Transcript Gases

Gases
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
Next
• Expand when heated
Properties
that determine physical behavior
of a gas
Amount
• Mass or moles
Volume
lxwxh
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
N/cm2
Pa
kPa
Torr
mmHg
lb/in2
Newton per square meter
Newton per square centimeter
Pascal
kilopascal
Torr
millimeters of mercury
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 values
dwater = 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
Measurement of Gas Pressure
Mercury
Barometer
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
b.
1.50 atm x 101.325 kPa
1 atm
mmHg
1.50 atm x 760 mmHg
1 atm
= 152 kPa
= 1140 mmHg
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 Laws
Boyle’s Law
Charles’ Law
Gay-Lussac’s Law
Combined Gas Law
Temperature & Gas Laws
Three 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 = V 2
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
T1
or
= P2V2
T2
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
V1 = 12.0 L
T1 = 300 K
P2 = ?
V2 = 3.5 L
T2 = 325 K
Equation:
P1V1
T1
or
= P2V2
T2
P1V1T2 = P2V2T1
(760 Torr)(12.0 L)(325 K) =
( P2)(3.5 L)(300 K)
P2 = 2.8 x 10³ Torr
Law of Combining
Volumes
Gay-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 Law
Volume & 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.
b.
Equal
Volumes
equal number of
molecules
Equal number of –
moleculess
equal
volumes
Avogadro’s Law
At a fixed temperature and pressure, the
volume of a gas is directly proportional to
the amount of gas.
V=c·n
V = 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 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
Ideal Gas Law
Molar Mass
Density
Molar Mass Determination
•
n = number of moles
•
Moles = mass of sample
Molar mass
Ideal Gas Equation: n = PV
RT
Substituting: m = PV
M RT
M = mRT
PV
m
M
Example: Molar Mass
The 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
Reactions
Ideal Gas Law
&
Balanced 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 of
Partial Pressure
Mixture 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 =
ntot
Pa =
Ptot
Va
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 ---- >
1 mol CH4
16.0 g CH4
= 0.563 mol CH4
nCH4 = 9.00 g CH4 x
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 =
Ptot
VO2
Vtot
0.281 atm
=
(0.281 + 0.841) atm =
VO2
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 Water
A 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
3 mol O2
= 0.811 g KClO3
123 g KCLO3
1 mol 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
Theory 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
•
u 2 = 3RT
particles velocities.
• The square root of u 2 is called the root mean square velocity.
• The unit of measurement is m/s.
M
Units: R = 8.3145 J K-1mol-1
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 volume
c) Mixing with a sample of Ar at the same
temperature
Answers
a. Increases
b. No effect
c. 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) Gases
Ideal vs. Nonideal (Real) Gases
Ideal
• High pressurecompressible, 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
Behave nonideally at
• High temperatures
• Low pressures
• Low temperatures
• High pressures
van der Waals Equation
van der Waals Equation
Equation 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/pu
bbooks/hillchem3/medialib/media_po
rtfolio/04.html
• Chemistry, 7th ed. Brown, LeMay &
Bursten, Prentice Hall