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Ch. 11 Properties of Gases
Brady & Senese, 5th Ed
Index
• 10.1. Familiar properties of gases can be explained at the molecular
level
• 10.2. Pressure is a measured property of gases
• 10.3. The gas laws summarize experimental observations
• 10.4. Gas volumes are used in solving stoichiometry problems
• 10.5. The ideal gas law relates P, V, T, and the number of moles of
gas, n
• 10.6. In a mixture each gas exerts its own partial pressure
• 10.7. Effusion and diffusion in gases leads to Graham's law
• 10.8. The kinetic molecular theory explains the gas laws
• 10.9. Real gases don't obey the ideal gas law perfectly
10.1 Familiar properties of gases can be explained at the molecular level
2
Properties of Gases
• What is the shape of the air in a balloon?
 Gases have an indefinite shape
• What is the volume of the gas in the balloon?
 They have an indefinite volume
10.1 Familiar properties of gases can be explained at the molecular level
3
How Does a Molecular Model Explain This?
• gases completely fill their
containers:
 Gases are in constant random motion
• gases have low density and are easy
to compress
 gas molecules are very far apart
• gases are easy to expand
 gas molecules don’t attract one
another strongly
10.1 Familiar properties of gases can be explained at the molecular level
4
What Is Pressure?
• The force of the collisions of the gas distributed over
the surface area of the container walls; P=force/area
 units : 1 atmosphere (atm) = 760 mm Hg (torr) =
1.01325(105) Pascal (Pa) = 14.7 psi=1013 millibar (mb)
• measured with a barometer*
 P=g×d×h
 d=density of the liquid
 g= gravitational acceleration
 h=height of the column supported
• Why use Mercury?
10.2 Pressure is a measured property of gases
5
Learning Check: Pressure Units
Convert 675 mm Hg to atm.
Known: 675 mmHg
Conversion factor?
Unknown: atm
760 mmHg = 1 atm
 1 atm 
675 mm Hg x 
  .888 atm
 760 mmHg 
10.2 Pressure is a measured property of gases
6
Force of Collisions
Learning Check: P 
Area
What happens to gas pressure when you raise the
temperature?
If the container can expand In a rigid walled container
in response to the force
No change in pressure is
observed because the area
increased.
Pressure increases
because the faster moving
molecules hit the walls of
the container with greater
force
10.3 The gas laws summarize experimental observations
7
Learning Check
Force of Collisions
P
Area
What happens to gas pressure when you increase the number
of molecules in the container?
If a container can expand
No pressure change
is observed.
In a rigid walled container
pressure increases
because more molecules
hit the walls of the
container, thus exert a
greater force on the
container
10.3 The gas laws summarize experimental observations
8
Absolute Zero
Temperature at which
there is zero molecular
motion
10.3 The gas laws summarize experimental observations
9
Ideal Gases
• Their behavior is predicted by the gas laws
• There are NO ideal gases
• However, most gases behave ideally at atmospheric P
or lower and room T or higher
• You need to know when they are not useful
10.3 The gas laws summarize experimental observations
10
Combined Gas Law:
•
•
•
•
•
1
P
V
Boyle’s Law
T V
Charles’ Law
Gay-Lussac’s Law T  P
Thus combining this information
And therefore, for any 2 conditions:
T
P
V
P1V1 P2V2

T1
T2
10.3 The gas laws summarize experimental observations
11
P1V1
P2 V2

T1
T2
Used for calculating the effects of changing
conditions
Combined Gas Law
 works if the Temperature is in Kelvin, but P and V
can be any units so long as the units cancel
• Learning Check:
• If a sample of air occupies 500. mL at STP*, what
is the volume at 85 °C and 560 torr?
760 torr  500. mL
560torr  V2

273.15 K
358.15K
890 mL
*Standard Temperature (273.15K) and Pressure (1 atm)
10.3 The gas laws summarize experimental observations
12
Learning Check
A sample of oxygen gas occupies 500.0 mL at 722
torr and –25 ºC. Calculate the temperature in ºC
if the gas has a volume of 2.53 L at 491 mm Hg.
.
P1V1
PV
 2 2
T1
T2
722 torr  500.0 mL
491torr  2530mL

248.15 K
T2
T2=853.90 K
T2=581 °C
10.3 The gas laws summarize experimental observations
13
Learning Check
A sample of helium gas occupies 500.0 mL at 1.21
atm Calculate the volume of the gas if the
pressure is reduced to 491 torr
P1V1
P2 V2

T1
T2
1.21 atm  500.0 mL  0.64605 atm  V2
936 mL
10.3 The gas laws summarize experimental observations
14
Your Turn!
22.4 L of He at 25 ºC are heated to 200.ºC. What is
the resulting volume?
A. 22.4 L
B. 179 L
C. 43.1 L
D. not enough information given
10.3 The gas laws summarize experimental observations
15
Molar Volume
• The volume of one mole of any gas at STP is 22.4
L.
 Identity of the gas doesn’t matter
 Molar mass of the gas doesn’t matter
• Corollary: equal volumes of any gas contain the
same number of particles as long as the T and P
are the same
10.4 Gas volumes can be used in solving stoichiometry problems
16
Learning Check:
How many liters of N2(g) at 1.00 atm and 25.0 °C are
produced by the decomposition of 150. g of NaN3?
2NaN3(s)  2Na(s) + 3N2(g)
150.g NaN3 1 mol NaN3
3 mol N2


1
65.0099 g 2mol NaN3
6.9220 mol N2
 22.4 L  77.53L at STP
1 mol at STP
1
V1 V2
V1 T2

; V2 
T1 T2
T1
77.53 L  298.15K
V2 
 84.6L
273.15K
10.4 Gas volumes can be used in solving stoichiometry problems
17
Bringing It Together
•
•
•
•
•
Avogadro: n directly proportional to V
Boyle: P indirectly proportional to V
Charles: T directly proportional to V
Gay-Lussac: T directly proportional to P
Combining these variables into one equation results
in the Ideal Gas Law.
 R is the constant of proportionality (the “ideal” or
“universal” gas constant) its value is 0.082057
L•atm/mol•K
nT
V
R
P
10.5 The ideal gas law relates P, V, T, and the number of moles of gas, n
18
Ideal Gas Law
• Used to describe a sample of gas under one
set of conditions
• The units have to be:




PV = nRT
P in atm or torr
V in L
n in mol
T in K
• R = 0.082057 L•atm/mol•K = 62.36 L•torr/mol•K
10.5 The ideal gas law relates P, V, T, and the number of moles of gas, n
19
Your Turn!
12.2 g of Ne are placed into a 5.0 L flask at 25 ºC.
What is the pressure of the gas?
A. 3.0 atm
B. 60. atm
C. 0.25 atm
D. None of these
10.5 The ideal gas law relates P, V, T, and the number of moles of gas, n
20
Gas Density
• The number of moles may be related to both the mass
(m) of the gas sample and the molar mass (MM) of the
gas involved
• Thus we may rewrite the Ideal Gas Law as
m
PV 
RT
MM
• Further, since d=m/V, we can rewrite the equation in
terms of density
m
PMM  RT  dRT
V
10.5 The ideal gas law relates P, V, T, and the number of moles of gas, n
21
Learning Check
What is the molar mass of a gas with a density of 6.7
g/L at -73.ºC and a pressure of 36.7 psi?
PxMM  dxRxT
6.7g 0.082057 L  atm
2.4983 atm  MM 

 200.15K
L
mol  K
44 g/mol = MM
10.5 The ideal gas law relates P, V, T, and the number of moles of gas, n
22
Learning Check
What is the density of NO2 at 200 ˚C and 600. torr?
PxMM  dxRxT
46.006 g
0.082057 L  atm
0.789atm 
 d
 473.15K
mol
mol  K
0.9 g/L
10.5 The ideal gas law relates P, V, T, and the number of moles of gas, n
23
Your Turn!
What is the density of Helium gas at 35 ºC and 1.2
atm?
A. 5.1 g/L
B. 0.20 g/L
C. 2.34 g/L
D. None of these
10.5 The ideal gas law relates P, V, T, and the number of moles of gas, n
24
Learning Check
A sample of fluorine gas occupies 275 mL at 945
torr and 72 ºC. What is the mass of the sample?
m
PV 
RT
PV = nRT
MM
mol
0.082057 L  atm
1.2434 atm  0.275 L  m

 345.15K
37.997g
mol  K
0.459 g = mass
10.5 The ideal gas law relates P, V, T, and the number of moles of gas, n
25
Learning Check
Determine the molecular weight of a gas if 1.053 g
of the gas occupies a volume of 1.000L at 25 °C
and 752 mm Hg (The Dumas Method)
PV = nRT
m
PV 
RT
MM
1.053g 0.082057 L  atm
0.98947 atm  1.000 L 

 298.15K
MM
mol  K
26.0 g/mol = mass
10.5 The ideal gas law relates P, V, T, and the number of moles of gas, n
26
Your Turn!
What is the molar mass of a sample of gas if 2.22 g
occupies a volume of 5.0 L a 35 ºC and 769 mm
Hg?
A. 1.3 g/mol
B. 0.015 g/mol
C. 0.090 g/mol
D. None of these =11 g/mol
10.5 The ideal gas law relates P, V, T, and the number of moles of gas, n
27
Dalton’s Law
The partial pressure of a gas is the pressure that the
gas would exert if it were in the container by
itself
10.6 In a mixture each gas exerts its own partial pressure
28
Learning Check:
Pump 520 mm Hg N2 and 250 mm Hg O2 into an
empty gas cylinder. What is the overall pressure
of the mixture?
Pt=520 mm Hg + 250 mm Hg=770 mm Hg
PTotal = P1 + P2 + P3 + ….
10.6 In a mixture each gas exerts its own partial pressure
29
Learning Check:
32.5 mL of Hydrogen gas is collected over water at 25 ºC
and 755 torr. What is the pressure of dry hydrogen gas?
VP25ºC = 23.76 mmHg)
Correct Pt to find the Pdry gas:
755-23.76 torr=731.24 torr
731 torr = Phydrogen
PTotal = P1 + P2 + P3 + ….
10.6 In a mixture each gas exerts its own partial pressure
30
Mole Fraction, X
• Each gas molecule contributes a
fraction of the total pressure
 Xa=the mole fraction of substance “a”
 na =the moles of component “a”
 nt= the total number of moles of gas in
the mixture
• Application: The partial pressure
contributed by the component gas “a”
is a fraction of the total pressure
10.6 In a mixture each gas exerts its own partial pressure
na
Xa 
nt
pa  X apt
31
Learning Check:
What is the mole fraction of N2 in the atmosphere?
1.000atm Air = .7808 atm N2+ .2095 atm O2+
.0093 atm Ar + .00036 atm CO2
na
Xa 
nt
.7808 = Xnitrogen
10.6 In a mixture each gas exerts its own partial pressure
32
Diffusion vs. Effusion
•
•
When the
partition is
removed, blue
molecules
diffuse to mix
The molecules
effuse through
a pinhole into
a vacuum
10.7 Effusion and diffusion in gases leads to Graham’s Law
33
Graham’s Law Of Effusion
• At the same temperature, particles of the different
gases will have the same average kinetic energy
• The greater the molecular mass of the gas, the
slower its velocity to achieve this kinetic energy
• Gases with lower molecular masses have higher
velocities and gases with higher molecular masses
have lower velocities at the same temperature
10.7 Effusion and diffusion in gases leads to Graham’s Law
34
Your Turn!
The average kinetic energy of all gas molecules
is the same at the same temperature.
Compared to lighter atoms at the same
temperature, heavier atoms on average:
A. move faster
B. move slower
C. move at the same average velocity
10.7 Effusion and diffusion in gases leads to Graham’s Law
35
Your Turn!
Three balloons are filled with equal volumes of the
gases: CH4, H2, and He. After 5 hours the balloons
look like the picture.
• Which is the He balloon?
A
B
10.7 Effusion and diffusion in gases leads to Graham’s Law
C
36