Chapter 10 Gases CHEMISTRY The Central Science 9th
Chapter 10 Gases CHEMISTRY The Central Science 9th Edition David P. White OCTOBER 5, 2010 HOMEWORK PRESSURE CONVERSIONS
P 433 Q 9, 17, 21 READ SECTION 10.3 P 404 Characteristics of Gases Gases are highly compressible and occupy the full volume of their containers. When a gas is subjected to pressure, its volume decreases. Gases always form homogeneous mixtures with other
gases. Gas molecules only occupy about 0.1 % of the volume of their containers. Have extremely low densities. Pressure Pressure is the force acting on an object per unit area: F P
A Gravity exerts a force on the earths atmosphere A column of air 1 m2 in cross section exerts a force of 105 N. The pressure of a 1 m2 column of air is 100 kPa. Atmospheric pressure is the weight of air per unit of area. F= m . a
mass of air column = 10,000 kg acceleration due to gravity at the surface of Earth = 9.8 m/s^2 Atmosphere Pressure and the Barometer
Atmosphere Pressure and the Barometer SI Units: 1 N = 1 kg.m/s2; 1 Pa = 1 N/m2. Atmospheric pressure is measured with a barometer. If a vacated tube is inserted into a container of mercury open to the atmosphere, the mercury will rise 760 mm up
the tube. Standard atmospheric pressure is the pressure required to support 760 mm of Hg in a column. Standard Pressure Normal atmospheric pressure at sea level. It is equal to 1.00 atm
760 torr (760 mm Hg) 101.325 kPa Units of Pressure 1 atm = 760 mmHg = 760 torr = 1.01325 10^5 Pa = 101.325 kPa.
Pascals 1 Pa = 1 N/m2 Bar 1 bar = 105 Pa = 100 kPa Examples Convert the following pressures: 1. 658.2 mm Hg to kPa 2. 1.85 atm to torr
3. 337.3 kPa to atm Examples Convert the following pressures: 1. 658.2 mm Hg to kPa 87.75 kPa 2. 1.85 atm to torr
1410 torr 3. 337.3 kPa to atm 3.329 atm Manometer Used to measure the difference in pressure between
atmospheric pressure and that of a gas in a vessel. Pressure and the Manometer The pressures of gases not open to the atmosphere are measured in manometers. A manometer consists of a bulb of gas attached to a U-tube containing Hg:
If Pgas < Patm then Pgas = Patm.- Ph If Pgas > Patm then Pgas = Patm + Ph. Example The mercury in a manometer is 46 mm higher on the open end than on the gas bulb end. If atmospheric pressure is 102.2 kPa, what is the pressure of the gas in the bulb? Example The mercury in a manometer is 46 mm
higher on the open end than on the gas bulb end. If atmospheric pressure is 102.2 kPa, what is the pressure of the gas in the bulb? 813 mm Hg 108 kPa The Gas Laws
The Pressure-Volume Relationship: Boyles Law Weather balloons are used as a practical consequence to the relationship between pressure and volume of a gas. As the weather balloon ascends, the volume increases.
As the weather balloon gets further from the earths surface, the atmospheric pressure decreases. Boyles Law: the volume of a fixed quantity of gas is inversely proportional to its pressure (assuming all other variables are unchanged). Boyle used a manometer to carry out the experiment. Boyles Law The volume of a fixed quantity of gas at
constant temperature is inversely proportional to the pressure. The Pressure-Volume Relationship: Boyles Law Mathematically: 1 V constant P
PV constant A plot of V versus P is a hyperbola. T constant. Similarly, a plot of V versus 1/P must be a straight line passing through the origin. P1V1 P2V2 As P and V are
inversely proportional A plot of V versus P results in a curve. Since PV = k V = k (1/P) This means a plot of V versus 1/P will be
a straight line. Examples 1. A gas that occupies 2.84 L has a pressure of 88.6 kPa. What would be the pressure of the gas sample if it only occupied 1.66 L (assuming the same temperature)? 2. A 10.0-L sample of argon gas has a pressure of
0.885 atm. At what volume would the sample have a pressure of 6.72 atm? Examples 1. A gas that occupies 2.84 L has a pressure of 88.6 kPa. What would be the pressure of the gas sample if it only occupied 1.66 L (assuming the same temperature)? 152 kPa
2. A 10.0-L sample of argon gas has a pressure of 0.885 atm. At what volume would the sample have a pressure of 6.72 atm? 1.32 L OCTOBER 6 GAS LAWS Section 3-4
Boyles Law Charless Law Avogadros Law The ideal gas equation HOMEWORK : PAGE 434 10.23 TO 10.41
41 IS A CHALLENGE QUESTION ODD ONLY The Temperature-Volume Relationship: Charless Law Charless Law: the volume of a fixed quantity of gas at constant pressure increases as the temperature increases (assuming pressure constant).
If we express T in Kelvin degrees, P constant V constant T V1 V2 T1 T2 V constant
T A plot of V versus T is a straight line. When T is measured in C, the intercept on the temperature axis is -273.15C. We define absolute zero, 0 K = -273.15C. Note the amount of gas and pressure remain constant.
Charless Law The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. i.e., V
=k T A plot of V versus T will be a straight line. Examples (remember to change T to K to use Charles Law!!!) 1. A gas sample occupies a volume of 1.89 L at 25C. What would be the volume of the sample at
75C? 2. A sample of carbon dioxide in a 5.00-L container has a temperature of 56.9C. At what temperature will this sample of CO2 occupy a volume of 3.00 L? Examples 1. A gas sample occupies a volume of 1.89 L at
25C. What would be the volume of the sample at 75C? 2.21 L 2. A sample of carbon dioxide in a 5.00-L container has a temperature of 56.9C. At what temperature will this sample of CO2 occupy a volume of 3.00 L? 198 K or -75C
The Quantity-Volume Relationship: Avogadros Law Gay-Lussacs Law of combining volumes: at a given temperature and pressure, the volumes of gases which react are ratios of small whole numbers. Avogadros Hypothesis: equal volumes of gas at the same temperature and pressure will contain the same number of
molecules. Avogadros Law: the volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas. Mathematically: n1 n2 V constant n
V1 V2 We can show that 22.4 L of any gas at 0C contain 6.02 1023 gas molecules. The Ideal Gas Equation Consider the three gas laws. 1 V (constant n, T ) Boyles Law:
P Charless Law: V T (constant n, P ) Avogadros Law: V n (constant P, T ) We can combine these into a general gas law: nT V
P The Ideal Gas Equation If R is the constant of proportionality (called the gas constant), then nT V R
P The ideal gas equation is: PV nRT R = 0.08206 Latm/molK = 8.314 J/molK We define STP (standard temperature and pressure) = 0C, 273.15 K, 1 atm. Volume of 1 mol of gas at STP is: PV nRT
nRT 1 mol 0.08206 Latm/molK 273.15 K V 22.41 L P 1.000 atm Examples 1. What is the volume of a 15.0-g sample of neon
gas at 130.0 kPa and 39C? 14.8 L 2. At what temperature will 2.85 moles of hydrogen gas occupy a volume of 0.58 L at 1.00 atm? 2.5 K or -270.7C Relating the Ideal-Gas Equation and the Gas Laws If PV = nRT and n and T are constant, then PV = constant and we have Boyles law.
Other laws can be generated similarly. In general, if we have a gas under two sets of conditions, then P1V1 P2V2 n1T1 n2T2 Examples
1. A gas sample occupies a volume of 0.685 L at 38C and 0.775 atm. What will be the temperature of the sample if it occupies 0.125 L at 1.25 atm? 91.6 K or -181.6C 2. A sample of nitrogen gas in a 2.00 L container has a pressure of 1800.6 mmHg at 25C. What would be the pressure on the sample in a 1.00 L container at 125C?
4810 mm Hg OCTOBER 7 Section 10-5 Ideal Gas Equation Problems Gas Densities and Molar Mass Volumes of gases in chemical reactions HW 38 ideal gas equation
43-45-46 concepts 47-49-50 density and molar mass 52-54 to 56 Stoichiometry with gases Further Applications of the Ideal-Gas Equation Gas Densities and Molar Mass Density has units of mass over volume. Rearranging the ideal-gas equation with M as molar mass
we get PV nRT n P V RT nM PM d
V RT Examples 1. What is the density of nitrogen gas at 35C and 1.50 atm? Examples 1. What is the density of nitrogen gas at 35C and
1.50 atm? 1.67 g/L The molar mass of a gas can be determined as follows: dRT OR mRT M M P
PV The cat paws throw dirt over the P Volumes of Gases in Chemical Reactions The ideal-gas equation relates P, V, and T to number of moles of gas. The n can then be used in stoichiometric calculations. Examples 1. 4.83 g of a gas occupy a 1.50 L flask at 25C and
112.4 kPa. What is the molar mass of the gas? 71.0 g/mol STOICHIOMETRY PROBLEMS If an air bag has a volume of 36 L and is filled up with Nitrogen at a pressure of
1.15 atm at a temperature of 26C, how many grams of NaN3 must be decomposed 2 Na N3 (s) 2 Na (s) + 3 N2 (73.2 g) How many L of NH3 at 850 0C and 5
atm are required to react wih 32 g of O2? 4NH3+ 5 O2 ->4NO + 6 H2 O (14.7 L ) October 8 Section 10-6 Gas Mixtures and Partial Pressures
Partial Pressures and Mole Fraction Collecting Gases over Water HW page 436 Q 55 and 56 collecting gas over water 10:57 to 67 odd numbers (red) Gas Mixtures and Partial Pressures Since gas molecules are so far apart, we can assume they behave independently.
Daltons Law: in a gas mixture the total pressure is given by the sum of partial pressures of each component: Ptotal P1 P2 P3 Each gas obeys the ideal gas equation: RT Pi ni V
Combining the equations RT Ptotal n1 n2 n3 V Partial Pressures and Mole Fractions
Let ni be the number of moles of gas i exerting a partial pressure Pi, then Pi i Ptotal where i is the mole fraction (ni/nt). Examples: 1. Hydrogen gas is added to a 2.00-L flask at a pressure of 5.6 atm. Oxygen gas is added until the total pressure in the flask measures 8.4 atm.
What is the mole fraction of hydrogen in the flask? What the pp of Oxygen? 2 .35 moles of argon and 2.75 moles of neon are placed in a 15.0-L tank at 35C. What is the total pressure in the flask? What is the pressure exerted by neon? Examples: 1. Hydrogen gas is added to a 2.00-L flask at a
pressure of 5.6 atm. Oxygen gas is added until the total pressure in the flask measures 8.4 atm. What is the mole fraction of hydrogen in the flask? 0.67 2.8 atm 2. 1.35 moles of argon and 2.75 moles of neon are
placed in a 15.0-L tank at 35C. What is the total pressure in the flask? What is the pressure exerted by neon? Ptot = 6.91 atm Collecting Gases over Water It is common to synthesize gases and collect them by displacing a volume of water. To calculate the amount of gas produced, we need to correct for the partial pressure of the water: Ptotal Pgas Pwater
Collecting Gases over Water Example - 150.82 mL of an unknown gas is collected over water at 27C and 1.032 atm. The mass of the gas is 0.1644 g. What is the molar mass of the gas? The vapor pressure of water at 27C is 26.74 torr
OCTOBER 12 Section 10-7 and 10-8 Kinetic-Molecular Theory Application to the Gas Laws Molecular Effusion Grahams Law Diffusion and Mean Free Path HW P 437 69 to 79 odd only and 80 LABORATORY BOOK MUST GET BY WED
FOR THURSDAY MUST STUDY LAB # 8 DETERMINING THE MOLAR MASS OF A GAS AND DO PRELAB QUESTIONS PEARSON SITE DO REGISTER ONLY 7 PEOPLE REGISTERED Kinetic Molecular Theory Theory developed to explain gas behavior. Theory of moving molecules.
Assumptions: 1.-Gases consist of a large number of molecules in continuous, random motion. 2.-Volume of individual molecules negligible compared to volume of container. 3.-Intermolecular forces (forces between gas molecules) are negligible (no attraction or repulsion between molecules)
4.- Energy can be transferred between molecules during collisions, but AVERAGE kinetic energy DOES NOT CHANGE at constant temperature (When molecules collide one speeds up the other
slows down) 5.-Average kinetic energy of molecules is proportional to the ABSOLUTE temperature. At a given temperature molecules of all gases have same average KE Kinetic molecular theory gives us an
understanding of pressure and temperature on the molecular level. Pressure of a gas results from the number of collisions per unit time on the walls of container. Magnitude of pressure given by how often and how hard the
molecules strike against the walls. Temperature MOLECULAR MOTION INCREASES WITH TEMPERATURE Molecules at 1 T at a given moment have a wide range of speed At higher temperature a larger fraction of molecules have a
higher speed. Gas molecules have an average kinetic energy. Each molecule has a different energy. If 2 different gases are at same T their molecules have the same Average Kinetic Energy. If T increases their motion increases too. MOLECULAR MOTION INCREASES WITH T
Individual molecules move at different speeds but the average kinetic energy is one at each temperature. When particles collide, the momentum is conserved. That means that one particle will move faster but the other will slow down to conserve the total amount of energy in the system. The diagram shows the distribution of molecular speeds for a sample of gas at different temperatures.
Increasing T increases both the most probable speed ( curve maximum) and rms speed (u) root mean square At higher T, a larger fraction of molecules move a greater speeds. The peak shows the most probable speed ( speed of the largest number of molecules. 400m/s is approximately 1000miles/hour KE = m x speed2 Root mean square is an statistical expression that is close to the mean.
Example , the average between 4 different speeds V1,v2,v3, v4 vav = (v1+v2+v3+v4) u= As kinetic energy increases, the velocity of the gas molecules increases. Root mean square speed, u, is the speed of a gas molecule having average kinetic energy. For one molecule Average kinetic energy, , is related to root mean square
speed: 2 1 mu 2 m is the mass of the molecules u is the root mean square speed
T is proportional to the speeds of the particles. Application to Gas Laws As volume increases at constant temperature, the average kinetic energy of the gas remains constant. Therefore, u is constant. However, volume increases so the gas molecules have to travel further to hit the walls of the container. Therefore, pressure decreases. If temperature increases at constant volume, the average
kinetic energy of the gas molecules increases. Therefore, there are more collisions with the container walls and the pressure increases. Effusion The escape of gas molecules through a tiny hole.
Diffusion The spread of one substance throughout a space or throughout a second substance. Spread of a gas.
Molecular Effusion and Diffusion As kinetic energy increases,(T) the velocity of the gas molecules increases. Average kinetic energy of a gas is related to its mass:
1 mu 2 (KE) 2 Consider two gases at the same temperature: both have same KE therefore the lighter gas molecules have to have higher speeds (rms speed u) than the heavier gas. Mathematically: (M is molecular mass)
3RT u M The lower the molar mass, M, the higher the speed of the molecules (rms speed). Calculate the rms speed for O2 molecules at 25 0 C
Convert mass to kg units of speed m/s answer 482 m/s approximately 1100 miles /hour Grahams Law of Effusion As kinetic energy increases, the velocity of the gas
molecules increases. Effusion is the escape of a gas through a tiny hole (a balloon will deflate over time due to effusion). The rate of effusion can be quantified. Consider two gases with molar masses M1 and M2, the
relative rate of effusion is given by: r1 M2 r2 M1 Only those molecules that hit the small hole will escape through it. Therefore, the higher the rms the more likelihood of a gas molecule hitting the hole.
Consider two gases with molar masses M1 and M2, the relative rate of effusion is given by: 3RT r1 u1 M2 M1
3RT r2 u2 M1 M2 Only those molecules that hit the small hole will escape through it. Therefore, the higher the rms the more likelihood of a gas molecule hitting the hole.
Examples For each pair of gases, determine which will effuse faster, and by how much it will be faster. 1. CH4 and Xe 2. Cl2 and N2 3. F2 and He Examples For each pair of gases, determine which will effuse faster, and by how much it will be
faster. CH4 and Xe 2.8607 1. Cl2 and N2 1.59095
2. F2 and He 3.08027
Diffusion and Mean Free Path Diffusion of a gas is the spread of the gas through space. Diffusion is faster for light gas molecules. Diffusion is significantly slower than rms speed (consider someone opening a perfume bottle: it takes while to detect the odor but rms speed at 25C is about 1150 mi/hr). Diffusion is slowed by gas molecules colliding with each
other. Average distance of a gas molecule between collisions is called mean free path. At sea level, mean free path is about 6 10-6 cm. Tetrafluoroethylene C2F4 effuses through a barrier at a rate of 4.6 x 10 -6 mol/h. An unknown gas, consisting only of boron and hydrogen, effuses at a rate
of 5.8 x 10 -6 mol/h under the same conditions. What is the molar mass of the unknown gas? Use Grahams Law M= 63g/mol OCTOBER 14 SECTION 10-9 REAL GASES DEVIATIONS FROM IDEAL BEHAVIOR
VAN DER WAALS EQUATION HW 10.8 10.81,85,95,97 Real Gases: Deviations from Ideal Behavior From the ideal gas equation, we have PV n
RT For 1 mol of gas, PV/RT = 1 for all pressures. BUT REAL GASES DO NOT BEHAVE IDEALLY SPECIALLY AT HIGH PRESSURES!!! In a real gas, PV/RT varies from 1 significantly. The higher the pressure the more the deviation from ideal behavior (For P<10 atm we could use ideal-gas equation) As the pressure on a gas increases, the molecules are
forced closer together. As the molecules get closer together, the volume of the gas molecules begin to be significant, and the volume of the gases is greater that what is predicted by the ideal gas equation. Also at the higher the pressure, the attraction between gas particles begins to manifest and less particles hit the walls of the container, therefore there is a reduction in the Pressure due to the molecular attractions
From the ideal gas equation, we have PV n RT As temperature increases, the gases behave more ideally. The assumptions in kinetic molecular theory show where ideal gas behavior breaks down: the molecules of a gas have finite
volume; molecules of a gas do attract each other. As the gas molecules get closer together, the smaller the intermolecular distance. The smaller the distance between gas molecules, the more likely attractive forces will develop between the molecules.
Therefore, the less the gas resembles and ideal gas. As temperature increases, the gas molecules move faster and further apart. Also, higher temperatures mean more energy available to break intermolecular forces. Then an increase in T helps the gases behave more ideally Real Gases: Deviations
from Ideal Behavior Therefore, the higher the temperature, the more ideal the gas. The T increase helps the molecules break the forces of attractions. The van der Waals Equation
We add two terms to the ideal gas equation: one to correct for volume of molecules and the other to correct for intermolecular attractions The correction terms generate the van der Waals equation: nRT n 2a P 2
V nb V where a and b are empirical constants. Real Gases: Deviations from Ideal Behavior The van der Waals Equation nRT n 2a P
V nb V 2 Corrects for molecular volume Corrects for molecular attraction General form of the van der Waals equation:
2 n P a V nb nRT 2 V
The constant b adjusts the volume, since gas particles do have volume the total volume that the particles have to move in is less than the whole container. That is why b is substracting V, and the units are L/mol ( V-n b)
The constant a accounts for the attraction between molecules which increases with the square of the number of molecules per unit of Volume. Units for a L2 atm/mol2 It indicates HOW STRONGLY THE MOLECULES ATTRACT EACH OTHER!!! If 1.00 mol of an ideal gas is confined to a 2.4 L at 0.0 C,
it would exert a pressure of 1.00 atm. Use the Van Der Waals equation and the contants for Cl2 to estimate the pressure exerted by 1 mol of Cl2 in 22.4 L at 0.0 C Monday October 18 Test on unit 10. One period 2 problems and multiple choice. Friday October 15. Experiment # 8 Determining the Molar Volume of a Gas. MUST BRING LAB NOTEBOOK (a marble book)
AND LAB BOOK KNOW THE PROCEDURE!!! You have to know what are you going to do BEFORE walking into the lab room.
Statistics of Fingerprints . Dakota Boyd, Dustin Short, Elizabeth Lee, John Huppenthal, Shelby Proft, Wacey Teller. History of Fingerprinting. Originally used paper and ink fingerprints. Fingerprints were matched using trained individuals.
[email protected] June 2006 MRST Parton distributions 1. Uncertainties on partons I: experimental errors Uncertainties due to errors on experimental data fitted obtained using both the Hessian and Lagrange multiplier approaches.
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