Chapter 12: Gases

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Chemistry-140 Lecture 30

• Chapter 12 Gases

• Chapter Highlights
• pressure measurements
• concept of STP
• gas laws (Charles, Boyles, Avogadros)
• ideal gas law
• applications of ideal gas law
• partial pressures
• kinetic theory of gases
• diffusion effusion (Grahams law)
• van der Waals equation

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Chemistry-140 Lecture 30

• Definition of a Gas

• Gas a substance that expands to fill its
  container and attains the container's shape
• is highly compressible
• usually nonmetallic
• simple molecular formula
• low molar mass

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Chemistry-140 Lecture 30

• Characteristics of a Gas

• Only substances that are gaseous under normal
  conditions of temperature and pressure are called
  gases
• A substance that is normally a liquid or solid is
  called a vapour in the gas state. example water
  vapour
• Gases form homogeneous mixtures regardless of the
  amounts and characteristics of the components.

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Chemistry-140 Lecture 30

• Pressure

• Pressure the force a gas exerts on the walls of
  its vessel per unit area P F/Area .
• Newtons are the SI units of force. (1 N 1
  kg-m/s2)
• Pascals are the SI units of pressure. (1 Pa 1
  N/m2)
• Atmospheric pressure the gas pressure most
  commonly measured. The mass of a column of
  atmosphere 1 m2 in cross-sectional area and
  extending to the top of the atmosphere exerts a
  force of 1.01 x 105 N

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Chemistry-140 Lecture 30
Barometric Pressure

• Atmospheric pressure is measured with a barometer
  and is called barometric pressure

• The standard pressure, (1 atmosphere, atm), is
  the pressure required to support a mercury column
  to a height of 760 mm 760 Torr
• 1.01325 x 105 Pa
• 101.325 kPa

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Chemistry-140 Lecture 30

• Gas Laws

• Four variables can adequately describe a gas
  sample
• T Temperature (generally expressed in Kelvin)
• n amount of material (generally in moles)
• P pressure (atmospheres is most common)
• V volume (litres is most common)

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Chemistry-140 Lecture 30

• Boyles Law The Pressure-Volume Relationship

• The volume of a fixed amount of gas (n) at
  constant temperature (T) is inversely
  proportional to the pressure of the gas PV
  constant (at constant T and n).
• Comparing the gas at different pressures

P1V1 P2V2 P3V3 .
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Chemistry-140 Lecture 30

• Charles Law The Temperature-Volume Relationship

• The volume of a fixed amount of gas (n) at
  constant pressure (P) is directly proportional to
  the temperature of the gas constant
• Comparing the gas at different pressures

V2
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Chemistry-140 Lecture 30

• Avogadros Law The Quantity-Volume Relationship

• The volume of a gas at constant pressure and
  temperature is directly proportional to the
  amount of gas present, expressed in moles
  constant
• Comparing two gas samples

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Chemistry-140 Lecture 30

• The Ideal Gas Equation

• Boyle's law V R (_at_ constant n, T)
• Charles's law V R T (_at_ constant n, P)
• Avogadro's law V R n (_at_ constant P, T)
• This can combine to give a more general law V R
  

or
PV nRT
where R the gas constant
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Chemistry-140 Lecture 30

• An Ideal Gas

• An ideal gas one that can be described by the
  ideal-gas equation PV nRT
• The usual value of the ideal gas constant is
• R 0.08206 L-atm/K-mol
• Temperature must be expressed in Kelvin and the
  units of volume and pressure must match the units
  of R
• (also note that R can have the value 8.3145
  J/K-mol)

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Chemistry-140 Lecture 30

• STP Standard Temperature Pressure

• STP The conditions 0.00 C (273.15 K) and 1
  atm are referred to as standard temperature and
  pressure, STP. At STP, the volume of 1 mol of an
  ideal gas is 22.41 L. this known as the molar
  volume of a gas at STP

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Chemistry-140 Lecture 30

• Applications of the Ideal Gas Equation

Question (similar to example 12.6) Calcium
carbonate, CaCO3(s), decomposes to CaO(s) and
CO2(g). A sample of CaCO3(s) is decomposed and
the CO2(g) collected in a 250 mL flask. After
decomposition, the gas has a pressure of 1.3 atm
at a temperature of 31oC. How many moles of
CO2(g) were generated?
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Chemistry-140 Lecture 30

• Applications of the Ideal Gas Equation

Answer Step 1 Identify the unknown quantity
and known quantities. Use units
consistent with R. n ? R 0.0821
L-atm/mol-K P 1.3 atm V 250 mL
0.250 L T 31 oC (31 273) K 304 K
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Chemistry-140 Lecture 30

• Applications of the Ideal Gas Equation

Answer Step 2 Rearrange the ideal gas
equation and solve for n. n

0.013 mol CO2(g)
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Chemistry-140 Lecture 30

• Applications of the Ideal Gas Equation

Question (similar to example 12.4) A sample
of argon gas is confined to a 1.00 L tank at
27.0oC. The pressure in the tank is 4.15 atm.
The gas is allowed to expand into a larger
vessel. Upon expansion, the temperature drops to
15.0oC and the pressure drops to 655 Torr. What
is the final volume of the gas?
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Chemistry-140 Lecture 30

• Applications of the Ideal Gas Equation

Answer Step 1 Identify the unknown quantity
and tabulate the known quantities in units
consistent with those in R. Notice that
we are missing both V and n in the final state!!
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Chemistry-140 Lecture 30

• Applications of the Ideal Gas Equation

Answer Step 2 Since the number of moles of
gas does not change we can calculate the number
of moles initially present and know how many were
present in the final state n

0.168 mol Ar(g)
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Chemistry-140 Lecture 30

• Applications of the Ideal Gas Equation

Answer Step 3 We can then use this to
calculate the final volume. V2

4.62 L Ar(g)
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Chemistry-140 Lecture 30

• Applications of the Ideal Gas Equation

Question The gas pressure in an aerosol can
is 1.5 atm at 25oC. Assuming that the gas obeys
the ideal-gas equation what would the pressure be
if the can was heated to 450oC?
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Chemistry-140 Lecture 30

• Applications of the Ideal Gas Equation

Answer Step 1 Identify the unknown quantity
and tabulate the known quantities in units
consistent with those in R.
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Chemistry-140 Lecture 30

• Applications of the Ideal Gas Equation

Answer Step 2 Since the can is a closed
container, the volume and of moles cannot
change V1 V2 and n1 n2. Therefore
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Chemistry-140 Lecture 30

• Applications of the Ideal Gas Equation

Answer Step 3 Rearrange and
calculate P2 P2


3.6 atm
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Chemistry-140 Lecture 31

• Chapter 12 Gases

• Chapter Highlights
• pressure measurements
• concept of STP
• gas laws (Charles, Boyles, Avogadros)
• ideal gas law
• applications of ideal gas law
• partial pressures
• kinetic theory of gases
• diffusion effusion (Grahams law)
• van der Waals equation

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Chemistry-140 Lecture 31

• Molar Mass Gas Density

• Recall that
• d (density) m(mass)/V(volume)
• n(moles) m(mass)/M(molar mass)
• So if we write the ideal-gas equation as
  
• substitute for n or
  
• thus d and M

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Chemistry-140 Lecture 31

• Molar Mass Gas Density

Question (similar to excercise 12.7) What is
the density of carbon tetrachloride, CCl4, vapour
at 714 Torr and 125oC?
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Chemistry-140 Lecture 31

• Molar Mass Gas Density

Answer Step 1 Recall that d We
need the molar mass of CCl4 M(CCl4) (12.0
4(35.5)) 154 g/mol d

4.43 g/L
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Chemistry-140 Lecture 31

• Molar Mass Gas Density

Question (no corresponding example) A flask is
evacuated and found to weigh 134.567 g. It is
filled to a pressure of 735 Torr at 31OC with a
gas of unknown molar mass and then reweighed
137.456 g. The flask is then filled with water
and weighed again 1067.9 g. What is the molar
mass of the unknown gas? (density of water at
31oC is 0.997 g/cm3)
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Chemistry-140 Lecture 31

• Molar Mass Gas Density

Answer Step 1 We need to know the volume of
the flask. We are given the mass of water when
the flask is filled, so we can use the density of
water to calculate the volume of the flask.
V 936 cm3
0.936 L
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Chemistry-140 Lecture 31

• Molar Mass Gas Density

Answer Step 2 Since we now know the volume of
the flask and the mass of gas is easily
calculated, we can obtain the gas density and use
this to get the molar mass. m(gas) (137.456
g - 134.567 g) d
M
2.889 g
3.09 g/L
79.7 g/mol
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• Gas Mixtures Partial Pressure

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• Gas Mixtures Partial Pressure

Pi XiPt
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Chemistry-140 Lecture 31

• Partial Pressures

Question (similar to example 12.1) A gaseous
mixture made from 6.00 g O2 and 9.00 g CH4 is
placed in a 15.0 L vessel at 0 oC. What is the
partial pressure of each gas, and what is the
total pressure in the vessel?
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Chemistry-140 Lecture 31

• Partial Pressures

Answer Step 1 Since each gas behaves
independently, we can calculate the pressure that
each would exert if the other were not
present. Convert masses to moles n(O2)
(6.00 g O2)
n(CH4) (9.00 g CH4)

0.188 mol O2
0.563 mol CH4
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Chemistry-140 Lecture 31

• Partial Pressures

Answer Step 2 Use the ideal-gas equation to
calculate the partial pressure of each gas.
P(O2)
P(CH4)

0.281 atm
0.841 atm
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Chemistry-140 Lecture 31

• Partial Pressures

Answer Step 3 We can now calculate the total
pressure from Pt P1 P2 P3 ...
Pt (0.281 atm) (0.841
atm) Pt
1.12 atm
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Chemistry-140 Lecture 31

• Partial Pressures

• Gases are often generated in the laboratory.
  Either as a product of a reaction or as a gaseous
  reactant to be used in a chemical reaction.

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Chemistry-140 Lecture 31

• Partial Pressures

Question The industrial synthesis of nitric
acid involves the reaction of nitrogen dioxide
gas with water. 3 NO2(g) H2O(l)
2 HNO3(aq) NO(g) How many moles of
HNO3 can be prepared using 450 L of NO2(g) at a
pressure of 5.00 atm and a temperature of 295 K?

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Chemistry-140 Lecture 31

• Partial Pressures

Answer Step 1 Use the ideal-gas equation to
determine the moles of NO2. n

92.9 mol
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• Partial Pressures

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Trapped Gases

• A common way to trap and measure the gas formed
  is a technique called displacement.

The gas that is collected is saturated with water
vapour. Total pressure inside the jar is then
Ptotal Pgas Pwater
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Chemistry-140 Lecture 31

• Trapped Gases

Question (similar to example 12.12) A sample
of KClO3 is partially decomposed producing O2
that is collected over water. The volume
collected is 0.250 L at 26 oC and 765 Torr total
pressure. 2 KClO3(s) 2 KCl(s)
3 O2(g) Knowing the vapour pressure of
water is 25 Torr at 26 oC, calculate how many
grams of KClO3 decomposed.
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Chemistry-140 Lecture 31

• Trapped Gases

Answer Step 1 We know V and T but not P(O2).
This can be determined from P(O2)
(765 Torr) - (25 Torr)
n(O2)

740 Torr
9.92 x 10-3 mol O2
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• Trapped Gases

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Chemistry-140 Lecture 32

• Chapter 12 Gases

• Chapter Highlights
• pressure measurements
• concept of STP
• gas laws (Charles, Boyles, Avogadros)
• ideal gas law
• applications of ideal gas law
• partial pressures
• kinetic theory of gases
• diffusion effusion (Grahams law)
• van der Waals equation

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Chemistry-140 Lecture 32

• Kinetic Molecular Theory

• The ideal-gas equation explains how gases
  behave.
• Kinetic-molecular theory
• Explains why ideal gases behave as they do.

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Chemistry-140 Lecture 32

• Kinetic Molecular Theory

• Gases consist of large numbers of molecules in
  continuous, random motion.
• The volume of all the molecules is negligible
  compared to the total volume in which the gas is
  contained.
• Attractive and repulsive interactions among gas
  molecules are negligible.

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• Kinetic Molecular Theory

• The collisions are elastic. Energy is
  transferred between molecules during collisions.
• The average kinetic energy is proportional to the
  absolute temperature.

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• Molecular Speeds

• Although kinetic energy at a given temperature is
  the same for all gases, the molecular speeds are
  different. At a constant kinetic energy, as the
  molecular mass increases, the molecular speed
  decreases.

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• Kinetic Molecular Theory Explains...

• Pressure is caused by gas molecules bombarding
  the container walls. The total force of these
  collisions depends on the number of collisions
  the average force per collision.

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• Kinetic Molecular Theory Explains...

• A temperature increase at constant volume gives
  molecules a higher kinetic energy and therefore
  higher speeds. Because of the increased speeds,
  more collisions occur and the pressure exerted by
  the gas increases.
• P is proportional to T (_at_ constant n, V)

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Chemistry-140 Lecture 32

• Kinetic Molecular Theory Explains...

• Boyles Law A volume increase at constant
  temperature is such that there are fewer
  molecules per unit volume and therefore fewer
  collisions. As a result the pressure exerted by
  the gas decreases.
• P is proportional to 1/V (_at_ constant n, T)

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Chemistry-140 Lecture 32

• Kinetic Molecular Theory Explains...

• P is proportional to
• (impulse imparted per collision) x (rate of
  collisions)
• impulse imparted per collision depends on
• momentum of the molecule (mass) x (speed) mu
• rate of collision is proportional to
• number of molecules per unit volume (n/V) and
  their speed (u)

P is proportional to (mu)(n/V)(u)
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Chemistry-140 Lecture 32

• Kinetic Molecular Theory Explains...

• P is proportional to
• Since KE 1/2mu2 and KE is
  proportional to T
• Then mu2 is proportional to T
• and P is proportional to
• Ideal Gas Equation P

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Chemistry-140 Lecture 32

• RMS Speed

• The root-mean-square speed (rms speed) of a gas,
  , is given by Maxwells Equation

As temperature increases, the rms speed of the
gas increases. As molecular mass increases, the
rms speed of the gas decreases.
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• Molecular Speed (Boltzmann Distribution)

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Chemistry-140 Lecture 32

• Molecular Speed (Boltzmann Distribution)

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Chemistry-140 Lecture 32

• Molecular Speeds

Exercise 12.13 Calculate the rms speed of a
N2 molecule at 25oC.
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• Molecular Speeds

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• Molecular Diffusion Effusion

• Diffusion The ability of a gas to disperse
  itself throughout a vessel. An example of
  diffusion would be odors spreading throughout a
  building.
• Rate of diffusion depends on the mean free path
  The average distance travelled by a molecule
  between collisions.

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Chemistry-140 Lecture 32

• Molecular Diffusion Effusion

• Effusion The ability of a gas to escape a vessel
  through a tiny hole.
• Graham's law of effusion the relative rates of
  effusion (r1 and r2) of two gases under identical
  conditions are inversely proportional to the
  square roots of their molar masses (M1 and M2)

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• Grahams Law of Molecular Effusion

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• Molecular Effusion

Question An unknown gas composed of diatomic
molecules effuses at a rate that is only 0.355
times that of O2(g) at the same temperature.
What is the identity of the unknown gas?
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Chemistry-140 Lecture 32

• Molecular Effusion

Answer Use Graham's Law. Let r(X2) and M(X2)
be the rate of effusion and molar mass of the
unknown diatomic gas.


0.355
Since molecule is diatomic AW(X) 254/2 127.
Molecule is I2.
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Chemistry-140 Lecture 32

• Nonideal Behaviour Real Gases

• Ideal gas behaviour assumes 1) negligible volume
  for the gas molecule and
• 2) no interactions between molecules.
• BUT.

P(observed) lt P(ideal)
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Chemistry-140 Lecture 32

• The van der Waals Equation

correction for intermolecular forces
correction for molecular volume
a and b van der Waals constants for the
particular gas
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Chemistry-140 Lecture 32

• The van der Waals Equation

Question (similar to exercise 12.15) Determine
the pressure of 8.00 mol of Cl2(g) in a 4.00 L
tank at 27 oC using both the ideal gas equation
and van der Waals equation. (For Cl2(g) a 6.49
atm L2/mol2 and b 0.0562 L/mol)
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• The van der Waals Equation

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• The van der Waals Equation

Answer van der Waals equation
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• Textbook Questions From Chapter 12
• Gas Laws 14, 18, 20, 23
• Ideal Gas Equation 26, 32, 34, 38, 44, 46, 50
• Gas Mixtures 52, 55, 58
• Kinetic Molecular Theory 60, 64, 68
• Non-Ideal Gases 70
• General Questions 82

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• Final Exam
• Monday December 10th, 2001
• 1200 Noon

St. Denis Centre Field House