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Understanding Gas Laws: Boyle's, Charles', and Avogadro's Principles

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This lesson focuses on key gas laws including Boyle's Law, Charles' Law, and Avogadro's Law, explaining their principles and applications. Students will learn how to apply these laws to calculate changes in volume, pressure, and temperature under various conditions. Activities include demonstrations to showcase these gas relationships, as well as practice problems that reinforce learning through calculations involving direct and inverse proportions. By the end of the lesson, students will be able to predict the behavior of gases in response to changes in environmental conditions.

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Understanding Gas Laws: Boyle's, Charles', and Avogadro's Principles

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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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