- 1. GASES
- 2. GASES Kinetic Molecular Theory of Gases -The volume of an individual gas molecule is negligible compared to the volume of the container holding the gas. This means that individual gas molecules, with virtually no volume of their own, are extremely far apart and most of the container is “empty” space -There are neither attractive nor repulsive forces between gas molecules -Gas molecules have high translational energy. They move randomly in all directions, in straight lines -When gas molecules collide with each other or with a container wall, the collisions are perfectly elastic. This means that when gas molecules collide, somewhat like billiard balls, there is no loss of kinetic energy -The average kinetic energy of gas molecules is directly related to the temperature. The greater the temperature, the greater the average kinetic energy.
- 3. GASES Gas molecules have high translational energy. They move randomly in all directions, in straight lines
- 4. GASES Pressure Volume Temperature ALL related to one another. If you change one, you change the other.
- 5. GASES: Pressure Pressure (P) is measured in kilopascals (kPa) Pressure = Force Area The unit Pa is the same as N/m2 For example: The standard atmospheric pressure at 0ºC is 101.3kPa How to measure pressure in pascals (Pa)? Converting common units: 760 mm Hg = 760 Torr = 1 atm = 101.3 kPa
- 6. GASES: Volume Volume (V) is measured in Litres (L)
- 7. GASES: Temperature Temperature (T) is measured in Kelvin (K) These are the same degrees as ºC, but: 0ºC = 273.15K To convert Celcius to Kelvin, use the following formula: T (in K) = ºC + 273
- 8. GASES: Relationships Boyle’s Law: Pressure and volume are inversely proportional PiVi = PfVf Charles’ Law: Volume and temperature are directly proportional Vi = Vf Ti Tf Gay-Lussac’s Law: Pressure and temperature are directly proportional Pi = Pf Ti Tf (assuming constant temperature) (assuming constant pressure) (assuming constant volume)
- 9. GASES: Relationships P 1 V If a sample of gas at initial conditions has an increase of pressure applied to it, its volume decreases proportionally Boyle’s Law: Pressure and volume are inversely proportional PiVi = PfVf
- 10. GASES: Relationships Boyle’s Law: Pressure and volume are inversely proportional PiVi = PfVf
- 11. GASES: Relationships Charles’ Law: Volume and temperature are directly proportional Vi = Vf Ti Tf
- 12. GASES: Relationships Gay-Lussac’s Law: Pressure and temperature are directly proportional Pi = Pf Ti Tf
- 13. GASES: Relationships COMBINED GAS LAW PiVi = PfVf Ti Tf Since pressure, volume, and temperature are all related, they can all be combined together:
- 14. GASES STANDARD TEMPERATURE and PRESSURE (STP) Pressure = 101.3 kPa Temperature = 273K (0°C) STANDARD AMBIENT TEMPERATURE and PRESSURE (SATP) Pressure = 100 kPa Temperature = 298K (25°C) One of two conditions will be used for gas calculations:
- 15. GASES: Calculations Sandra is having a birthday party on a mild winter’s day. The weather changes and a higher pressure (103.0 kPa) cold front (-25°C) rushes into town. The original air temperature was -2°C and the pressure was 100.8 kPa. What will happen to the volume of the 4.2 L balloons that were tied to the front of the house? Given: Pi = 100.8 kPa Pf = 103.0 kPa Vi = 4.2 L Vf = ? Ti = -2°C = 271K Tf = -25°C = 248K PiVi = PfVf Ti Tf (100.8 kPa) (4.2 L) = (103.0 kPa) Vf (271 K) (248K) Vf = 3.76 L Therefore the volume of the balloons will decrease to 3.8 L
- 16. GASES: Calculations Practice: Page 457 #19-21
- 17. GASES: Dalton’s Law Dalton’s Law of Partial Pressures: The total pressure of a mixture of gases is the sum of the pressures of each of the individual gases Ptotal = P1 + P2 + P3 + P4 + P5 +… + Pn Practice: Page 460 #22-23
- 18. GASES: Ideal Gas Law Gay-Lussac: Mole ratios are the same as volume ratios Avogadro’s Hypothesis: Equal volumes of all ideal gases at the same temperature and pressure contain the same number of molecules
- 19. GASES: Ideal Gas Law For example: 2H2(g) + O2(g) 2H2O(g) 2 mol + 1 mol 2 mol 2 volumes + 1 volume 2 volume So if you react 2L of hydrogen gas with 1L of oxygen gas, you will get… 2L of water vapour!!!
- 20. GASES: Ideal Gas Law Ideal Gas Law formula: (Most IMPORTANT formula) PV = nRT Pressure (kPa) Volume (L) Number of moles (mol) Universal gas constant 8.314 kPa∙L mol∙K Temperature (K)
- 21. GASES: CALCULATIONS Sulfuric acid reacts with iron metal to produce gas and an iron compound. What volume of gas is produced when excess sulfuric acid reacts with 40.0g of iron at 18°C and 100.3 kPa? Given: mFe= 40.0g T = 18.0°C = 291K P = 100.3 kPa STEP 1: Write the balanced chemical equation Fe(s) + H2SO4(aq) H2(g) + FeSO4(aq) n = m/M = (40.0g) / (55.85g/mol) = 0.716 mol Fe STEP 2: Use molar ratios to solve for amount of product made (stoichiometry!) 1 mol H2 = x 1 mol Fe 0.716 mol Fe x = 0.716 mol H2 STEP 3: Use the ideal gas law to solve for the volume
- 22. GASES: CALCULATIONS Sulfuric acid reacts with iron metal to produce gas and an iron compound. What volume of gas is produced when excess sulfuric acid reacts with 40.0g of iron at 18°C and 100.3 kPa? Given: mFe= 40.0g T = 18.0°C = 291K P = 100.3 kPa STEP 3: Use the ideal gas law to solve for the volume PV = nRT V = nRT P = (0.716 mol x 8.314 kPa∙L/mol∙K x 291 K) (100.3 kPa) = 17.3 L Therefore 17.3 L of hydrogen gas are produced Practice: Page 506 #30-34
- 23. GASES: CALCULATIONS A student reacts magnesium with excess dilute hydrochloric acid to produce hydrogen gas. She uses 0.15g of magnesium metal. What volume of dry hydrogen does she collect over water at 28°C and 101.8 kPA? Given: mMg= 0.15g T = 28.0°C = 301K P = 98.0 kPa ? Pressure of water vapour at 28°C = 3.78 kPa (from page 596, Table 1) Ptotal = PH2O + PH2 (101.8 kPa) = 3.78 kPa + PH2 Dalton’s Law of Partial Pressures PH2 = 98.0 kPa
- 24. GASES: CALCULATIONS A student reacts magnesium with excess dilute hydrochloric acid to produce hydrogen has. She uses 0.15g of magnesium metal. What volume of dry hydrogen does she collect over water at 28°C and 101.8 kPA? Given: mMg= 0.15g T = 28.0°C = 301K P = 98.0 kPa STEP 1: Write the balanced chemical equation Mg(s) + 2HCl(aq) MgCl2(ag) + H2(g) n = m/M = (0.15g) / (24.31g/mol) = 0.0062 mol STEP 2: Use molar ratios to solve for amount of product made (stoichiometry!) 1 mol H2 = x 1 mol Mg 0.0062 mol Mg x = 0.0062 mol H2 STEP 3: Use the ideal gas law to solve for the volume
- 25. GASES: CALCULATIONS STEP 3: Use the ideal gas law to solve for the volume PV = nRT V = nRT P = (0.0062 mol x 8.314 kPa∙L/mol∙K x 301 K) (98.0 kPa) = 0.16 L Therefore the student collects 0.16 L of dry hydrogen Practice: Page 511 #37-39 A student reacts magnesium with excess dilute hydrochloric acid to produce hydrogen gas. She uses 0.15g of magnesium metal. What volume of dry hydrogen does she collect over water at 28°C and 101.8 kPA? Given: mMg= 0.15g T = 28.0°C = 301K P = 101.8 kPa