THE-GAS-LAWS-COMPLETE LESSON POWERPOINT!

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1. The Gas Laws

2. PRESSURE , VOLUME , TEMPERATURE are measurable properties of gases that are related to each other. If one of these variables changed, there is a corresponding change in other variables depending on its relationship. Gas law equations can be derived whenever ne of these variables is altered.

3. G AS PRESSURE – described as a force acting on a specific area. P = Force over(divide) Area Has units of atm Atmosphere Mm Hg Millimeter mercury Torr Torr lb /in^2 Pound per square inch kPa Kilopascals 1 atm = 760 mm hg = 760 torr = 101.325 kPa

4. Volume – three dimensional space occupied by a gas. Ml Milliliter L Liter m^3 Cubic meter Cm^3 Cubic centimeter 1 L = 1000 ml, 1 ml = 1 cm^3

5. Temperature – measure of the warmth or coldness of a body Measure of the average kinetic energy of the particles in an object Kinetic energy is the movement of the particle. High KE, faster movement F Fahrenheit C Celsius K kelvin

6. Quantity – measured in moles (mol) 1 mol = 6.022 x 10^23 units of a substance

7. MAIN GAS LAWS Boyle’s Law Charles’ Law Gay-Lussac’s Law These laws are products of various experiments that are done many centuries ago. The Ideal Gas Law can be used to describe the relationship between variables used by the gas laws.

8. BOYLE’S LAW: Volume-Pressure Relationship

9. ROBERT BOYLE An Anglo-Irish chemist. Investigated the relationship between pressure and volume of a gas using a J-shaped tube apparatus which was closed to one end. He then proposed that , the volume of a given mass of gas held at constant temperature is inversely proportional to its pressure.

10. Graph that shows the relationship of Pressure and Volume Schematic Illustration of Boyle’s Law

11. Boyle has shown us that as the volume decreases, the pressure increases, which pertains to an inverse relationship. This relationship is more commonly called Boyle’s Law. This relationship can be expressed in a mathematical way: V ∝ where ∝ means “ proportional to” . To change symbol ∝ to an equal sign, a proportionality constant k, is introduced. Hence, V = ( ) or more simply: PV =  

12. For a given sample of gas under two different conditions at a constant temperature, the product of pressure and volume is constant , thus, it is written as follows: = where and are the initial pressure and volume , while, and are the final pressure and volume of the same amount of gas at the same temperature.  

13. =   P1 = Initial Pressure V1 = initial Volume P2 = Final Pressure V2 = Final volume

14. =   A gas occupies 12.3 liters at a pressure of 40.0 mm hg. What is the volume when the pressure is increased to 120.0 mmhg ? 1. GIVEN: V1 = 12.3 L P1 = 40.0 mm hg P2 = 120.0 mm hg V2 = ???

15. =   1. GIVEN: V1 = 12.3 L P1 = 40.0 mm hg P2 = 120.0 mm hg V2 = ??? P1V1 = P2V2 (40.O mmhg ) (12.3L) = (120.0mmhg) V2 492 mmhgL = (120.0mmhg) V2 = V2   4.1 L = V2

16. =   If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm? GIVEN: 2. V1 = 3.60 L P1 = 1 atm P2 = 2.50 atm V2 = ???

17. =   GIVEN: 2. V1 = 3.60 L P1 = 1 atm P2 = 2.50 atm V2 = ??? P1V1 = P2V2 (1 atm) (3.6 L) = (2.5 atm) V2 3.6 atm(L) = (2.5 atm) V2 = V2   1.44 L = V2

18. =   Convert 350.0 mL at 740.0 mmHg to its new volume at standard pressure. GIVEN: 3. V1 = 350 mL P1 = 740 mmHg P2 = standard pressure which is 760mm hg V2 = ???

19. =   GIVEN: 3. V1 = 350 mL P1 = 740 mmHg P2 = 760mm hg V2 = ??? P1V1 = P2V2 (740 mmhg ) (350 mL) = (760mmhg) V2 259,000mmhg(mL) = (760mmhg) V2 = V2   340.79 mL = V2

20. CHARLES’S LAW: Volume-Temperature Relationship

21. JACQUES ALEXANDRE CHARLES A French Physicist performed an experiment in a balloon , hot water and cold water more than 100 years later after Boyle’s experiment about the volume-pressure relationship. He then proposed that “the Kelvin temperature and the volume of a gas are directly related when there is no change in pressure of a gas ”.

22. Graph that shows the relationship of Volume and Temperature Schematic Illustration of Charles’ Law

23. From his experiment, Charles has shown us that as the volume increases and/or decreases, the temperature manifests the same increase and/or decreases, which pertains to a direct relationship. This relationship is more commonly called Charles’ Law .

24. This relationship can be expressed in a mathematical way: V ∝ T, or more simply, = k, where K is proportional constant. For a given sample of gas under two different conditions at a constant pressure, the equation can be written as: = where and are the initial temperature in Kelvin and Volume, while, , also in Kelvin, and are the final temperature and volume of the same amount of gas at the same pressure.  

25. =   A syringe contains 56.11 mL of gas at 311 K. Determine the volume that the gas will occupy if the temperature is increased to 400 K . 1. GIVEN: V1 = 56.11 mL T1 = 311 K T2 = 400 K V2 = ???

26. =   1. GIVEN: V1 = 56.11 mL T1 = 311 K T2 = 400 K V2 = ??? =   V2 (311 K) = 56.11 mL (400 K) V2 (311 K) = 22,444 ml K =   V2 = 72.17 mL V2T1 = V1T2 GIVEN: V1 = 56.11 mL T1 = 311 K T2 = 400 K

27. =   If 540 mL of nitrogen at 0.00 °C is heated to a temperature of 100 °C, what will be the new volume of the gas? 2. GIVEN: V1 = 540 mL T1 = 0.00 °C T2 = 100 °C WE HAVE TO CONVERT CELCIUS TO KELVIN FIRST V2 = ???

28. =   If 540 mL of nitrogen at 0.00 °C is heated to a temperature of 100 °C, what will be the new volume of the gas? 2. GIVEN: V1 = 540 mL T1 = 0.00 °C T2 = 100 °C V2 = ??? T1= 0 C + 273 = 273 K T2= 100 C + 273 = 373 K

29. =   2. GIVEN: V1 = 540 mL T1 = 273 K T2 = 373 K V2 = ??? =   V2 (273 K) = 540 mL (373 K) V2 (272 K) = 201,420 ml K =   V2 = 737.8 mL V2T1 = V1T2

30. =   At constant pressure, Neon gas contracts from 2.00 L to 0.75 L. The initial temperature is 24 C. Find the final temperature 3. GIVEN: V1 = 2 L T1 = 24 C T2 = ??? WE HAVE TO CONVERT CELCIUS TO KELVIN FIRST V2 = 0.75 L

31. =   At constant pressure, Neon gas contracts from 2.00 L to 0.75 L. The initial temperature is 24 C. Find the final temperature 3. GIVEN: V1 = 2 L T1 = 24 C T2 = ??? V2 = 0.75 L T1= 24 C + 273 = 297 K GIVEN: V1 = 2 L GIVEN:

32. =   3. =   0.75 L (297 K) = 2 L (T2) 222.75 L K = 2 L (T2) =   111.38 K = T2 V2T1 = V1T2 T1 = 297 K T2 = ??? V2 = 0.75 L V1 = 2 L GIVEN:

33. GAY-LUSSAC’S LAW: Temperature-Pressure Relationship

34. JOSEPH-LOUIS GAY-LUSSAC - A French Chemist and Physicist. - He proved that one of the postulates in Kinetic Molecular Theory is the effect of temperature on the motion of gas particles. - Year 1802, he conducted and experiment and discovered the relationship between the Temperature and Pressure. He found out that the pressure of the pressure of a gas increased or decreased proportionally with a change in temperature. He summarized his findings in his proposed Gay-Lussac’s Law which states that “the pressure of a fixed amount of a gas is directly proportional to the absolute temperature (Kelvin).

35. Graph that shows the relationship of Temperature and Pressure Schematic Illustration of Gay-Lussac’s Law

36. It can be mathematically expressed as follows P ∝ T (n, V constant). Or dividing both sides of the equation by T , hence, P, = k or = K. For any two sets of pressure and temperature, at constant volume, the equation can be stated as: =  

37. =   Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C. 1. GIVEN: P1 = 1 atm T1 = 20 C T2 = 30 C P2 = ??? WE HAVE TO CONVERT CELCIUS TO KELVIN FIRST

38. =   Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C. 1. GIVEN: P1 = 1 atm T1 = 20 C T2 = 30 C P2 = ??? T1= 20 C + 273 = 293 K T2= 30 C + 273 = 303 K

39. =   1. =   P2(293 K) = 1 atm (303 K) P2 (293 K)= 303 atm K =   P2 = 1.03 atm P2T1 = P1T2 GIVEN: P1 = 1 atm T1 = 293 K T2 = 303 K P2 = ???

40. =   A gas has a pressure of 699.0 mmHg at 40.0 °C. What is the temperature at standard pressure? 2. GIVEN: P1 = 699.0 mmhg T1 = 40 C T2 = ? P2 = standard pressure WE HAVE TO CONVERT CELCIUS TO KELVIN FIRST

41. =   A gas has a pressure of 699.0 mmHg at 40.0 °C. What is the temperature at standard pressure? 2. GIVEN: P1 = 699.0 mmhg T1 = 40 C T2 = ? P2 = standard pressure T1= 40 C + 273 = 313 K Standard pressure is 760 mmhg

42. =   2. =   760(313 K) = 699 (T2) 237,880 = 699 (T2) =   340.31 K = T2 P2T1 = P1T2 GIVEN: P1 = 699.0 mmhg T1 = 313K T2 = ? P2 = 760 mmhg

43. COMBINED GAS LAWS: Volume-Pressure-Temperature Relationship

44. The three properties of a gas (pressure, volume and temperature) usually change at once under experimental condition. The four possible variations are as follows: 1. Both T and P cause an increase in V. 2. Both T and P cause a decrease in V. 3. T causes an increase in V and P cause a decrease in V. 4. T causes a decrease in V and P cause an increase in V.

45. The three gas laws (Boyle’s, Charles’ and Gay-Lussac’s laws) discussed earlier can be combined and may be expressed into one equation, = K , and expressed as the Combined Gas Law or the general gas law. Combined Gas Laws allows you to directly solve for the changes in pressure, volume, or temperature. Scientists generally follow the customary reference point of gases which is 0˚C and 1 atm pressure by international agreement. This reference point refers to the Standard Temperature and Pressure or (STP). The temperature can also be expressed as 273 K and standard pressure as 760 torr,760 mmHg and 76 cmHg .  

46. The equation can be expressed as follows: BOYLE’S LAW : V ∝ CHARLES’S LAW: V ∝ T COMBINED GAS LAW: V ∝ or PV ∝ T or PV = constant Thus, = k In a constant n, or mole, this can be derived as: =  

47. =   A gas has a volume of 800.0 mL at −23.0 °C and 300.0 torr. What would the volume of the gas be at 227.0 °C and 600.0 torr of pressure? 1. GIVEN: P1 = 300 torr T1 = -23 C V1 = 800 mL P2 = 600 torr T2 = 227 C V2 = ? WE HAVE TO CONVERT CELCIUS TO KELVIN FIRST

48. =   A gas has a volume of 800.0 mL at −23.0 °C and 300.0 torr. What would the volume of the gas be at 227.0 °C and 600.0 torr of pressure? 1. GIVEN: P1 = 300 torr T1 = -23 C V1 = 800 mL P2 = 600 torr T2 = 227 C V2 = ? T1= -23 C + 273 = 250 K T2= 227 C + 273 = 500 K

49. =   1. =   =   120,000,000 = 150,000 (V2) =   800 mL = V2 =   GIVEN: T2 = 227 C V2 = ? P1 = 300 torr T1 = -23 C V1 = 800 mL P2 = 600 torr

50. =   A balloon of air now occupies 10.0 L at 25.0 °C and 1.00 atm. What temperature was it initially, if it occupied 9.40 L and was in a freezer with a pressure of 0.939 atm? 2. GIVEN: P1 = 0.939 atm T1 = ? V1 = 9.40 L P2 = 1 atm T2 = 25 C V2 = 10 L WE HAVE TO CONVERT CELCIUS TO KELVIN FIRST T1= 25 C + 273 = 298 K

51. =   2. =   =   = 2,630.3268   =   T1 = 263.03 K or -10.03 C =   GIVEN: P1 = 0.939 atm T1 = ? V1 = 9.40 L P2 = 1 atm T2 = 298 K V2 = 10 L

52. =   Suppose you have a sample of gas at 303K in a container with a volume of 2L and pressure of 760mmHg. The sample shifts to a temperature of 340 K and the volume increases slightly to 2.1L. What is the pressure of the sample now? 3. GIVEN: P1 = 760 mmhg T1 = 303 K V1 = 2 L P2 = ? T2 = 340 K V2 = 2.1 L

53. =   3. =   =   = 516,800   =   P2 = 812.20 mmhg =   GIVEN: P1 = 760 mmhg T1 = 303 K V1 = 2 L P2 = ? T2 = 340 K V2 = 2.1 L

54. Avogadro’s law, =   Amadeo Avogadro  was an Italian physicist who stated, in 1811, that the volume of any gas is proportional to the number of molecules of gas (measured in  Moles  – symbol  mol ). In other words if the amount of gas increases, then so does its volume. 

55. =   The initial amount of dry ice (gas) is 1 mol, and occupies a volume of 2 L. What is the volume if we increase the amount of dry ice (gas) to 2 mol? 1. GIVEN: n1 = 1 mol V1 = 2 L n2 = 2 mol V2 = ? n1 = 1 mol V1 = 2 L

56. =   1. GIVEN: =   V2 (1) = 2 (2) V2 = 4 V2 = 4 L V2n1 = V1n2 n2 = 2 mol V2 = ? n1 = 1 mol V1 = 2 L

57. =   5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)? 2. GIVEN: n1 = 0.965 mol V1 = 5 L n2 = 1.80 mol V2 = ?

58. =   2. GIVEN: =   V2 (0.965) = 5 (1.8) V2 (0.965) = 9 V2n1 = V1n2 n1 = 0.965 mol V1 = 5 L n2 = 1.80 mol V2 = ? =   V2 = 9.33 L

59. The Ideal Gas Law

60. Aside from pressure, volume and temperature, moles may also be combined to the three properties of gas establish the relationship among the four variables affecting the behavior of gases. Boyle’s, Charles’s, Gay-Lussac’s and Avogadro’s Law when combined constitute the ideal Gas Law or ideal gas equation.

61. The ideal Gas is based on the experimental measurements of the physical properties of gases: temperature, pressure, volume and number of moles. Ideal Gas equation can be expressed as: PV = nRT

62. Where R is the proportionality constant at STP and is equal to 0.0821 L . atm/mol . K and referred as the ideal gas constant. This tells that the volume of a gas varies directly with the number of moles and absolute temperature and inversely proportionality with pressure. PV = nRT

63.   What pressure is required to contain 0.023 moles of nitrogen gas in a 4.20 L container at a temperature of 20 C. 1. GIVEN: P = ? V = 4.20 L n = 0.023 moles T = 20 + 273 = 293 K R = 0.0821 ( L atm/mol K )

64. 1. GIVEN: P (4.20) = 0.5532719 =   P = 0.13 atm     P (4.20)   P = ? V = 4.20 L n = 0.023 moles T = 20 + 273 = 293 K R = 0.0821 (L atm/mol K)

65.   Oxygen gas is collected at a pressure of 123 kPa in a container which has a volume of 10.0 L. What temperature must be maintained on 0.500 moles of this gas in order to maintain this pressure? Express the temperature in degree Celsius. 2. GIVEN: P = 123 kPa V = 10.0 L n = 0.500 moles T = ? R = 0.0821 (L atm/mol K) 1 atm = 760 mm hg = 760 torr = 101.325 kPa

66. 2. GIVEN: 12.1 = 0.04105 (T) =   294.76 K or 21.76 C = T     1.21 (10)   P = 121 atm V = 10.0 L n = 0.500 moles T = ? R = 0.0821 (L atm/mol K)

67. Thank you! 