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Chapter 5 The Gas Laws. Pressure Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions.

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Presentation on theme: "Chapter 5 The Gas Laws. Pressure Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions."— Presentation transcript:

1 Chapter 5 The Gas Laws

2 Pressure

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

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

5 Manometer n Column of mercury to measure pressure. n h is how much lower the pressure is than outside.

6 Manometer n h is how much higher the gas pressure is than the atmosphere. h Gas

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

8 Gas Laws

9 Boyle’s Law n Pressure and volume are inversely related at constant temperature. n PV= k n As one goes up, the other goes down. n P 1 V 1 = P 2 V 2

10 V P (at constant T)

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

12 PV P (at constant T) CO 2 O2O2 22.41 L atm

13 Examples n 20.5 L of nitrogen at 25ºC and 742 torr are compressed to 9.8 atm at constant T. What is the new volume?

14 Charle’s Law n Volume of a gas varies directly with the absolute temperature at constant pressure. n V = kT (if T is in Kelvin) n V 1 = V 2 T 1 = T 2

15 V (L) T (ºC) He H2OH2O CH 4 H2H2 -273.15ºC

16 Examples n What would the final volume be if 247 mL of gas at 22ºC is heated to 98ºC, if the pressure is held constant?

17 Avogadro's Law n At constant temperature and pressure, the volume of gas is directly related to the number of moles. n V = k n (n is the number of moles) n V 1 = V 2 n 1 n 2

18 Gay- Lussac Law n At constant volume, pressure and absolute temperature are directly related. n P = k T n P 1 = P 2 T 1 = T 2

19 Examples n A deodorant can has a volume of 175 mL and a pressure of 3.8 atm at 22ºC. What would the pressure be if the can was heated to 100.ºC?

20 Ideal Gas

21 Ideal Gas Law n PV = nRT n V = 22.41 L at 1 atm, 0ºC, n = 1 mole, what is R? n R is the ideal gas constant. n R = 0.08206 L atm/ mol K n Tells you about a gas NOW. n The other laws tell you about a gas when it changes.

22 Gas Constant

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

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

25 Examples n A 47.3 L container containing 1.62 mol of He is heated until the pressure reaches 1.85 atm. What is the temperature?

26 n Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC What is the mass of Kr?

27 n A sample of gas has a volume of 4.18 L at 29ºC and 732 torr. What would its volume be at 24.8ºC and 756 torr?

28 Gas Density and Molar Mass n D = m/V Let M stand for molar mass Let M stand for molar mass M = m/n M = m/n n n= PV/RT M = m PV/RT M = m PV/RT M = mRT = m RT = DRT PV V P P M = mRT = m RT = DRT PV V P P

29 Examples n What is the density of ammonia at 23ºC and 735 torr?

30 n A compound has the empirical formula CHCl. A 256 mL flask at 100.ºC and 750 torr contains.80 g of the gaseous compound. What is the molecular formula?

31 Gases and Stoichiometry n Reactions happen in moles n At Standard Temperature and Pressure (STP, 0ºC and 1 atm) 1 mole of gas occuppies 22.42 L. n If not at STP, use the ideal gas law to calculate moles of reactant or volume of product.

32 Examples n Mercury can be achieved by the following reaction n What volume of oxygen gas can be produced from 4.10 g of mercury (II) oxide at STP?

33 Examples n Using the following reaction n calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L of carbon dioxide at 25ºC and 2.00 atm.

34 n 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?

35 Examples n Consider the following reaction What volume of NO at 1.0 atm and 1000ºC can be produced from 10.0 L of NH 3 and excess O 2 at the same temperature and pressure?

36 n What volume of O 2 measured at STP will be consumed when 10.0 kg NH 3 is reacted?

37 The Same reaction n What mass of H 2 O will be produced from 65.0 L of O 2 and 75.0 L of NH 3 both measured at STP?

38 n What mass of NO is produced from 500. L of NH 3 at 250.0ºC and 3.00 atm?

39 Partial Pressure

40 Dalton’s Law n The total pressure in a container is the sum of the pressure each gas would exert if it were alone in the container. n The total pressure is the sum of the partial pressures. n P Total = P 1 + P 2 + P 3 + P 4 + P 5... n For each P = nRT/V

41 Dalton's Law n P Total = n 1 RT + n 2 RT + n 3 RT +... V V V n In the same container R, T and V are the same. n P Total = (n 1 + n 2 + n 3 +...)RT V n P Total = (n Total )RT V

42 The mole fraction n Ratio of moles of the substance to the total moles. symbol is Greek letter chi  symbol is Greek letter chi     = n 1 = P 1 n Total P Total    = n 1 = P 1 n Total P Total

43 Examples n The partial pressure of nitrogen in air is 592 torr. Air pressure is 752 torr, what is the mole fraction of nitrogen?

44 n What is the partial pressure of nitrogen if the container holding the air is compressed to 5.25 atm?

45 Examples 3.50 L O 2 1.50 L N 2 2.70 atm n When these valves are opened, what is each partial pressure and the total pressure? 4.00 L CH 4 4.58 atm 0.752 atm

46 Vapor Pressure n Water evaporates! n When that water evaporates, the vapor has a pressure. n Gases are often collected over water so the vapor pressure of water must be subtracted from the total pressure. n It must be given.

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48 Example n N 2 O can be produced by the following reaction what volume of N 2 O collected over water at a total pressure of 94 kPa and 22ºC can be produced from 2.6 g of NH 4 NO 3 ? ( the vapor pressure of water at 22ºC is 21 torr)

49 Kinetic Molecular Theory (KMT)

50 Kinetic Molecular Theory n Theory tells why the things happen. n explains why ideal gases behave the way they do. n Assumptions that simplify the theory.  The particles are so small compared to distance between them, we can ignore their volume.  The particles are in constant motion and their collisions cause pressure.

51 Kinetic Molecular Theory  The particles do not affect each other, neither attracting or repelling.  The average kinetic energy is proportional to the Kelvin temperature. n Appendix 2 shows the derivation of the ideal gas law and the definition of temperature. n We need the formula KE = 1/2 mv 2

52 Root Mean Squared Velocity n (KE) avg = N A (1/2 mu 2 ) n (KE) avg = 3/2 RT Where M is the molar mass in kg/mole, and R has the units 8.3145 J/Kmol. n The velocity will be in m/s

53 Example n Calculate the root mean square velocity of carbon dioxide at 25ºC. n Calculate the root mean square velocity of hydrogen at 25ºC. n Calculate the root mean square velocity of chlorine at 25ºC.

54 Range of velocities n The average distance a molecule travels before colliding with another is called the mean free path and is small (near 10 -7 ) n Temperature is an average. There are molecules of many speeds in the average. n Shown on a graph called a velocity distribution

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57 Velocity n Average increases as temperature increases. n Spread increases as temperature increases.

58 Effusion n Passage of gas through a small hole, into a vacuum. n The effusion rate measures how fast this happens. n Graham’s Law the rate of effusion is inversely proportional to the square root of the mass of its particles.

59 Effusion n Passage of gas through a small hole, into a vacuum. n The effusion rate measures how fast this happens. n Graham’s Law the rate of effusion is inversely proportional to the square root of the mass of its particles.

60 Examples n A compound effuses through a porous cylinder 3.20 time faster than helium. What is it’s molar mass?

61 n If 0.00251 mol of NH 3 effuse through a hole in 2.47 min, how much HCl would effuse in the same time?

62 n A sample of N 2 effuses through a hole in 38 seconds. what must be the molecular weight of gas that effuses in 55 seconds under identical conditions?

63 Diffusion n The spreading of a gas through a room. n Slow considering molecules move at 100’s of meters per second. n Collisions with other molecules slow down diffusions. n Best estimate is Graham’s Law.

64 Real Gases n Real molecules do take up space and they do interact with each other (especially polar molecules). n Need to add correction factors to the ideal gas law to account for these.

65 Volume Correction n The actual volume free to move in is less because of particle size. n More molecules will have more effect. n Corrected volume V’ = V - nb n b is a constant that differs for each gas. n P’ = nRT (V-nb)

66 Pressure correction n Because the molecules are attracted to each other, the pressure on the container will be less than ideal n depends on the number of molecules per liter. n since two molecules interact, the effect must be squared.

67 Pressure correction n Because the molecules are attracted to each other, the pressure on the container will be less than ideal n depends on the number of molecules per liter. n since two molecules interact, the effect must be squared. P observed = P’ - a 2 () V n

68 Altogether n P obs = nRT - a n 2 V-nb V n Called the Van der Wall’s equation if rearranged n Corrected Corrected Pressure Volume ()

69 Where does it come from n a and b are determined by experiment. n Different for each gas. n Bigger molecules have larger b. n a depends on both size and polarity. n once given, plug and chug.

70 Example n Calculate the pressure exerted by 0.5000 mol Cl 2 in a 1.000 L container at 25.0ºC n Using the ideal gas law. n Van der Waal’s equation –a = 6.49 atm L 2 /mol 2 –b = 0.0562 L/mol


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