Gases
Chapter 5
1
Gas Properties

Four properties determine the physical behavior of
any gas:
Amount of gas
 Gas pressure
 Gas volume
 Gas temperature

2
Gas pressure

Gas molecules
exert a force on the
walls of their
container when
they collide with it
3
Gas pressure
Gas pressure can support a column of liquid
 Pliquid = g•h•d

g = acceleration due to the force of gravity (constant)
 h = height of the liquid column
 d = density of the liquid

4
Atmospheric
pressure

Torricelli barometer



In the closed tube, the liquid
falls until the pressure exerted
by the column of liquid just
balances the pressure exerted
by the atmosphere.
Patmosphere = Pliquid = ghd
Patmosphere  liquid height
Standard atmospheric
pressure (1 atm) is
760 mm Hg
5
Units for pressure

In this course we usually convert to atm
6
Gas pressure

Pliquid = g•h•d
Pressure exerted by a column
of liquid is proportional to the
height of the column and the
density of the liquid
 Container shape and volume
do not affect pressure

7
Example

A barometer filled with perchloroethylene
(d = 1.62 g/cm3) has a liquid height of 6.38 m.
What is this pressure in mm Hg (d = 13.6 g/cm3)?
P = ghd = g hpce dpce = g hHg dHg
 hpce dpce = hHg dHg
 hHg = hpce d pce = (6.38 m)(1.62 g/cm3) = 0.760 m
dHg
13.6 g/cm3
 hHg = 760 mm Hg

8
Gas pressure

A manometer compares the pressure of a gas in a
container to the atmospheric pressure
9
Gas Laws: Boyle
In 1662, Robert Boyle
discovered the first of
the simple gas laws
 PV = constant

For a fixed amount of gas at
constant temperature, gas pressure
and gas volume are inversely
proportional
10
Gas Laws: Charles

In 1787, Jacques Charles discovered a relationship
between gas volume and gas temperature:
• relationship between volume
and temperature is always linear
• all gases reach V = 0 at same
temperature, –273.15 °C
volume (mL)
• this temperature is
ABSOLUTE ZERO
temperature (°C)
12
A temperature scale for gases:
the Kelvin scale
A new temperature scale was invented: the Kelvin
or absolute temperature scale
 K = °C + 273.15
 Zero Kelvins = absolute zero

13
Gas laws: Charles

Using the Kelvin scale, Charles’ results is

For a fixed amount of gas at constant pressure, gas
volume and gas temperature are directly proportional
V
 constant
T

A similar relationship was found for pressure and
temperature:
P
 constant
T
14
Standard conditions for gases

Certain conditions of pressure and temperature
have been chosen as standard conditions for gases
Standard temperature is 273.15 K (0 °C)
 Standard pressure is exactly 1 atm (760 mm Hg)


These conditions are referred to as STP
(standard temperature and pressure)
16
Gas laws: Avogadro

In 1811, Avogadro proposed that equal volumes of
gases at the same temperature and pressure contain
equal numbers of particles.

At constant temperature and pressure, gas volume is
directly proportional to the number of moles of gas
V
 constant
n
 Standard molar volume: at STP, one mole of gas
occupies 22.4 L
17
Putting it all together:
Ideal Gas Equation

Combining Boyle’s Law, Charles’ Law, and
Avogadro’s Law give one equation that includes all
four gas variables:
PV
R
nT
or PV  nRT
R is the ideal or universal gas constant
 R = 0.08206 atm L/mol K

19
Using the Ideal Gas Equation

Ideal gas equation may be expressed two ways:

One set of conditions: ideal gas law
PV  nRT

Two sets of conditions: general gas equation
P1V1 P2V2

n1T1 n2T2
20
Ideal Gas Equation and molar mass

Solving for molar mass (M)
PV  nRT
m
n
M
mRT
PV 
M
mRT
M
PV
22
Ideal Gas Equation and gas density
mRT
PV  nRT 
M
mRT
P
VM
d
dRT
P
M
m
V
MP
d
RT
24
Gas density
Gas density depends directly
on pressure and inversely on
temperature
 Gas density is directly
proportional to molar mass

MP
d
RT
25
Mixtures of Gases
Ideal gas law applies to pure gases and to mixtures
 In a gas mixture, each gas occupies the entire
container volume, at its own pressure
 The pressure contributed by a gas in a mixture is
the partial pressure of that gas
 Ptotal = PA + PB
(Dalton’s Law of Partial Pressures)

27
Mixtures of
Gases

When a gas is collected over water, it is always
“wet” (mixed with water vapor).
Ptotal = Pbarometric = Pgas + Pwater vapor
 Example: If 35.5 mL of H2 are collected over water at
26 °C and a barometric pressure of 755 mm Hg, what is
the pressure of the H2 gas? The water vapor pressure at
26 °C is 25.2 mm Hg.

28
Gas mixtures

The mole fraction represents the contribution of
each gas to the total number of moles.

XA = mole fraction of A
nA
XA 
ntotal
29
Gas Mixtures

For gas mixtures,
nA
PA
VA


ntotal Ptotal Vtotal
mole fraction
equals
pressure fraction
equals
volume fraction
Each gas occupies
the entire container.
The volume fraction describes
the % composition by volume.
32
Gases in Chemical Reactions
PV
To convert gas volume into moles for
n
RT
stoichiometry, use the ideal gas equation:
 If both substances in the problem are gases, at the
same T and P, gas volume ratios = mole ratios.

n2

n1
P2 V2
RT2
P1V1
RT1
V2

V1
P2 = P1 and T2 = T1
34
A Model for Gas Behavior
Gas laws describe what gases do, but not why.
 Kinetic Molecular Theory of Gases (KMT) is the
model that explains gas behavior.

developed by Maxwell & Boltzmann in the mid-1800s
 based on the concept of an ideal or perfect gas

36
Ideal gas





Composed of tiny particles in constant, random, straight-line motion
Gas molecules are point masses, so gas volume is just the empty
space between the molecules
Molecules collide with each other and with the walls of their
container
The molecules are completely independent of each other, with no
attractive or repulsive forces between them.
Individual molecules may gain or lose energy during collisions, but
the total energy of the gas sample depends only on the absolute
temperature.
37
Molecular collisions and pressure

Force of molecular collisions depends on
collision frequency
 molecule kinetic energy, ek

1 2
ek  mu
2
 ek
depends on molecule mass m and molecule speed u
 molecules move at various speeds in all directions
38
Molecular speed
Molecules move at various speeds
 Imagine 3 cars going 40 mph, 50 mph, and 60 mph

Mean speed = u = (40 + 50 + 60) ÷ 3 = 50 mph
 Mean square speed (average of speeds squared)
u2 = (402 + 502 + 602) ÷ 3 = 2567 m2/hr2
 Root mean square speed
urms = √2567 m2/hr2 = 50.7 mph

39
Distribution of molecule speeds
40
The basic equation
of KMT

1N
2
P
mu
3V
Combining collision frequency, molecule kinetic
energy, and the distribution of molecule speeds
gives the basic equation of KMT
P = gas pressure and V = gas volume
 N = number of molecules
 m = molecule mass
 u2 = mean square molecule speed (average of speeds squared)

41
Combine the Equations of
KMT and Ideal Gas
1N
2
P
mu
3V
If n = 1,
N = NAAvogadro’s number
and
PV = RT
PV  nRT
1 NA
2
P
mu
3 V
PV  N A m u  RT
1
3
2
42
Combine the Equations of
KMT and Ideal Gas
PV  N A m u  RT
1
3
2
N A mu  3RT  M u
2
2
NA x m (Avogadro’s number x mass of one molecule)
= mass of one mole of molecules (molar mass M)
43
Combine the Equations of
KMT and Ideal Gas
3RT  M u
2
u  urms 
2
3RT
M
We can calculate the root mean square speed
from temperature and molar mass
44
Calculating root mean square speed

To calculate root mean square speed from
temperature and molar mass:
Units must agree!
 Speed is in m/s, so

urms 
3RT
M
R must be 8.3145 J/mol K
 M must be in kg per mole, because Joule = kg m2 / s2


Speed is inversely related to molar mass: light
molecules are faster, heavy molecules are slower
45
Interpreting temperature

Combine the KMT and ideal gas equations again
Again assume n=1, so N = NA and PV = RT
PV  NAmu   NA mu
2
1
3
PV  N
2
3
 mu
1
A 2
2
2
3

1
2
2
3
2
N A e k  RT
R
ek 
T  constant T
NA
3
2
47
Interpreting
temperature
R
ek 
T
NA
3
2
Absolute (Kelvin)
temperature is
directly
proportional to
average
molecular kinetic
energy
 At T = 0, ek = 0

48
Diffusion and Effusion

Diffusion (a) is
migration or mixing
due to random
molecular motion

Effusion (b) is escape
of gas molecules
through a tiny hole
49
Rates of diffusion/effusion

The rate of diffusion or effusion is directly
proportional to molecular speed:
rat e of effusion of A (urms)A
3RT MA



rat e of effusion of B (urms)B
3RT MB

MB
MA
The rates of diffusion/effusion of two different
gases are inversely proportional to the square roots
of their molar masses (Graham’s Law)
50
Using Graham’s Law
rat e of effusion of A (urms)A


rat e of effusion of B (urms)B

MB
MA
Graham’s Law applies to relative rates, speeds,
amounts of gas effused in a given time, or
distances traveled in a given time.
51
Using Graham’s Law with times
Graham’s law can be confusing when applied to
times
rat
e
A
n
/t
M
A
A
B
rate = amount of gas (n)


time (t)
rat e B nB/t B
MA

For same amount s of A and B
tB

tA
MB
MA
52
Use common sense
with Graham’s Law

When you compare two gases, the lighter gas
escapes at a greater rate
 has a greater root mean square speed
 can effuse a larger amount in a given time
 can travel farther in a given time
 needs less time for a given amount to escape or travel


Make sure your answer reflects this reality!
53
Reality Check

Ideal gas molecules

constant, random,
straight-line motion
point masses

independent of each other

gain / lose energy during
collisions, but total energy
depends only on T ek)

Real gas molecules
same
are NOT points – molecules
have volume; Vreal gas > Videal gas
are NOT independent –
molecules are attracted to
each other, so Preal gas < Pideal gas
same (some energy may be
absorbed in molecular
)
56
Real gas corrections

For a real gas,
a corrects for attractions between gas molecules, which
tend to decrease the force and/or frequency of collisions
(so Preal < Pideal)
 b corrects for the actual volume of each gas molecule,
which increases the amount of space the gas occupies
(so Vreal > Videal)


The values of a and b depend on the type of gas
57
An equation for real gases:
the van der Waals equation
2

n
a

P 
V – nb  nRT  P V
re al
2
re al
ideal ideal

V 
Add correction to Preal to make it equal to Pideal,
because intermolecular attractions decrease real pressure
Subtract correction to Vreal to make it equal to Videal,
because molecular volume increases real volume
58
When do I need
the van der Waals equation?

Deviations from ideality become significant when
molecules are close together (high pressure)
 molecules are slow (low temperature)

}
non-ideal
conditions
At low pressure and high temperature, real gases
tend to behave ideally
 At high pressure and low temperature, real gases
do not tend to behave ideally

59