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Gas Laws: Pressure, Volume, and Temperature Relations

Discover the fundamental principles of the gas laws - Boyle’s Law, Charles' Law, Avogadro’s Law, Gay-Lussac’s Law, and the Ideal Gas Law. Learn how pressure, volume, and temperature are interconnected in the behavior of gases, and master the calculations to predict gas properties accurately.

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Gas Laws: Pressure, Volume, and Temperature Relations

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  1. Chapter 5 The Gas Laws pp

  2. 5.1 Pressure • Force per unit area. • Gas molecules fill container. • Molecules move around and hit sides. • Collisions are the force. • Container has the area. • Measured with a barometer.

  3. Barometer Vacuum • The pressure of the atmosphere at sea level will hold a column of mercury 760 mm Hg. • 1 atm = 760 mm Hg 760 mm Hg 1 atm Pressure

  4. Manometer • Column of mercury to measure pressure. • h is how much lower the pressure is than outside. h Gas

  5. Manometer • h is how much higher the gas pressure is than the atmosphere. h Gas

  6. Units of pressure • 1 atmosphere = 760 mm Hg • 1 mm Hg = 1 torr • 1 atm = 101,325 Pascals = 101.325 kPa • Can make conversion factors from these. • What is 724 mm Hg in kPa? . . . • 96.5 kPa • In torr? • 724 torr • In atm? • 0.952 atm.

  7. 5.2 The Gas Laws of Boyle, Charles, and Avogadro • Boyle’s Law • Pressure and volume are inversely related at constant temperature. • PV= k • As one goes up, the other goes down. • P1V1 = P2 V2 • Graphically

  8. 1 atm • As the pressure on a gas increases 4 Liters

  9. As the pressure on a gas increases the volume decreases • Pressure and volume are inversely related 2 atm 2 Liters

  10. A Boyle’s Law Relationship

  11. V P (at constant T)

  12. Slope = k V 1/P (at constant T)

  13. Ne 22.41 L atm O2 PV CO2 P (at constant T)

  14. Example pp • 20.5 L of nitrogen at 25.0ºC and 742 torr are compressed to 9.80 atm at constant T. What is the new volume? Steps . . . • P1V1 = P2 V2 • (20.5 L)(742 torr) = (9.80 atm x 760 torr/atm)(x) • x = 2.04 L

  15. You try it! • 30.6 mL of CO2 at 740 torr is expanded at constant temperature to 750 mL. What is the final pressure in kPa? Ans. . . • 4.02 kPa. Steps . . . • P1V1 = P2 V2 • (30.6 mL)(740 torr) =(750 mL) (x torr) • x = 30.2 torr • (30 torr)(101.325 kPa/ 760 torr) = 4.02 kPa

  16. Charles’ Law • Volume of a gas varies directly with the absolute temperature at constant pressure. • V = kT (if T is in Kelvin) • V1 = V2 T1 = T2 • Graphically

  17. Tr 51A Charles’ Law Relationship

  18. He CH4 H2O V (L) H2 -273.15ºC T (ºC)

  19. Example pp • What would the final volume be if 247 mL of gas at 22ºC is heated to 98ºC , if the pressure is held constant? Steps. • V1/T1 = V2/T2 • (247 mL)/(22 + 273) K = (x mL)/(273 + 98) K • x = 311 ml

  20. Examples (do this) • At what temperature would 40.5 L of gas at 23.4ºC have a volume of 81.0 L at constant pressure? a) 173 ºC b) 280 ºC c) 320 ºC d) 593 ºC • 320. ºC (change K to ºC!)

  21. Avogadro's Law • At constant temperature and pressure, the volume of gas is directly related to the number of moles. • V = k n (n is the number of moles) • V1 = V2 n1 = n2

  22. Figure 5.10 p. 195Balloons Holding 1.0 L of Gas at 25º C and 1 atm. Each contains 0.041 mol of gas, or 2.5 x 1022 molecules, even though their masses are different (equal numbers, different masses).

  23. Gay- Lussac Law • At constant volume, pressure and absolute temperature are directly related. • P = k T • P1 = P2 T1 = T2

  24. Tr 52A Gay-Lussac’s Law Relationship

  25. Combined Gas Law pp • If the moles of gas remain constant, use this formula and cancel out the other things that don’t change. • P1 V1 = P2 V2. T1 T2

  26. Examples pp • A deodorant can has volume of 175 mL & pressure of 3.8 atm at 22ºC. What pressure if heated to 100.ºC? • V is constant, so (P1V1)/T1 = (P2V2/T2) • (3.8)/(295) = (x)/373 • x = 4.8 atm

  27. Examples • A deodorant can has volume of 175 mL & pressure of 3.8 atm at 22ºC. What volume of gas could the can release at 22ºC and 743 torr? Answer . . . • 680 mLSteps. . . • (3.8)(175)/(273 + 22) = (743/760)(x)/(273 + 22)

  28. 5.3 Ideal Gas Law pp • PV = nRT • V = 22.41 L at 1 atm, 0ºC, n = 1 mole, what is R? • R is the ideal gas constant. • R = 0.08206 L atm/ mol K • R also = 62.396 L torr/k mol or 8.3145 J/k mol • Tells you about a gas NOW. • The other laws tell you about a gas when it changes.

  29. Ideal Gas Law • An equation of state. • Independent of how you end up where you are at. Does not depend on the path. • Given 3 measurements you can determine the fourth. • An Empirical Equation - based on experimental evidence.

  30. Ideal Gas Law • A hypothetical substance - the ideal gas • Think of it as a limit. • Gases only approach ideal behavior at low pressure (< 1 atm) and high temperature. • Use the laws anyway, unless told to do otherwise. • They give good estimates.

  31. Examples pp • A 47.3 L container with 1.62 mol of He heated until pressure reaches 1.85 atm. What is the temperature? Steps . . . • PV = nRT • T = (1.85)(47.3)/(.0821)(1.62)=658 K or 385 ºC

  32. Examples pp • Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC. What is the mass of Kr? Ans. . . • 546 g Steps . . . • n = PV/RT = 6.51 mol • Since M = m/mol = m/n, m = M • n • Get M from periodic table • m = 83.8 • 6.51 = 546 g

  33. Examples • A gas is 4.18 L at 29 ºC & 732 torr. What volume at 24.8ºC & 756 torr? Ans. . . • Use combined law. Why? . . . • The conditions are changing • x = 3.99 L (can convert torr to atm when calculating, but pressure units cancel out so don’t need to for this problem).

  34. 5.4 Gases and Stoichiometry • Reactions happen in moles • At Standard Temperature and Pressure (STP, 0ºC and 1 atm) 1 mole of gas occupies 22.41 L. • If not at STP, use the ideal gas law to calculate moles of reactant or volume of product.

  35. Figure 5.11A Mole of Any Gas Occupies a Volume of Approximately 22.4 L at STP How many particles are in each box? 6.022 x 1023 particles

  36. Examples pp • Can get mercury by the following: What volume oxygen gas produced from 4.10 g of mercury (II) oxide (M = 217 g/mol) at STP? Steps. • Balance first! 2HgO  2Hg + O2 • Since at STP can then use stoich • 4.10 g HgO • 1 mol HgO/217g HgO • 1 mol O2/2 mol HgO • 22.4 L O2/1 mol O2 • = 0.212 L or 212 mL

  37. Examples pp • Can get mercury by the following: What volume oxygen gas can be produced from 4.10 g of mercury (II) oxide at 400.ºC & 740. torr? (Hint: conditions are changed, so what equation do you then use?) . . . • 1st, find molar mass of mercury(II) oxide. • 2nd, balance reactions & spider under STP • 3rd, then use the combined law to convert. • (1)(212)/273 = (740/760)(x)/(273 + 400) • 0.537 L

  38. Examples • Using the following reaction calculate the mass of sodium hydrogen carbonate (M = 84.01) that produces 2.87 L of carbon dioxide at 25ºC & 2.00 atm. • Not at STP so use PV=nRT; ans. . . • 19.7 g

  39. Examples • Using the following reaction • If 27 L of gas are produced at 26ºC and 745 torr when 2.6 L of HCl are added what is the concentration of HCl? • Use stoich to get moles HCl, then mole ratio  moles CO2, then moles/L = concentration. Or, use Chas law to correct to 24.2 L, then stoich to get 1.08 mol HCl = 1.08 mole CO2; so concentration is 0.415 M

  40. Examples • Consider the following reaction What volume NO at 1.0 atm & 1000ºC produced from 10.0 L of NH3 & xs O2 at same temperature & pressure? • T & P constant & cancel out so volume is proportional to moles = 10 L

  41. You try it! • Consider the following reaction What volume of O2 at STP will be consumed when 10.0 kg NH3 is reacted? (Use “rounded” molar masses). Ans. . . • watch Kg! Ans. Rounded to 16 500 L (from calc ans of 16 470)

  42. The Same reaction with some twists pp • What mass of H2O produced from 65.0 L O2 & 75.0 L of NH3 measured at STP? . . . Ans. • As before, but must find LR = O2 ans. 313 g • What volume of NO would be produced? • Above + mole ratio. Ans. = 52 L • What mass of NO is produced from 500. L of NH3 at 250.0ºC and 3.00 atm? Ans. • non-STP so use combined gas law to get to STP (783 L), then “spider” Ans. = 895 g

  43. Gas Density and Molar Mass • D = m/V • n= grams of gas/molar mass or m/molar mass • Let M stand for molar mass • M = m/n • n= PV/RT • M = m PV/RT • M = mRT = mR T = DRT PV V P P

  44. Finding M or D from ideal gas law • M = m (PV/RT) • M =m RT V P • Since m/V = density (D) • M = DRT = molar mass P • Density varies directly with M and P • Density varies indirectly with T

  45. Quick Example pp • What mass of ethene gas, C2H4, is in a 15.0 L tank at 4.40 atm & 305 K? (steps) • PV = mRT/M and M = 28 g/mol • m = PVM/RT • m = (4.40 atm)(15.0 L)(28 g/mol) (0.0821 L atm/mol K)(305 K) • m = 73.8 g C2H4

  46. Another way to solve it (quick) pp • What mass of ethene gas, C2H4, is in a 15.0 L tank at 4.40 atm & 305 K? (steps) • PV = nRT, solve for n • n = PV/RT • n = (4.40 atm)(15.0 L) (0.0821 L atm/mol K)(305 K) • n = 2.64 mol C2H4 • C2H4 (M = 28 g/mol) • Answer is 73.8 g C2H4

  47. Quick Example • A gas has a density of 3.36 g/L at 14 ºC and 1.09 atm. What is its molar mass? • Use M = DRT/P to get an answer of . . . • 72.6 g/mol. Calculation . . . • (3.36)(0.0821)(14 + 273)/(1.09) = 72.6 g/mol

  48. AP Examples • Density of ammonia at 23ºC & 735 torr? Ans. • D = MP/RT = 0.68 g/L

  49. AP Examples • A compound has empirical formula CHCl. A 256 mL flask at 100.ºC & 750 torr contains 0.80 g of the gaseous compound. What is the molecular formula? Steps. . . • What do you need to get the molecular formula? . . . • The molecular mass, M. • How do you get this? • M = g/mol. You are given the grams, so find the number of moles using . . . • PV = nRT and solve for n . . . Try it! . . .

  50. AP Examples cont. • EF of CHCl. 0.80 g are in 256 mL flask at 100.ºC & 750 torr. What is the MF? • n = 0.0083 from n = PV/RT • What’s M? . . . • M  97 from 0.80 g/0.0083 moles = g/mol • How do you find molecular formula from EF and MF? • Find how many Efs are in the MF (CHCl) then multiply the subscripts of the EF by that number. Try it! . . . • 97/49  2, so formula is C2H2Cl2

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