Ch. 5 Gases
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Elements that exist as gases at
0
25 C
and 1 atm
Physical Characteristics of Gases
•
•
•
•
•
Gases adopt the shape of their containers (fluidity).
Gases adopt the volume of their containers (Diffusion/compressibility).
Gases will mix evenly and completely when in the same volume
Gases have much lower densities than liquids and solids.
Gas volume & pressure change with temperature.
WF6 gas: 13 g/L
11x heavier than air, but still
>75x less dense than water
SF6 Blooper;
www.youtube.com/watch?v=V2FR6-gEwjU
Properties of Gases
DIFFUSION - Uniform spreading of gas molecules
EFFUSION - Movement of gas through small hole
FLUIDITY - Ability to flow and take shape of their
container (liquids and gases)
Pressure =
Force
Area
Barometer
vacuum
Decreased Area,
Increased Pressure
Increased Area,
Decreased Pressure
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa
Hg(l)
TedEd: The history of the barometer
www.youtube.com/watch?v=EkDhlzA-lwI
Atmospheric pressure:
Dependant on elevation, temperature,
weather
10 miles
4 miles
Sea level
0.2 atm
0.5 atm
1 atm
Veritasium: World's Longest Vertical Straw
www.youtube.com/watch?v=HUmZrtiXDik
Pressure of a gas
Pressure = the collision of gas particles with a surface;
force per unit area
As number of collisions
increase, pressure increases
As force of collisions
increase, pressure increases
TedEd: How heavy is air?
www.youtube.com/watch?v=VDf00z8sMFw
Manometers Used to Measure Gas Pressures
closed-tube
For below 1 atm pressures
open-tube
For above atm pressures
Pressure conversion Example 1
What is the pressure in atmospheres if the barometer reading is 688
mmHg (torr)?
We need to know:
Kinetic Molecular Theory Summary
1. Gas particles separated from each other by large
distances
2. Gas particles are in constant motion in random
directions, and they frequently collide.
3. Negligible intermolecular forces
4. Kinetic energy is proportional to temperature
(gases at the same temperature will have the
same average KE)
Be able to cite each and know each of their implications
Kinetic Molecular Theory of Gases
1. A gas is composed of molecules that are separated
from each other by large distances far greater than
their own dimensions. The molecules can be
considered to possess mass but have insignificant
volume.
• Very low density (particles per volume)
• Compressibility of gases (plenty of space to get closer)
Kinetic Molecular Theory of Gases
2. Gas molecules are in constant motion in random
directions, and they frequently collide with one
another. Collisions among molecules are perfectly
elastic (don’t lose energy).
• Explains diffusion/effusion
property of gases
• Adopts the shape of container
Kinetic Molecular Theory of Gases
3. Gas molecules exert neither attractive nor
repulsive forces on one another.
(liquid)
“Negligible intermolecular forces”
Inter = “between”
Molecules too far apart to effect each other
• Explains why gases mix completely
regardless of identity
These bonds only occur
in liquids and solids
Kinetic Molecular Theory of Gases
4. The average Kinetic energy (KE) of the molecules
is proportional to the temperature of the gas
KE = ½ mv2 ; v2 is average velocity
• As KE goes up, molecule velocity goes up.
• Therefore, as Temp ↑, Velocity ↑
• Faster particles leads to more collisions (pressure)
*Any two gases at the same temperature will
have the same average KE, regardless of size.
Any two gases at the same temperature will have the
same average KE, regardless of size.
KE = ½
2
mv
We can compare a two gas mixture with relative speed
at an arbitrary KE value of 5 at given temperature.
Helium (4 g/mol) vs Chlorine gas (Cl2; 71 g/mol)
He: 5 = ½ 4*v2
v = 1.6
Cl2: 5 = ½ 71*v2
v = 0.38
He = 1.6
= 4.2
Cl2 0.38
At this temperature, Helium is moving
4.2 times faster than chlorine gas.
Same amount of energy can move a
smaller mass faster.
Gas diffusion is the gradual mixing of molecules of one gas
with molecules of another by virtue of their kinetic properties.
Because two gases have the same KE at the same
temp (KE = ½*m*v2). We can relate them to
determine an unknown gas’s mass using a standard
gas’s diffusion rate.
v1
M2
=
v2
M1
Graham’s law of diffusion
NH4Cl(s)
random
molecular path
NH3
HCl
17 g/mol
36 g/mol
88
Gas effusion is the process by which gas under pressure
escapes from one compartment of a container to another
by passing through a small opening.
Distance/time
v1
v2
=
(d/t1)
(d/t2)
rate units
m/s
Smaller effuses faster
=
t2
t1
=
time units
sec or min
M2
M1
Example: Effusion of a gas
An unknown hydrocarbon (CxHy) gas
takes 1.50 min to effuse our setup.
It takes an equal volume of Br2(g)
vapor 4.73 min to effuse through the
same barrier.
Calculate the molar mass of the
unknown gas, and suggest what this
gas might be.
t2
t1
=
M2
M1
Gas effusion. Gas molecules move
from a high-pressure region (left) to a
low-pressure one through a pinhole.
Example: Effusion solution
Solution From the molar mass of Br2, we write
Because the molar mass of carbon is 12.01 g and that of
hydrogen is 1.008 g, the gas is methane (CH4).
Jimmy Fallon-Kevin Delaney Makes a Cloud
www.youtube.com/watch?v=iQWtZd8jM3g
Apparatus for Studying Molecular
Speed Distribution and average molecular speed
After numerous hits, the molecular deposition will eventually
become visible. Density of each region is measured
Speed of sound (340 m/s; Mach 1)
-280 °F
80 °F
As Temp ↑
Speed ↑
Fastest flight speed (2,020 m/s; Mach 6)
Accompanied by a sonic boom
Ruben’s Tube; www.youtube.com/watch?v=BbPgy4sHYTw
~800 °F
Speed distributions of different
gases at the same temperature
80 °F
Speed distributions for N2
gas molecules at different
temperatures
vrms
As Mass↓
Speed ↑
MM
82
Average gas molecule speed (vrms)
Total kinetic energy of a mole of gas equals 3/2RT
NA•(1/2mv2) = 3/2•R•T
R = Gas constant
(we’ll explain later)
MM (molar mass) = NAm
R = 8.314 J/K · mol
vrms
MM
1 Joule = 1 kg m2/s2
Because our constant (R) uses Joules, which uses kg,
we must express Molar mass in kg. We must also use
Kelvin for temperature.
Example: Speed of a gas
Calculate and compare the root-mean-square speeds of
helium atoms and nitrogen molecules in m/s at 25°C.
vrms
R = 8.314 J/K · mol
MM
1 Joule = 1 kg m2/s2
The molar mass of He is 4.003 g/mol, or 0.004 kg/mol.
The temperature is 25°C , but needs to be expressed as 298 K
Example: Speed of a gas Solution
vrms
MM
Considering 1 J = 1 kg m2/s2, the rest of the units cancel out
At 25°C, helium travels on
average ~3000 mph
Example: Speed of a gas Solution
The procedure is the same for N2, the molar mass of which is 28.02
g/mol, or 2.802 × 10−2 kg/mol so that we write
Escape velocity is the speed where an object’s KE is equal to the
gravitational potential energy. The speed needed to “break-free” of
Earth’s gravity is ~ 11,000 m/s
Earth’s atmosphere has low abundance of H2 & He because the
molecules are light and travel fast enough to escape the Earth’s pull.
Crash Course: Passing Gases
www.youtube.com/watch?v=TLRZAFU_9Kg
Three physical properties can describe
a sample of gas
• Volume
• Pressure
They are each interconnected
with each other – if one changes,
the others must change with it.
• Temperature
There are several scientific gas laws that define
the behavior of gases with these parameters
Apparatus for Studying the Relationship Between
Pressure and Volume of a Gas
As P increases
V decreases
Doubling
Pressure
Tripling
Pressure
Boyle’s Law: Pressure-Volume relationship
Constant Temp. & Moles of gas
“∝” = proportional
P ∝ 1/V
“Inversely proportional”
The product of Pressure
and Volume at any point on
the line is always the same.
P x V = constant (k1)
P1 x V1 = P2 x V2
Boyle’s Law (Pressure-Volume)
P1 x V1 = P2 x V2
• As Volume decreases, Collisions
become more frequent in smaller space.
• More collisions = more pressure
As Volume ↓; Pressure ↑ or
As Pressure ↓; Volume ↑
Boyle’s Law Practice: P1 x V1 = P2 x V2
Units must be equal on both sides
• If we have 5 L of Nitrogen gas (N2) at 2 atm, what is the
volume if we apply 4 times the pressure?
2 atm x 5 L = 8 atm x V2
V2 = 1.25 L
• Helium gas is found in a container with a volume of 3L
at 4 atm. If the container double in size, what is the
new pressure?
4 atm x 3 L = P2 x 6 L
P2 = 2 atm
Variation in Gas Volume with Temperature
at Constant Pressure
KMT explains this: as
molecules travel faster (at
higher T), Volume would
have to increase to
maintain the same
Pressure
As T increases
V increases
Charles’s Law
V∝T
Variation of Volume with Temp.
(Constant Pressure)
proportional
V =k
2
T
Temperature must be
in Kelvin
K = 0C + 273
Also, P ∝ T
(Constant V)
We must use the Kelvin Scale in all gas calculations
Kelvin is an absolute scale
°C and °F are relative scales
20 °C → 100 °C
NOT a 400% increase
Quadrupling the temperature
in °C does not quadruple the
thermal energy its measuring.
293 K → 373 K
only a 27% increase in
thermal energy
First Determination of Absolute zero
As Temp ↑; Volume ↑
V∝T
V1/T1 = V2 /T2
T (K) = t (0C) + 273.15
Data stops here
from condensation
Lord Kelvin realized each line extrapolated to the same point
• To a theoretically lowest attainable staring temperature
• He identified -273 0C as absolute zero
Charles’s Law Practice:
Units must be equal on both sides, must use Kelvins
• If we have 0.8 mL of Oxygen gas (O2) at 30 °C, what is the
volume if we heat the gas to 90 °C?
0.8 mL
303 K
=
V2
363 K
V2 = 0.96 mL
• Air is found at 1 atm, 25 °C in a fixed volume. If the pressure
increases 5x, what is the new temperature?
1 atm
273 K
=
5 atm
T2
T2 = 1,365 K
~ 2,000 ° F
Veritasium: Fire Syringe
www.youtube.com/watch?v=4qe1Ueifekg
Avogadro’s Law (Moles of gas)
Constant temperature
Constant pressure
V  number of moles (n)
V = constant x n First proposed by Avogadro, 1811: Gas
V∝n
volume does not depend on molecule
size, only number of particles
or
P1
P2
4 volumes
→
2 volumes
Mass is conserved, but volume is NOT
Avogadro’s Law
Combustion of hydrocarbons (like octane) drive the pistons in
combustion engines from the expansion of gases (CO2 and H2O vapor)
• Guns/Rockets/Bombs all function based on a reaction
that produces gases that expand.
N2(g)
+
TNT(s) → CO(g)
+
H2O(g)
Reactions: Movie Explosions Explained
www.youtube.com/watch?v=3e839o4YGao
Candle wax demo
Paraffin wax: a mixture of long hydrocarbons
Remember: Combustion of a hydrocarbon gives CO2 and H2O
2C20H42(s) + 61O2(g) → 40CO2(g) + 42H2O(l)
Conversion of available oxygen gas to CO2 produces less moles
of gas, so there’s a decrease in pressure.
Less pressure inside, causes atmospheric pressure to push the
water up (just like mercury in a barometer)
You can watch it done here if missed in class:
https://www.youtube.com/watch?v=0WGOpSpuDYQ
Ideal Gas Equation
Boyle’s law: P  1 (at constant n and T)
V
Charles’s law: V  T (at constant n and P)
+
Avogadro’s law: V  n (at constant P and T)
Combining them all we see:
nT
nT
V = constant x
=R
P
P
PV = nRT
V  nT
P
R is the gas constant
Ideal Gas Equation: assumes
negligible intermolecular
forces between particles
The conditions 0 0C and 1 atm are called standard
temperature and pressure (STP). (SATP 25 °C & 1 atm)`
Experiments show that at STP, 1 mole of an ideal gas occupies
22.414 L. (regardless of molecular identity)
PV = nRT
PV
R=
nT
(1 atm)(22.414L)
=
(1 mol)(273.15 K)
Gas constant
R = 0.082057 (L • atm) / (mol • K)
Must use these units
Remember, K = C° + 273; 1 atm = 760 mmHg
Example: Ideal Gas #1
Sulfur hexafluoride (SF6) is a colorless and odorless gas.
Due to its lack of chemical reactivity, it is used as an
insulator in electronic equipment.
Calculate the pressure (in atm) exerted by 265.9 grams of
the gas in a steel vessel of volume 9,500 mL at 32°C.
*Convert grams to moles with MM
PV = nRT
(265.9 g) / (146.1 g/mol) = 1.82 mol
*Add 273 to °C to use Kelvin/ Convert mL to L
(1.82 mol)(0.0821 L•atm/K•mol)(32 + 273 K)
nRT
P=
=
V
9.5 L
Crash Course: The Ideal Gas Law
www.youtube.com/watch?v=BxUS1K7xu30
= 4.8 atm
Example: Ideal gas #2
Calculate the volume (in L) occupied
by 7.40 g of NH3 at STP
*This relation can only be
used at STP conditions
Or we could use the ideal gas equation where 7.40 g NH3 =
0.435 moles of NH3, and then applying V = nRT/P.
V = (0.435 mol)(0.082057)(273.15 K)
1 atm
= 9.74 L
Combined Ideal Gas Equation
Can be used when gas sample conditions change
PV = nRT
R=
(Before change)
R=
(After change)
The R’s are the same so we can set them equal
Sometimes called the
Modifed gas law
Can be used in place of each prior gas law used alone
Example: Combined Gas Law #1
Argon is an inert gas used in light bulbs to stop
the vaporization of the tungsten filament.
Electric light bulbs are
usually filled with argon.
A certain light bulb containing argon at 1.20 atm
and 18°C is heated to 85°C at constant volume.
Calculate its final pressure (in atm).
n1 = n2 because bulb is sealed
V1 = V2 because bulb volume does not expand
which is Charles’ law
Example: Combined Gas Law #1 Solution
Next we write
Initial Conditions
Final Conditions
P1 = 1.20 atm
P2 = ?
T1 = (18 + 273) K = 291 K T2 = (85 + 273) K = 358 K
The final pressure is given by
Check At constant volume, the pressure of a given amount of gas is
directly proportional to its absolute temperature. Therefore the increase
in pressure is reasonable.
Example: Combined Gas Law #2
A small bubble rises from the bottom of a lake, where the
temperature and pressure are 8°C and 6.4 atm, to the
water’s surface, where the temperature is 25°C and the
pressure is 1.0 atm.
Calculate the final volume (in mL) of the bubble if its initial
volume was 2.1 mL.
We can remove n
because it’s constant
in this problem
Example: Combined Gas Law #2 Solution
The given information is summarized:
Initial Conditions
P1 = 6.4 atm
V1 = 2.1 mL
T1 = (8 + 273) K = 281 K
Rearranging
Final Conditions
P2 = 1.0 atm
V2 = ?
T2 = (25 + 273) K = 298 K
Practice: Combined Gas Law #3
An inflated He balloon at sea level (1.0 atm) with
a V = 7.1 L (basketball) is allowed to rise to a
height of 8.8 km (Mt. Everest), where the pressure
is about 0.33 atm. The change in temperature
drops from 20°C to -30°C.
What is the final volume of the balloon?
n1 = n2 because
balloon is sealed
Dalton’s Law of Partial Pressures
Individual gas pressures are cumulative regardless of chemical
identity (negligible intermolecular forces)
P1
P2
Ptotal = P1 + P2
V and T are constant
Dalton’s Law
+
V and T are constant
=
Consider a case in which two gases, A
and B, are in a container of volume V.
nART
PA =
V
nBRT
PB =
V
PT = PA + P B
nA is the number of moles of A
nB is the number of moles of B
nA
XA =
nA + nB
P A = XA PT
Pi = X i PT
nB
XB =
nA + nB
PB = X B PT
ni
mole fraction (Xi ) =
nT
Example: Dalton’s Law
A mixture of gases contains 4.46 moles of neon (Ne),
0.74 mole of argon (Ar), and 2.15 moles of xenon (Xe).
Calculate the partial pressures of the gases if the total
pressure is 2.00 atm at a certain temperature.
XNe=
4.46 mol Ne = 0.607
7.35 mol Total
Example: Dalton’s Law Solution
Check Make sure that the sum of the partial pressures is equal to the
given total pressure; that is, (1.21 + 0.20 + 0.586) atm = 2.00 atm.
More Practice: Ptotal = 0.78 atm of 1.2 mol CO2 & 3.4 mol O2
Finding Molar Mass by Gas Density
PV = nRT
m
so, n = M
n P
=
V RT
&
M=
m
n
(grams)
(mole)
n is moles of gas
m is the mass of the gas in grams
M is the molar mass of the gas
Density (d) Calculations
PM
m
d=
=
V
RT
Molar Mass (M) of a Gaseous Substance
dRT
M=
P
d is the density of the
gas in g/L
Example: Molar mass to Density #1
Calculate the density of carbon dioxide (CO2) in
grams per liter (g/L) at 0.990 atm and 55°C.
PM
d=
RT
We will use T = 273 + 55 = 328 K and
44.01 g/mol for the molar mass of CO2
PM
d=
RT
Example: Gas Density to Molar mass #2
A chemist has synthesized a greenish-yellow gaseous
compound of chlorine and oxygen and finds that its density
is 7.71 g/L at 36°C and 2.88 atm.
Calculate the molar mass of the compound
To find the molecular formula we’ll then use the ideal gas law to
determine how many moles exist in 1 L of this gas
Example: Gas Density to Molar mass #2 Solution
From the given density we know there are 7.71 g of the gas in 1 L
The number of moles of the gas in this volume can be obtained from the
ideal gas equation:
Therefore, the molar mass is given by
Example: Gas Identity Problem #3
The chemical analysis of a gaseous compound showed that
it contained 33.0% Silicon and 67.0% Fluorine by mass.
At 35°C, 0.210 L exerted a pressure of 1.70 atm.
If the mass of 0.210 L of the compound was 2.38 g, find
the molecular formula of the compound.
First determine Empirical Formula:
÷1.17 = 1 mol Si
÷1.17 = 3 mol F
Empirical Formula: SiF3
MM = 85.1 g
With volume and mass we
can find gas density to
solve for the Molar Mass
dRT
M=
P
Example: Gas Identity Solution #3
Density of Gas = 2.38g ÷ 0.210 L = 11.33 g/L
dRT
M=
P
=
(11.33 g/L)(0.0821)(308 K)
1.70 atm
Empirical Formula:
SiF3
MM = 85.1 g
Si2F6
MM = 170.2 g
= 169 g/mol
169 g
~ 2x as massive
85.1 g
Gas Stoichiometry
Similar to gravimetric analysis of solid samples
Example: Gas Stoichiometry #1
Sodium azide (NaN3) is used in some automobile air
bags. The impact triggers the decomposition of NaN3:
The N2 gas produced quickly inflates the bag.
Calculate the N2 volume generated at 80°C and 823
mmHg by the decomposition of 60.0 g of NaN3.
2 mol NaN3  3 mol N2
Airbag Deploying in Slow Mo - The Slow Mo Guys
www.youtube.com/watch?v=KRcajZHc6Yk
Example: Gas Stoichiometry #2
Calculate the volume of O2 (in liters) required
for the complete combustion of 7.64 L of
acetylene (C2H2) measured at STP.
5 mol O2  2 mol C2H2
Avogadro’s Law: gas volume is independent
of its identity:
5 Liters O2  2 Liters C2H2
The reaction of calcium
carbide (CaC2) with
water produces acetylene
(C2H2), a flammable gas.
Collecting a Gas over Water
P T = P O 2 + P H 2O
2KClO3 (s)
2KCl (s) + 3O2 (g)
71
Vapor of Water and Temperature
Vapor pressure increases until equal to
atmospheric pressure then boiling occurs.
72
Example
5.15
Oxygen gas generated by the decomposition
of KClO3 is collected as shown
2KClO3 (s)
2KCl (s) + 3O2 (g)
The volume of oxygen collected at 24°C and atmospheric pressure of
762 mmHg is 128 mL.
Calculate the mass (in grams) of oxygen gas obtained.
The pressure of the water vapor at 24°C is 22 mmHg.
Example
5.15
Therefore,
From the ideal gas equation we can determine moles of O2:
To use R, we must convert to from mmHg to atm and C to K
n = PV
RT
(0.974 atm)(0.128 L) = 0.00511 moles O
=
2
(0.0821)(297 K)
= 0.00511 moles O2 = 0.164 grams O2
Chemistry in Action:
Scuba Diving and the Gas Laws
Depth (ft)
Pressure
(atm)
0
1
33
2
66
3
D
(Boyle’s Law)
P
V
Ascending too fast can cause the “bends” - Decompression
sickness (gas bubbles enlarging in the blood stream)
TedEd: The effects of underwater pressure on the body
www.youtube.com/watch?v=_cj8AtODiHc
Non-Ideal Gas: Effect of intermolecular forces
on the pressure exerted by a gas.
1) At High Pressure, density
increases, and intermolecular forces
are no longer negligible.
2) Molecules slow down at Low
Temperature and lowers KE to
overcome attractive forces
“Attractive forces “lessen” the
force exerted on the walls
Deviations from Ideal Behavior
1 mole of ideal gas
PV = nRT
PV
n = RT = 1.0
Repulsive Forces
Attractive Forces
Most gases act “ideally” below ~ 5 atm
Van der Waals Gas equation
For non-ideal gas
corrected
pressure
}
}
an2
( P + V 2 ) (V – nb) = nRT
corrected
volume
a correlates to molecule attraction
b “roughly” correlates to
molecule size
*Experimentally determined
Example
5.18
Given that 3.50 moles of NH3 occupy 5.20 L at 47°C, calculate
the pressure of the gas (in atm) using
(a) the ideal gas equation
(b) the van der Waals equation
Example
5.18
(b) It is convenient to first calculate the correction terms in Equation
(5.18) separately. From Table 5.4, we have
a = 4.17 atm · L2/mol2
b = 0.0371 L/mol
so that the correction terms for pressure and volume are
Example
5.18 Solution
Finally, substituting these values in the van der Waals equation:
The value is 1.5 atm lower than when using the ideal gas equation.
This makes sense as we expect the pressure to be reduced by
added intermolecular forces.
Crash Course: Real Gases
www.youtube.com/watch?v=GIPrsWuSkQc