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Day 1

Day 1. I CAN… Understand and apply Boyle ’ s Law Understand and apply Charles ’ Law Observe and explain demos using gas laws. Gas Laws. 1 torr = 1mm Hg 1 atm = 760 mm Hg. Convert 687 torrs into atmospheres .

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Day 1

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  1. Day 1 • I CAN… • Understand and apply Boyle’s Law • Understand and apply Charles’ Law • Observe and explain demos using gas laws

  2. Gas Laws 1 torr = 1mm Hg 1 atm = 760 mm Hg Convert 687 torrs into atmospheres. If temperature is constant, what happens to volume (V) as pressure (P) increases? What if P decreases?

  3. ANSWERS F A • Pressure (P) is COLLISIONS and is force per unit area 2. 687 torr x 1 atm = 0.904 atm 760 torr • If T is constant… as P increases, V decreases! • If T is constant… as P decreases, V increases! P V P V

  4. Properties of Gases Gases can be described using the following four variables: V = volume of the gas (liters, L) P = pressure (atmospheres, atm) T = temperature (Kelvin, K) n = amount (moles, mol) UNITS

  5. Inverse and DirectProportions Indirect Direct

  6. Inverse / Indirect Relationship • What variables did you observe to have an indirect relationship yesterday with the simulation?

  7. Direct Relationship • What variables did you observe to have a direct relationship yesterday with the simulation?

  8. INVERSE PROPORTIONS P • As one variable goes up, the other goes down! • Produces a curved graph… V T constant

  9. Demo Time  • Vacuum pump

  10. Inversely ProportionalEx: Boyle’s Law(V as a function of P)

  11. Boyle's Law P1 x V1=P2xV2 P may change and V may change, but their product stays the same! Multiplying the two variables equals a constant.

  12. Boyle’s Law A gas filled syringe has a volume of 150mL and a starting pressure of 0.947atm. If the pressure is increased to 0.987 atm what is the new volume? Given: P1= 0.947 atm V1= 150mL= 0.15L P2= 0.987 atm V2 = ???? P1V1=P2V2 V2 = 0.947 atm (.15L) = 0.987 atm

  13. Demo Time  • Liquid nitrogen

  14. Temperature and Volume

  15. Directly ProportionalEx: Charles’ Law(V as a function of T)

  16. T V DIRECTPROPORTIONS • As one variable goes up, so does the other! • Produces a straight line graph… • Dividing the one variable by the other equals a constant. V1 V2 T1 T2 = YES!!!

  17. Charles’ Law The volume of a balloon is 2.45 L when at a temperature of 273 K. What happens to the volume when the temperature is raised to 325K? Given: V1= 2.45L T1= 273 K V2= ? T2 = 325 K V2 = 2.45 L (325K) = 273 K V1 V2 T1 T2 =

  18. Charles’ Law • Temperature MUST be in Kelvins! • Kelvins is the unit of temperature that gives the direct relationship to energy and the other units. Celsius does not! • 0°C = 273 K • So… what is 23°C in K? • 23 + 273 = 296 K

  19. Stop here for today! • Let’s practice 

  20. Demo Time  • Candle

  21. Practice #1 • A sample of helium gas has a pressure of 3.54 atm in a container with a volume of 23.1 L. This sample is transferred to a new container and the pressure is measured to be 1.87 atm. What is the volume of the new container? Assume constant temperature. 43.7 L

  22. Practice #2 • A 2.45 L sample of nitrogen gas is collected at 273 K and heated to 325 K. Calculate the volume of the nitrogen gas at 325 K. Assume constant pressure. 2.92 L

  23. Day 2 • I CAN… • Understand and apply Gay-Loussac’s Law • Understand and apply Avogadro’s Law • Understand and apply the Combined Gas Law

  24. Pressure depends on Temp Today’s temp: 35°F Today’s temp: 85°F Pressure Gauge Pressure Gauge

  25. P1 Gay-Loussac’s Law

  26. Gay Loussac’s law A bike tire has a pressure of 0.987 atm at a temperature of 25°C. What temperature would bring the pressure down to 0.795 atm? Given: P1= 0.987 atm T1= 298 K P2= 0.795 atm T2 = ??? T2 = 0.795atm (298K) = 0.98.7atm

  27. Avogadro’s Law: V1 V2 n1 n2 =

  28. Avogadro’s Law A 59.5 L cylinder has 2.55 moles of hydrogen gas. More hydrogen is added so that there are now 7.83 moles, what is the new volume? Given: V1= 59.5 L n1= 2.55 mol V2= ??? n2 = 7.83 mol V2 = 59.5 L (7.83mol) = 2.55 mol V1 V2 n1 n2 =

  29. A little review • Combined Gas Law • Which is a direct relationship? • Which is inverse? P1V1=P2V2 n1T1 n2T2

  30. Examples • A 15 L cylinder of gas at 4.8 atm pressure at 25ºC is heated to 75ºC and compressed to 17 atm. What is the new volume? 4.9 L

  31. Examples • If 6.2 L of gas at 723 mm Hg at 21ºC is compressed to 2.2 L at 4117 mm Hg, what is the temperature of the gas? 590K

  32. When measured at STP, a quantity of gas has a volume of 500 L. What volume will it occupy at 0 oC and 93.3 kPa? P1 = 101.3 kPa T1 = 273 K V1 = 500 L P2 = 93.3 kPa T2 = 0 oC + 273 = 273 K V2 = ? V2 = 542.9 L

  33. 1  If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas? V2=30 L Combined

  34. A gas has a temperature of 14 0C, and a volume of 4.5 liters. If the temperature is raised to 29 0C and the pressure is not changed, what is the new volume of the gas? V2 = 4.7 L Charles’

  35. A sample of gas under a pressure of 720 mm Hg has a volume of 300. mL. The pressure is changed to 800. mmHg. What volume will the gas then occupy? V2 =0.270 L Boyle’s

  36. If I have 2.9 L of gas at a pressure of 5.0 atm and a temperature of 50 0C, what will be the temperature of the gas if I decrease the volume of the gas to 2.4 L and decrease the pressure to 3.0 atm? T2= 160 K Combined

  37. Tonights HW problem + 21, 29, 43 A gas that has a volume of 28 liters, a temperature of 45 0C, and an unknown pressure has its volume increased to 34 liters and its temperature decreased to 35 0C. If I measure the pressure after the change to be 2.0 atm, what was the original pressure of the gas? P1= 2.5 atm

  38. Day 3 • I CAN… • Understand and apply the Ideal Gas Law • Explain how ideal gases and real gases differ in their behavior

  39. PV = R n T PV =nRT Ideal gas law • IF WE COMBINE ALL OF THE LAWS TOGETHER INCLUDING AVOGADRO’S LAW MENTIONED EARLIER WE GET: WHERE R IS THE UNIVERSAL GAS CONSTANT NORMALLY WRITTEN AS

  40. Ideal Gas Law PV = nRT P = pressure (in atm!!) V = volume (in L!!) n = number of moles R = universal gas constant = T = temperature (in K!!) 0.08206 L atm K mol

  41. Argon is an inert gas used in lightbulbs to slow the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm, temperature of 18.0 0C, and a volume of 0.500L. How many moles of argon are present in the lamp? Grams? PV = nRT Given: P = 1.20 atm T = 18.0°C R = 0.08206 L atm K mol V = 0.500L Need to rearrange to solve for n: PV = n RT (1.20 atm)(0.500L) = n (0.08206 L*atm/mol*K)(291K) =0.0251 mol Ar

  42. L•atm 0.08206 mol•K 9.45g 26g/mol What volume does 9.45g of C2H2 occupy at STP? • 1atm • P • R • ? • V • T • 273K • n • = .3635 mol C2H2

  43. L•atm mol•K (0.08206 ) V =nRT P (273K) (.3635mol) V = (1 atm) V = 8.14L

  44. Volume of gas under STP • L  Mol under STP conditions: • 22.4 L/mol of any gas 0.03635 mol C2H2x 1 mol= 8.14 L C2H2 22.4 L

  45. If I have an unknown quantity of oxygen gas held at a temperature of 1195 K in a container with a volume of 25.0 liters and a pressure of 560.0 atm, how many grams of oxygen gas do I have? • n = 143 moles • m = 4580 g O2

  46. Using PV=nRT A camping stove propane tank holds 3000 g of C3H8. How large would a container have to be to hold the same amount of propane gas at 25 ºC and a pressure of 303 kPa?

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