Chapter 5: Gases
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
5.10
5.11
5.12
Early Experiments
The Gas Laws of Boyle, Charles, and Avogadro
The Ideal Gas Law
Gas Stoichiometry
Dalton’s Law of Partial Pressures
The Kinetic Molecular Theory of Gases
Effusion and Diffusion
Collisions of Gas Particles with the Container Walls
Intermolecular Collisions
Real Gases (first 3 paragraphs, pgs 171 - 172)
Characteristics of Several Real Gases
Chemistry in the Atmosphere
1
Substances that are Gases under Normal
Conditions
Substance
Helium
Neon
Argon
Hydrogen
Nitrogen
Nitrogen Monoxide
Oxygen
Hydrogen Chloride
Ozone
Ammonia
Methane
Formula
MM (g/mol)
He
Ne
Ar
H2
N2
NO
O2
HCl
O3
NH3
CH4
4.0
20.2
39.9
2.0
28.0
30.0
32.0
36.5
48.0
17.0
16.0
2
Some Important Industrial Gases
Name - Formula
Origin and use
Methane (CH4)
Natural deposits; domestic fuel
Ammonia (NH3)
From N2 + H2 ; fertilizers, explosives
Chlorine (Cl2)
Electrolysis of seawater; bleaching
and disinfecting
Oxygen (O2)
Liquefied air; steelmaking
Ethylene (C2H4 )
High-temperature decomposition of
natural gas; plastics
3
Important Characteristics of Gases
1) Gases are highly compressible
An external force compresses the gas sample and decreases its
volume; removing the external force allows the gas volume to
increase.
2) Gases are thermally expandable
When a gas sample is heated, its volume increases; when it is
cooled its volume decreases.
3) Gases have low viscosity
Gases flow much easier than liquids or solids.
4) Most gases have low densities
Gas densities are on the order of grams per liter ,whereas liquids
and solids are grams per cm3 (mL), 1000 times greater.
5) Gases are infinitely miscible
Gases mix in any proportion (air is a mixture of many gases).4
Pressure of the Atmosphere
• Called “atmospheric pressure”
– the force exerted on earth’s surface by the gases in air
– the force exerted upon us by the atmosphere above us
• The force per unit area of these gases or a measure of
the weight of the atmosphere pressing down upon us.
Pressure =
Force
Area
• Can be measured using a barometer – a device that
can weigh the atmosphere above us
5
Figure 5.1:
A Torricellian barometer.
Density of Mercury = 13.6 g/cm3
760 mm column of 1 cm2 area weighs
= 76 cm x 1 cm2 x 13.6 g/cm3
= 1030 g = 1.03 kg = 2.28 lbs
P = force / area
= 2.28 pounds / cm2
= 14.7 pounds / in2
= 1.00 atm
6
Common Units of Pressure
Unit
Atmospheric Pressure
Pascal (Pa) = N/m2;
kilopascal (kPa)
bar
1.01325 x 105 Pa
101.325 kPa
1.01325 bar
Scientific Field Used
SI unit; physics,
Chemistry
Meteorology, Chemistry
atmosphere (atm)
1 atm
Chemistry
millimeters of mercury
(mmHg), also called
“torr”
760 mmHg
760 torr
Chemistry, medicine,
biology
pounds per square inch
(psi or lb/in2)
14.7 lb/in2
Engineering
7
Converting Units of Pressure
Problem: A chemist collects a sample of carbon dioxide from the
decomposition of limestone (CaCO3) in a closed-end manometer
(i.e., vacuum is the reference pressure), the height of the mercury is
341.6 mm Hg. Calculate the CO2 pressure in torr, atmospheres, and
kiloPascals.
Plan: The pressure is in mm Hg, so we use the conversion factors from
the table on the previous slide to find the pressure in the other units.
Solution:
PCO2
converting from mmHg to torr:
760 torr
(torr) = 341.6 mm Hg x
= 341.6 torr
760 mm Hg
converting from torr to atm:
1 atm
PCO2( atm) = 341.6 torr x
= 0.4495 atm
760 torr
converting from atm to kPa:
PCO2(kPa) = 0.4495 atm x
101.325 kPa
= 45.54 kPa
1 atm
8
Measuring Gas Pressure Inside a Container
Figure 5.2:
simple
manometer
Pgas = Patm - h
Pgas = Patm + h
9
Boyle: “pressure/volume relation”
By adding more mercury
to the J tube, Boyle
increased the pressure of
the gas trapped at the
closed end.
PV= constant
10
Figure 5.4: Plotting Boyle’s data from Table 5.1
11
Boyle’s Law: P & V of a gas sample inversely
proportional when n and T are constant
Pressure is inversely proportional to volume:
P=
k
V
or
V=k
P
or
PV = k
“Change of Conditions Problems”:
if n and T are constant
P1V1 = k
P2V2 = k
then
P 1 V 1 = P2 V 2
12
Decreasing the pressure in the
marshmallow vessel will cause the
marshmallows to...
PV = k
1. Get bigger
2. Get smaller
3. Stay the same
Marshmallow Demo
13
Applying Boyles Law to Gas Problems
Problem: A gas sample at a pressure of 1.23 atm has a volume of
15.8 cm3, what will be the volume in L if the pressure is increased to
3.16 atm? (Do you expect volume to increase or decrease?)
Solution:
P1 = 1.23 atm
P2 = 3.16 atm
3
V1 = 15.8 cm
V2 = unknown
T and n remain constant
1cm3 = 1 mL
V1 (mL)
P1V1 = P2V2
1 mL
1L
V1 = 15.8 cm3 x
x
3
1 cm 1000mL
V1 (cm3)
1000mL = 1L
= 0.0158 L
V1 (L)
x P1/P2
V2 = V1 x
P1
1.23 atm
= 0.0158 L x
= 0.00615 L
P2
3.16 atm
V2 (L)
14
Figure 5.7:
Plots of V versus T (°C) for several gases.
15
Figure 5.8:
Plots of V versus T as in Fig. 5.7 except that here
the Kelvin scale is used for temperature.
16
Charles Law: V & T proportional
At constant pressure and for a fixed amount (# moles) of gas:
volume is directly proportional to temperature
V= constant x Absolute Temperature
V=bxT
“Change of Conditions” problems:
Since V/T = b
or
V1 / T1 = V2 / T2
or:
T1
V1
=
T2
V2
T1 = V1 x
T2
V2
Temperatures must be expressed in Kelvin!
17
According to Charles’ law if the
temperature of a gas increases what
happens to its volume?
V/T = b
1. It decreases
2. It increases
3. It stays the same
Charles Law Demo
18
A sample of carbon monoxide, a poisonous
gas, occupies 3.20 L at 125 °C. At what
temperature (°C) will the gas occupy 1.54 L if
the pressure remains constant?
T1
V1
1.
2.
3.
4.
=
T2
V2
192 °C
60 °C
-81 °C
-213 °C
19
Charles Law Problem
A sample of carbon monoxide, a poisonous gas, occupies
3.20 L at 125 oC. Calculate the temperature (oC) at which the
gas will occupy 1.54 L if the pressure remains constant.
V1 = 3.20 L
T1 = 125oC = 398 K
V2 = 1.54 L
T2 = ?
(volume decreases, so what will happen to T?)
T2 = 398 K x 1.54 L
3.20 L
T2 = T1 x ( V2 / V1)
T2 = 192 K
oC
= 192 K
= K - 273.15 = 192 - 273
= -81oC
oC
20
AVOGADRO
volume/molecules
“for gas at constant temperature and
pressure, the VOLUME is directly
proportional to the number of
MOLES”
Avogadro’s # = 6.022 x 1023 particles/mole
21
Which gas would have the largest number
of molecules per mole?
1.
2.
3.
4.
5.
6.
7.
CO2
CH4
He
O2
Xe
H2
All are equal
22
Avogadro’s Law - Moles and Volume
The volume of gas is directly proportional to the amount of gas
(in moles), when measured at the same P and T:
V∝n
or
V = constant x n
V=axn
n1 = initial moles of gas
n2 = final moles of gas
n1 =
V1
n2
V2
V1 = initial volume of gas
V2 = final volume of gas
or
n1 = n2 x V1
V2
23
Avogadro’s Law: Volume and Amount of Gas
Problem:
Sulfur hexafluoride is a gas used to trace pollutant plumes in the
atmosphere. If the volume of 2.67 g of SF6 at 1.143 atm and
28.5 °C is 2.93 m3, what will be the mass of SF6 in a container
with a volume of 543.9 m3 at 1.143 atm and 28.5 °C?
Avogadro's Law
Temp and
Pressure don’t
change
Calculate moles of
SF6
n1 n 2
=
V1 V2
Convert to moles
after volume change
Convert back
to mass of SF6
24
If the volume of 2.67 g of sulfur hexafluoride
(SF6) at 1.143 atm and 28.5 °C is 2.93 m3, how
many moles of SF6 will be in a container with a
volume of 543.9 m3 at 1.143 atm and 28.5 °C?
n1 n 2
=
V1 V2
1.
2.
3.
4.
0.0183 mol
0.0381 mol
3.40 mol
7.08 mol
25
Avogadro’s Law: Volume and Amount of Gas
Problem: Sulfur hexafluoride is a gas used to trace pollutant plumes in
the atmosphere, if the volume of 2.67 g of SF6 at 1.143 atm and 28.5 oC
is 2.93 m3, what will be the mass of SF6 in a container with a volume of
543.9 m3 at 1.143 atm and 28.5 oC?
Solution: Molar mass SF6 = 146.07 g/mol
2.67g SF6
= 0.0183 mol SF6
146.07g SF6/mol
n2 = n1 x
V2
V1
= 0.0183 mol SF6 x
543.9 m3
= 3.40 mol SF6
2.93 m3
mass SF6 = 3.40 mol SF6 x 146.07 g SF6 / mol = 496 g SF6
26
GASES
•
•
•
•
Torricelli 1608-1647
Boyle 1627-1691
Charles 1746-1823
Avogadro 1776-1856
“air exerts pressure”
“pressure/volume”
“volume/temperature”
“volume/molecules”
IDEAL GAS LAW:
PV=nRT
Knowledge of pressure, volume and temperature
completely define the state of a gas since the fourth
property can be determined from the ideal gas equation.
27
IDEAL GAS LAW
The ideal gas equation combines the Boyle’s, Charles’, and
Avogadro’s laws into one easy-to-remember law:
PV=nRT
n = number of moles of gas in volume V
R = Ideal gas constant
R = 0.08206 L·atm / (mol·K) = 0.0821 L·atm·mol-1·K-1
An ideal gas is one for which both the volume of the
molecules and forces between the molecules are so small
that they have insignificant effect on its P-V-T behavior.
28
29
Variations on the Ideal Gas Equation
During chemical and physical processes, any of the four variables
in the ideal gas equation may be fixed.
Thus, PV=nRT can be rearranged for the fixed variables:
– for a fixed amount (n) at constant temperature (T)
PV = nRT = constant
Boyle’s Law
– for a fixed amount (n) at constant pressure (P)
V/T = nR/P = constant
Charles’ Law
– for a fixed pressure (P) and temperature (T)
V = n (RT/P) or V ∝ n
Avogadro’s Law
REMEMBER…rearrange PV = nRT as needed
For example, for constant volume (V):
when amount (n) constant…P/T = nR/V = constant
when T constant…P = n (RT/V) so P ∝ n
30
Standard Temperature and Pressure (STP)
A set of standard conditions have been chosen to make it easier to
quantify gas amounts (i.e., “liters at STP”).
Standard Temperature = 0 °C = 273.15 K
Standard Pressure = 1 atmosphere = 760 mm Hg
At these “standard” conditions, if you have 1.0 mole of any ideal gas, it
will occupy a “standard molar volume”: V = nRT/P
Standard Molar Volume = 22.414 liters = 22.4 L or L/mol
31
32
Values of R in Different Units
L · atm
R* = 0.08206 mol · K
*
R = 62.36
L · torr
mol · K
R = 8.314
kPa · dm3
mol · K
R = 8.314
J#
mol · K
Most calculations in this course use values of R to 3 significant figures.
#
J is the abbreviation for joule, the SI unit of energy. The joule is a
derived unit composed of the base units kg·m2/s2
33
Gas Law: Solving for Pressure
Problem: Calculate the pressure in a 87.5 L container filled with
5.038 kg of xenon at a temperature of 18.8 °C.
Plan:
Volume is in
L
Convert kg of
Xe to moles
Ideal Gas Law
Check value for R
Pressure in
atm
Convert temp
to K
34
What is the pressure in a 87.5 L container
filled with 5.038 kg of xenon at a
temperature of 18.8 °C?
PV=nRT
R = 0.08206 L*atm/mol*K
1.
2.
3.
4.
1.64 atm
10.5 atm
25.5 atm
676 atm
35
Gas Law: Solving for Pressure
Problem: Calculate the pressure in a 87.5 L container filled with
5.038 kg of xenon at a temperature of 18.8 °C.
Solution:
nXe =
5038 g Xe
= 38.37014471 mol Xe
131.3 g Xe / mol
T = 18.8 oC + 273.15 K = 291.95 K
PV = nRT
P=
so P = nRT
V
(38.37 mol )(0.0821 L·atm)(291.95 K)
87.5 L (mol·K)
= 10.5 atm
36
Ammonia Density Problem
Calculate the density of ammonia gas (NH3) in grams per liter at
752 mm Hg and 55 °C.
density = d = mass per unit volume = g / L
P = 752 mm Hg x (1 atm/ 760 mm Hg) =0.989 atm
T = 55 °C + 273.15 = 328.15 K
PV = nRT so…
( 0.989 atm)
P
n/V =
=
R xT
( 0.08206 L·atm / mol·K) ( 328.15 K)
n/V =
0.03674 mol/L
density = 0.03674 mol/L x (17.03 g/mol NH3)
d = 0.626 g / L (more than 1000x less dense than water!)
37
Ammonia Density Problem – a different approach
Calculate the density of ammonia gas (NH3) in grams per liter at
752 mm Hg and 55 °C.
density =d = mass per unit volume = g / L
P = 752 mm Hg x (1 atm/ 760 mm Hg) =0.989 atm
T = 55 oC + 273.15 = 328.15 K
n = mass / Molar Mass = m / MM
PV = nRT
so
n= m = P· V
MM
R·T
therefore,
d=
P x MM
=
R xT
d = m/V = (P·MM)/(RT)
( 0.989 atm) ( 17.03 g/mol)
( 0.08206 L·atm/mol·K) ( 328 K)
d = 0.626 g / L
38
Applying the Temperature - Pressure
Relationship (constant V and n)
Problem: A copper tank is filled with compressed gas to a pressure of
4.28 atm at a temperature of -17.68 °C . What will be the pressure if the
temperature is raised to 95.6 °C?
Plan: The volume of the tank and number of moles are not changed so we
only have to deal with the temperature change. Convert temperature to
Kelvin and use the T ratio and initial pressure to calculate the new P.
Solution:
T1 = -17.68 oC + 273.15 K = 255.47 K
T2 = 95.6 oC + 273.15 K = 368.8 K
P1
P
= 2 = nR/V
T1
T2
P2 = P1 x
T2
=?
T1
P2 = 4.28 atm x
368.8 K
255.47 K
= 6.18 atm
39
Change of Conditions
A gas sample in the laboratory has a volume of 45.9 L at
25 °C and a pressure of 743 mm Hg. If the temperature is
increased to 155 oC by compressing the gas to a new volume of
3.10 mL, what is the pressure?
Plan:
Convert all units to
K, L, atm
Constant moles
Set up “before”
and “after”
equations
Solve for new
pressure
40
Change of Conditions
P1= 743 mm Hg x 1 atm/ 760 mm Hg = 0.978 atm
V1 = 45.9 L V2 = 3.10 mL = 0.00310 L
T1 = 25 °C + 273.15 = 298.15 K
T2 = 155 °C + 273.15 = 428.15 K
P1 x V1
T1
=
P2 x V2
T2
( 0.978 atm) ( 45.9 L)
( 298.15 K)
= nR = constant
=
P2 (0.00310 L)
(428.15 K)
( 428.15.15 K) ( 0.978 atm) ( 45.9 L)
= 2.08 x 104 atm
P2 =
41
( 298 K) ( 0.00310 L)
Example: Molar Mass of a gas from its
mass and P, V, and T
Problem: A sample of natural gas is collected at 25.0 °C in a
250.0 mL flask. If the sample had a mass of 0.118 g at a pressure of
550.0 torr, what is the molecular weight of the gas?
Plan:
Convert all units to
K, L, atm
Calculate mass of vapor
Use ideal gas law to
calculate moles
Use mass / mole
relationship to calculate
molar mass
42
Example: Molar Mass of a gas from its
mass and P, V, and T
Problem: A sample of natural gas is collected at 25.0 °C in a
250.0 mL flask. If the sample had a mass of 0.118 g at a pressure of
550.0 torr, what is the molecular weight of the gas?
Plan: Use the ideal gas law to calculate n, then calculate the molar
mass.
Solution:
1mm Hg
1.00 atm
P = 550.0 torr x
x
= 0.724 atm
1 torr
760 mm Hg
V = 250.0 mL x
1.00 L
= 0.250 L
1000 mL
T = 25.0 °C + 273.15 K = 298.2 K
(0.724 atm)(0.250 L)
n=
(0.0821 L atm/mol K) (298.2 K)
n=
PV
RT
= 0.007393 mol
MM = mass / n = 0.118 g / 0.007393 mol = 16.0 g/mol
43
Problem: A volatile liquid is placed in a 590.0 mL flask and heated
until all of the liquid is gone. Only vapor fills the flask at a
temperature of 100.0 °C and a pressure of 736 mm Hg. If the mass of
the empty flask was 148.375 g and the mass of the flask containing
the vapor is 149.457 g, what is the molar mass of the liquid?
Plan: Use the gas law to calculate the moles, then the moles to
calculate the molar mass.
PV=nRT
R = 0.08206 L*atm/mol*K
1.
2.
3.
4.
0.076 g/mol
0.186 g/mol
1.082 g/mol
58.02 g/mol
44
Molar Mass of a Gas (like lab #4)
Problem: A volatile liquid is placed in a 590.0 mL flask and heated until
all of the liquid is gone. Only vapor fills the flask at a temperature of
100.0 °C and a pressure of 736 mm Hg. If the mass of the empty flask was
148.375 g and the mass of the flask containing the vapor is 149.457 g,
what is the molar mass of the liquid?
Plan: Use the gas law to calculate the moles, then the moles to calculate
the molar mass.
Solution:
1 atm
pressure = 736 mm Hg x
= 0.9684 atm
760 mm Hg
n = (PV/RT) =
( 0.9684 atm)(0.590 L)
(0.0821 L atm/mol K)(373.2 K)
= 0.0186 mol
mass = 149.457g - 148.375g = 1.082 g
M = mass / n = 1.082 g / 0.0186 mol = 58.02 g/mol
note: the compound is acetone C3H6O = MM = 58 g mol !
Chemical Equations and
Calculations
Atoms (Molecules)
Avogadro’s
Number
6.022 x 1023 mol-1
Reactants
Molarity
moles / L
Solutions
Moles
Molar
Mass
45
Mass
g/mol
Products
PV = nRT
Gases
46
Gas Reaction Problem
Problem: Calculate the volume of carbon dioxide produced by the
thermal decomposition of 36.8 g of calcium carbonate at 715 mm Hg and
37 °C, according to the reaction below:
CaCO3(s)
CaO(s) + CO2 (g)
Plan:
Use reaction to determine
moles of CO2 formed
Convert CaCO3 mass to
moles
Using MM of CaCO3
Use ideal gas equation to
calculate V
Convert pressures and temps
into appropriate units
Kelvin and atm
47
Gas Reaction Problem
Problem: Calculate the volume of carbon dioxide produced by the
thermal decomposition of 36.8 g of calcium carbonate at 715 mm Hg and
37 °C, according to the reaction below:
CaCO3(s)
CaO(s) + CO2 (g)
Solution:
1 mol CaCO
36.8g x 100.1g CaCO3 = 0.368 mol CaCO3
3
Moles of CO2 = mol CaCO3 x
V=
nRT
P
=
1 mol CO2
1 mol CaCO3
(0.368 mol CO2)(0.08206 L atm/mol K)(310.15K)
(715 mmHg x
atm
)
760 mm Hg
V = 9.96 L
48
Gas Law Stoichiometry - I - NH3 + HCl
Problem: A partition separating two containers is removed and the gases
are allowed to mix and react. The first container with a volume of 2.79 L
contains ammonia gas at a pressure of 0.776 atm and a temperature of
18.7 °C. The second, with a volume of 1.16 L, contains HCl gas at a
pressure of 0.932 atm and a temperature of 18.7 °C. What mass of solid
ammonium chloride will be formed, what is left in the container, and
what is the final pressure?
Plan: This is a limiting reactant problem, so we must calculate the moles
of each reactant using the gas law to determine the limiting reagent. Then
we can calculate the mass of product and determine what is left in the
combined volume of the container and the final P.
Solution:
Equation:
NH3 (g) + HCl(g)
NH4Cl(s)
T (in Kelvin) = 18.7 oC + 273.15 = 291.9 K
49
Gas Law Stoichiometry - II - NH3 + HCl
ni =
PiV
RT
(0.776 atm) (2.79 L)
nNH3 =
= 0.0903 mol NH3 ¸ 1
(0.0821 L atm/mol K) (291.9 K)
nHCl =
(0.932 atm) (1.16 L)
= 0.0451 mol HCl ¸ 1
(0.0821 L atm/mol K) (291.9 K)
limiting
reactant
Therefore, the product will be 0.0451 mol NH4Cl or 2.28 g NH4Cl
Ammonia remaining = 0.0903 mol - 0.0451 mol = 0.0452 mol NH3
V = 1.16 L + 2.79 L = 3.95 L
PNH3=
nNH3RT (0.0452 mol) (0.0821 L atm/mol K) (291.9 K)
=
=0.274 atm
V
(3.95 L)
50
Gas Mixtures: Dalton’s Law of Partial Pressures
Ptotal = P1+P2+P3…
n1RT + n2RT + n3RT
V
V
V
RT
Ptotal = (n1+n2+n3…)
V
Ideal gas equation applies to a mixture as a whole and to each gas
species individually (i):
PtotalV = ntotalRT
Ptotal =
Ptotal is what is measured by pressure gauges.
PiV = niRT for each species, i. Pi harder to measure.
51
Dalton’s Law of Partial Pressures
A 2.00 L flask contains 3.00 g of CO2 and 0.10 g of helium at a
temperature of 17.0 °C.
What are the partial pressures of each gas and the total pressure in
the flask?
T = 17 °C + 273 = 290 K
nCO2 = 3.00 g CO2 x
PCO2 = nCO2RT/V
PCO2 =
mol CO2
= 0.0682 mol CO2
44.01 g CO2
( 0.0682 mol CO2) ( 0.08206 L·atm/mol·K) ( 290 K)
PCO2 = 0.811 atm
(2.00 L)
52
Dalton’s Law of Partial Pressures
= 0.025 mol He
nHe = 0.10 g He x mol He
4.003 g He
PHe = nHeRT/V
PHe =
(0.025 mol) ( 0.08206 L atm / mol K) ( 290 K )
( 2.00 L )
PHe = 0.30 atm
PTotal = PCO2 + PHe = 0.811 atm + 0.30 atm
PTotal = 1.11 atm
53
Mole Fraction of a Species
χi = ni
ntotal
ntotal =n1+n2+n3…
Σ χi = χ1+χ2+χ3…= 1.00000
PiV = niRT for each i and
Pi =
niRT =
V
Σ Pi = P1+P2+P3…= Ptotal
ni
.
ntotal (ntotalRT/V) = χi Ptotal
Partial pressure of each gas is equal to its mole
fraction multiplied by the total pressure.
54
Figure 5.9: Partial pressure of each gas in a mixture of
gases depends on the number of moles of that gas (ni).
Defined as Pi = (ni/ntotal) Ptotal
55
Example Problem Using Mole Fractions
A mixture of gases contains 4.46 mol Ne, 0.74 mol Ar and 2.15
mol Xe. What are the partial pressures of the gases if the total
pressure is 2.00 atm ?
Plan:
Calculate total
number of
moles
Calculate individual
mole fractions
Add up all mole
values
Divide individual moles
by total moles
Calculate
partial pressures
Multiply mole
fractions by total
pressure
56
A mixture of gases contains 4.46 mol Ne,
0.74 mol Ar and 2.15 mol Xe. What is the
partial pressure of the neon if the total
pressure is 2.00 atm ?
1.
2.
3.
4.
0.202 atm
0.586 atm
1.21 atm
4.46 atm
57
Example Problem Using Mole Fractions
A mixture of gases contains 4.46 mol Ne, 0.74 mol Ar and 2.15 mol Xe.
What are the partial pressures of the gases if the total pressure is 2.00 atm ?
Total # moles = 4.46 + 0.74 + 2.15 = 7.35 mol = ntotal
xNe = nNe / ntotal = 4.46 mol Ne / 7.35 mol = 0.607
PNe = xNe PTotal = 0.607 (2.00 atm) = 1.21 atm Ne
xAr = 0.74 mol Ar / 7.35 mol = 0.101
PAr = xAr PTotal = 0.101 (2.00 atm) = 0.202 atm Ar
xXe = 2.15 mol Xe / 7.35 mol = 0.293
PXe = xXe PTotal = 0.293 (2.00 atm) = 0.586 atm Xe
58
Figure 5.10: Production of O2(g) by
thermal decomposition of KCIO3:
It is mixed with vapor pressure of water.
59
Saturation
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. .
. .
.
60
Collection of Hydrogen Gas Over Water
Vapor pressure - I
2 HCl(aq) + Zn(s)
ZnCl2 (aq) + H2 (g)
Calculate the mass of hydrogen gas collected over water if
156 mL of gas is collected at 20oC and 769 mm Hg.
PTotal = PH2 + PH2O
PH2 = PTotal - PH2O
FROM TABLE
PH2 = 769 mm Hg - 17.5 mm Hg
= 752 mm Hg
T = 20oC + 273 = 293 K
PH2 = (752 mm Hg )(1 atm/ 760 mm Hg) = 0.989 atm
V = 0.156 L
61
Collection of Hydrogen Gas Over Water
Vapor Pressure - II
PH2V = nH2RT
nH 2 =
nH2 = PH2V / RT
(0.989 atm) (0.156 L)
(0.0821 L atm/mol K) (293 K)
nH2 = 0.00641 mol
Mass H2 = 0.00641 mol x (2.01 g H2 / mol H2)
= 0.0129 g H2
62
KINETIC MOLECULAR THEORY
of GASES (abbreviated)
is a model that tries to explain the properties of ideal
gases and speculates about behavior of individual
gas particles.
This model assumes:
• gas molecules are small (negligible volume)
• gas molecules are moving
• molecules do not attract or repel each other
• average kinetic energy of gas molecules is
temperature dependent regardless of identity
63
Postulates of Kinetic Molecular Theory (KMT)
I
The particles are so small compared to the distances between them
that the volume of the individual particles can be assumed to be
negligible (zero).
II
The particles are in constant motion. The collisions of the particles
with the walls of the container are the cause of the pressure exerted
by the gas.
III The particles are assumed to exert no forces on each other; they are
assumed to neither attract nor repel each other.
IV The average kinetic energy of a collection of gas particles assumed
to be directly proportional to the Kelvin temperature of the gas.
64
Relating KMT to PV = nRT
Boyle’s Law (P vs. V at constant n and T)
P = (nRT)
constant
1
V
KMT: Decrease in volume
means gas particles will hit
the container walls more
often.
P vs T at constant n and V
P=
( nRV )T
constant
KMT: Increase in temperature
means the speed of gas particles
increases; particles will hit the
walls with greater force and
frequency.
65
Relating KMT to PV = nRT
Charles’ Law (V vs. T at constant n and P)
( nRP )T
V=
constant
KMT: Increase in temperature means
the speed of gas particles increases;
they will hit the walls with greater
force and frequency. P can only be
constant if V increases to offset
increased particle speeds.
Avogadro’s Law (V vs. n at constant T and P)
( RTP )n
V=
constant
KMT: Increase in number of
particles at constant T would
increase the P if V held constant;
to keep P constant, V must
increase
66
Relating KMT to PV = nRT
Dalton’s Law (Mixture of Gases)
Ptotal = P1 + P2 + P3…
KMT: assumes all gas particles
are independent of each other
and that the particle volumes
are negligible
→ identity of gases does not
matter when determining
pressure (only the number of
particles)
67
Which of the following statements is true?
1.
2.
3.
4.
A gas particle that is
behaving ideally has no
mass.
The observed pressure of a
gas depends on the size of
the gas molecule.
The gas particles of a
theoretical ideal gas would
not collide with each other.
Carbon dioxide will behave
more ideally than helium at
the same conditions of
pressure and temperature.
68
Looking into the Kinetic Molecular Theory
(applying principles of physics)
The kinetic energy associated with the velocity of molecules:
KE = (1/2) mass x velocity2
KE = (1/2) m x u2
The temperature of matter is a measure of its thermal energy,
such that T is proportional to the average KE (velocity (u) is
proportional to √T).
Gas molecules collide with the walls and, on average, simply reverse
their momentum in the direction perpendicular
to the wall. This gives rise to the pressure (force per unit area)
on the wall.
Experiment:
PV = RT
n
Theory:
PV ∝ T
n
69
Velocity and Energy
• Average Kinetic Energy of a gas particle
KEavg = ½ mu2
• Average Kinetic Energy for a mole of gas particles:
KEavg = NA (½ mu2 )
• Applying definitions of velocity, momentum, force, and pressure, we can
derive the following expression for pressure:
3[
P = 2
• Substituting…
n NA (½ mu2 )
V
P =
2
3
[
]
n (KEavg)
V
]OR
PV = 2 (KE )
avg
n
3
• From KMT, T (in K) indicates KEavg of the particles
PV
n
∝
T
70
Velocity and Energy
• Combining
PV = RT
n
= 2
3 (KEavg)
KEavg =
3 RT
2
• Kelvin temperature is an index of the random motions of the
particles of a gas → higher temperature means greater motion
• KEavg is independent of gas identity!
• Energy units are “joules” (kg·m2/s2), so use the gas constant with
the appropriate units: 8.314 J/mol·K
71
Velocity and Energy
• Average velocity of particles from KMT is a special kind of
average…root mean square velocity:
urms =√(u2)
• From:
= 3
2 RT
KEavg = NA (½ mu2 )
½ mu2 = 3 RT
2 NA
u2 = 3RT/(mNA)
= 3RT/M
(since m = M/NA)
√(u2) = urms = √(3RT/M)
M is molar mass in kg/mol
72
Probable path of a single gas particle in a box.
Actual path affected by:
• collisions (mean free path: average distance a particle travels
between collisions)
• temperature
73
Fig 5.15
Individual molecules
exhibit different
velocities.
Plot of O2 molecules
reveals that AVERAGE
velocity is the velocity
shared by most
molecules.
74
Figure 5.16:
A plot of the relative
number of N2
molecules that have a
given velocity at three
temperatures.
Average velocities of
molecules are
TEMPERATURE
and MOLAR MASS
dependent:
urms = √(3RT/M)
75
Based on the graph, which of the following gas
molecules, at 0 °C and 1 atm, would you expect to
have the highest average velocity?
1.
2.
3.
4.
5.
He
Cl2
CH4
NH3
All are equal
76
urms = √(3RT/M)
@ 273K
At same T, heavier gases go slower, as 1/√M
At a given temperature,
all gases have the same:
77
He, O2, H2 , Cl2 ...
molecular kinetic energy
distributions
Or
average molecular kinetic
energy
for ANY type of gas
molecules !
KEavg = 3/2 RT, independent of M!
78
Three Ways to Describe Typical Velocity
for Gas Particles
“Root mean square velocity”
urms =√(u2) = √(3RT/M)
Using the Maxwell-Boltzmann distribution:
(kB and R related by NA)
“Most probable velocity”
ump = √(2kBT/m) = √(2RT/M)
“Average velocity”
uavg = √(8kBT/πm) = √(8RT/πM)
M is molar mass in kg
79
(must use 8.314 J/mol K for R; remember that 1 J = 1
kg·m2/s2)
Molecular Mass and Molecular Speeds - I
Problem: Calculate the molecular speeds (urms ) of the molecules of
hydrogen, methane, and carbon dioxide at 300K.
Plan: Use the equations for the average kinetic energy of a molecule and
the relationship between kinetic energy and velocity to calculate the
average speeds of the three molecules.
Solution: for hydrogen, H2 = 2.016 g/mol
KEmolecule =
3 R
8.314 J/mol·K
x T = 1.5 x
x 300K
2 NA
6.022 x 1023 molecules/mol
KEmolecule = 6.213x10 - 21 J/molecule = 1/2 mu2
m=
2.016 x 10 -3 kg/mole
= 3.348x10- 27 kg/molecule
6.022 x 1023 molecules/mole
6.213x10 - 21
KEmolecule = 1/2 mu2
J/molecule = 1.674x10 - 27 kg/molecule (u2)
urms = 1.926 x 103 (J/kg)1/2 = 1.926 x 103 m/s = 1,926 m/s
80
What are the molecular speeds (urms ) of the molecules of
methane (CH4) and carbon dioxide at 300K?
KEmolecule = 6.213x10 - 21 J/molecule = 1/2 mu2
1.
2.
3.
4.
5.
6.
2.78x10-11 m/s,
1.68x10-11 m/s
15.29 m/s, 9.22 m/s
21.62 m/s, 13.04 m/s
483.5 m/s, 291.6 m/s
683.8 m/s , 412.3 m/s
4.68x105 m/s,
1.70x105 m/s
81
Molecular Mass and Molecular Speeds - II
Root mean square velocity = urms = √(3RT/M)
√(u2) = √(3RT/M)
R = 8.314 J/mol K = 8.314 kg·m2/(s2·mol·K)
Note: 1 J = 1 kg·m2/s2
For methane , CH4: M = 16.04 g/mole = 16.04 x 10 - 3 kg/mole
√(u2) = √ [(3 x 8.314 kg·m2/(s2·mol·K) x 300 K)/ 16.04 x 10 - 3 kg/mole]
urms = 683.8 m/s
For carbon dioxide , CO2 = 44.01 g/mole
= 44.01 x 10 - 3 kg/mole
√(u2) = √ [(3 x 8.314 kg·m2/(s2·mol·K) x 300 K)/ 44.01 x 10 - 3 kg/mole]
urms = 412.3 m/s
82
Molecular Mass and Molecular Speeds - III
Molecule
H2
Molecular
Mass (g/mol)
2.016
Kinetic Energy
(J/molecule)
Velocity
(m/s)
µ 1/√M
6.213 x 10 - 21
1,926
CH4
CO2
16.04
6.213 x 10 - 21
44.01
6.213 x 10 - 21
683.8
412.3
83
DIFFUSION and EFFUSION
Effusion - A gas escaping from a container into a
vacuum. There are no other (or very few)
molecules available for collisions.
Diffusion - One gas mixing with another gas, or
gases, where the molecules are colliding with each
other and exchanging energy between molecules.
84
EFFUSION
“rate of effusion of a gas is inversely proportional to
the square root of the mass of its particles”
“relative rates of effusion of two gases are given by the
inverse ratio of square roots of masses of gas
particles”
GRAHAM (1805-1869)
Heavy molecules are moving slowly
85
Effusion of gas into an
evacuated chamber
86
Effusion
• Grahams Law
– Rate of effusion is inversely proportional to
the square root of the mass of the gas.
• For two gases, the relative rate of effusion
is:
Rate of effusion gas 1
Rate of effusion gas 2
=
M2
M1
87
DIFFUSION
• diffusion is the rate of gas mixing
• diffusion velocities are proportional to effusion
velocities
• diffusion velocities are much slower than
effusion velocities due to collisions with gas
molecules present in the existing space
88
Diffusion Rates: proportional to
average velocity or to 1/√M
dNH3 distance traveled by NH3 uavg(NH3)
= distance traveled by HCl u =
dHCl
avg(HCl)
√MHCl
=
√MNH3
Rate1/Rate2 = uavg 1/ uavg 2 = (M2/M1)1/2
HCl = 36.46 g/mol
NH3 = 17.03 g/mol
RateNH3 = ( 36.46 / 17.03 )1/ 2 x RateHCl
RateNH3 = 1.463 x RateHCl
89
Figure 5.19: (a) demonstration of the
relative diffusion rates of NH3 and HCl
molecules through air.
90
Figure 5.19: (b) When HCl (g) and
NH3 (g) meet in the tube, a white ring
of NH4Cl (s) forms.
91
Collisions with the Container Walls
“the pressure a gas exerts is caused by the collisions of its
particles with the walls of the container”
“collision frequency of gas molecules depends on
temperature and pressure”
In order to compute collision frequency (ZA, collisions per
second) we need to consider:
• average velocity of the gas particles
• size of the area being considered
• number of particles per unit of volume
92
Intermolecular Collisions
“collision frequency of gas molecules depends on
temperature and pressure”
“the effects of the collisions must somehow ‘cancel out’
relative to the properties P, V, and T of an ideal gas”
(KMT says there’s no interaction)
In order to compute collision frequency (Z) we need to
consider:
•
•
•
•
speed of one particle
particle diameter
number of particles per unit of volume
motion of the other particles
93
IDEAL GASES vs REAL GASES
→ An ideal gas is a hypothetical concept.
→ No gas follows the ideal gas law exactly.
• Many come close at low P and/or high T
→ Ideal behavior is approached by real gases under
certain conditions.
→ KMT: a simple model explains ideal behavior,
albeit by making very drastic assumptions
• no inter-particle interactions; zero volume for particles
A model is an approximation and will inevitably
fail! We can learn a lot from the failures.
94