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Gas Laws

Gas Laws. BOYLE. CHARLES. AVOGADRO. GAY-LUSSAC. Push!. Consider This. What happens to the volume of a gas when you increase the pressure? (e.g. Press a syringe that is stoppered). Push!. Consider This. What happens to the volume of a gas when you increase

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Gas Laws

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  1. Gas Laws BOYLE CHARLES AVOGADRO GAY-LUSSAC

  2. Push! Consider This What happens to the volume of a gas when you increase the pressure? (e.g. Press a syringe that is stoppered)

  3. Push! Consider This What happens to the volume of a gas when you increase the pressure? (e.g. Press a syringe that is stoppered)

  4. Consider This What happens to the Volume of a Gas When you Increase the Pressure? (e.g. Press a syringe that is stoppered) Why? There is lots of space between gas particles. Therefore, gases are compressible!

  5. Let’s investigate the relationship between pressure and volume if the quantity of gas and temperature are held constant. 100 kPa

  6. 100 kPa 100 50 Volume = 50 L

  7. 200 kPa 100 50 200 25 Volume = 25 L

  8. 400 kPa 100 50 200 25 400 12.5 Volume = 12.5 L

  9. 800 kPa What is the mathematical relationship between P and V? P x V = constant 100 50 5000 5000 5000 5000 200 25 400 12.5 800 6.25 Volume = 6.25 L

  10. Boyle’s Law In the 17th Century Robert Boyle described this property as, “the spring of air”. Boyle showed that when temperature and amount of gas were constant then: P  1/V OR: PV = k

  11. Boyle’s Law For a fixed quantity of gas at a constant temperature, the volume and pressure are inversely proportional. Boyle showed that when temperature and amount of gas were constant then: P  1/V OR: PV = k

  12. Who Cares? Scuba Divers! At sea level air pressure = 100 kPa At 10 m deep in water pressure = 200 kPa At 20 m deep = 300 kPa At 30 m deep = 400 kPa SCUBA provides air at the same pressure

  13. A Scuba Diver goes to a depth of 90 m and takes a breath of 3 L volume from her tank. Suddenly! A Dolphin lunges at the diver and takes the SCUBA! The Diver holds her breath and quickly returns to the surface. What will the volume of air in the diver’s lungs be at the surface (100 kPa)? What will happen to the diver?

  14. A Scuba Diver goes to a depth of 90 m and takes a breath of 3 L volume from her tank. Assuming that T is constant we can use Boyle’s Law: PV = k At 90 m: k = P1V1 At surface: k = P2V2 Therefore! P1V1 = P2V2 (1000 kPa)(3 L) = 100 kPa(V2) V2 = 300 L Yikes Exploding lungs

  15. Consider This! Two balloons are filled with equal volumes of air. What happens to the Volume of each if one is heated and the other is frozen?

  16. HEATED BALLOON 50oC, V=1.18 L FROZEN BALLOON

  17. HEATED BALLOON 60oC, V=1.22 L 40oC, V=1.14 L FROZEN BALLOON

  18. HEATED BALLOON 70oC, V=1.25 L 30oC, V=1.11 L FROZEN BALLOON

  19. HEATED BALLOON 80oC, V=1.29 L 20oC, V=1.07 L FROZEN BALLOON

  20. HEATED BALLOON 90oC, V=1.32 L 10oC, V=1.04 L FROZEN BALLOON

  21. HEATED BALLOON 100oC, V=1.36 L 0oC, V=1.00 L FROZEN BALLOON

  22. To study this relationship let’s look at this data in a table. Graph this data using temperature as the independent variable

  23. If this line is extended backwards the volume of 0 L of gas is found to be -273 oC

  24. -273.15 oC is known as absolute zero. When using gas laws, temperature must be expressed using a temperature scale where 0 is -273.15oC. This is called the Kelvin scale of absolute temperature. 0 K = -273.15 oC When changing oC to K simply add 273.15 What is 12.3oC in K? 12.3 + 273.15 = 285.45 = 285.5 K

  25. Now let’s look at this table with temperatures in Kelvin (K). Can you spot a mathematical relationship between T and V.

  26. V/T in Kelvin is a constant. 0.0037 0.0037 0.0037 0.0037 0.0037 0.0037 0.0037 0.0037 0.0037 0.0037 0.0037

  27. Charles’ Law In the 17th Century Jacques Charles examined the relationship between Temperature and Volume Charles showed that when Pressure and amount of gas were constant then: V  T OR: V/T = k

  28. Charles’ Law For a fixed quantity of gas at a constant pressure, absolute temperature and volume are directly proportional. Charles showed that when Pressure and amount of gas were constant then: V  T OR: V/T = k

  29. Combined Gas Law V1 V2 T1 T2 V2P2 V1P1 T1 T2 If = V1P1 V2P2 and = then = or V1P1 T2 =V2P2 T1 V2T1 V2T1 What does P2 = ?

  30. RECAP Boyle’s Law At Constant T and n: VP = k Charles’ Law At Constant P and n: V/T = k Combined Gas Law For a fixed quantity of gas: VP/T = k

  31. V1P1 V2P2 T1 T2 V1P1 = V2P2 If 12.5 L of a gas at a pressure of 125 kPa is placed in an elastic container at 15oC what volume would it occupy if the pressure is increased to 145 kPa? Given: V1 = 12.5 L P1 = 125 kPa T1 = 15oC V2 = ? P2 = 145 kPa T2 = 15oC = Since T1 = T2 cancel them to get (12.5 L)(125 kPa) = V2(145 kPa) V2 = (12.5 L)(125 kPa)/145 kPa V2 = 10.8 L

  32. 125 kPa 12.5 L @ 125 kPa Does this answer make sense?

  33. 165 kPa Does this answer make sense?

  34. 165 kPa Does this answer make sense?

  35. 165 kPa Does this answer make sense? Yes, as the pressure increases at constant temperature the volume decreases. V reduced to 10.8 L

  36. V1P1 V2P2 T1 T2 V1/T1 = V2 /T2 If 15.6 L of a gas at a pressure of 165 kPa is placed in an elastic container at 15oC what volume would it occupy if the temperature is increased to 98oC? Given: V1 = 15.6 L P1 = 165 kPa T1 = 15oC V2 = ? T2 = 98oC P2 = 165 kPa = Since P1 = P2 cancel them to get 288 K (15.6 L)/(288 K) = V2/(371 K) V2 = (15.6 L)(371 K) / 288K V2 = 20.1 L 371 K

  37. If 5.3 L of a gas at a pressure of 75 kPa is placed in an elastic container at 24oC what volume would it occupy if the temperature is increased to 62oC and pressure to 155 kPa? V1P1 V2P2 T1 T2 Given: V1 = 5.3 L P1 = 75 kPa T1 = 24oC V2 = ? T2 = 62oC P2 = 155 kPa V1P1T2 =V2P2 T1 = OR (5.3 L)(75 kPa)(335 K) 297 K V2 = (155 kPa)(297K) 335 K V2= 2.9 L

  38. Ideal Gas Law An Ideal Gas is a hypothetical gas that obeys all the gas laws perfectly under all conditions. PV = nRT Where n is the number of moles of gas, P is pressure in kPa, R = 8.313 kPaL/molK and T is temperature in K. Find the mass of helium gas which would be introduced into a 0.95 L container to produce a pressure of 125 kPa at 25oC.

  39. Find the volume occupied by 25 g of chlorine gas at SATP. 4.2 g of propane gas is introduced into a 325 mL container at 45oC. What is the pressure of the container. Propane is C3H8. What is the density of NH3 gas at STP if 1.0 mol of this gas occupies 22.4 L. At what temperature does methane gas have a density of 1.2 mg/L if its pressure is 65 kPa. Methane is CH4.

  40. Dalton's Law of Partial Pressure

  41. This gas exerts a pressure of 100 kPa inside this container. If another gas is injected into the same container and it exerts a pressure of 70 kPa what is the total pressure in the container? 170 kPa

  42. This gas exerts a pressure of 100 kPa inside this container. 170 kPa If another gas is injected into the same container and it exerts a pressure of 70 kPa what is the total pressure in the container?

  43. This gas exerts a pressure of 100 kPa inside this container. 170 kPa If another gas is injected into the same container and it exerts a pressure of 70 kPa what is the total pressure in the container?

  44. The total pressure of a gas mixture is the sum of the partial pressures of each of the gases in the mixture. Example If a container of air has a pressure of 100 kPa and the % of N2 in the container is 78%, % of O2 is 21%, what are the partial pressures of each of these gases inside this container. PN2= 78 kPa, PO2= 21 kPa

  45. What would the total pressure be if the gas in container 1 was injected into container 2. 1 5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 2

  46. The total pressure is the sum of the pressures of gas 1 and gas 2.Since gas 1 changed volume and temperature its pressure changed. 1 5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 2

  47. V1=5.0 L, V2=6.0 L, T1=300 K, T2 = 400 K P1=125 kPa, P2=? V1P1T2=V2P2T1 1 5.0 L x 125 kPa x 400 K P2 = = 139 kPa 6.0 L x 300 K 5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 2

  48. Total pressure is 155 kPa + 139 kPa = 294 kPa = 2.9 x 102 kPa 1 5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 139 kPa 2

  49. Find the total pressure in container 1 if the gas in container 2 is injected into container 1. 1 5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 2

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