1. Gas Laws AP/IB Chem 1

2. Scheffler Properties of Gases Variable volume and shape Expand to occupy volume available Volume, Pressure, Temperature, and the number of moles present are interrelated Can be easily compressed Exert pressure on whatever surrounds them Easily diffuse into one another 2

3. Scheffler Mercury Barometer Used to define and measure atmospheric pressure On the average at sea level the column of mercury rises to a height of about 760 mm. This quantity is equal to 1 atmosphere It is also known as standard atmospheric pressure 3

4. Scheffler Barometer The mercury barometer was the basis for defining pressure, but it is difficult to use or to transport Furthermore Mercury is very toxic and seldom used anymore Most barometers are now aneroid barometers or electronic pressure sensors, 4

5. Scheffler Pressure Units & Conversions The above represent some of the more common units for measuring pressure. The standard SI unit is the Pascal or kilopascal. (kPa) The US Weather Bureaus commonly report atmospheric pressures in inches of mercury. Pounds per square inch or PSI is widely used in the United States. Most other countries use only the metric system. 5

6. Scheffler Standard Temperature and Pressure Standard Temperature and Pressure or STP = 0 o C or 273.15 Kelvin and a pressure of 1 atmosphere. This is used as a reference point when comparing quantities of gases Gases are seldom measured at exactly these conditions. We need to be able to compute the volume at various temperature and pressures

7. Scheffler = 340 kPa If the pressure of helium gas in a balloon has a volume of 4.00 dm 3 at 210 kPa, what will the pressure be at 2.50 dm 3 ? Note that dm 3 = Liter and cm 3 = mL also KPa is kilo Pascal unit of pressure P 1 V 1 = P 2 V 2 (see Boyle’s law ) (210 kPa) (4.00 dm 3 ) = P 2 (2.50 dm 3 ) P 2 = (210 kPa) (4.00 dm 3 ) (2.50 dm 3 ) Sample Problem 1: 7

8. Scheffler Boyle’s Law According to Boyle’s Law the pressure and volume of a gas are inversely proportional at constant pressure. PV = constant. P 1 V 1 = P 2 V 2 8

9. Scheffler Boyle’s Law A graph of pressure and volume gives an inverse function A graph of pressure and the reciprocal of volume gives a straight line 9

10. Scheffler Standard Temperature and Pressure (STP) The volume of a gas varies with temperature and pressure. Therefore it is helpful to have a convenient reference point at which to compare gases. For this purpose standard temperature and pressure are defined as: Temperature = 0 o C 273 K Pressure = 1 atmosphere = 760 torr = 101.3 kPa This point is often called STP 10

11. Scheffler Charles’ Law According to Charles’ Law the volume of a gas is proportional to the Kelvin temperature as long as the pressure is constant V = kT V 1 = T 1 V2 T2V2 T2 Note: The temperature for gas laws must always be expressed in Kelvin where Kelvin = o C +273.15 (or 273 to 3 significant digits) 11

12. Scheffler Charles’ Law A graph of temperature and volume yields a straight line. Where this line crosses the x axis (x intercept) is defined as absolute zero 12

13. Scheffler Sample Problem 2 A gas sample at 40 o C occupies a volume of 2.32 dm 3. If the temperature is increased to 75 o C, what will be the final volume? 2.58 dm 3 13 V 1 = V 2 T 1 T 2 Convert temperatures to Kelvin. 40 o C = 313K 75 o C = 348K 2.32 dm 3 = V 2 313 K 348K (313K)( V 2 ) = (2.32 dm 3 ) (348K) V 2 =

14. Scheffler Gay-Lussac’s Law Gay-Lussac’s Law defines the relationship between pressure and temperature of a gas. The pressure and temperature of a gas are directly proportional P 1 = P 2 T1T1 T2T2 14

15. Scheffler Sample Problem 3: The pressure of a gas in a tank is 3.20 atm at 22 o C. If the temperature rises to 60 o C, what will be the pressure in the tank? 3.6 atm 15 P 1 = P 2 T 1 T 2 Convert temperatures to Kelvin. 22 o C = 295K 60 o C = 333K 3.20 atm = P 2 295 K 333K (295K)( P 2 ) = (3.20 atm)(333K) P 2 =

16. Scheffler The Combined Gas Law 1. If the amount of the gas is constant, then Boyle’s Charles’ and Gay-Lussac’s Laws can be combined into one relationship 2. P 1 V 1 = P 2 V 2 T2T2 T1T1 16

17. Scheffler Sample Problem 4: A gas at 110 kPa and 30 o C fills a container at 2.0 dm 3. If the temperature rises to 80 o C and the pressure increases to 440 kPa, what is the new volume? V 2 = 0.58 dm 3 17 P 1 V 1 = P 2 V 2 T 1 T 2 Convert temperatures to Kelvin. 30 o C = 303K 80 o C = 353K V 2 = V 1 P 1 T 2 P 2 T 1 = (2.0 dm 3 ) (110 kPa ) (353K) (440 kPa ) (303 K)

18. Scheffler Advogadro’s Law Equal volumes of a gas under the same temperature and pressure contain the same number of particles. If the temperature and pressure are constant the volume of a gas is proportional to the number of moles of gas present V = constant * n where n is the number of moles of gas V/n = constant V 1 /n 1 = constant = V 2 /n 2 V 1 /n 1 = V 2 /n 2 18

19. Scheffler Universal Gas Equation Based on the previous laws there are four factors that define the quantity of gas: Volume, Pressure, Kelvin Temperature, and the number of moles of gas present (n). Putting these all together: PV nT = Constant = R The proportionality constant R is known as the universal gas constant 19

20. Scheffler Universal Gas Equation The Universal gas equation is usually written as PV = nRT Where P = pressure V = volume T = Kelvin Temperature n = number of moles The numerical value of R depends on the pressure unit (and perhaps the energy unit) Some common values of R include: R = 62.36 dm 3 torr mol -1 K -1 = 0.0821 dm 3 atm mol -1 K -1 = 8.314 dm 3 kPa mol -1 K-1 20

21. Scheffler Sample Problem 5 Example: What volume will 25.0 g O 2 occupy at 20 o C and a pressure of 0.880 atmospheres? : V = (0.781 mol)(0.08205 dm -3 atm mol -1 K -1 )(293K) 0.880 atm V = 21.3 dm 3 (25.0 g) n = ----------------- = 0.781 mol (32.0 g mol -1 ) V =? P = 0.880 atm; T = (20 + 273)K = 293K R = 0.08205 dm -3 atm mol -1 K -1 PV = nRT so V = nRT/P Data Formula Calculation Answer 21

22. Scheffler Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L Universal Gas Equation –Alternate Forms 22

23. Scheffler A 2.10 dm 3 vessel contains 4.65 g of a gas at 1.00 atmospheres and 27.0 o C. What is the molar mass of the gas? dRT P M = d = m V 4.65 g 2.10 dm 3 = = 2.21 g dm 3 M = 2.21 g dm 3 1 atm x 0.0821 x 300.15 K dm 3 atm molK M = 54.6 g/mol Sample Problem 6 23

24. Scheffler Dalton’s Law of Partial Pressures The total pressure of a mixture of gases is equal to the sum of the pressures of the individual gases (partial pressures). P T = P 1 + P 2 + P 3 + P 4 +.... whereP T = total pressure P 1 = partial pressure of gas 1 P 2 = partial pressure of gas 2 P 3 = partial pressure of gas 3 P 4 = partial pressure of gas 4 24

25. Scheffler Dalton’s Law of Partial Pressures Applies to a mixture of gases Very useful correction when collecting gases over water since they inevitably contain some water vapor. 25

26. Scheffler Sample Problem 7 Henrietta Minkelspurg generates Hydrogen gas and collected it over water. If the volume of the gas is 250 cm 3 and the barometric pressure is 765.0 torr at 25 o C, what is the pressure of the “dry” hydrogen gas at STP? (P H2O = 23.8 torr at 25 o C) 26

27. Scheffler Sample Problem 8 Henrietta Minkelspurg generated Hydrogen gas and collects it over water. If the volume of the gas is 250 cm 3 and the barometric pressure is 765 torr at 25 o C, what is the volume of the “dry” hydrogen gas at STP? From the previous calculation the adjusted pressure is 742.2 torr V 2 = (250 cm3)(742.2 torr)(273K) (298K)(760.torr) V 2 = 223.7 cm 3 P 1 = P H2 = 742.2 torr; P 2 = Std Pressure = 760 torr T 1 = 298K; T 2 = 273K; V 1 = 250 cm 3 ; V 2 = ? (V 1 P 1 /T 1 ) = (V 2 P 2 /T 2 ) therefore V 2 = (V 1 P 1 T 2 )/(T 1 P 2 ) 27

28. Scheffler Kinetic Molecular Theory Matter consists of particles (atoms or molecules) that are in continuous, random, rapid motion The Volume occupied by the particles has a negligibly small effect on their behavior Collisions between particles are elastic Attractive forces between particles have a negligible effect on their behavior Gases have no fixed volume or shape, but take the volume and shape of the container The average kinetic energy of the particles is proportional to their Kelvin temperature 28

29. Scheffler Maxwell-Boltzman Distribution Molecules are in constant motion Not all particles have the same energy The average kinetic energy is related to the temperature An increase in temperature spreads out the distribution and the mean speed is shifted upward 29

30. Scheffler The distribution of speeds for nitrogen gas molecules at three different temperatures The distribution of speeds of three different gases at the same temperature u rms = 3RT M  Velocity of a Gas 30

31. Scheffler Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. NH 3 17.0 g/mol HCl 36.5 g/mol NH 4 Cl Diffusion 31

32. Scheffler DIFFUSION AND EFFUSION Diffusion is the gradual mixing of molecules of different gases. Diffusion is the gradual mixing of molecules of different gases. Effusion is the movement of molecules through a small hole into an empty container. Effusion is the movement of molecules through a small hole into an empty container. 32

33. Scheffler Graham’s Law Graham’s law governs effusion and diffusion of gas molecules. Graham’s law governs effusion and diffusion of gas molecules. KE=1/2 mv 2 Thomas Graham, 1805-1869. Professor in Glasgow and London. The rate of effusion is inversely proportional to its molar mass. 33

34. Scheffler Sample Problem 9 1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850 o C according to the equation: 4 NH 3 (g) + 5 O 2 (g)  4 NO(g) + 6 H 2 O(g) Using Graham's Law, what is the ratio of the effusion rates of NH 3 (g) to O 2 (g)? 34

35. Scheffler Sample Problem 10 What is the rate of effusion for H 2 if 15.00 cm 3 of CO 2 takes 4.55 sec to effuse out of a container? Rate for CO 2 = 15.00 cm 3 /4.55 s = 3.30 cm 3 /s 35

36. Scheffler Sample Problem 11 What is the molar mass of gas X if it effuses 0.876 times as rapidly as N 2 (g)? 36

37. Scheffler Ideal Gases v Real Gases Ideal gases are gases that obey the Kinetic Molecular Theory perfectly. The gas laws apply to ideal gases, but in reality there is no perfectly ideal gas. Under normal conditions of temperature and pressure many real gases approximate ideal gases. Under more extreme conditions more polar gases show deviations from ideal behavior. 37

38. Scheffler In an Ideal Gas --- The particles (atoms or molecules) in continuous, random, rapid motion. The particles collide with no loss of momentum The volume occupied by the particles is essentially zero when compared to the volume of the container The particles are neither attracted to each other nor repelled The average kinetic energy of the particles is proportional to their Kelvin temperature At normal temperatures and pressures gases closely approximate idea behavior 38

39. Scheffler Real Gases These deviations occur because Real gases do not actually have zero volume Polar gas particles do attract if compressed For ideal gases the product of pressure and volume is constant. Real gases deviate somewhat as shown by the graph pressure vs. the ratio of observed volume to ideal volume below. 39

40. Scheffler van der Waals Equation (P + n 2 a/V 2 )(V - nb) = nRT The van der Waals equation shown below includes corrections added to the universal gas law to account for these deviations from ideal behavior wherea => attractive forces between molecules b => residual volume or molecules The van der Waals constants for some elements are shown below Substance a (dm 6 atm mol -2 ) b (dm 3 mol -1 ) He0.03410.02370 CH 4 2.250.0428 H 2 O5.460.0305 CO 2 3.590.0437 40

41. Scheffler Sample Problem 12 What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 0.0821 x 310.15 K dm 3 atm molK 1.00 atm = = 4.76 dm 3 41