Gases
Chapter 10
(very loosely)
Jespersen 7th Edition
Dr. C. Yau
Spring 2015
1
Properties of Gases: Macroscopic View
• Gases can be compressed
• Gases exert a pressure.
• The pressure of a gas depends on how
much gas is confined.
• Gases fill completely any container
into which they are placed.
• Gases mix freely and quickly with
each other.
2
Kinetic Molecular Theory (KMT) of Gases:
View at the Particulate Level
1. A gas is made of an extremely large number of
very tiny particles that are in constant, random
motion.
2. The gas particles themselves occupy a net
volume so small in relation to the volume of their
container that their contribution to the total
volume can be ignored.
3. The particles often collide in perfectly elastic
collisions with themselves and with the walls of
the container.
3
Kinetic Molecular Theory of Gases:
View at the Particulate Level (cont'd.)
4. The particles move in straight lines neither
attracting nor repelling each other.
5. The pressure of a gas is due to the frequency
and force of collisions of the particles with the
sides of the container.
6. The particles within a sample of gas have
different kinetic energy (KE), but for samples at
a given temperature, the average kinetic
energy is the same. It is affected only by
temperature.
4
The Ideal Gas Law
This is the law for the ideal gas.
It's the gas that is ideal, not the law.
There are "real gases" that do not follow this law.
KNOW THIS WELL!
PV = nRT where P = pressure of gas in atm
V = volume of gas in L
n = # moles of gas
T = temp. of gas in Kelvin
R = 0.08206 atm L mol-1 K-1 or  atm  L 


(no need to memorize R)
mol

K


5
Parameters of a Gas
To describe a gas, one must specify the P, V and
T as they are all interdependent on each other.
Pressure:
Units: 1 atm = 760 mmHg = 760 torr
1 mmHg = 1 torr
What has P to do with mm and mercury?
The barometer measures the P of the
atmosphere. It is constructed with a tube of
mercury, and the height of the mercury column
is measured in millimeters, hence, mm Hg.
6
Barometric Pressure (Pbarom or Patm)
The column of Hg is affected
by the pressure of the
atmosphere pressing down
on the reservoir of Hg.
Which day has the higher
atmospheric pressure?
A taller column means a
higher Patm.
Day 1 has the higher
atmospheric pressure (or
barometric pressure).
Day 1
Day 2
752 mm Hg means a
higher P than 750 mm Hg.
7
Parameters of a Gas: P, V, T (cont'd.)
Pressure:
The manometer measures the P of a gas sample.
Temperature:
• The Celsius scale was arbitrarily based on the fp of
water to be zero.
• The behavior of gases cannot be based on
something arbitrary. Instead it is based on the
absolute zero being the lowest temp possible.
• The Kelvin scale: TK = ToC + 273.15 KNOW THIS!
• If P is not changed, V of gas decreases with T.
Theoretically when V = zero, T is at lowest (zero K).
• KE of gas particles decreases with T.
• At 0 K, theoretically gas particles have zero KE
(motionless).
8
Parameters of a Gas: P, V, T (cont'd.)
Volume: Varies with P and T. This makes it
difficult to compare gases when measured at
different P and T...hence the definition of
STP:
STP (Standard Temperature & Pressure)
KNOW THIS WELL!
STP means "standard temperature &
pressure):
0oC (273.15 K)
and 1 atm (760 mm Hg)
9
Molar Volume at STP
For ease of comparison, we usually give the
volume of a gas at STP.
For example, which gas sample contains more
molecules of gas?
Sample A:
3.45 L of gas at 758 mmHg and 38.4oC
Or
Sample B:
1.22 L of gas at 798 mmHg and 17.3oC?
Sample A has a larger V, but it is at a lower P and
higher T. It is difficult to compare the two
samples without doing some calculations. 10
Molar Volume at STP
"Molar volume" refers to the V of one mole of gas.
Molar Volume of any ideal gas at STP = 22.4 L
(Not an exact number, but 3 sig. fig.)
Do not forget this is ONLY at STP, for ideal gas.
For an ideal gas, since the gas particles have
negligible volume and do not sense the
presence of neighboring particles (no attraction
nor repulsion) the volume does NOT depend on
the type of gas we have.
11
Gas Calculations
TWO types of Gas Problems dealing with PV=nRT:
1) One set of parameters given: Use PV=nRT
2) Two sets of parameters given:
Rearrange equation so only R is on one side of
eqn:
PV
=R
nT
Since R is a constant, this means that under any
conditions, (PV)/(nT) must always equal to each
other:
P1 V1
P2 V2
=
n1 T1
n1 T2
12
Example 1 (from handout):
What is the volume of one mole of gas at STP?
First identify which type of problem it is.
Example 2
What does "molar volume of a gas" refer to?
Example 3
Sulfur hexafluoride SF6 is a colorless,
odorless, very unreactive gas. Calculate
the pressure (in atm) exerted by 1.82
moles of the gas in a steel vessel of
volume 5.43 L at 69.5C.
Ans. 9.42 atm
13
Example 4
A gas at 772 mmHg and 35.0C occupies a
volume of 6.85 L. Calculate its volume at
STP.
Ans. 6.17 L
Example 5
If the volume of a gas is 5.0 L at 758 torr,
what is the volume when the pressure is
increased to 1.23 atm. Assume the
temperature remains constant.
Ans. 4.1 L
14
Example 6:
A 2.0 L balloon filled with nitrogen at 18°C at a
picnic. By noon, the temperature has gone up
to 28°C. What is the volume of the balloon at
that point? Assume the atmospheric pressure
has not changed throughout the day.
Ans. 2.1 L
15
Example 7:
Sodium azide (NaN3) is used in some automobile
air bags. The impact of a collision triggers the
decomposition of NaN3 as follows:
2NaN3 (s)  2Na (s) + 3N2 (g)
The nitrogen gas produced quickly inflates the
bag between the driver and the windshield.
Calculate the volume of N2 generated at 80.0C
and 823 mmHg by the decomposition of 60.0 g
of NaN3.
Ans. 37.1 L
16
How would you rearrange PV=nRT
to give you the density of the gas?
How do you normally calculate the density of an
object?
Which part of PV = nRT would give us density?
Which part of the eqn would give us molar mass
(MM)?
Rearrange the equation to calculate MM.
17
D = mass/volume = g/V where g= mass, V=volume
There is no term for mass(g) in PV=nRT but we
know MM = g/n (where g= mass and n = moles)
Rearrange to n = g/MM
Thus
PV =
n R T
Becomes PV = (g/MM)RT
g
P MM
Density = =
V
R T
Significance? How is density related to molar mass?
Next, rearrange the equation to calculate MM.
18
Example 9:
The volume of a gas was determined to be 200.0
mL at 99oC and 733 mmHg. It has a mass of
0.970 g. What is its molar mass?
Is it chloroform (CHCl3) or carbon tetrachloride?
Ans. CCl4
19
Effusion & Diffusion of Gases
Effusion is the gradual movement of gas particles
through a very tiny hole into a vacuum.
Diffusion is the spontaneous mixing of the
particles of one gas with those of another.
20
Graham's Law of Effusion:
The rate of effusion (r) is inversely proportional to
the square root of its density (d).
rA
=
rB
dB
dA
dB
MB
and
=
where M = molar mass
dA
MA
The density of a gas is directly proportional to its
molar mass.
Thus, rate of effusion is inversely proportional
to the square root of its molar mass.
rA
=
rB
MB
MA
21
Graham's Law of Effusion
Rate of effusion is inversely proportional to
the square root of its molar mass.
rA

rB
MB
MA
where r = rate of effusion
M = molar mass
The heavier gas effuses slower at a given T.
Example: Which effuses faster, Ne or Xe?
How much faster?
22
Example: Which effuses faster, Ne or Xe?
How much faster?
Ne is lighter (M = 20.2 g/mol)
Xe is heavier (M = 131 g/mol)
Ne should effuse faster than Xe.
rNe

rXe
M Xe
= ?
M Ne
Ans. Ne effuses _____ times faster than Xe.
23
If a molecule B is 4 times heavier than
A, how much faster would A be
effusing?
Ans. A is 2X as fast.
How does T affect the rate of effusion?
24
Dalton's Law of Partial Pressures
This law concerns mixtures of gases that do not
react with each other. Each gas exerts its own
partial pressure.
In a mixture of gases (such as A and B), the total
pressure is the sum of the partial pressures of
each gas.
PTOT = p
A
+p
B
n A RTA
n B RTB
p A=
and p B =
VA
VB
Note: Since they are in the same container, T and
V must be the same. R is also the same.
25
Collecting Gases Over Water
Fig.10.11
p.491
26
Table 10.2
27
Example 10.10 (p. 492)
A sample of oxygen is collected over water at
15oC and a pressure of 738 torr. Its volume
is 316 mL.
(a) What is the partial pressure of the oxygen?
(b) What would be its volume when dry at
STP?
Ans. 725 torr, 286 mL at STP.
Practice Exercises 23 & 24 p. 493
28
Dalton's Law of Partial Pressures (cont'd.)
pA  X A PTOTAL
nA
where X A = mole fraction =
nA  nB
(Partial pressure of A is a fraction of the total pressure.)
Rearrangement of equation gives us…
pA
PTOTAL
nA

nA  nB
What this tells us is that the fraction in pressures is
equal to the fraction in moles, the mole fraction.
REMEMBER: This is only for mixtures of gases. 29
Example 10.11 p. 495
Suppose a mixture of oxygen and nitrogen is
prepared in which there are 0.200 mol O2
and 0.500 mol N2. If the total pressure of
the mixture is 745 torr, what are the partial
pressures of the two gases in the mixture?
Ans. 213 torr and 532 torr
Practice Exercises 25 & 26 p. 496
30
Real Gas vs. Ideal Gas
A real gas differs from an ideal gas:
1) Gas particles do have volume.
2) There are interactions between particles in a
gas (attraction and repulsion).
These are noticeable only under these conditions:
1) At slightly higher P or VERY low T (near
liquefaction) particles are moving very slowly
and can "socialize." EFFECT: Actual V smaller
than ideal. P is lower than ideal.
2) At VERY high P, particles are forced VERY
closely together and their volume becomes
significant. EFFECT: Actual P is higher than
ideal.)
31
Real Gas vs. Ideal Gas
Due to molecules having sig. size.
Due to attraction between molecules of a real gas.
You do not need to memorize this equation or do
any calculations with it. Know when and how real
gases deviate in the Ideal Gas Law (previous slide).32
Real vs. Ideal Gases
This is a graph for 1 mole of
gas (n = 1 mole). For the
Ideal Gas, PV = n RT
therefore(PV)/(RT) = 1.0 mol
at any pressure.
We see that all gases at normal conditions (P = 1 atm) behave
like an ideal gas, (PV)/(RT) = 1.0.
At slightly higher than 1.0 atm (but below 10 atm), the graphs for
real gases dip below 1.0. This can be explained by real gases
having intermolecular attraction.
As the P rises further, the graphs rise above 1.0. This can be
explained by real gases having significant volume.
33