2 P1 V1n2 P1 V1= P2 V2 n1T1 n2T2 = (2.0 atm)(0.25 L)(1.5 mol)(298K) (1.1 atm)(0.5 L)(4.0 mol) T2= 101.59 K – 273 = -171.41 ˚C
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