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Gas Laws

Gas Laws. Boyle’s Law Charles’s Law Gay-Lussac’s Law The Combined Gas Law The Ideal Gas Law. The Pressure-Volume Relationship (Boyle’s Law) Anglo-Irish chemist : Robert Boyle (1627-1691). For a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure

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Gas Laws

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  1. Gas Laws Boyle’s Law Charles’s Law Gay-Lussac’s Law The Combined Gas Law The Ideal Gas Law

  2. The Pressure-Volume Relationship (Boyle’s Law)Anglo-Irish chemist : Robert Boyle (1627-1691) For a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure P1 x V1 = P2 x V2 The graph of an inverse relationship is a downward curve

  3. Boyle’s Law – Sample problem • A high altitude balloon contains 30.0 L of helium gas at 103 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant.) • 1. AnalyzeList the knowns and unknown • P1 = 103kPa V1 = 30.0 L P2 = 25.0 kPa • Unknown: V2 = ? L • Use the known values and Boyle’s Law expression to calculate the unknown value

  4. Boyle’s Law – Sample problem • 2. CalculateSolve for the unknown. • Rearrange the expression for Boyle’s Law to isolate V2. • V1 x P1 • V2 = P2 • Substitute the known values for P1, V1, and P2 into the equation and solve. V2 = 30.0 L X 103 kPa 25.0 kPa V2 = 1.24 x 102 L

  5. Boyle’s Law – Sample problem • 3. EvaluateDoes the result make sense? Using kinetic theory, a decrease in pressure at constant temperature must correspond to a proportional increase in volume. The calculated result agrees with both the kinetic theory and the pressure-temperature relationship. Also, the units have canceled correctly and the answer is expressed to the proper number of significant figures.

  6. The Temperature-Volume Relationship (Charles’s Law)French Physicist & balloonist: Jacques Charles (1746-1823) The volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant. V1 = V2 T1 T2 The graph of a relationship such as Charles’s law that is a direct proportion, is a straight line.

  7. Charles’s Law – Sample • A balloon inflated in a room at 24 0C has a volume of 4.00 L. The balloon is then heated to a temperature of 58 0C. What is the new volume if the pressure remains constant? • 1. AnalyzeList the knowns and unknown • V1 = 4.00 L T1 = 24 0C T2 = 58 0C • Unknown: V2 = ? L • Use the known values and Charles’s Law expression to calculate the unknown value

  8. Charles’s Law - Sample • 2. CalculateSolve for the unknown. Because the gas laws will be applied, express the temperatures in kelvins. T1 = 24 0C + 273 = 297 K T2 = 58 0C + 273 = 331 K Rearrange the expression for Charles’s law to isolate V2. V2 = V1 x T2 T1

  9. Charles’s Law - Sample Step 2 continued……… Substitute the known values for T1, V1, and T2 into the equation and solve V2 = 4.00 L x 331 K 297 K = 4.46 L

  10. Charles’s Law - Sample • 3. EvaluateDoes the result make sense? From kinetic theory, the volume should increase with an increase in temperature (at constant pressure). This result agrees with the kinetic theory and Charles’s law. The volume does increase with increasing temperature.

  11. The Temperature-Pressure Relationship (Gay-Lussac’s Law)French Chemist: Joseph Gay-Lussac (1778-1850) The pressure of a gas is directly proportional to the Kelvin temperature if the volume remains constant. P1 = P2 T1 T2

  12. Gay-Lussac’s Law - Sample The gas left in a used aerosol can is at a pressure of 103 kPa at 25 0C. If this can is thrown into a fire, what is the pressure of the gas when its temperature reaches 928 0C? (Calculating the answer to this problem will show you why it is dangerous to dispose of aerosol cans in a fire. Most aerosol cans carry warnings on their labels that clearly say not to incinerate (burn) or to store above a certain temperature.)

  13. Gay-Lussac’s Law - Sample 1. AnalyzeList the knowns and unknown Known: P1 = 103 kPa, T1 = 25 0C, T2 = 928 0C Unknown: P2 = ? kPa Use the known values and Gay-Lussac’s law expression to calculate the unknown. Remember, because this problem involves temperatures and a gas law, the temperatures must be expressed in kelvins.

  14. Gay-Lussac’s Law - Sample • 2. CalculateSolve for the unknown. First convert degrees Celsius to kelvins. T1 = 25 0C + 273 = 298 K T2 = 928 0C + 273 = 1201 K Rearrange Gay-Lussac’s law to isolate P2. P2 = P1 x T2 T1

  15. Gay-Lussac’s Law - Sample • Step 2 continued……….. Substitute the known values for P1, T2, and T1 into the equation and solve P2 = 103 kPa x 1201 K 298 K = 4.15 x 102 kPa

  16. Gay-Lussac’s Law - Sample • 3. EvaluateDoes the result make sense? From the kinetic theory, one would expect the increase in temperature of a gas to produce an increase in pressure if the volume remains constant. The calculated value does show such an increase.

  17. The Combined Gas Law • A single expression (the combined gas law) combines the three gas laws as follows: P1 x V1 = P2 x V2 T1 T2 This expression provides a useful way for you to do calculations in which none of the variables are constant.

  18. Combined Gas Law - Sample • The volume of a gas-filled balloon is 30.0 L at 40 0C and 153 kPa. What volume will the balloon have at standard temperature (273 K) and pressure (standard pressure =101.3 kPa) STP?

  19. Combined Gas Law - Sample 1. AnalyzeList the knowns and unknown Known: P1 = 153 kPa, T1 = 40 0C, V1 = 30.0 L P2, = 101.3 kPa, T2 = 273 K Unknown: V2 = ? L Use the known values and the combined gas law expression to calculate the unknown. Remember, because this problem involves temperatures and a gas law, the temperatures must be expressed in kelvin units.

  20. Combined Gas Law - Sample • 2. CalculateSolve for the unknown. First convert degrees Celsius to kelvins. T1 = 40 0C + 273 = 313 K T2 = standard temperature in K (given) Rearrange the combined gas law to isolate V2. V2 = V1 x P1 x T2 P2 x T1

  21. Ideal Gas Law - Sample • You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas (N2)(g) to a final pressure of 2.00 x 104 kPa at 28 0C. How many moles of N2(g) does the cylinder contain?

  22. Combined Gas Law - Sample Step 2 continued……… Substitute the known values into the equation and solve V2 = 30.0 L x 153 kPa x 273 K 101.3 kPa x 313 K = 39.5 L

  23. Combined Gas Law - Sample • 3. EvaluateDoes the result make sense? Temperature decreases; therefore, the temperature ratio is less than 1 (273K/313K). The pressure decreases, so the pressure ratio is greater than 1 (153 kPa/101.3 kPa). Recalculate by multiplying the initial volume of gas by these two ratios, and you will see the result is the same.

  24. Assignment • Calculations using Gas Law formulas • Page 335 # 10, 11 • Page 337 # 12, 13 • Page 338 # 14 • Page 339 # 15 • Page 340 # 16 , 17 • SHOW YOUR WORK, with units

  25. The Ideal Gas Law This law has an advantage over the combined gas law – it permits you to solve for the number of moles of a contained gas when P, V, and T are known. R = P x V P1 x V1= P2 x V2 T x n T1 x n1 T2 x n2 or…. P x V = n x R x T

  26. Ideal gas law values When T, P, and V of a gas are known, you can use the ideal gas law to calculate the number of moles of the gas. • n = number of moles • R = 8.31 (L x kPa) / (K x mol) • (R is the ideal gas law constant) • P = pressure • V = volume • T = temperature • STP = 22.4 L / I mole

  27. The Ideal Gas Law - sample 1. AnalyzeList the knowns and unknown Known: P = 2.00 x 104 kPa T = 28 0C V = 20.0 L Unknown: n = ? Moles of N2(g)

  28. The Ideal Gas Law - sample 2. CalculateSolve for the unknown. First – convert temperature to kelvin units 28 0C + 273 = 301 K Rearrange the ideal gas law to isolate n n = P x V R x T

  29. The Ideal Gas Law - Sample • Step 2 continued…………… Substitute the known quantities, P, V, R, and T into the equation and solve n = (2.00 x 104 kPa) x 20.0 L L x kPa 8.31 K x mol x 301 K n = 160 mol N2(g)

  30. The Ideal Gas Law - sample • 3. EvaluateDoes the result make sense? The gas is at a high pressure, but the volume is not large. This means that a large number of moles of gas must be compressed into the volume. The large answer is thus reasonable, and the units have canceled correctly.

  31. Avogadro's hypothesis • Equal volumes of gases at the same temperature and pressure contain equal numbers of particles.

  32. Avogadro’s - sample • Determine the volume (in liters) occupied by 0.202 mol of a gas at standard temperature and pressure (STP).

  33. Avogadro’s - sample AnalyzeList the knowns and unknown Known: n = 0.202 mol Unknown: V = ? L The number of moles of a gas at STP is a constant value. Convert from moles to the volume (V). moles  volume STP Conversion factor: 22.4 L / 1mol

  34. Avogadro’s - Sample 2. CalculateSolve for the unknown. Multiplying the known value by the conversion factor yields: V = 0.202 mol X 22.4 L 1 mol V = 4.52 L

  35. Avogadro’s - sample 3. EvaluateDoes the result make sense? Because 1 mol of the gas occupies 22.4 L at STP, 0.202 mol of the gas should occupy about one-fifth of that volume, or about 4.5 L.

  36. Dalton’s Law – Partial Pressure At constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases. Ptotal = P1 + P2 + P3 + ….. Pn

  37. Dalton’s Law - sample Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2) at 101.3 kPa of total pressure if the partial pressures of nitrogen, carbon dioxide, an other gases are 79.10 kPa, 0.040 kPa, and 0.94 kPa respectively.

  38. Dalton’s Law - sample 1. AnalyzeList the knowns and unknown Known: PN2 = 79.10 kPa PCO2 = 0.040 kPa Pothers = 0.94 kPa Ptotal = 101.30 kPa Unknown: PO2 = ? kPa Use the known values and Dalton’s law of partial pressures to calculate the unknown value.

  39. Dalton’s Law - sample 2. CalculateSolve for the unknown. Rearrange the expressions for Dalton’s Law to isolate PO2. Substitute the values for the partial pressure and solve the equation. PO2 = Ptotal – (PN2 + PCO2 + Pothers) =101.30 kPa – (79.10 kPa + 0.040 kPa + 0.94 kPa) = 21.22 kPa of Oxygen

  40. Dalton’s Law - sample 3. EvaluateDoes the result make sense? The partial pressure of oxygen must be smaller than that of nitrogen because Ptotal is only 101.30 kPa. The other partial pressures are small, so an answer of 21.22 kPa seems reasonable.

  41. Assignment • “Gas Laws – Math 12-5” • Page 342 - # 22 • Page 348 - # 31, 33 • Page 351 - # 37 • SHOW your work and units in the answers

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