Chapter 5
Gases and Gas Laws
Chem 110
Chap 5 Gas Laws
Slide 1
An Overview of the Physical States of Matter
Distinguishing gases from liquids and solids.
•  Gas volume changes significantly with pressure.
–  Solid and liquid volumes are not greatly affected by pressure.
•  Gas volume changes significantly with temperature.
–  Gases expand when heated and shrink when cooled.
–  The volume change is 50 to 100 times greater for gases than for liquids
and solids.
•  Gases flow very freely.
•  Gases have relatively low densities.
•  Gases form a solution in any proportions.
–  Gases are freely miscible with each other.
Chap 5 Gas Laws
Slide 2
1
The Gaseous State
Can we discuss the various states of matter from a molecular
standpoint?
–  Yes! Kinetic Molecular Theory (KMT) is a powerful
concept in discussing just how molecules move, interact,
and react!
•  Gases are a good place to start:
–  Least dense state of matter;
–  Intermolecular forces (IMFs) are minimum
Gases can be very easily described using FOUR variables:
Volume (V); Pressure (P); amount (n); and Temperature (T).
Chap 5 Gas Laws
Slide 3
Volume
No need to describe in depth as concept of volume is straightforward. Use liter to quantify (L).
A gas will occupy COMPLETELY whatever volume container it is
placed in:
Chap 5 Gas Laws
Slide 4
2
Pressure
•  More difficult to define and describe.
•  P = force per unit area gas exerts on the walls of its container;
Each time wall gets hit the
pressure changes.
Measured using a barometer:
P=
F ma mg mg mgh
=
=
=
=
= dgh
A
A
A V
V
h
d = density; g = 9.807 m/sec2
dHg = 13.59 g/cm3
Chap 5 Gas Laws
Slide 5
Pressure units
Pressure is measured is a variety of units:
Pascal (Pa): 1 kg m-1sec-2
Atmosphere (atm): 101.325 kPa
Bar (bar): 1.013 atm
Torr (torr) or mm Hg: 1.316 x 10-3 atm
•  Typically, only use pascals if SI units exclusively are needed
(commonly needed in energy calculations).
•  Atm and torr are most commonly used (Pa is too small)
•  1 atm = 760 torr = 760 mm Hg
Chap 5 Gas Laws
Slide 6
3
Chap 5 Gas Laws
Slide 7
Questions
• 
• 
• 
• 
How does pressure vary with volume?
How does pressure vary with temperature?
How does pressure vary with amount?
How does energy of a gas vary with temperature?
Let’s begin by looking at some of the basic early experiments
regarding gases (scientific method).
Chap 5 Gas Laws
Slide 8
4
The Gas Laws
•  The gas laws describe the physical behavior of gases in terms
of 4 variables:
– 
– 
– 
– 
pressure (P)
temperature (T)
volume (V)
amount (number of moles, n)
•  An ideal gas is a gas that exhibits linear relationships among
these variables.
•  No ideal gas actually exists, but most simple gases behave
nearly ideally at ordinary temperatures and pressures.
Chap 5 Gas Laws
Slide 9
Boyle’s Law
Boyle: studied how P and V were related.
•  J-tube experiment
(a) When Δh = 0, Pint = Pext = 1 atm
(b) Add Hg thru open end, volume of
confined gas decreases;
•  New pressure goes up as follows:
Pnew = Pinitial (atm) +
Chap 5 Gas Laws
h (mmHg)
760 mmHg/atm
Slide 10
5
Boyles Law: Results
T2
x
x
x
x
P
x
x
x
x
x
x
x x
x
x
x
x
x
x
x
T1
• 
• 
• 
• 
• 
PV = constant at a fixed T
P = c (1/V)
c = 22.414 L-atm (!)
At low P, satisfied exactly;
At 1 atm small corrections
needed;
•  At high P (> 50 atm)
substantial corrections req’d;
1/V
P1 V2
=
P2 V1
PV
1 1 = PV
2 2
Chap 5 Gas Laws
Slide 11
Charles Law
•  How does T (which is NOT a mechanical property) affect gases?
•  Charles: at low P all gases expand by the same relative amount
if Ti and Tf are the same for each gas (unlike liquids and solids)
V ∝ t at fixed P and n
Extrapolation leads to
absolute zero and Kelvin
scale (T);
Charles law is simplified if
T (in K) used (true linear
relationship)
Since T (in K) = t + 273.15
Chap 5 Gas Laws
T1 V1
=
T2 V 2
Slide 12
6
Avogadro’s Law
Avogadro came up with the idea of ‘combining volumes’.
3 L of hydrogen gas and 1 L of nitrogen gas produces 2 L of ammonia gas
3H 2 ( g ) + N 2 ( g ) ⎯⎯
→ 2 NH 3 ( g )
“Equal volumes of different gases at the same temperature and
pressure contain equal numbers of particles”.
V ∝ n at fixed P and T
V/n = constant
Chap 5 Gas Laws
Slide 13
Combined Gas Law
What we know from the “ABC”s of Gas Laws:
•  V ∝ 1/P at fixed n and T (PV = const)
•  V ∝ T at fixed P and n (V/T = const)
•  V ∝ n at fixed P and T (V/n = const)
So:
PV
PV
1 1
= 2 2
n1T1 n2T2
and
PV
1 1
= constant = R
n1T1
R = 0.082058 L atm mol –1 K –1 = 8.3145 J mol –1 K –1
Chap 5 Gas Laws
Slide 14
7
Ideal Gas Law
•  PV = nRT
•  R is the universal gas constant
•  IGL is an equation of state: defining any three states (such as
P, n, and T) allows one to calculate the fourth and final state
(here, V);
•  Works well as long as one deals with ideal gases, or gases
behaving ideally;
•  Best ideal gas: monatomic, low pressure, high temperature;
most gases are real and deviate from ideality (but not much).
Chap 5 Gas Laws
Slide 15
Molar Volume
•  Avogadro’s hypothesis: equal volumes of gases have equal
numbers of ‘particles’.
•  Evaluate IGL in terms of volume of a gas and moles:
V RT
=
= Vm
n
P
Vm is the Molar Volume
Note: independent of gas identity!
•  At SATP (standard ambient temperature and pressure):
 Vm = 24.79 L·mol–1 (@ 298.15 K and 1 bar exactly)
•  At STP (standard temperature and pressure):
 Vm = 22.41 L·mol–1 (@ 273.15 K and 1 atm exactly)
Chap 5 Gas Laws
Slide 16
8
Molar Volume Comparisons
At STP
(0 °C and 1 atm)
STP vs. SATP
(0 °C vs. 25 °C);
(1 bar vs. 1 atm)
Chap 5 Gas Laws
Slide 17
Molar Mass from the Ideal Gas Law
n=
m
M
M=
=
PV
RT
mRT
PV
Chap 5 Gas Laws
Slide 18
9
Density
•  Gases are the least dense of all matter; typical to measure and
report in grams per liter of gas;
•  IGL can be used to derive expression for density and how it
relates to a gas:
mass m
m
MM =
= ; so n =
mol
n
MM
m
PV = nRT =
RT
MM
So, the density of a gas can be
⎛ m ⎞ RT
RT
P=⎜ ⎟
=d
used to determine the molar
MM
⎝ V ⎠ MM
mass and thus the identity of
P( MM )
an unknown gas.
d=
RT
Chap 5 Gas Laws
Slide 19
Gas Law Problems-1
A sample of SO2 (g) at 80°C is confined to a 3.0 L
container and the pressure determined to be 4.5 atm; how
many grams of SO2 are in the container?
Since you are being asked about ‘physical’ amounts of the gas, n
must be determined first; since PV = nRT then
n=
PV
(4.5 atm)(3.0 L)
=
= 0.466 mol
RT (0.08206 L ⋅ atm ⋅ mol –1 ⋅ K –1 )(353 K)
Since n = m/MM and SO2 has a MM=64 g/mol,
n(MM) = m and thus m = 29.8 g of SO2
Chap 5 Gas Laws
Slide 20
10
Gas Law Problems-2
An ideal gas at a pressure of 2.50 atm is cooled at constant
volume until the pressure drops to 1.00 atm. If the gas is initially
at 100. °C, what will the final temperature be?
In this two-state problem, n,V and R are all constant,
so P/T = nR/V = constant for both states; therefore
P1 P2
P T (1.00 atm)(373 K)
= ; so T2 = 2 1 =
T1 T2
P1
2.50 atm
Solving, we get T2 = 149 K (or -124 °C)
Chap 5 Gas Laws
Slide 21
Gas Law Problems-3
Into a 500. mL container is placed 1.35 g of an unknown gas
and the container is placed into a 25°C constant temperature
bath. If the pressure of the gas is determined to be 600. torr,
what is the molecular weight of the gas?
Since P(MM) = mRT = (m/V)RT we have all the information we
need to solve this problem (remember to convert P to atm and T
to kelvins):
mRT (1.35 g)(0.08206 L ⋅ atm ⋅ mol –1 ⋅ K –1 )(298 K)
MM =
=
PV
(0.789 atm)(0.500 L)
Solving, we get MM = 83.7 g/mol
Chap 5 Gas Laws
Slide 22
11
Sample Problem 5.8
PROBLEM:
Finding the Molar Mass of a Volatile Liquid
An organic chemist isolates a colorless liquid from a petroleum
sample. She places the liquid in a preweighed flask and puts the
flask in boiling water, causing the liquid to vaporize and fill the
flask with gas. She closes the flask and reweighs it. She obtains
the following data:
Volume (V) of flask = 213 mL
mass of flask + gas = 78.416 g
T = 100.0°C
P = 754 torr
mass of flask = 77.834 g
Calculate the molar mass of the liquid.
PLAN: The variables V, T and P are given. We find the mass of the gas by
subtracting the mass of the flask from the mass of the flask with
the gas in it, and use this information to calculate M.
Chap 5 Gas Laws
Slide 23
Sample Problem 5.8
SOLUTION
:
V = 213 mL x
P = 754 torr x
m of gas = (78.416 - 77.834) = 0.582 g
1L
= 0.213 L
103 mL
1 atm
760 torr
M=
mRT
PV
T = 100.0°C + 273.15 = 373.2 K
= 0.992 atm
atm·L
x 373 K
mol·K
0.213 L x 0.992 atm
0.582 g x 0.0821
=
Chap 5 Gas Laws
= 84.4 g/mol
Slide 24
12
Gas Mixtures
•  Most gases are found as mixtures; for example, dry air is a
mixture (w/w) of N2 (75.52%), O2 (23.14%), CO2 (0.05%),
and Ar (1.29%). Does this affect gas behavior?
•  At low pressures (1-2 atm and lower) a mixture of gases that
do not react with one another behaves like a single pure gas
and can therefore be treated as one in using the IGL.
•  Dalton’s Law of Partial Pressures: the total pressure of a
mixture of gases is the sum of the partial pressures of its
components.
Ptot = PA + PB + + Pi = ∑ Pi
Chap 5 Gas Laws
Slide 25
Law of Partial Pressures
Some uses and modifications:
a)  Humid gas calculations
Phumid air = Pdry air + Pwater vapor
b)  Pressure in terms of mole composition:
n RT nB RT
n RT
Ptot = PA + PB + + Pi = A
+
+ + i
V
V
V
RT
Ptot = (nA + nB + + ni )
V
Ptot
P
n
RT
=
= tot ; so PA = A Ptot
V
(nA + nB + + ni ) ntot
ntot
define mole fraction, xi =
ni
n
then Pi = i Ptot = xi Ptot
ntot
ntot
Chap 5 Gas Laws
Slide 26
13
Table 5.2 Vapor Pressure of Water (Pwater) at Different T
T(0C)
0
5
10
12
14
16
18
20
22
24
26
28
30
35
P
(torr)
H2O
T(0C)
4.6
6.5
9.2
10.5
12.0
13.6
15.5
17.5
19.8
22.4
25.2
28.3
31.8
42.2
40
45
50
55
60
65
70
75
80
85
90
95
100
P
(torr)
H2O
55.3
71.9
92.5
118.0
149.4
187.5
233.7
289.1
355.1
433.6
525.8
633.9
760.0
Chap 5 Gas Laws
Slide 27
Kinetic Molecular Theory
The empirical ideal gas law suggests that we should be able to
develop a model that explains gas behavior at the molecular
level.
What exactly is occurring at the microscopic level?
–  How do individual particles exert force to give pressure?
–  Why are gases compressible? And what happens on the
molecular level?
–  Why does the law of partial pressures work?
–  What is the temperature effect on molecules?
–  Why does Avogadro’s law work? Shouldn’t smaller
molecules occupy less space than larger ones? What
about pressure considerations?
Chap 5 Gas Laws
Slide 28
14
Molecular Motion in Gases
•  Gas laws suggest a lot about gases, but not much about their motion
(except that they don’t interact much and their speed seems to increase with
temperature). Are all gas processes IGL- derivable?
•  Two processes are not:
Diffusion: gradual dispersal of one gas thru another (equal pressure
process).
Effusion: escape of a gas thru a pinhole (or porous barrier) from a region
of high pressure to one of lower pressure.
Chap 5 Gas Laws
Slide 29
Effusion and Size
Effusion and diffusion, while different processes, can be explained using similar
approaches; effusion is easier to treat quantitatively.
Graham: at constant T, the rate of effusion is inversely proportional to the
square root of the molar mass:
1
Rate of effusion ∝
∝ Average speed
MM
This is Grahams Law of Effusion
For two gases, we can say the following:
effusion Rate A
effusion RateB
effusion Time A
effusion TimeB
=
=
NA
MM B
NB
MM A
=
MM B
MM A
MM A
MM B
Chap 5 Gas Laws
Slide 30
15
Effusion and Temperature
Effusion experiments at different temperatures reveals that the rate of effusion
of a given gas increases as the temperature is raised; specifically:
effusion Rate T2
effusion Rate T1
=
T2
T1
Therefore:
Rate of effusion ∝ Average speed ∝
T
And so:
Average speed ∝
T
MM
Chap 5 Gas Laws
Slide 31
Kinetic Molecular Theory: Assumptions
1)  A pure gas consists of a large number of identical molecules
separated by distances that are great compared to their size
(point-charge assumption).
2)  Gas molecules are constantly moving in random directions
with a distribution of speeds in straight-line trajectories.
3)  Gas molecules exert no attractive or repulsive forces on one
another between collisions, so between collisions they move in
straight lines with constant velocities.
4)  Since gas molecules are much, much smaller than the volume
they occupy, collisions between two molecules are rare (or,
rare compared with collisions with the walls).
5)  Collisions of molecules with the walls are elastic: no energy is
lost during a collision
Chap 5 Gas Laws
Slide 32
16
Consequences of KMT
Out of a rigorous mathematical treatment the following results are obtained:
1.  EK = (3/2)RT; thus, the temperature of a gas is a measure of its average
kinetic energy, independent of the identity of the gas;
2.  The distribution of molecular speeds can be represented by the MaxwellBoltzmann speed distribution:
3 2
⎛ m ⎞ 2 – ( mu / 2 k T )
f (u) = 4π ⎜
ue
⎝ 2π k T ⎟⎠
2
B
B
Once at thermal equilibrium, this distribution will persist indefinitely.
3.  Since f(u) is a probability function, three specific speeds can be identified:
ump =
2RT
MM
; uave =
8RT
π ( MM )
; and urms =
3RT
MM
Chap 5 Gas Laws
Slide 33
Maxwell-Boltzmann Distribution
ump = most probable
speed; f(u) is at its
maximum;
uav = average speed;
urms = root mean square
speed; this can be
thought of as the
‘typical’ speed used
to describe a gas. It
is also equal to
ump : uav :urms = 1.000 : 1.128: 1.225
2( E K )
MM
Chap 5 Gas Laws
Slide 34
17
Applications of KMT
• 
Zw: the rate of collisions of gas molecules with a section of the wall with area A:
Zw =
1⎛ N⎞
⎜
4⎝ V
⎟⎠ (uave ) A
(Graham’s law of effusion explained by this)
• 
Z1: the frequency of collisions with other molecules; assumes molecules are approx.
spherical and sweep out a cylinder in its flight (d = molecular diameter):
⎛ N ⎞ 2 π RT
d
⎝ V ⎟⎠
MM
Z1 = 4 ⎜
• 
λ: the mean free path; distance traveled between collisions:
λ=
uave
=
Z1
1
2(π )(d 2 )( N V )
Chap 5 Gas Laws
Slide 35
Molecular Questions, KMT Answers
Envoking KMT to answer the questions raised earlier regarding
what is occurring at the molecular level:
Pressure: collisions with container cause this; anything that
causes them to ⇑ will cause ⇑ P;
Compressibility: ⇓ space and molecules hit walls more often: ⇑
P;
Partial Pressures: Simply an ⇑ in the number of molecules;
more molecules to collide with walls (assume NO IMF’s!)
Temp Effect: ⇑ energy and molecules move faster and hit walls
more often: ⇑ P;
Avogadro’s: ⇑ molecules, ⇑ collisions; ⇑ V until collisions per
surface area gets to where it was before addition
Chap 5 Gas Laws
Slide 36
18
Real Gases
•  The IGL is an equation of state, one that simply relates all states to one
another (P, V, T, and n); how do real gases differ?
•  Deviations are revealed in the compressibility factor, z:
PV
z=
for 1 mol of a gas
RT
25 C
N2
Chap 5 Gas Laws
Slide 37
van der Waals Equation of State
2
2
⎛
n ⎞
nRT
n
⎜⎝ P + a V 2 ⎟⎠ (V − nb) = nRT or P = (V − nb) – a V 2
Modifications take into account that forces do exist between molecules and that these
are repulsive at short distances and attractive at long distances; that gas molecules do
indeed have molecular volumes; and that attractions occur between pairs of molecules.
Effective volume = (V-nb); b is a constant that describes the volume excluded per mole
of molecules (units are L per mol);
Pressure attractive correction = a(n2/V2); a is a positive constant that depends on the
attractive forces of the gases involved.
Now z can be written as follows:
z=
1
1 − (nb V )
–
an
RTV
If a and b = 0, z = 1 (ideal gas)
If a is small and b is appreciable, z > 1
If b is small and a is appreciable, z < 1)
Chap 5 Gas Laws
Slide 38
19
Van der Waals constants a and b
Chap 5 Gas Laws
Slide 39
Table 5.4 Van der Waals Constants for Some Common Gases
a
Gas
He
Ne
Ar
Kr
Xe
H2
N2
O2
Cl2
CH4
CO
CO2
NH3
H2O
atm·L2
mol2
0.034
0.211
1.35
2.32
4.19
0.244
1.39
1.36
6.49
2.25
1.45
3.59
4.17
5.46
Chap 5 Gas Laws
b
L
mol
0.0237
0.0171
0.0322
0.0398
0.0511
0.0266
0.0391
0.0318
0.0562
0.0428
0.0395
0.0427
0.0371
0.0305
Slide 40
20
Chap 5 Gas Laws
Slide 41
End Chapter 5
Gas Laws
Chap 5 Gas Laws
Slide 42
21