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Gas Laws Day 2

Gas Laws Day 2. Gas Law Foldable. Fold the left and right to the middle. Cut along solid lines (but only to the crack!). 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 ….

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Gas Laws Day 2

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  1. Gas LawsDay 2

  2. Gas Law Foldable • Fold the left and right to the middle. • Cut along solid lines (but only to the crack!)

  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. 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.

  7. The Combined Gas Law P1 V1 = P2 V2 n1T1 n2T2

  8. 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

  9. 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

  10. 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

  11. *** 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

  12. 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 if T1 = T2

  13. Inverse Indirect

  14. 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.

  15. 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.

  16. 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

  17. P1 V1= P2 V2 n1T1 n2T2 P1V1 = P2V2 V2 V2 P2 = (3.5 atm)(5 L) 25 L P2 = 0.7 atm

  18. Charles’ Law: Volume vs Temperature At constant pressure, the volume of a fixed mass of gas varies directly with the Kelvin temperature V1 = V2 if P1=P2 T1 T2

  19. Direct

  20. Charles’ Law Graph Temp decreases Vol decreases.

  21. Temperature and Volume with constant Pressure Changing the temperature but requiring the pressure to stay the same causes the volume to increase.

  22. 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.

  23. If 22.4 L of oxygen is heated from 23˚C to 50 ˚C, what is its new volume? V1 = 22.4 L T1 = 296 K V2 = ? T2 = 323 K Example

  24. P1V1= P2V2 n1T1n2T2 V1 = V2 T1 T2 V2 = V1 T2 T1 V2 = (22.4 L)(296 K) = 20.5 L (323 K)

  25. 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.

  26. 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

  27. Avogadro’s Law • As the number of moles increases, the volume expands to make room for the additional gas

  28. 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 P1V1= P2V2 n1T1 n2T2 V1 = 4.65 L

  29. V1 = V2 n1 n2 V2 = V1n2 n1 V2 = (4.65 L)(3.75 mol) = 21.79 L (0.8 mol) (20 L)

  30. Whew! My brain hurts!!!

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