The Gas Phase
characteristics of gases:
Chapter 5:
The Gaseous
State
◆
◆
fluid
◆
highly compressible
◆
force
pressure, P = ––––––
area
The pressure exerted by a gas is the
result of the collisions of particles
with the walls of the container.
◆
◆
low P - result of infrequent
collisions of little force
high P - result of frequent
collisions of greater force
expand to fill container in both shape and
volume
◆
◆
Pressure
least dense form of matter
particles in constant, random, and chaotic
motion
P,V, T, and n are all interdependent relationships defined by the gas laws
Measurement of Atmospheric Pressure
The barometer was invented by Torricelli in the 17th
century to measure atmospheric pressure.
Measurement of Pressure Using a Manometer
Units of Pressure
notes:
Pascal is the SI unit of pressure
1 torr = 1 mmHg so 1 atm = 760 torr
Boyle’s Law
At constant temperature for a given sample of gas,
volume is inversely proportional to pressure.
P ∝ 1/V at constant n & T
or P = constant (1/V)
so
PV = constant
Boyle’s Law
example:
A 10.0 L volume of He gas exerts a pressure of
0.75 atm. What pressure is exerted by this
sample of He if the volume occupied is
decreased to 8.0 L at constant temperature?
pick 2 points, define coordinates:
at point 1: 1/V1 = constant•P1 ! P1V1 = constant
at point 2: 1/V2 = constant•P2 ! P2V2 = constant
P1V1 = P2V2
Charles’s Law
For a given sample of gas at constant pressure,
volume is directly proportional to its absolute
temperature.
or
V ∝ T at constant n & P
V = constant • T
so V⁄T = constant
V1/T1 = V2/T2
Avogadro’s Law
Molar Volume of a Gas
example:
A chemical reaction produces O2 (g) that
occupies 4.38 dm3 at 19ºC. What volume will
be occupied by this amount of O2 (g) at 25ºC
and constant pressure?
Equal volumes of any two gases at the same T and P
contain the same number of particles.
V∝n
or V = constant • n
V⁄n = constant
V1/n1 = V2/n2
molar volume of a gas - volume occupied per mol of gas
◆ units L/mol
◆ dependent on T and P
◆ independent of identify of the gas
Standard Temperature and Pressure (STP)
STP is defined as T = 0ºC (273 K) and P = 1 atm
At STP:
molar volume of a gas = 22.41 L/mol
or 1 mol gas = 22.41 L
example:
Calculate the volume occupied by 12.5 g C2H2 (g)
at STP.
example:
2.50 mol N2 (g) occupies a volume of 3.68 L at
some T and P. If an additional 1.00 mol N2 (g) is
added at constant T and P, determine the volume
occupied by the sample.
Given:
P • V = constant
V = constant • T
V = constant • n
Ideal Gas Law:
PV = nRT
example:
0.75 mol of a gas in a 2.5 L container exerts a
pressure of 490 torr. Determine the temperature
of this system.
example:
Determine the number of N2 molecules in a 3.65 L
sample at 20ºC and 1.20 atm.
Relationship Between
Pressure, Temperature, Density & Molar Mass
mass
molar mass, M = ––––––
mol
and:
so:
m
n = ––––
M
PV = nRT
m
PV = –––
M RT
m
PM = ––– RT
V
or:
PM = dRT
example:
Determine the density (in g/L) of AsH3 (g) at 725
torr and 35ºC.
example:
Determine the molar mass of an unknown gas with
density of 0.400 g/L at 190ºC and 60.0 mmHg.
example:
Determine the density (in g/L) of PH3 (g) at STP.
example:
Determine the molar mass of an unknown gas with
density of 1.96 g/L at STP.
Stoichiometry Calculations for Reactions
Involving Gas Phase Reactants and Products
◆
this typically involves a both a stoichiometry
calculation and an ideal gas law calculation
example:
2 KClO3 (s) " 2 KCl (s) + 3 O2 (g)
If 0.0100 mol potassium chlorate is consumed,
determine the volume of O2 (g) produced
a. at STP
Consider the following chemical reaction that has
been used in automobile air bags to quickly
generate N2 (g) to cause inflation of the air bag:
6 NaN3 (s) + Fe2O3 (s) ! 3 Na2O (s) + 2 Fe (s) + 9 N2 (g)
Calculate the mass of NaN3 required to produce
75.0 L N2 (g) at 25ºC and 748 mmHg.
b. at 30ºC and 1.04 atm
Another Perspective on Homogeneous
Gas-Phase Reactions:
Comparing Molar Volumes . . .
Consider the molar volumes of the following:
1 mol NaCl (s) at 25ºC & 1 atm:
27cm3
1 mol ethanol (l) at 25ºC & 1 atm:
1 mol O2 (g) at 25ºC & 1 atm:
example:
59 mL
24.5 L
now consider the following reaction:
4 C3H5(NO3)3 (l) ! 6 N2 (g) + O2 (g) + 12 CO2 (g) + 10 H2O (g)
example:
The following volumes of CO and H2 are
combined at constant T and P and allowed to
react:
CO (g) + 2 H2 (g) " CH3OH (g)
3.25 L
10.0 L
yield in L?
Another Perspective on Homogeneous
Gas-Phase Reactions:
Dalton’s Law of Partial Pressures:
Describing Nonreactive Gas Mixtures
or alternatively . . .
The following pressures of CO and H2 are
combined at constant T and V and allowed
to react:
CO (g) + 2 H2 (g) " CH3OH (g)
3.25 atm 5.00 atm
P after reaction?
Dalton’s Law of Partial Pressures:
Describing Nonreactive Gas Mixtures
1. Each gas in a mixture of nonreactive gases
behaves independently with respect to pressure.
2. The pressure exerted by each component in a
mixture is the partial pressure of the component.
Consider a mixture of 3 gases A, B, and C:
PAV = nART
PBV = nBRT
PCV = nCRT
PtotV = ntotRT
Ptot = PA + PB + PC
ntot = nA + nB + nC
3. The total pressure of the mixture is equal to the
sum of the partial pressures of the components.
Ptot = ΣPA + PB
A gas mixture’s composition can be defined in terms
of mole fraction, !:
4. a. Each component in the mixture follows the
ideal gas law.
mol A
! A = ––––––––;
mol total
b. The gas mixture follows the ideal gas law.
mol B
!B = ––––––––;
mol total
!A + !B + !C = 1
mol C
!C = ––––––––
mol total
Relationship Between PA, Ptot, χA
for a mixture of nonreactive gases A and B at
constant T and V:
Ptot = (RT/V)ntot
PA = (RT/V)nA
PB = (RT/V)nB
so:
!
!
!
Ptot/ntot = constant
PA/nA = constant
PB/nB = constant
Ptot
PA
––––
=
––––
ntot
nA
!
PA
nA = !A
–––
= –––
n
tot
Ptot
PB
Ptot = ––––
––––
nB
ntot
!
PB = –––
nB = !B
–––
n
tot
Ptot
Kinetic Molecular Theory (KMT)
◆
theoretical model to explain observed behavior of gases
1. Gases are composed of
particles that are so small
relative to the distances
between them that the volume
of the individual particles can
be assumed to be negligible.
2. Particles of a gas are in
constant, random, and chaotic
motion. The collisions of the
particles with the walls of the
container are the cause of the
pressure exerted by the gas.
example:
A 10.0 L flask contains 1.031 g O2 and some amount
of CO2 at 18ºC. For this system, Ptot = 0.108 atm.
Determine PCO2, !CO2, and !O2.
example:
At 20ºC a 1.00 L flask contains 0.241 mol He (!He =
0.743) and some amount of Ne.
Determine PHe, Ptot, and nNe.
Kinetic Molecular Theory (KMT) continued
3. Particles are assumed to exert no force on
each other; they neither attract nor repel each
other.
4. The average kinetic energy of a collection of
gas particles is directly proportional to the
temperature of the gas.
The kinetic energy of particles in a gas sample
is dependent only on the temperature of the
sample.
kinetic energy = 3/2(RT)
Focus on Pressure
Pressure exerted by a gas phase sample is the result
of collisions of particles with the walls of the
container.
◆ affected primarily by the frequency & force of
particle collisions
frequency of collisions:
force of collisions:
affected by
◆
◆
affected by
concentration, C
C = n/V
◆
particle speed, u
u ∝T
directly proportional
to T
related to speed (u)
and molar mass (M)
◆ How will changing V, T, and n affect the
frequency & force of particle collisions?
P ∝ frequency
P ∝ force
Using KMT to Explain Relationships:
P ∝ 1/V at constant n, T
kinetic energy of
the particles
frequency ∝ C
force ∝ u
frequency ∝ u
force ∝ M
Determining Molecular Speeds
for a sample of gas at some constant P & V:
u = √(3RT)/M
P∝ n at constant V, T
u = rms speed or
“average” molecular
speed
M in kg/mol
T in Kelvin
R = 8.314 kg•m2/s2•K•mol
note:
u ∝ √T
and
u ∝ 1/√M
Maxwell’s Distribution of Molecular Speeds
◆
◆
u ∝ 1/√M
the greater the molar mass,
the slower the rms speed
example:
Determine the temperature (in K) at which
xenon atoms have u = 382 m/s.
in samples of lower molar
mass, there is a wider
distribution of rms speeds
◆
◆
u ∝ √T
the higher the temperature, the
faster the rms speed
in samples at lower
temperatures, there is a more
narrow distribution of rms
speeds
Comparing Relative Speeds of 2 Gases at Same T:
example:
Determine the rms speed (in m/s) of F2
molecules at 100 K.
Diffusion and Effusion of Gases
consider 2 gases A and B:
uA = √3RT/MA
and
Diffusion:
the process by which one
gas disperses through
another gas to occupy the
space uniformly
uB = √3RT/MB
if 3RT = constant, then:
uA/uB = √MB/MA
example:
Determine the relative speeds of N2 and
Cl2 molecules at the same temperature.
Effusion:
the process by which gas phase
particles escape through small
holes or pores;
movement is from a region of
higher P to a region of lower P
Graham’s Law of Effusion:
(also applies to diffusion)
The rate of effusion of a gas is inversely proportional
to the square root of its molar mass.
rateeff ∝ 1/√M
∆V
rateeff = ––––
∆t
consider 2 gases A and B at constant T:
rateA/rateB = √MB/MA
Real Gases: Deviation from Ideal Behavior
example:
3.52 s is required for 25.0 mL of He (g) to
effuse. Determine the rate of effusion of O2 (g)
under identical experimental conditions.
example:
An unknown gas takes 2.55 times longer than
does Ne (g) to effuse under identical
experimental conditions. Determine the molar
mass of the unknown gas.
◆ The behavior of real gas
samples deviates from ideal gas
behavior because of
intermolecular forces of
attraction and repulsion
between particles.
◆ The deviation from ideal
gas behavior is most
significant under
conditions of very high P
and very low T.
Van der Waals Equation:
an2 )(V – nb) = nRT
(P + –––
V2
1 mole of an ideal gas
occupying a volume of 22.41 L at 0ºC
would exert a pressure of 1 atm.
Treating SO2 as a real gas, determine the
pressure exerted (in atm) by a 1 mole sample in
22.41 L at 0ºC.
For SO2, a = 6.865 L2•atm/mol2
b = 0.05679 L/mol