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1 Unit 2- Simple Particle Energy and States The Gas Laws.

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Presentation on theme: "1 Unit 2- Simple Particle Energy and States The Gas Laws."— Presentation transcript:

1 1 Unit 2- Simple Particle Energy and States The Gas Laws

2 2 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.

3 3 Atmospheric Pressure n results from the gravitational pull on the particles in atmosphere n decreases w/ increased elevation b/c air around earth thins out as elevation increases n barometers- measure atmospheric pressure

4 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 5 Manometer Gas h n Column of mercury to measure pressure. n h is how much lower the pressure is than outside. n when gas pressure is less than atmospheric pressure, mercury is pushed toward the gas reservoir. n What is doing the “pushing?”

6 6 Manometer n When the gas pressure is greater than atmospheric pressure, the mercury is pushed toward the open end. n h is how much higher the gas pressure is than the atmosphere. h Gas

7 7 Manometers n The difference between the heights of the mercury on each side of the tube is a measure of the pressure of the gas. n when the atmospheric pressure is greater the difference is subtracted from the atmospheric pressure to calculate gas pressure.(Pr) n when the atmospheric pressure is less than the gas pressure the difference is added to the atomospheric pressure to calculate gas pressure.

8 8 Manometer Problems

9 9 Units of pressure n 1 atmosphere = 760 mm Hg n 1 mm Hg = 1 torr n 1 atm = 101,235 Pascals = 101.325 kPa n Can make conversion factors from these.

10 10 Pressure Conversions Sample Problems: 1. What is 724 mm Hg in kPa? 724 mmHg x _________ = _________kPa n in torr? 724 mm Hg x _________= _________ torr n in atm? 724 mmHg x __________ = _______ atm

11 11 Pressure Conversions 1. What is 1.5 atm ‘s in kPa’s? 2. What is 115.2 kPa in mmHg?

12 12 The Gas Laws n 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 n Graphically

13 13 V P (at constant T)

14 14 V 1/P (at constant T) Slope = k

15 15 PV P (at constant T) CO 2 O2O2 22.41 L atm

16 16 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? PVTn InitialFinalEffect

17 17 Example n 30.6 mL of carbon dioxide at 740 torr is expanded at constant temperature to 750 mL. What is the final pressure in kPa?

18 18 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 n Graphically

19 19 V (L) T (ºC) He H2OH2O CH 4 H2H2 -273.15ºC

20 20 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?

21 21 Examples n At what temperature would 40.5 L of gas at 23.4ºC have a volume of 81.0 L at constant pressure?

22 22 Avogadro's Law n Avagadro’s 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

23 23 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

24 24 Combined Gas Law n If the moles of gas remains constant, use this formula and cancel out the other things that don’t change. n P 1 V 1 = P 2 V 2. T 1 T 2

25 25 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? n What volume of gas could the can release at 22ºC and 743 torr?

26 26 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 = 0.08306 L atm/ mol K n Tells you about a gas is NOW. n The other laws tell you about a gas when it changes.

27 27 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.

28 28 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.

29 29 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? 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? 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?

30 30 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

31 31 Examples n What is the density of ammonia at 23ºC and 735 torr? 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 empirical formula?

32 32 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.

33 33 Examples n Mercury can be achieved by the following reaction What volume of oxygen gas can be produced from 4.10 g of mercury (II) oxide at STP? n At 400.ºC and 740 torr?

34 34 Examples n Using the following reaction calaculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L of carbon dioxide at 25ºC and 2.00 atm. 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 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 temperture and pressure? n What volume of O 2 measured at STP will be consumed when 10.0 kg NH 3 is reacted?

36 36 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? n What volume Of NO would be produced? n What mass of NO is produced from 500. L of NH3 at 250.0ºC and 3.00 atm?

37 37 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

38 38 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

39 39 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

40 40 Examples n The partial pressure of nitrogen in air is 592 torr. Air pressure is 752 torr, what is the mole fraction of nitrogen? n What is the partial pressure of nitrogen if the container holding the air is compressed to 5.25 atm?

41 41 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

42 42 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.

43 43 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)

44 44 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, but don’t work in real gases.  The particles are so small we can ignore their volume.  The particles are in constant motion and their collisions cause pressure.

45 45 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

46 46 What it tells us n (KE) avg = 3/2 RT n This the meaning of temperature. n u is the particle velocity. n u is the average particle velocity. n u 2 is the average particle velocity squared. the root mean square velocity is  u 2 = u rms the root mean square velocity is  u 2 = u rms

47 47 Combine these two equations n (KE) avg = N A (1/2 mu 2 ) n (KE) avg = 3/2 RT

48 48 Combine these two equations 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

49 49 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.

50 50 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

51 51 number of particles Molecular Velocity 273 K

52 52 number of particles Molecular Velocity 273 K 1273 K

53 53 number of particles Molecular Velocity 273 K 1273 K

54 54 Velocity n Average increases as temperature increases. n Spread increases as temperature increases.

55 55 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.

56 56 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.

57 57 Deriving n The rate of effusion should be proportional to u rms n Effusion Rate 1 = u rms 1 Effusion Rate 2 = u rms 2

58 58 Deriving n The rate of effusion should be proportional to u rms n Effusion Rate 1 = u rms 1 Effusion Rate 2 = u rms 2

59 59 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.

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

61 61 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.

62 62 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.

63 63 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)

64 64 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.

65 65 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

66 66 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 ()

67 67 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.

68 68 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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