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
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b) Collisions of randomly moving particles with the walls of the container exert a force per unit area that we perceive as gas pressure
• 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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• Pressure–Volume Law (Boyle’s Law):
Boyle’s Law
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Pressure1
Volume
V1P1 k15
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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Example 1: Boyle’s Law
• Temperature–Volume Law (Charles’ Law):
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Charles’ Law
V T
V1
T1=k1
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Charles’ Law
Example 2: Charles’ Law
A sample of CO2(g) at 35C has a volume of 8.56 x10-4 L. What would the resulting volume be if we increased the temperature to 85C? (at constant moles and pressure)
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• The Volume–Amount Law (Avogadro’s Law):
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Avogadro’s Law
nV
11
1 knV
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Avogadro’s Law
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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We can combine Boyle’s and charles’ law to come up with the combined gas law
Use Kelvins for temp, any pressure, any volume
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1
11 T
VP
T
VP
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Combined Gas Law
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.
• 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
TRnVP
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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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• Density and Molar Mass Calculations:
• You can calculate the density or molar mass (MM) of a gas. The density of a gas is usually very low under atmospheric conditions.
TR
MMP
V
MMnd
volume
mass
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The Ideal Gas Law
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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• In a mixture of gases the total pressure, Ptot, is the
sum of the partial pressures of the gases:
• Dalton’s law allows us to work with mixtures of
gases.
• T has to be in K
• V has to be in L
nV
RTP total
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Dalton’s Law of Partial Pressures
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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• For a two-component system, the moles of components A and B can be represented by the mole fractions (XA and XB).
1
BA
BA
BB
BA
AA
XX
nn
nX
nn
nXMole fraction is related to
the total pressure by:
totii PXP
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Dalton’s Law of Partial Pressures
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)30
Example 15:
All of the mole fractions of elements in a given compound must add up to?
1. 100
2. 1
3. 50
4. 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 33
Kinetic Molecular Theory
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• Average Kinetic Energy (KE) is given
by:
KE 1
2mu2
MM
RT
mN
RTu
A
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U = average speed of a gas particle
R = 8.314 J/K mol
m = mass in kg
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.
MM
RTu
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Taking square root of both sides gives the equation
MM
RTurms
3
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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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• Maxwell speed distribution curves.
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Kinetic Molecular Theory
• 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.
• For two gases at same temperature and pressure:
MM
RTRate rms
3 u
1
2
1
2
2
1
MM
MM
MM
MM
Rate
Rate
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Graham’s Law
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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• 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.
Behavior of Real Gases
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•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.
nRTbnVV
naP –
2
2
IntermolecularAttractions
ExcludedVolume
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Behavior of Real Gases
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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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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Example 21: Ideal Vs. Van Der Waals
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