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

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

4. ANSWERS 1.Pressure (P) is COLLISIONS and is force per unit area 2. 687 torr x 1 atm = 0.904 atm 760 torr 3.If T is constant… as P increases, V decreases! 4.If T is constant… as P decreases, V increases! F A

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

6. Inverse and Direct Proportions

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

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

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

10. Demo Time Vacuum pump

12. Inversely Proportional Ex: Boyle’s Law (V as a function of P) P (torr)V (mL) 1004560 2002280 3001520 4001140 500912 600760 700651 800570 900507 1000456 1100414 1200380 1300351 1400326 1500304

13. P 1 x V 1 = P 2 x V 2 P may change and V may change, but their product stays the same! Multiplying the two variables equals a constant.

14. 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: P 1 = 0.947 atm V 1 = 150mL= 0.15L P 2 = 0.987 atm V 2 = ???? P 1 V 1 =P 2 V 2 V 2 = 0.947 atm (.15L) = 0.987 atm

15. Demo Time Liquid nitrogen

16. Temperature and Volume

17. Directly Proportional Ex: Charles’ Law (V as a function of T) T (K)V (mL) 00 50100 200 150300 200400 250500 300600 350700 400800 450900 5001000 5501100 6001200 6501300 7001400

18. DIRECT PROPORTIONS As one variable goes up, so does the other! Produces a straight line graph… Dividing the one variable by the other equals a constant. T V V 1 V 2 T 1 T 2 = YES!!!

19. 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: V 1 = 2.45L T 1 = 273 K V 2 = ? T 2 = 325 K V 1 V 2 T 1 T 2 = V 2 = 2.45 L (325K) = 273 K

20. 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

21. Stop here for today! Let’s practice

22. Demo Time Candle

23. 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

24. 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

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

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

27. Gay-Loussac’s Law P 1

28. 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: P 1 = 0.987 atm T 1 = 298 K P 2 = 0.795 atm T 2 = ??? T 2 = 0.795atm (298K) = 0.98.7atm

29. Avogadro’s Law: V 1 V 2 n 1 n 2 =

30. 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: V 1 = 59.5 L n 1 = 2.55 mol V 2 = ??? n 2 = 7.83 mol V 1 V 2 n 1 n 2 = V 2 = 59.5 L (7.83mol) = 2.55 mol

31. P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2 P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2  Combined Gas Law Which is a direct relationship? Which is inverse?  Combined Gas Law Which is a direct relationship? Which is inverse? A little review

32. 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? 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

33. 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? 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

34. When measured at STP, a quantity of gas has a volume of 500 L. What volume will it occupy at 0 o C and 93.3 kPa? P 1 = 101.3 kPa T 1 = 273 K V 1 = 500 L P 2 = 93.3 kPa T 2 = 0 o C + 273 = 273 K V 2 = ? V 2 = 542.9 L

35. Demo Break Demo Break Set up POE Set up POE BalloonBalloon Liquid nitrogenLiquid nitrogen Vacuum pumpVacuum pump Predict and explain in terms of the four variables used to describe gases: volume (V), pressure (P), temperature (T), and moles (n) Predict and explain in terms of the four variables used to describe gases: volume (V), pressure (P), temperature (T), and moles (n)

36. Group Work Time Put all of your names on your paper Put all of your names on your paper Each person does the work out Each person does the work out I will collect one sheet from each group to grade for a worksheet grade I will collect one sheet from each group to grade for a worksheet grade You won’t know whose I collect until the end of class! You won’t know whose I collect until the end of class! Work together and make sure that each person understand the problem. Work together and make sure that each person understand the problem. Homework Check Homework Check

37. 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? V 2 =30 L Combined 1

38. A gas has a temperature of 14 0 C, and a volume of 4.5 liters. If the temperature is raised to 29 0 C and the pressure is not changed, what is the new volume of the gas? V 2 = V 2 = 4.7 L Charles’

39. 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? V 2 = 0.270 L Boyle’s

40. If I have 2.9 L of gas at a pressure of 5.0 atm and a temperature of 50 0 C, 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? T 2 = 160 K Combined

41. Tonights HW problem + 21, 29, 43 A gas that has a volume of 28 liters, a temperature of 45 0 C, and an unknown pressure has its volume increased to 34 liters and its temperature decreased to 35 0 C. If I measure the pressure after the change to be 2.0 atm, what was the original pressure of the gas? P 1 = P 1 = 2.5 atm

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

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

44. 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 K mol

46. 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 0 C, and a volume of 0.500L. How many moles of argon are present in the lamp? Grams? PV = nRT Need to rearrange to solve for n: PV = n RT Given: P = 1.20 atm T = 18.0°C R = 0.08206 L atm K mol V = 0.500L (1.20 atm)(0.500L) = n (0.08206 L*atm/mol*K) (291K) = 0.0251 mol Ar

47. What volume does 9.45g of C 2 H 2 occupy at STP? What volume does 9.45g of C 2 H 2 occupy at STP? P  V  T T  T T  1atm ? ? 273K R  n n  n n  =.3635 mol C 2 H 2 0.08206 Latm molK 9.45 g 26 g/mol

48. V = nRT P V = nRT P (1 atm) V V (.3635 mol ) (273K) V = 8.14L = = (0.08206 ) Latm molK Latm molK

49. Volume of gas under STP  L  Mol under STP conditions:  22.4 L/mol of any gas 0.03635 mol C 2 H 2 x 1 mol = 8.14 L C 2 H 2 22.4 L 22.4 L

50.   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 O 2

51. Using PV=nRT A camping stove propane tank holds 3000 g of C 3 H 8. 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?