gases gases. characteristics of gases expand to fill and assume the shape of their container next
TRANSCRIPT
GasesGases GasesGases
Characteristics of Gases
• Expand to fill and assume the shape of their container
Next
• Diffuse into one another and mix in all proportions.
• Particles move from an area of high concentration to an area of low concentration.
• Next
Invisible
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• Expand when heated
PropertiesPropertiesPropertiesProperties
that determine physical that determine physical behavior of a gasbehavior of a gas
Amount• Mass or moles
Volume l x w x h πr²h
Temperature
Pressure
The molecules in a gas are in constant motion. The purple balls represent gaseous atoms that collide with each other and the walls of the container. Gas molecules are in constant motion. "Pressure" is a measure of the collisions of the atoms with the container.
Pressure
• Force per unit area
• Equation: P = F/A
F = force A = area Next
Force• Pushing or pulling on something
Formulas For Surface Area
• Square: 6s²• rectangle: 2ab + 2bc + 2ac• Cylinder: 2πr²+ 2πrh• sphere: 4πr²
Units of Measurement
• N/m2 Newton per square meter• N/cm2 Newton per square centimeter• Pa Pascal• kPa kilopascal• Torr Torr• mmHg millimeters of mercury• lb/in2 pound per square inch
Calculating Pressure Using P = F/A
Suppose that a woman weighing 135 lb and wearing high-heeled shoes momentarily places all her weight on the heel of one foot. If the area of the heel is 0.50 in²., calculate the pressure exerted on the underlying surface.
P = F/A = 135 lb/0.50 in² P = 270 lb/in²
Liquid Pressure
The pressure depends height of liquid column and density of liquid.
Ρ = dgh = density x acceleration due to gravity x
height
Calculating Pressure Exerted by a Liquid
Calculate the height of a mercury column, in feet, that is equal to a column of water that is 110 ft. high.
Known valuesdwater = 1.0 g/cm3 dmercury = 13.6 g/cm3 g = 32 ft/s2
Ρ = dgh Pwater = Pmercury
dgh = dgh
(1.0 g/cm3)(32ft/s2)(110 ft) = (13.6 g/cm3)(32 ft/s2)(h) h = 8.1 ft. Hg
Barometer• Barometer is a device
used to measure the pressure exerted by the atmosphere.
• Height of mercury varies with atmospheric conditions and with altitude.
Aneroid barometer
Mercury Barometer
Measurement of Gas Pressure
Standard Atmosphere (atm)
• Pressure exerted by a mercury column of exactly 760 mm in height
• the density of Hg = 13.5951 g/cm3 (0oC)
• The acceleration due to gravity (g) is 9.80665 m/s2 exactly.
1 atm =
• 760 mmHg• 760 Torr• 14.696 lb/in2
• 101.325 kPa
Check textbook for more values
Converting Pressure to an Equivalent Pressure
A gas is at a pressure of 1.50 atm. Convert this pressure to
a. Kilopascals 1 atm = 101.325 kPa
1.50 atm x 101.325 kPa = 152 kPa 1 atmb. mmHg
1.50 atm x 760 mmHg = 1140 mmHg 1 atm
Manometers
• Used to compare the gas pressure with the barometric pressure.
Next
Types of Manometers
• Closed-end manometer The gas pressure is equal to the
difference in height (Dh) of the mercury column in the two arms of the manometer
Closed-end Manometer
Open-end Manometer
The difference in mercury levels (Dh) between the two arms of the manometer gives the difference between barometric pressure and the gas pressure
Three Possible Relationships
1. Heights of mercury in both columns are equal if gas pressure and atmospheric pressure are equal.
Pgas = Pbar
2. Gas pressure is greater than the barometric pressure.
∆P > 0
Pgas = Pbar + ∆P
3. Gas pressure is less than the barometric pressure.
∆P < 0
Pgas = Pbar + ∆P
Standard Temperature & Pressure (STP)
Temperature = 0ºC Pressure = 1 atm
The Simple Gas LawsThe Simple Gas LawsThe Simple Gas LawsThe Simple Gas LawsBoyle’s LawBoyle’s LawCharles’ LawCharles’ Law
Gay-Lussac’s LawGay-Lussac’s LawCombined Gas LawCombined Gas Law
Temperature & Gas LawsThree temperature scales Fahrenheit (ºF) Celsius (ºC) Kelvin K (no degree symbol)
Always use K when performing calculations with the gas law equations.
K = (ºC) + 273 Example: ºC = 20º K = 293 Next
Absolute Zero of Temperature
Temperature at which the volume of a hypothetical gas becomes zero
-273.14 ºC Next
Absolute or Kelvin Scale
Temperature scale that has -273.15 ºC as its zero.
Temperature interval of one Kelvin equals one degree celsius.
1 K = 1 ºC
Boyle’s Law
For a fixed amount of gas at constant temperature, gas volume is inversely proportional to gas pressure.
P1V1 = P2V2
Charles’ Law
The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin (absolute) temperature.
V1 = V2
T1 T2
or V1T2 = V2T1
Example . Charles’ Law
A 4.50-L sample of gas is warmed at constant pressure from 300 K to
350 K. What will its final volume be?
Given: V1 = 4.50 L T1 = 300. K T2 = 350. K V2 = ?
Equation: V1 = V2
T1 T2
or V1T2 = V2T1
(4.50 L)(350. K) = V2 (300. K)
V2 = 5.25 L
Gay-Lussac’s Law
The pressure of a sample of gas is directly proportional to the absolute temperature when volume remains constant.
P1 = P2
T1 T2
or P1T2 = P2T1
On the next slide (43)
The amount of gas and its volume are the same in either case, but if the gas in the ice bath (0 ºC) exerts a pressure of 1 atm, the gas in the boiling-water bath (100 ºC) exerts a pressure of 1.37 atm. The frequency and the force of the molecular collisions with the container walls are greater at the higher temperature.
Combined Gas Law
Pressure and volume are inversely proportional to each other and directly proportional to temperature.
P1V1 = P2V2
T1 T2
or P1V1T2 = P2V2T1
Example. Combined Gas Law
A sample of gas is pumped from a 12.0 L vessel at
27ºC and 760 Torr pressure to a 3.5-L vessel at 52ºC. What is the final pressure?
Given:P1 = 760 Torr P2 = ?V1 = 12.0 L V2 = 3.5 LT1 = 300 K T2 = 325 K
Equation:
P1V1 = P2V2 T1 T2
or P1V1T2 = P2V2T1
(760 Torr)(12.0 L)(325 K) = ( P2)(3.5 L)(300 K)
P2 = 2.8 x 10³ Torr
Law of Combining Law of Combining VolumesVolumes
Law of Combining Law of Combining VolumesVolumes
Gay-Lussac’s LawGay-Lussac’s Law
Law of Combining Volumes
Gases react in volumes that are related as small whole numbers.
The small whole numbers are the stoichiometric coefficients.
Avogadro’s LawAvogadro’s LawAvogadro’s LawAvogadro’s Law
Volume & MolesVolume & Moles
Avogadro’s Explanation of Gay-Lussac’s Law
When the gases are measured at the same temperature and pressure, each of the identical flasks contains the same number of molecules. Notice how the combining ratio: 2 volumes H2 to 1 volume O2 to 2 volumes H2O leads to a result in which all the atoms present initially are accounted for in the product.
Avogadro’s Hypothesis
Different gases compared at same temperature and pressure
a. Equal - equal number of Volumes molecules
b. Equal number of – equal moleculess volumes
Avogadro’s LawAt a fixed temperature and pressure, the
volume of a gas is directly proportional to the amount of gas.
V = c · nV = volume c = constant n= # of moles
Doubling the number of moles will cause the volume to double if T and P are constant.
More STP Values For Gases
• 1 mol of a gas = 22.4 L
• Number of molecules contained in 22.4 L of a gas is 6.022 x 1023
A Molar Volume of Gas
Source: Photo by James Scherer. © Houghton Mifflin Company. All rights reserved.
Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law
Equation
Includes all four gas variables:• Volume• Pressure• Temperature• Amount of gas Next
PV = nRT• Gas that obeys this equation if said to be an ideal gas (or perfect gas).
• No gas exactly follows the ideal gas law, although many gases come very close at low pressure and/or high temperatures.
• Ideal gas constant, R, is R = PV nT = 1 atm x 22.4 L 1 mol x 273.15 K
R = 0.082058 L·atm/mol· K
Applications of the Applications of the Ideal Gas LawIdeal Gas Law
Applications of the Applications of the Ideal Gas LawIdeal Gas Law
Molar MassMolar MassDensityDensity
Molar Mass Determination
• n = number of moles
• Moles = mass of sample m Molar mass M
Ideal Gas Equation: n = PV RT Substituting: m = PV M RT
M = mRT PV
Example: Molar MassThe density of carbon tetrachloride vapor at 714 torr and
125ºC is 4.43 g/L. What is its molar mass?
M = dRT P = (4.43 g/L)(0.0821 L-atm/mol-k)(398 K) (714 torr x 1 atm/760 torr)
M = 154 g/mol
Gas Density (d)
• d = m/V so V = m/d
• Ideal Gas Equation: PV = nRT Substituting P m = mRT d M
“m” cancels out d = PM RT
Finding the Vapor Density of a Substance
Example. Vapor Density
The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressue is 1.6 Earth atm. Assuming ideal behavior, calcuate the density of Titan’s atmosphere.
d = PM RT
= (1.6)(1 atm)(28.6 g/mol) (0.0821 L-atm/mol-K)(95 K)
d = 5.9 g/L
Gases in Chemical Gases in Chemical ReactionsReactions
Gases in Chemical Gases in Chemical ReactionsReactions
Ideal Gas Law Ideal Gas Law & &
Balanced Chemical EquationBalanced Chemical Equation
Example. Reaction Stoichiometry
How many moles of nitric acid can be prepared using 450 L of NO2 at a pressure of 5.00 atm and a temperature of 295 K?
Step 1: Use Ideal Gas law equation n = PV (5.00 atm)(450 L) RT (0.0821 L-atm/mol-K)(295 K)
n = 92.9 mol Next ---- >
Step 2 3NO2 (g) + H2O --- > 2HNO3(aq) + NO(g)
92.9 mol NO2 x 2 mol HNO3
3 mol NO2
= 61.9 mol HNO3
Dalton’s Law ofDalton’s Law of Partial Pressure Partial PressureDalton’s Law ofDalton’s Law of Partial Pressure Partial Pressure
Mixture of GasesMixture of Gases
Total Pressure
The total pressure of a mixture of gases is the sum of the partial pressures of the components of the mixture.
Ptotal = PA + PB + ……
Total Pressure: Mixture of Gases
Total Volume
Vtotal = VA + VB + …….
The expression for percent by volume
(VA / Vtotal) x 100 %
Total Volume: Mixture of Gases
Mole Fraction of Compound in Mixture
• Fraction of all molecules in the mixture contributed by that component.
• Sum of all the mole fractions in a mixture is 1.
• Expression for mole fraction of a substance in terms of P and V
na = Pa = Va
ntot Ptot Vtot
Total Volume: Mixture of Gases
Example: Gas Mixtures & Partial Pressure
A gaseous mixture made from 6.00 g O2 and 9.00 g CH4 is placed in a 15.0 L vessel at 0ºC. What is the partial pressure of each gas, and what is the total pressure in the vessel?
Step 1: nO2 = 6.00 g O2 x 1 mol O2
32 g O2
= 0.188 mol O2
Next ---- >
nCH4 = 9.00 g CH4 x 1 mol CH4
16.0 g CH4
= 0.563 mol CH4
Step 2: Calculate pressure exerted by each
PO2 = nRT V = (0.188 mol O2)(0.0821 L-atm/mol-K)(273 K) 15.0 L = 0.281 atm
PCH4 = (0.563 mol)(0.0821 L-atm/mol-K)(273 K) 15.0 L = 0.841 atm
Step 3: Add pressures
Ptotal = PO2 + PCH4
= 0.281 atm + 0.841 atm
Ptotal = 1.122 atm
Mole fraction: na/ntotal
nO2 = 0.188 mol (0.188 + 0.563) mol
nO2 = 0.250
nCH4 = 0.563 mol (0.188 + 0.563) mol
nCH4 = 0.750
0.250 + 0.750 = 1.000
Volume of each Gas
PO2 = VO2
Ptot Vtot
0.281 atm = VO2
(0.281 + 0.841) atm = 15 L
VO2 = 3.8 L
VCH4 = 15 L – 3.8 L = 11 L
Gas Collected Over Water Wet Gas: mixture of the desired gas and
water vapor
Pbar = Pgas + Pwater
Pgas = Pbar – Pwater
Next
Pbar = Pgas + Pwater
Pgas = Pbar - Pwater
Gas Collected Over WaterA sample of KClO3 is partially
decomposed, producing O2 gas that is collected over water. The volume of gas collected is 0.250 L at 26ºC and 765 torr total pressure.
2KClO3(s) ----- > 2 KCl(s) + 3O2(g)
Next ---- >
a. How many moles of O2 are collected?
Step 1: Calculate pressure of O2 in mixture
PO2 = Ptot – Pwater vapor
= 765 torr – 25 torr
PO2 = 740 torr
Next ---- >
nO2 = PV
RT
= (740 torr)(1atm/760 torr)(0.250 L) (0.0821 L-atm/mol-K)(299 K)
nO2 = 9.92 x 10-3 mol
Next ----- >
b. How many grams of KClO3 were decomposed?
(9.92 x 10-3 mol O2) x 2 mol KClO3 x 123 g KCLO3
3 mol O2 1 mol KClO3
= 0.811 g KClO3
c. When dry, what volume would the collected O2 gas occupy at the same temperature and pressure?
Remove water vapor. P2 = 765 torr Temperature is the same.
P1V1 = P2V2
V2 = (740 torr)(0.250 L) (765 torr)
V2 = 0.242 L
Kinetic-Molecular Kinetic-Molecular Theory of GasesTheory of Gases
Kinetic-Molecular Kinetic-Molecular Theory of GasesTheory of Gases
Assumptions
• The molecules of gases are in rapid random motion.
• Their average velocities (speeds) are proportional to the absolute (Kelvin) temperature.
• At the same temperature, the average kinetic energies of the molecules of different gases are equal.
Next
Relationship Between Temperature and Average
Kinetic Energy
Higher temperature means greater motion
(KE)av = 3 RT 2
Obtained from PV = RT = 2 (KE)av
n 3
PV = RT = 2 (KE)av = ½ mv² n 3
Apply to all gases whether alone or mixed with other gases
1. Constant V a. T changes – KE changes b. T increases – pressure increases c. KE changes – speed of molecules changes
2. Constant temperature a. V increases, P decreases b. KE doesn’t change – T didn’t change c. Speed of molecules doesn’t change-KE didn’t change
Root-Mean Square Speed (u 2)
• u 2 represents the average of the squares of the particles velocities.• The square root of u 2 is called the root mean square velocity. • The unit of measurement is m/s.
• u 2 = 3RT Units: R = 8.3145 J K-1mol-1
M or 8.3145 kgm2/s2 K-1mol-1
Joule = kg·m2/s2
Joule is a unit of measurement for energy.
Example
How is the root-mean square speed (rms) of F2 molecules in a gas sample changed by
a) An increase in temperature.b) An increase in volumec) Mixing with a sample of Ar at the same
temperature
Answers
a. Increasesb. No effectc. No effect
Example. Root-Mean Square Speed
Calculate the root mean square speed for the atoms in a sample of helium gas at 25ºC.
Given:T = 25ºC = 298 K
Known:R = 8.3145 J/K·mol or 8.3145 kgm2/s2 / K·mol
MHe : Change g to kg 4.00 g/mol = 4.00 x 10-3 kg/mol
Equation: u 2 = 3RT M
u 2= 3 (8.3145 kgm2/s2 /K·mol)(298K)
4.00 x 10-3 kg/mol
u = 1.36 x 103 m/s
Effusion
The escape of gas molecules from their container through a tiny orifice or pin hole.
Model of Gaseous Effusion
Diffusion
Diffusion is the moving or mixing of molecules of different substances as a result of random molecular motion.
Factors That Influence Effusion and Diffusion
• The greater the temperature the faster the ga will move or vise versa/
• If the mass increases, the kinetic energy will also increase if the speed remains constant.
• If the mass increases, the speed decreases if the kinetic energy remains constant.
Example. Rate of Effusion
Calculate the ratio of the effusion rates of hydrogen gas (H2) and uranium hexafluoride (UF6), a gas used in the enrichment process fuel for nuclear reactors.
Known:Molar Masses H2 = 2.016 g/mol UF6 = 352.02 g/mol
(Rate of effusion)² = MU compound
MH gas
= 352.02 2.016
Rate of effusion = 13.21
Graham’s Law
• The rate of effusion (or diffusion) of two different gases are inversely proportional to the square roots of their molar masses.
• (rate of effusion of A)2 = MB
• (rate of effusion of B)2 MA
Ratios of Rates the square root of two molar
masses is also equal to the ratio
• Molecular speeds• Effusion times• Distance traveled by molecules• Amount of gas effused
Based on Assumptions of Kinetic Molecular Theory
• Rates of diffusion of different gases are inversely proportional to the square roots of their densities.
• Rates of diffusion are inversely proportional to the square roots of their molar masses.
Nonideal (Real) GasesNonideal (Real) GasesNonideal (Real) GasesNonideal (Real) Gases
Ideal vs. Nonideal (Real) Gases
Ideal
• High pressure- compressible, volume approaches zero
• Force of collisions with container wall is great
Nonideal
• High pressure – molecules are practically incompressible
• Force of collision with container wall is less due to attractive force among the molecules.
Behavior of Gases
Behave ideally at
• High temperatures
• Low pressures
Behave nonideally at
• Low temperatures• High pressures
van der Waals van der Waals EquationEquation
van der Waals van der Waals EquationEquation
van der Waals EquationEquation corrects for volume and intermolecular forces
(P + n²a/V²)(V-nb) = nRT
• n²a/V² = related to intermolecular forces of attraction
• n²a/V² is added to P = measured pressure is lower than expected
• a & b have specific values for particular gases
• V - nb = free volume within the gas
Intermolecular Force of Attraction
Attractive forces of the orange molecules for the purple molecule cause the purple molecule to exert less force when it collides with the wall than it would if these attractive forces did not exist.
Source of Sketches/problems
• http://cwx.prenhall.com/bookbind/pubbooks/hillchem3/medialib/media_portfolio/04.html
• Chemistry, 7th ed. Brown, LeMay & Bursten, Prentice Hall