Chapter 5: Gases
Renee Y. Becker
Valencia Community College
CHM 1045
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a) Gas is a large collection of particles moving at random
throughout a volume
b) Collisions of randomly moving particles with the
walls of the container exert a force per unit area that we
perceive as gas pressure
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• Units of pressure: atmosphere (atm)
Pa (N/m2, 101,325 Pa = 1 atm)
Torr (760 Torr = 1 atm)
bar (1.01325 bar = 1 atm)
mm Hg (760 mm Hg = 1 atm)
lb/in2 (14.696 lb/in2 = 1 atm)
in Hg (29.921 in Hg = 1 atm)
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Boyle’s Law
• Pressure–Volume Law (Boyle’s Law):
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Boyle’s Law
• Pressure–Volume Law (Boyle’s Law):
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Volume 
Pressure
• The volume of a fixed amount of gas maintained at
constant temperature is inversely proportional to
the gas pressure.
• Can use any units for P & V
V1P1  k1
P 1 V 1 = P2 V 2
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Example 1: Boyle’s Law
A sample of argon gas has a volume of
14.5 L at 1.56 atm of pressure. What
would the pressure be if the gas was
compressed to 10.5 L? (at constant
temperature and moles of gas)
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Charles’ Law
• Temperature–Volume Law (Charles’ Law):
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Charles’ Law
• Temperature–Volume Law (Charles’ Law):
V  T
• The volume of a fixed amount of gas at constant
pressure is directly proportional to the Kelvin
temperature of the gas.
• Use Kelvins for temperature!!
V1
=k
1
T1
V1 = V2
T1
T2
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Example 2: Charles’ Law
A sample of CO2(g) at 35C has a volume of
8.56 x10-4 L. What would the resulting
volume be if we increased the temperature
to 85C? (at constant moles and
pressure)
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Avogadro’s Law
• The Volume–Amount Law (Avogadro’s
Law):
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Avogadro’s Law
• The Volume–Amount Law (Avogadro’s Law):
V n
• At constant pressure and temperature, the volume
of a gas is directly proportional to the number of
moles of the gas present.
• Use any volume and moles
V1
 k1
n1
V1 = V2
n1
n2
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Example 3: Avogadro’s Law
6.53 moles of O2(g) has a volume of 146 L. If
we decreased the number of moles of
oxygen to 3.94 moles what would be the
resulting volume? (constant pressure and
temperature)
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Combined Gas Law
We can combine Boyle’s and charles’ law to
come up with the combined gas law
P1  V1
P2  V2

T1
T2
Use Kelvins for temp, any pressure, any volume
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Example 4:Combined Gas Law
Oxygen gas is normally sold in 49.0 L steel
containers at a pressure of 150.0 atm.
What volume would the gas occupy if the
pressure was reduced to 1.02 atm and the
temperature raised from 20oC to 35oC?
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Example 5: Gas Laws
An inflated balloon with a volume of 0.55 L at
sea level, where the pressure is 1.0 atm, is
allowed to rise to a height of 6.5 km, where
the pressure is about 0.40 atm. Assuming
that the temperature remains constant, what
is the final volume of the balloon?
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The Ideal Gas Law
• Ideal gases obey an equation incorporating the laws
of Charles, Boyle, and Avogadro.
P V  n  R  T
• 1 mole of an ideal gas occupies 22.414 L at STP
• STP conditions are 273.15 K and 1 atm pressure
• The gas constant R = 0.08206 L·atm·K–1·mol–1
– P has to be in atm
– V has to be in L
– T has to be in K
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Example 6: Ideal Gas Law
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.
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Example 7: Ideal Gas Law
What is the volume (in liters) occupied by
7.40 g of CO2 at STP?
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The Ideal Gas Law
• Density and Molar Mass Calculations:
mass
n  MM P  MM
d


volume
V
R T
• You can calculate the density or molar mass
(MM) of a gas. The density of a gas is
usually very low under atmospheric
conditions.
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Example 8: Density & MM
What is the molar mass of a gas with a density
of 1.342 g/L at STP?
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Example 9: Density & MM
What is the density of uranium hexafluoride,
UF6, (MM = 352 g/mol) under conditions of
STP?
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Example 10: Density & MM
The density of a gaseous compound is 3.38
g/L at 40°C and 1.97 atm. What is its
molar mass?
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Dalton’s Law of Partial Pressures
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Dalton’s Law of Partial Pressures
• In a mixture of gases the total pressure, Ptot, is the
sum of the partial pressures of the gases:
RT
P total 
V
n
• Dalton’s law allows us to work with mixtures of
gases.
• T has to be in K
• V has to be in L
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Example 11: Dalton
Exactly 2.0 moles of Ne and 3.0 moles of
Ar were placed in a 40.0 L container at
25oC. What are the partial pressures of
each gas and the total pressure?
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Example 12: Dalton
A sample of natural gas contains 6.25
moles of methane (CH4), 0.500 moles of
ethane (C2H6), and 0.100 moles of
propane (C3H8). If the total pressure of
the gas is 1.50 atm, what are the partial
pressures of the gases?
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Dalton’s Law of Partial Pressures
• For a two-component system, the moles of
components A and B can be represented by
the mole fractions (XA and XB).
Mole fraction is related to
the total pressure by:
Pi  X i Ptot
nA
XA 
nA  nB
nB
XB 
nA  nB
X A  X B 1
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Example 13: Mole Fraction
What is the mole fraction of each
component in a mixture of 12.45 g of
H2, 60.67 g of N2, and 2.38 g of NH3?
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Example 14: Partial Pressure
On a humid day in summer, the mole
fraction of gaseous H2O (water vapor) in
the air at 25°C can be as high as
0.0287. Assuming a total pressure of
0.977 atm, what is the partial pressure
(in atm) of H2O in the air?
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Gas Stoichiometry & Example
• In gas stoichiometry, for a constant
temperature and pressure, volume is
proportional to moles.
Example: Assuming no change in temperature
and pressure, calculate the volume of O2 (in
liters) required for the complete combustion
of 14.9 L of butane (C4H10):
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(l)
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Example 15:
All of the mole fractions of elements in a
given compound must add up to?
1.
2.
3.
4.
100
1
50
2
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Example 16:
Hydrogen gas, H2, can be prepared by
letting zinc metal react with aqueous
HCl. How many liters of H2 can be
prepared at 742 mm Hg and 15oC if
25.5 g of zinc (MM = 65.4 g/mol) was
allowed to react?
Zn(s) + 2 HCl(aq)  H2(g) + ZnCl2(aq)
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Kinetic Molecular Theory
• This theory presents physical properties of gases in
terms of the motion of individual molecules.
• Average Kinetic Energy  Kelvin Temperature
• Gas molecules are points separated by a great
distance
• Particle volume is negligible compared to gas
volume
• Gas molecules are in rapid random motion
• Gas collisions are perfectly elastic
• Gas molecules experience no attraction or
repulsion
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Kinetic Molecular Theory
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• Average Kinetic Energy (KE) is given
by:
1
2
KE  mu
2
U = average speed of a
gas particle
R = 8.314 J/K mol
m = mass in kg
3RT
3RT
u

mNA
MM
MM = molar mass, in
kg/mol
NA = 6.022 x 1023
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• The Root–Mean–Square Speed: is a
measure of the average molecular
speed.
3RT
u 
MM
2
3RT
urms 
MM
Taking square root of both
sides gives the equation
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Example 17:
Calculate the root–mean–square speeds
of helium atoms and nitrogen molecules
in m/s at 25°C.
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Kinetic Molecular Theory
• Maxwell speed distribution curves.
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Graham’s Law
• Diffusion is the
mixing of different
gases by random
molecular motion
and collision.
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Graham’s Law
• Effusion is when
gas molecules
escape without
collision, through a
tiny hole into a
vacuum.
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Graham’s Law
• Graham’s Law: Rate of effusion is
proportional to its rms speed, urms.
3RT
Rate u rms 
MM
• For two gases at same temperature and
pressure:
Rate1

Rate2
MM 2
MM 1

MM 2
MM 1
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Example 18:
Under the same conditions, an unknown
gas diffuses 0.644 times as fast as
sulfur hexafluoride, SF6 (MM = 146
g/mol). What is the identity of the
unknown gas if it is also a hexafluoride?
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Example 19: Diffusion
• What are the relative rates of diffusion
of the three naturally occurring isotopes
of neon: 20Ne, 21Ne, and 22Ne?
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Behavior of Real Gases
• Deviations result from assumptions about
ideal gases.
1. Molecules in gaseous state do not exert
any force, either attractive or repulsive, on
one another.
2. Volume of the molecules is negligibly small
compared with that of the container.
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Behavior of Real Gases
• At higher pressures, particles are much
closer together and attractive forces become
more important than at lower pressures.
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Behavior of Real Gases
• The volume taken up by gas particles is
actually less important at lower pressures
than at higher pressure. As a result, the
volume at high pressure will be greater than
the ideal value.
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Behavior of Real Gases
• Corrections for non-ideality require van der
Waals equation.

n 
 P  a 2   V – n  b   nRT
V 

2
Intermolecular
Attractions
Excluded
Volume
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Example 20: Ideal Vs. Van Der Waals
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. (a = 4.17, b =
0.0371)
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Example 21: Ideal Vs. Van Der Waals
Assume that you have 0.500 mol of N2 in
a volume of 0.600 L at 300 K. Calculate
the pressure in atmospheres using both
the ideal gas law and the van der Waals
equation.
• For N2, a = 1.35 L2·atm mol–2, and b =
0.0387 L/mol.
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