Chapter 10
Gases
Barometers and Standard
Atmospheric Pressure
Barometers and Standard
Atmospheric Pressure
• Standard atmospheric
pressure defined as the
pressure sufficient to support a
mercury column of 760mm
(units of mmHg, or torr).
Barometers and Standard
Atmospheric Pressure
• Standard atmospheric
pressure defined as the
pressure sufficient to support a
mercury column of 760mm
(units of mmHg, or torr).
• Another unit was introduced to
simplify things, the atmosphere
(1 atm = 760 mmHg).
Barometers and Standard
Atmospheric Pressure
• Standard atmospheric
pressure defined as the
pressure sufficient to support a
mercury column of 760mm
(units of mmHg, or torr).
• Another unit was introduced to
simplify things, the atmosphere
(1 atm = 760 mmHg).
• 1 atm = 760 mmHg = 760 torr =
101.325 kPa (page 262).
STP
standard temperature and pressure
Standard temperature
0°C or 273 K
Standard pressure
1 atm (or equivalent)
– Pressure varies inversely
with volume
– Volume varies inversely with
pressure
“The volume of a sample of gas is inversely proportional
to its pressure, if temperature remains constant.”
Boyle’s Law
Boyle’s Law: Pressure & Volume
(Figure 10.6 (a) page 263)
P V 
P V 

Boyle’s Law: Pressure-Volume Relationships
A sample of air occupies 73.3 mL at 98.7 atm and 0 ºC.
What volume will the air occupy at 4.02 atm and 0 ºC?
1800 mL
Boyle’s Law: Pressure-Volume Relationships
A sample of helium occupies 535 mL at 988 mmHg and
25 °C. If the sample is transferred to a 1.05-L flask at
25 °C, what will be the gas pressure in the flask?
503 mm Hg
• Effects of temperature on a gas
• Volume varies directly with
Temperature
“The volume of a quantity of gas, held
at constant pressure, varies directly
with the Kelvin temperature.”
Charles’s Law
a
Charles Law: Volume and Temperature
(Figure 10.8 Page 266)
Charles’ Law and Absolute Zero
•Extrapolation to
zero volume gives
a temperature of
-273°C or 0 K
Charles’s Law: Temperature-Volume Relationships
A sample of oxygen gas occupies a volume of 2.10 L at 25 °C.
What volume will this sample occupy at 150 °C? (Assume no
change in pressure.)
2.98 L
Charles’s Law: Temperature-Volume Relationships
A sample of oxygen gas occupies a volume of 2.10 L at 25 °C.
At what Celsius temperature will the volume of oxygen
occupy 0.750 L? (Assume no change in pressure.)
-167°C
Pressure vs. Temperature
• Pressure varies directly with Temperature
• If the temperature of a fixed volume of gas
doubles its pressure doubles.
Pressure vs. Temperature
• The pressure exerted by a
gas is directly related to the
Kelvin temperature.
• V is constant.
Pressure vs. Temperature
Example
A gas has a pressure of 645 torr at 128°C. What is the
temperature in Celsius if the pressure increases to 1.50 atm?
Pi = 645 torr
Ti = 128°C + 273
= 401 K
Pf = 1.50 atm 760 torr = 1140 torr
1 atm
Tf = ?K
Solution
T2 = 401 K x 1140 torr = 709K
645 torr
709K - 273 = 436°C
Combined Gas Law Problem
A sample of helium gas has a volume of 0.180 L, a
pressure of 0.800 atm and a temperature of 29°C.
What is the new temperature(°C) of the gas at a
volume of 90.0 mL and a pressure of 3.20 atm?
Combined Gas Law Problem
A sample of helium gas has a volume of 0.180 L, a
pressure of 0.800 atm and a temperature of 29°C.
What is the new temperature(°C) of the gas at a
volume of 90.0 mL and a pressure of 3.20 atm?
302 K
x
3.20 atm
0.800 atm
x
90.0 mL = 604 K
180.0 mL
604 K - 273 = 331 °C
Combined Gas Law
• A 10.0 cm3 volume of gas measured 75.6 kPa
and 60.0C is to be corrected to correspond
to the volume it would occupy at STP.
6.12 cm3
Gay-Lussac’s Law
Gay-Lussac’s Law of combining volumes: at a given
temperature and pressure, the volumes of gases which
react are ratios of small whole numbers.
How many liters of steam can be
formed from 8.60L of oxygen gas?
17.2 L
How many liters of hydrogen gas
will react with 1L of nitrogen gas to
form ammonia gas?
3L H2
A
A
How many mL of hydrogen are needed to
produce 13.98 mL of ammonia?
20.97 ml NH3
Avogadro’s Law: Equal volumes of gases at the same
temperature and pressure contain the same number of particles.
The molar volume of a gas at STP = 22.4L
22.4 L
Ideal Gas
• An ideal gas is defined as one for which both the
volume of molecules and forces of attraction
between the molecules are so small that they have
no effect on the behavior of the gas.
Ideal Gas Equation
PV=nRT
R values
•
A
a
•
I
•R values for atm and kPa
on Page 272 in book.
Calculate the volume occupied by 0.845 mol of
nitrogen gas at a pressure of 1.37 atm and a
temperature of 315 K.
15.9 L
Find the pressure in millimeters of mercury of a
0.154 g sample of helium gas at 32°C and contained
in a 648 mL container.
1130 mm Hg
An experiment shows that a 113 mL gas
sample has a mass of 0.171 g at a pressure of
721 mm Hg and a temperature of 32°C. What is
the molar mass (molecular weight) of the gas?
40.0 g/mol
Can the ideal “gas” equation be used to
determine the molar mass of a liquid?
Homework
• Do the lab summary for “The Molecular
Mass of a Volatile Liquid”. It is due ____.
• Attempt the pre-lab for “The Molecular
Mass of a Volatile Liquid”. It is due ____.
Problem: A volatile liquid is placed in a flask whose volume is 590.0 ml
and allowed to boil until all of the liquid is gone, and only vapor fills the
flask at a temperature of 100.0 oC and 736 mm Hg pressure. If the mass
of the flask before and after the experiment was 148.375g and 149.457 g,
what is the molar mass of the liquid?
57.9 g/mol
What is the density of methane gas (natural
gas), CH4, at 125oC and 3.50 atm?
1.71 g/L
Calculate the density in g/L of
O2 gas at STP.
1.43 g/L
Dalton’s Law of Partial Pressure
• The total pressure in a container is the sum of the partial
pressures of all the gases in the container.
• In a gaseous mixture, a gas’s partial pressure is the one
the gas would exert if it were by itself in the container.
• Ptotal = P1 + P2 + P3
• Ptotal = 100 KPa + 250 KPa + 200 KPa = 550 KPa
Two 1.0 L containers, A and B, contain gases
with 2.0 atm and 4.0 atm, respectively. Both
gases are forced into Container B. Find the total
pressure of the gas mixture in B.
A
B
P
V
A
2.0 atm
1.0 L
B
4.0 atm
1.0 L
Vmixture
P
2.0 atm
1.0 L
4.0 atm
Total = 6.0 atm
Dalton’s Law Problem
• Air contains oxygen, nitrogen, carbon dioxide,
and trace amounts of other gases. What is the
partial pressure of oxygen at standard
conditions if the partial pressure of nitrogen,
carbon dioxide, and other gases are 79.1 KPa,
0.04 KPa, and 0.94 KPa respectively?
• Ptotal = PO + PN + PCO + POther gases
2
2
2
• 101.3 KPa = PO2 + 79.1 KPa + 0.04 KPa + 0.94KPa
• PO = 101.3 KPa – (79.1 KPa + 0.04 KPa + 0.94KPa)
2
• PO = 21.2 KPa
2
Two 1.0 L containers, A and B, contain gases with
2.0 atm and 4.0 atm, respectively. Both gases are
forced into Container Z (vol. 2.0 L). Find the total
pressure of mixture in Z.
A
B
Z
Two 1.0 L containers, A and B, contain gases with
2.0 atm and 4.0 atm, respectively. Both gases are
forced into Container Z (vol. 2.0 L). Find the total
pressure of mixture in Z.
A
A
B
Z
PX
VX
2.0 atm
1.0 L
VZ
PX,Z
1.0 atm
2.0 L
B
4.0 atm
1.0 L
2.0 atm
Total = 3.0 atm
Find total pressure of the gas mixture in Container Z.
A
1.3 L
3.2 atm
B
2.6 L
1.4 atm
C
Z
3.8 L
2.7 atm
2.3 L
X atm
Find total pressure of the gas mixture in Container Z.
A
B
1.3 L
3.2 atm
2.6 L
1.4 atm
PX
VX
A
3.2 atm
1.3 L
B
1.4 atm
2.6 L
C
2.7 atm
3.8 L
C
Z
3.8 L
2.7 atm
2.3 L
X atm
VZ
PX,Z
1.8 atm
2.3 L
1.6 atm
4.5 atm
Total = 7.9 atm
Dalton’s Law
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 422
Dalton’s Partial Pressures
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 421
Dalton’s Law of Partial Pressures
The mole ratio in a mixture of gases determines each
gas’s partial pressure.
Total pressure of mixture (3.0 mol He and 4.0 mol Ne) is
97.4 kPa.
Find partial pressure of each gas
Dalton’s Law of Partial Pressures
Total pressure of mixture (3.0 mol He and 4.0 mol Ne) is 97.4 kPa.
Find partial pressure of each gas.
PHe
3 mol He
97.4 kPa 

7 mol gas
41.7 kPa
PNe
4 mol Ne
97.4 kPa 

7 mol gas
55.7 kPa
80.0 g each of He, Ne, and Ar are in a container. The total
pressure is 780 mm Hg. Find each gas’s partial pressure.
80.0 g each of He, Ne, and Ar are in a container. The total
pressure is 780 mm Hg. Find each gas’s partial pressure.
 1mol 
80 g He 
  20 mol He
 4g 
 1mol 
80 g Ne 
  4 mol Ne
 20 g 
 1 mol 
80 g Ar 
  2 mol Ar
 40 g 
PHe = 20/26
of total
Total:
26 mol gas
PNe = 4/26
of total
PAr = 2/26
of total
PHe  600 mm Hg, PNe  120 mm Hg, PAr  60 mm Hg
Example: A student generates oxygen
gas and collects it over water. If the
volume of the gas is 245 mL and the
barometric pressure is 758.0 torr at 25oC,
what is the volume of the “dry” oxygen
gas at STP? (Pwater = 23.8 torr at 25oC)
PO2 = PT - Pwater = 758.0 torr - 23.8 torr = 734.2 torr
Find the molar mass of an unknown gas if a 0.16 g sample of
the gas is collected over water and equalized to a pressure of
781.7 torr and a volume of 90.0 mL at a temperature of 28°C .
Find the molar mass of an unknown gas if a 0.16 g sample of
the gas is collected over water and equalized to a pressure of
781.7 torr and a volume of 90.0 mL at a temperature of 28°C .
44 g/mol
Homework
• Do the AP sample problem (1999 Test
question #5) in notebook. It will be
included as part of your homework.
• Don’t forget the pre-lab and lab summary
for “The Molecular Mass of a Volatile
Liquid”.
Gas Diffusion and Effusion
Graham's Law:
governs the rate of effusion and diffusion of gas molecules.
“Stink” or “Die”
a
The Root Mean Square Speed
Fig. 10.17 Page 285
To use Graham’s Law, both gases must be at same temperature.
diffusion: particle movement from
high to low concentration
NET MOVEMENT
effusion: diffusion of gas particles
through an opening
For gases, rates of diffusion & effusion obey Graham’s law:
more massive = slow; less massive = fast
Gas Diffusion and Effusion
Graham's Law:
governs the rate of effusion and diffusion of gas molecules.
Rate of A
molar mass of B

Rate of B
molar mass of A
Rate of diffusion/effusion is inversely
proportional to its molar mass.
35
36
Br
Graham’s Law
79.904
Kr
83.80
Determine the relative rate of diffusion
for krypton and bromine.
The lightest gas is “Gas A” and the heavier gas is “Gas B”.
Relative rate means find the ratio “vA/vB”.
vA

vB
v Kr

v Br2
m Br2
m Kr
mB
mA
159.80g/mol
 1.381

83.80g/mol
Kr diffuses 1.381 times faster than Br2.
1
8
H
O
Graham’s Law
1.00794
15.9994
A molecule of oxygen gas has an average speed of 12.3
m/s at a given temp and pressure. What is the average
speed of hydrogen molecules at the same conditions?
vA

vB
mB
mA
vH 2
12.3 m/s

32.00 g/mol
2.02 g/mol
vH 2
vH 2
vO2

mO2
mH 2
12.3 m/s
 3.980
vH2  49.0m/s
1
8
H2
Graham’s Law
2.0
O
15.9994
An unknown gas diffuses 4.0 times faster than O2.
Find its molar mass.
The lightest gas is “Gas A” and the heavier gas is “Gas B”.
The ratio “vA/vB” is 4.0.
vA

vB
vA

v O2
mB
mA
mO2
mA
Square both
sides to get rid
of the square
root sign.

32.00 g/mol
g/mol 
32.00
 4.0 



m
A
A


32.00 g/mol
16 
mA
32.00 g/mol
mA 
 2.0 g/mol
16
2
Kinetic Molecular Theory
Theory developed to explain gas behavior.
• Theory of moving molecules.
• Assumptions:
– Gases consist of a large number of molecules in constant
random motion.
– Volume of individual molecules negligible compared to
volume of container.
– Intermolecular forces (forces between gas molecules)
negligible.
– Energy can be transferred between molecules, but total
kinetic energy is constant at constant temperature.
– Average kinetic energy of molecules is proportional to
temperature.
Kinetic Molecular Theory
Kinetic molecular theory gives us an
understanding of pressure and
temperature on the molecular level.
• Pressure of a gas results from the
number of collisions per unit time on the
walls of container.
• Magnitude of pressure given by how
often and how hard the molecules
strike.
• Gas molecules have an average kinetic
energy.
• Each molecule has a different energy.
Kinetic Molecular Theory
There is a spread of
individual energies
of gas molecules
in any sample of
gas.
As the temperature
increases, the average
kinetic energy of the
gas molecules
increases
Kinetic Molecular Theory
As kinetic energy increases, the velocity of the gas
molecules increases.
• Root mean square speed, u, is the speed of a gas
molecules having the certain average kinetic energy.
• Average kinetic energy, , is related to root mean
square speed, u:
2
1
  2 mu
a
Kinetic Molecular Theory
As kinetic energy increases, the velocity of the gas
molecules increases.
• Root mean square speed, u, is the speed of a gas
molecules having the certain average kinetic energy.
• Average kinetic energy, , is related to root mean
square speed, u:
2
1
  2 mu
a
How does this theory explain Boyles Law?
As the volume of a container of gas increases at
constant temperature, the gas molecules have to travel
further to hit the walls of the container. There are
fewer collisions by the gas molecules with the walls
of the container. Therefore, pressure decreases.
If temperature increases at constant volume, the average
kinetic energy of the gas molecules increases.
Therefore, there are more collisions with the
container walls and the pressure increases.
How does this theory explain Charles Law?
If temperature increases at constant volume, the average
kinetic energy of the gas molecules increases and they
speed up. Therefore, there are more frequent and
more forceful collisions with the container walls by
the gas molecules and the pressure increases.
Ideal Gases vs. Real Gases
• An ideal gas is an “imaginary gas” made
up of particles with negligible particle
volume and negligible attractive forces.
Ideal Gases vs. Real Gases
• In a “Real Gas” the molecules of a gas do
have volume and the molecules do attract
each other.
• Therefore anything that makes gas particles
more likely to stick together or stay close to
one another make them behave less ideally.
Real Gases: Deviations from Ideal Behavior
As the pressure on a gas increases, the molecules are forced
into a smaller volume.
• As the volume becomes
smaller, the molecules get
closer together, and a
greater fraction of the
occupied space is actually
taken up by gas molecules.
• Therefore, the higher the
pressure, the less the gas
resembles an ideal gas.
Real Gases: Deviations from Ideal Behavior
• The smaller the distance between gas
molecules, the more likely attractive
forces will develop between the
molecules.
• As temperature increases, the gas
molecules move faster and are
further apart.
• Also, higher temperatures mean
more energy available to break
intermolecular forces.
• Therefore,
the
higher
the
temperature, the more ideal the gas.
Real Gases and Ideal Behavior
• A real gas typically exhibits behavior closest to
“ideal gas” behavior at low pressures and high
temperatures.
Real Gases: The van der Waals equation
We add two terms to the ideal gas equation one to correct for
volume of molecules and the other to correct for
intermolecular attractions
The correction terms generate the van der Waals equation:
nRT n 2 a
P
 2
V  nb V
where a and b are empirical constants.
2 

n
 P  a V  nb   nRT
2 

V


a corrects for the effect of molecular attractions (van der
Waals forces), and b corrects for the molecular volume
Real Gases: The van der Waals equation
We add two terms to the ideal gas equation one to correct for
volume of molecules and the other to correct for intermolecular
attractions
The correction terms generate the van der Waals equation:
• You will not be required to solve this equation but you should
know its form and which variables need to be corrected.
2 

n
 P  a V  nb   nRT
2 

V


a corrects for the effect of molecular attractions (van der
Waals forces), and b corrects for the molecular volume