AP Notes Chapter 11
Properties Of Gases
Temperature

An indirect measure of the
average kinetic energy of a
collection of particles
KEavg = kT
Boltzman Plot
Pressure
Measure of the number of
collisions between gas
particles and a unit area of
the wall of the container
Pressure = force / unit area
Force/area
English system:
pounds/in2 (psi)
Metric system:
2
Newton/m (pascal)
Torricelli Barometer
1 atmosphere
pressure
h = 760 mm Hg
1 atm = 760 torr (mm Hg)
= 101.325 kPa
= 1.01325 bar
=14.70 psi
Patm
Manometer
h
Pgas
Patm
Manometer
h
Pgas
Volume
Total space of a container that
gases occupy due to the free
random motion of the gas
molecules
Relationship between
Volume & Pressure of
Gases
P-V
V
P (at constant T)
1
V
P
V
Slope = k
1/P (at constant T)
In mathematical terms:
y = mx + b
1
V m b
P
k
or V 
P
Boyle’s Law
Relationship between
Volume & Temperature
of Gases
V-T
In mathematical terms:
y = mx + b
V = mT + b
V  kT
Charles’ Law
Where T must be in
Kelvin (K) temperature
K = 0C + 273
Relationship between
Pressure &
Temperature of Gases
P-T
In mathematical terms:
y = mx + b
P = mT + b
P  kT
Gay-Lussac’s Law
Relationship between
Volume & Moles
of Gases
V-n
In mathematical terms:
y = mx + b
V = mn + b
V  kn
Avogadro’s Law
Avogadro’s Hypothesis
At constant temperature and
pressure, equal volumes of
gases contain equal number
of particles
3. Hydrogen gas [8.3 liters]
reacts in the presence of 2.5
liters of nitrogen gas at 370C
and 100 kPa. What volume
of ammonia is produced at
these same conditions?
Combined Gas Law
k
V
V  kT
P
P  kT
V  kn
PV

 a constant
nT
Ideal & Real
Gasses
Kinetic Molecular
Theory
1. Gases consist of small
particles that are far apart in
comparison to their own size.
These particles are
considered to be tiny points
occupying a negligible
volume compared to that of
their container.
Kinetic Molecular Theory
2. Molecules are in rapid and
random straight-line motion.
This motion can be
described by well-defined
and established laws of
motion.
Kinetic Molecular Theory
3. The collisions of molecules
with the walls of a container
or with other molecules are
perfectly elastic. That is, no
loss of energy occurs.
Kinetic Molecular Theory
4. There are no attractive
forces between molecules or
between molecules and the
walls with which they collide.
Kinetic Molecular Theory
5. At any particular instant, the
molecules in a given sample
of gas do not all possess the
same amount of energy.



Have 1 particle,
with mass m,
with velocity 
PARTICLE
IN THE
BOX
Consider the P exerted:
f
P
A
f
P 2

But:
f=?
But:
f = ma
where
change in velocity
a
time





Change in velocity =
(
2
a 
t
Thus, the pressure
exerted by one
particle on a wall
is:
 2 
(m) 
t 

P1 
2

But,
?
But,
d

t
and, the distance a
particle travels between
collisions with the same
wall is ?
2
 
t
or
2
t 

 2 
Substituting
(m) 
t 

into
P1 
2

we get:
 2 


(m)
2 



P1 
2

m

Simplifying: P 
1
3

2
 V
3
but,
m
 P1 
V
2
This represents the
pressure (P) that
one particle exerts
striking opposite
walls in the box.
Now assume the
box contains N
particles. Then, N/3
particles are
traveling between
opposite walls.
Thus,
the total pressure on
opposite walls is:
 N   m
P   
 3  V
2



1 2 1
 
3 3 2
Substitute & rearrange

 2  N  1 
2
P      m
 3  V  2 



 1
2
  m  KE
2
 2N 
 PV  
  KE
 3 
From classical physics
3
KE  kT
2
where k is the
Boltzman constant
R
k
N0
where
R = universal gas constant
N0 = Avogadro’s number
3 R
 KE  
T
2 N0
2N 3 R
 PV 
 
T
3 2 N0
 PV  nRT
Ideal Gas Equation
L  atm
R  0.08206
mol  K
L  kPa
 8.314
mol  K
L  torr
 62.36
mol  K
Note that
PV
R
nT
is similar to the
Combined Gas Law
derived earlier. PV
nT
C
Variations on
Ideal Gas Equation
mass
PV 
RT
MM
4. What is the molar
mass of methylamine
if 0.157 g of the gas
occupies 125 mL with
a pressure of 99.5 kPa
0
at 22 C?
Bromine
Variations on
Ideal Gas Equation
mass
PV 
RT
MM
mass
P  MM
density 

volume
RT
5. Calculate the
density of fluorine
gas at:
0
30 C
STP
and 725 torr.
Real Gas
Behavior
Ideal Gas
Equation
PV = nRT
N2
2.0
CH4
H2
PV 1.0
nRT
CO2
Ideal
gas
0
0
200 400 600 800 1000
P (atm)
“correct” for volume
of molecules
(V - b)
also “correct” for
attractive forces
between molecules
a 

P  2 
V 

van der Waals’ Equation
a 

 P  2 V  b   RT
V 

for 1 mole
van der Waals’ Equation

an 
 P  2 V  bn   nRT
V 

for n moles
2
from CRC Handbook
a*
He 0.03412
b*
0.02370
Ne 0.2107
0.01709
*when P(atm) & V(L)
from CRC Handbook
NH3
a*
4.170
b*
0.03707
H2O
5.464
0.03049
*when P(atm) & V(L)
from CRC Handbook
a*
20.39
b*
0.1383
C5H12 19.01
0.1460
CCl4
*when P(atm) & V(L)
Cl2 gas has a = 6.49, b = 0.0562
 For 8.0 mol Cl2 in a 4.0 L tank at 27oC.
 P (ideal) = nRT/V = 49.3 atm
 P (van der Waals) = 29.5 atm

T & P conditions
where a real gas
approximates
an ideal gas?
203 K
N2 gas
1.8
PV
nRT
293 K
1.4
673 K
1.0
Ideal
gas
0.6
0
200
400 600
P (atm)
800
T & P conditions
where a real gas
approximates
an ideal gas?
high temperature
low pressure
Gaseous
Molecular
Movement
Partial Pressure
pressure exerted by
each component in a
mixture of gases
this assumes that
NO interactions
occurs between
the molecules of gas
must conclude
1. each gas acts as if it is
in container alone
2. each gas collides with
the container wall as an
“event”
n
PT   Pi
i1
where n = # components
or
PT = P1 + P2 + P3 + ...
Pi V = ni R T
or
ni RT
Pi 
V
thus:
n1RT n2 RT
PT 

 ...
V
V
or
RT
PT 
 n1  n2  ...
V
therefore:
nT =  ni
and PT  sum of mols
of gas
Mole Fraction
ni
Xi 
nT
n1 n2 n3


 ...  1
n n
n
Since:
ni R T
Pi 
V
and
nT R T
PT 
V
Then
ni
Pi
 Xi 
nT
PT
and
Pi = Xi PT
 diffusion
is the gradual
mixing of molecules of
different gases.
 effusion
is the movement
of molecules through a
small hole into an empty
container.
rate of 
effusion
average
speed
r
or
r  c
rA c A

rB c B
rA
A

rB
B
But ...

1
KE  m 
2
2
3
 nRT
2
where
m
n
 m  n  MM
MM
thus
then

1
n  MM  
2
1
2
2
 3RT 


 MM 
RMS speed
3
 nRT
2
substituting:
rA
rB
 3 RT

MM A


 3 RT

 MM B






1
2
1
2
simplifying
rA

rB
MMB
MMA
Graham’s Law
NH3-HCl
d
r
t
if “d” is constant
d
rA
t A tB



d
rB
tA
tB
MMB
MM A
if “t” is constant
dA
rA
dA
t



dB
rB
dB
t
MM B
MM A
GAS LAW
STOICHIOMETRY
1. Ethanol, C2H5OH, is
often prepared by
fermentation of sugars
such as glucose,
C6H12O6, with carbon
dioxide as the other
product.
[A] What volume of CO2
is produced from 125 g
of glucose if the reaction
is 97.5% efficient?
[B] Ethanol can also be
made from ethylene,
C2H4 according to the
following chemical
system:
3 C2H4(g) + 2 H2SO4 
C2H5HSO4 + (C2H5)2SO4
then
C2H5HSO4 + (C2H5)2SO4 + 3H2O
 3C2H5OH + 2 H2SO4
What volume (mL) of 95%
ethanol is produced from
142.5 dm3 of C2H4? The
density of 95% ethanol is
0.8161 g/mL.
2. What is the final pressure
[kPa] if 1000. g uranium
reacts with sufficient
fluorine gas to produce
gaseous uranium
o
hexafluoride at 32 C in a
300. L container?
3. What mass of sodium
metal is needed to
produce 250 mL of
o
hydrogen gas at 24 C
and 740 Torr?