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Gases The Gas Laws Labs #18 Molar Mass of a Volatile Liquid #19 Calcium Carbonate Analysis: Molar Volume of Carbon Dioxide 1/22/20141.

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Presentation on theme: "Gases The Gas Laws Labs #18 Molar Mass of a Volatile Liquid #19 Calcium Carbonate Analysis: Molar Volume of Carbon Dioxide 1/22/20141."— Presentation transcript:

1 Gases The Gas Laws Labs #18 Molar Mass of a Volatile Liquid #19 Calcium Carbonate Analysis: Molar Volume of Carbon Dioxide 1/22/20141

2 Gases have general characteristics Expansion: Expand indefinitely to fill the space available to them Indefinite shape: Fill all parts of container evenly, so have no definite shape of their own Compressibility: Most compressible of states of matter Mixing: Two or more gases will mix evenly and completely when confined to same container Low density: Have much lower densities than liquids and solids (typically about 1/1000 those of liquids/solids) Pressure: Exert pressure on their surroundings 1/22/20142

3 Vapors Gas phase at temperature where same substance can also exist in liquid or solid state Below critical temperature of substance (vapor can be condensed by increasing pressure without reducing temperature) 1/22/20143

4 Pressure: force per unit area (P= f/a) Force generated by collisions of gas particle w/container walls Related to velocity of gas particles Total force = sum of forces of all collisions each second per unit area Force = m x acceleration Pressure dependent on Gas particle velocity Collision frequency Collision frequency dependent on Gas particle velocity Distance to container walls. Changing temperature changes collision force, as well as collision frequency Collision frequency changed by altering size of container Force of collisions is not affected 1/22/20144

5 5 (psi) (force of 1 newton exerted on one square meter of area) (29.92 Hg)

6 Manometers Used to measure pressure of enclosed gas U-tube partially filled with liquid, typically Hg One end connected to container of gas being measured Other end sealed with vacuum existing above liquid, or open to atmosphere 1/22/20146

7 Pressure is just difference between two levels (in mm of Hg)-indicates pressure of system attached to apparatus Gas connected to one arm Space above Hg in other arm is vacuum Liquid in tube falls to height (directly proportional to pressure exerted by gas in the tube) Since pressure of gas causes liquid levels to be different in height, it is this difference (h) that is measure of gas pressure in container P gas = P h If left to atmosphere, it measures atmospheric pressure-barometer 1/22/20147 Closed-end manometer

8 Used to measure pressure of gas in container Difference in Hg levels indicates pressure difference in gas pressure and atmospheric pressure Atmospheric pressure pushes mercury in one direction Gas in container pushes it in the other direction Two ends connected to gases at different pressures Closed end to gas in bulb (gas filled) Open end to atmosphere If pressure of gas higher than atmospheric, Hg level lower in arm connected to gas (P gas = P barometric + h) If pressure of gas lower than atmospheric, Hg level higher in arm connected to gas (P gas = P barometric - h) If levels are equal, gas at atmospheric pressure 1/22/20148 Open-end manometer

9 1/22/20149 If fluid other than mercury is used: Difference in heights of liquid levels inversely proportional to density of liquid and represents the pressure Greater density of liquid, smaller difference in height High density of mercury (13.6 g/mL) allows relatively small monometers to be built Readings must be corrected for relative densities of fluid used and of mercury (mm Hg = mm fluid) density fluid density Hg

10 1/22/201410

11 A sample of CH 4 is confined in a water manometer. The temperature of the system is 30.0 °C and the atmospheric pressure is kPa. What is the pressure of the methane gas, if the height of the water in the manometer is 30.0 mm higher on the confined gas side of the manometer than on the open to the atmosphere side. (Density of Hg is g/mL). 1/22/201411

12 1) Convert 30.0 mm of H 2 O to equivalent mm of mercury: (30.0 mm) (1.00 g/mL) = (x) ( g/mL) x = mm (I will carry some guard digits.) 2) Convert mmHg to kPa: mmHg x ( kPa/760.0 mmHg) = kPa 3) Determine pressure of enclosed wet CH 4 : At point A in the above graphic, we know this: P atmo. press. = P wet CH4 + P the 30.0 mm water column kPa = x kPa x = kPa 4) Determine pressure of dry CH 4 : From Dalton's Law, we know this: P wet CH4 = P dry CH4 + P water vapor (Water's vapor pressure at 30.0 °C is 31.8 mmHg. Convert it to kPa.) kPa = x kPa x = kPa Based on provided data, use three significant figures; so 94.2 kPa. 1/22/201412

13 Boyles Law Pressure-Volume Relationship Temperature/# molecules (n) constant 1/22/ X axis independent variable Y axis dependent variable

14 Inversely proportional (one goes up, other goes down) V 1/P or P = k/T where K is constant PV = constant P 1 V 1 = P 2 V 2 Gas that strictly obeys Boyles Law is ideal gas (holds precisely for gases at very low temperatures) 1/22/201414

15 Pressure applied vs. volume measured Shows inverse proportion If pressure doubled, volume decreased by ½ 1/22/201415

16 Pressure vs. inverse of volume Plot of V (P) against 1/P (1/V) gives straight line –Graph of equation P = k 1 x 1/V Same relationship holds whether gas is being expanded or contracted 1/22/201416

17 A gas which has a pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant pressure? P 1 = 1.3 atm V 1 = 27 L P 2 = 3.9 atm V 2 = ? P 1 V 1 = P 2 V atm (27 L) = 3.9 atm (X) = 9.0 L 1/22/201417

18 Charles Law Temperature-Volume Relationship Pressure/n constant 1/22/201418

19 Volume of gas directly proportional to temperature (T V), and extrapolates to zero at zero Kelvin (convert Celsius to Kelvin) 1/22/201419

20 Used to determine absolute zero From extrapolated line, determine T at which ideal gas would have zero volume Since ideal gases have infinitely small atoms, only contribution to volume of gas is pressure exerted by moving atoms bumping against walls of container If no volume, then no kinetic energy left Absolute zero is T at which all KE has been removed Does not mean all energy has been removed, merely all KE 1/22/

21 If T, V and vs. V = kT P is constant k = proportionality constant V/T = constant 1/22/201421

22 A gas at 30 o C and 1.00 atm occupies a volume of L. What volume will the gas occupy at 60.0 o C and 1.00 atm? V 1 = atm T 1 = 30 o C = 303 K V 2 = ? T 2 = 60.0 o C = 333 K V 1 /T 1 = V 2 /T /303 K = X/333 K = L 1/22/201422

23 Gay Lussacs Law Pressure-Temperature Relationship Volume/n constant 1/22/201423

24 1/22/ When two gases react, do so in volume ratios always expressed as small whole numbers When H burns in O, volume of H consumed is always exactly 2x volume of O Direct relationship between pressure and temperature (P T) P/T = constant P i /T i = P f /T f

25 Avogadros Law (at low pressures) Volume-Amount Relationship Temperature/Pressure constant 1/22/201425

26 1/22/ Equal volumes of gases, measured at same temperature and pressure, contain equal numbers of molecules Avogadro's law predicts directly proportional relation between # moles of gas and its volume Helped establish formulas of simple molecules when distinction between atoms and molecules was not clearly understood, particularly existence of diatomic molecules Once shown that equal volumes of hydrogen and oxygen do not combine in manner depicted in (1), became clear that these elements exist as diatomic molecules and that formula of water must be H 2 O rather than HO as previously thought

27 V of gas proportional to # of moles present (V n) At constant T/P, V of container must increase as moles of gas increase V/n = constant V i /n i = V f /n f 1/22/201427

28 A 5.20 L sample at 18.0oC and 2.00 atm pressure contains moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? V 1 = 5.20 L n 1 = mol V 2 = X n 2 = = 1.70 mol V 1 /n 1 = V 2 /n L/0.436 mol = X/1.70 mol = 20.3 L 1/22/201428

29 Ideal Gas Law (Holds closely at P < 1 atm) 1/22/ Generalization applicable to most gases, at pressures up to about 10 atm, and at temperatures above 0°C. Ideal gass behavior agrees with that predicted by ideal gas law.

30 Formulated from combination of Boyles law, Charless law, Gay-Lussacs law, and Avogadros principle Combined If proportionality constant called R Rearrange to form ideal gas equation 1/22/201430

31 1/22/ Experimentally observed relationship between these properties is called the ideal gas law: PV = nRT Pressure (P) in atm Volume (V) in L Absolute temperature (T) in K (Charles Law) Amount (number of moles, n) R (universal ideal gas constant) liter atm/mole K or L·atm·K –1 ·mol – liter kPa/mole K 8.31 J/mole K 8.31 V C/mole K 8.31 x g cm 2 /sec 2 mole K 6.24 x 10 4 L mm Hg/mol K 1.99 cal/mol K

32 1/22/ You can use ideal gas law equation for all problems Given 3 of 4 variables and calculate 4 th P i V i = n i RT i P f V f n f RT f Cancel all constants and R, make appropriate substitutions from given data to perform calculation

33 A sample containing moles of a gas at 12.0 o C occupies a volume of 12.9 L. What pressure does the gas exert? PV = nRT P (12.9 L) = (0.614 mol)( L atm/K mol)(285 K) = 1.11 atm 1/22/201433

34 1/22/201434

35 Homework: Read , pp Q pp , #29, 31-34, 44 1/22/201435

36 Standard Temperature and Pressure (STP) STP = 0°C (273K) and 1.00 atm pressure (760 mm Hg) One mole of gas at STP will occupy L Allows you to compare gases at STP to each other 1/22/201436

37 What volume will 1.18 mole of O 2 occupy at STP? PV = nRT (1atm)(X) = (1.18 mol)( L atm/K mol)(273 K) = 26.4 L Alternate way: At STP, 1 mole occupies 22.4 L V i /n i = V f /n f 22.4 L/1 mol = X/1.18 mol = 26.4 L 1/22/201437

38 A sample containing 15.0 g of dry ice, CO 2 (s), is put into a balloon and allowed to sublime according to the following equation: CO 2 (s) CO 2 (g) How big will the balloon be (what is the volume of the balloon) at 22 o C and 1.04 atm after all of the dry ice has sublimed? 15.0 g CO 2 1 mol CO 2 = mol CO g CO 2 PV = nRT (1.04 atm)(X) = (0.341 mol)( L atm/K mol)(295K) = 7.94 L 1/22/201438

39 0.500 L of H 2 (g) are reacted with L of O 2 (g) according to the equation 2H 2 (g) + O 2 (g) 2H 2 O(g). What volume will the H 2 O occupy at 1.00 atm and 350 o C? Find limiting reactant: L H 2 1 mol H 2 = mol H 2 / L L O 2 1 mol O 2 = mol O 2 / L PV = nRT (1atm)(X) = ( mol)( L atm/K mol)(623 K) = 1.14 L 1/22/201439

40 1/22/201440

41 Molar Mass Gives density which can be determined if P, T and molar mass are known Density directly proportional to molar mass Density increases as gas pressure increases Density decreases as temperature increases 1/22/201441

42 A gas at 34.0 o C and 1.75 atm has a density of 3.40 g/L. Calculate the molar mass (M.M.) of the gas. M.M. = dRT P M.M. = (3.40 g/L)( L atm/K mol)(307K) (1.75 atm) M.M. = 48.9 g/mol 1/22/201442

43 Example What are the expected densities of argon, neon and air at STP? PV = nRT PV = g (RT) molar mass Density = g/V = (Molar mass)P/RT density Ar = (39.95 g/mol)(1.00 atm)/( L- atm/mol-K)(273 K) = 1.78 g/L density Ne = g/L density air = 1.28 g/L molar mass = (0.80)(28 gN 2 /mol) + (0/20)(32 g O 2 /mol) = 28.8 g/mol 1/22/201443

44 Molar Volume (V/n) V = RT n P ( L-atm/mol-K)(273K) = 22.4 L/mol 1.00 atm 1/22/201444

45 Homework: Read 5.4, pp Q pp , #52, 56, 58, 60 1/22/201445

46 Daltons Partial Pressure 1/22/201446

47 When two gases are mixed together, gas particles tend to act independently of each other Each component gas of mixture of gases uniformly fills containing vessel Each component exerts same pressure as it would if it occupied that volume alone Total pressure of mixture is sum of individual pressures, called partial pressures, of each component P t = P A + P B + P C + … 1/22/201447

48 1/22/201448

49 1/22/ Since each component obeys ideal gas law (PV=nRT) and has same T and V, it follows that partial pressure of each gas in container is directly proportional to # moles of gas present P i /P t = n i /n t P i = n i /n t x P t P i = X i P t X i = mole fraction of gas component i

50 A volume of 2.0 L of He at 46 o C and 1.2 atm pressure was added to a vessel that contained 4.5 L of N 2 at STP. What is the total pressure and partial pressure of each gas at STP after the He is added? Find # moles of He at original conditions-will lead us to finding partial pressure of He at STP. PV = nRT (1.2 atm)(2.0 L) = n( L atm/k mol)(319 K) n = mol He When gases are combined under STP, partial pressure of He will change while N 2 will remain the same since it is already at STP. P He V = nRT (X)(4.5 L) = (0.091 mol)( L atm/K mol)(273 K) P = atm Total pressure = = 1.46 atm 1/22/201450

51 Mole fraction X i = n i = P i n total P total You can use either the number of moles or the pressure of each component of your system to evaluate the mole fraction 4.5 L N 2 1 mol = mol N L X N2 = mol = atm = mol atm X He = mol = atm = mol atm 1/22/201451

52 1/22/ A mixture of gases consists of 3.00 moles of helium, 4.00 moles of argon, and 1.00 moles of neon. The total pressure of the mixture is 1,200 torr. Calculate the mole fraction of each gas. n t = 3.00 mol mol mol = 8.00 mol X i = n i /n t X He = 3.00 mol/8.00 mol = X Ar = 4.00 mol/8.00 mol = X Ne = 1.00 mol/8.00 mol = Calculate the partial pressure of each gas. P i = X i P i P He = x 1,200 torr = 450. torr P Ar = x 1,200 torr = 600. torr P Ne = x 1,200 torr = 150. torr He-1.2 L, 0.63 atm, 16 o CNe-3.4 L, 2.8 atm, 16 o C

53 1/22/ Whenever gases are collected by displacement of water, total gas pressure is sum of partial pressure of collected gas and partial pressure of water vapor P t = P gas + P H2O Consequently, partial pressure of collected gas is P gas = P t – P H2O, where partial pressure of water depends on temperature and corresponds to vapor pressure of water

54 1/22/ When oxygen gas is collected over water at 30°C and the total pressure is 645 mm Hg: What is the partial pressure of oxygen? Given the vapor pressure of water at 30°C is 31.8 mm Hg. P t = P O2 + P H2O P O2 = P t – P H2O = 645 mm Hg – 31.8 mm Hg P O2 = 613 mm Hg What are the mole fractions of oxygen and water vapor? Recall that the partial pressures of O 2 and H 2 O are related to their mole fractions, P O2 = X O2 P t P H2O = X H2O P t X O2 = 613/645 = 0.950, and X H2O = 31.8/645 = Note also that the sum of the mole fractions is 1.0, within the number of significant figures, given: X O2 + X H2O = = 0.999

55 The vapor pressure of water in air at 28 o C is 28.3 torr. Calculate the mole fraction of water in a sample of air at 28 o C and 1.03 atm pressure. X H2O = P H2O = 28.3 torr = P air 783 torr 1/22/201455

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57 1/22/201457

58 The safety air bags in cars are inflated by nitrogen gas generated by the rapid decomposition of sodium azide. If an air bag has a volume of 36 L and is to be filled with nitrogen gas at a pressure of 1.15 atm at a temperature of 26.0 o C, how many grams of NaN 3 must be decomposed? 2NaN 3 (s) 2Na(s) + 3N 2 (g) Gas data mol N 2 mol NaN 3 g NaN 3 n = PV/RT = (1.15 atm)(36 L) = 1.7 mol N 2 ( L-atm/mol-K)(299K) 1.7 mol N 2 2 mol NaN g NaN 3 = 72 g NaN 3 3 mol N 2 1 mol NaN 3 1/22/201458

59 Homework: Read 5.5, pp Q pg. 235, #64, 66, 68, 70, 72 1/22/201459

60 Kinetic Molecular Theory of Gases Describes the properties of gases 1/22/201460

61 Main assumptions that explain behavior of ideal gas: Gases composed of tiny, invisible molecules that are widely separated from one another in empty space Gas molecule behave as independent particles Volume occupied by molecule considered negligible Gas molecules are in constant motion, continuous, random and straight-line motion Collisions of particles with walls of container cause pressure exerted by gas Molecules collide with one another, but collisions are perfectly elastic (no net loss of energy-exert no force on each other) Attractive forces between atoms and/or molecules in gas are negligible Average kinetic energy of collection of gas particles assumed to be directly proportional to Kelvin temperature of gas 1/22/201461

62 Temperature is measure of average kinetic energy of gas Equal #s of molecules of any gas have same average kinetic energy at same temperature Greater temperature, greater fraction of molecules moving at higher speeds (higher average kinetic energy of gas molecules) Individual molecules move at varying speeds Momentum conserved in each collision, but one colliding molecule might be deflected off at high speed while another is nearly stopped Result is that molecules at any instant have wide range of speeds 1/22/201462

63 At higher temperatures, greater fraction of molecules moving at higher speeds Average KE, e, related to root mean square (rms) speed u 4 molecules in gas sample have speeds of 3.0, 4.5, 5.2, and 8.3 m/s. Average speed rms Because mass of molecules does not increase, rms speed of molecules must increase with increasing temperature 1/22/201463

64 Average kinetic energy of single gas molecule KE = ½ mv 2 m = mass of molecule (kg) v = speed of molecule (meters/sec) KE measured in joules Average translational kinetic energy of any kind of molecule in ideal gas is given by: 1/22/201464

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67 1/22/ J = kg· m 2 /s 2

68 1/22/ Based on understanding of pressure, molecular meaning of gas laws can be appreciated as follows:

69 Boyles Law: inverse relationship between P & V: Kinetic molecular theory agrees If V of container decreased, gas particles strike walls of container more frequently, increasing pressure of gas If V of container increased, fewer impacts of gas molecules with wall per second, decreasing pressure 1/22/201469

70 Charles Law: direct relationship between T & V: Increasing temperature increases average kinetic energy and speed of gas molecules Increases force of each collision w/container wall Increases frequency of collisions w/container wall Both increase pressure If pressure of gas is to remain constant, volume must increase to decrease frequency of collisions with container walls If one of the walls is a movable piston, then the volume of the container will expand until balanced by the external force 1/22/201470

71 Gay-Lussacs Law: direct relationship between T & P Increase in temperature increases kinetic energy of gas particles Force of each collision increases Frequency of collisions with container walls increases When gas is heated in container with fixed volume, gas molecules impact more forcefully with wall, increasing pressure 1/22/201471

72 Avogadros Law As more gas molecules are added to the container, # of impacts per second with wall increases, pressure increases correspondingly If V of container is not fixed, then V will increase 1/22/201472

73 Grahams Law of Effusion Experiments show that molecules of a gas do not all move at same speed but are distributed over a range. 1/22/201473

74 Maxwell-Boltzmann distribution Probability distribution that forms basis of kinetic theory of gases Explains many fundamental gas properties, including pressure and diffusion 1/22/201474

75 1/22/ As temp. increases, curve flattens out (more molecules have higher kinetic energies and therefore higher average speeds) (speed) Highest point on curve marks most probable speed-some move with speeds much less than most probable speed, while others move with speeds much greater than this speed

76 Diffusion Spread of gas molecules throughout volume Much slower than average molecular speed because of collision between molecules Mean free path Average distance molecule travels between collisions Factors affecting it: Density (increasing density decreases MFP) Radius of molecule (increasing size increases collision frequency, reducing MFP) Pressure (increasing pressure increases collision frequency, reducing MFP) Temperature (increasing T increases collision frequency, but does not affect MFP) 1/22/201476

77 Effusion Escape of gas through small opening into vacuum Rate (in mol/s) proportional to average speed u More collisions gas has w/walls of container, higher probability it will hit pinhole and go through it 1/22/201477

78 1/22/ Average kinetic energy has same value for all gases at same temperature Gas molecules with higher molar masses will have slower average speeds, while lighter molecules will have higher average speeds Rates two gases effuse through a pinhole in a container are inversely related to the square roots of the molecular masses of gas particles Use equation to find relative rates of diffusion of gases, whether evacuated or not) All gases expand to fill container Heavier gases diffuse more slowly than lighter ones

79 1/22/ KE = cT = ½ mv 2 KE-average kinetic energy c = constant that is same for all gases T = temperature in Kelvin M = mass of gas v 2 = average of the square of the velocities of the gas molecules Since cT will be the same for all gases at the same temperature, the average KE of any two gases at the same T will be the same KE 1 = KE 2 ½ m 1 V 2 1 = ½ m 2 v 2 2 root mean square velocity = v rms = v 2 m 1 /m 2 = v rms2 /v rms1 Higher the T, higher v rms

80 1/22/201480

81 1/22/ Effusion: Diffusion:

82 How many times faster than He would NO 2 gas effuse? M NO2 = g/mol M He = g/mol M NO2 / M He = rate He /Rate NO /4.003 = So He would effuse 3.39 times faster as NO 2 1/22/201482

83 Real Gases (Nonideal Gases) Ideal gas law works very well for most gases, however, it does not work well for gases under high pressures or gases at very low temperature Conditions used to condense gases Generalized that any gas close to its BP (condensation point) will deviate significantly from ideal gas law Kinetic molecular theory makes two assumptions Gases have no volume They exhibit no attractive or repulsive forces 1/22/201483

84 1/22/ As a real gas is cooled and/or compressed, distances between particles decreases dramatically, and these real volumes and forces can no longer be ignored Cooling gas decreases average KE of its molecules All gases when cooled enough will condense to liquid Suggests that intermolecular forces of attraction exist between molecules of real gases Condensation occurs when average KE is not great enough for molecules to break away from intermolecular attractive forces When they stick together, there are fewer particles bouncing around and creating P, so real P in nonideal situation will be smaller than P predicted by ideal gas equation

85 1/22/201485

86 Van der Waals Equation Developed modification of ideal gas law to deal with nonideal behavior of real gases 1/22/201486

87 Real Gases Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important) 1/22/201487

88 1/22/ corrected pressure corrected volume P ideal V ideal P ideal V effective = nRT Van der Waals (real gases) P = pressure of gas (atm) V = volume of gas (L) n = #moles of gas (mol) T = absolute temperature (K) R = gas constant, L-atm/mol-K a = constant, different for each gas, that takes into account the attractive forces between molecules b = constant, different for each gas, that takes into account the volume of each molecule correction for attractive forcescorrection for small/finite volume

89 1/22/ Calculate the pressure exerted by mol of He in a L container at -25 o C Using the ideal gas law Using van der Waals equation PV = nRT (X)(0.2000L) = ( mol)() L atm/K mol)(248) = atm From table 5.3, a = atm L 2 /mol 2 and b = L mol (P x / )( – x ) = ( )(248)= atm

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91 Pneumatic trough used for collecting gas samples over water A gas (H 2, O 2 ) which is not soluble (or only slightly soluble) in water can be collected over water Pneumatic trough w/bottle full of water submerged Gas from reaction bubbles into bottle, displacing water Pressure of gas inside collecting bottle determined to solve ideal gas equation When gases are collected over water, there is some water vapor collected Pressure of water vapor depends on T only and is obtained from reference table Pressure of gas that was generated is calculated from Daltons law of partial pressure P gas = P atm - P water 1/22/201491

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93 Homework: Read , pp Q pp , #74, 76, 78, 80, 82, 84, 86, 93 Do one additional exercise and one challenge problem. Submit quizzes by to me: umdahl/ace/launch_ace.html?folder_path=/chemistry/book_con tent/ _zumdahl/ace&layer=act&src=ch05_ace1.xml umdahl/ace/launch_ace.html?folder_path=/chemistry/book_con tent/ _zumdahl/ace&layer=act&src=ch05_ace2.xml umdahl/ace/launch_ace.html?folder_path=/chemistry/book_con tent/ _zumdahl/ace&layer=act&src=ch05_ace3.xml 1/22/201493

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