Chemistry-140
Lecture 30
Chapter 12: Gases
 Chapter Highlights
 pressure measurements
 concept of STP
 gas laws (Charle’s, Boyle’s, Avogadro’s)
 ideal gas law
 applications of ideal gas law

partial pressures

kinetic theory of gases

diffusion & effusion (Graham’s law)

van der Waal’s equation
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
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.
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 crosssectional area and extending to the top of the atmosphere
exerts a force of 1.01 x 105 N
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
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)
Chemistry-140
Lecture 30
Boyle’s 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 ….
 PV

1 1
V2 = 

 P2 
Chemistry-140
Lecture 30
Charle’s Law: The Temperature-Volume Relationship
 The volume of a fixed amount of gas (n) at constant
pressure (P) is directly proportional to the temperature
V 
of the gas;   = constant
T 
 Comparing the gas at different pressures:
 V1 
 V2 
  =  
 T1 
 T2 
 V1T2 
V2 = 

 T1 
Chemistry-140
Lecture 30
Avogadro’s Law: The Quantity-Volume Relationship
 The volume of a gas at constant pressure and
temperature is directly proportional to the amount of
V 
gas present, expressed in moles;   = constant
 n
 Comparing two gas samples;
 V1 
 V2 
  =  
 n1 
 n2 
 V1 n2 
V2 = 

 n1 
Chemistry-140
Lecture 30
The Ideal Gas Equation
 1
 Boyle's law: V R   (@ constant n, T)
 P
 Charles's law: V R T (@ constant n, P)
 Avogadro's law V R n (@ constant P, T)
 nT 
 This can combine to give a more general law: V R 

 P 
or
PV = nRT
where R = the gas constant
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)
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
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?
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
Chemistry-140
Lecture 30
Applications of the Ideal Gas Equation
Answer:
Step 2: Rearrange the ideal gas equation and
solve for n.
 PV 
n = 

 RT 
1.3 atm0.250 L


= 

 0.0821 L atm / mol K 304 K  
=
0.013 mol CO2(g)
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?
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.
1: Initial
P
V
n
T
4.15 atm
1.00 L
?
300 K
?
?
288 K
2: Final 0.862 atm
Notice that we are missing both V and n in the final state!!
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
 PV

1 1
n =

 RT1 
4.15 atm1.00 L


= 




 0.0821 L atm / mol K 300 K 
=
0.168 mol Ar(g)
Chemistry-140
Lecture 30
Applications of the Ideal Gas Equation
Answer:
Step 3: We can then use this to calculate the final volume.
 nRT2 
V2 = 

 P2 
 0.168 mol 0.0821 L atm / mol K 288 K  

= 
0.862 atm


= 4.62 L Ar(g)
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?
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.
P
V
n
T
1: Initial
1.5 atm
V1
n1
298 K
2: Final
P2
V2
n2
723 K
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:
 n1 R 
 P1 

 =  
 T1 
 V1 
then
and
 n2 R 
 P2 

 =  
 V2 
 T2 
 P1 
 P2 
  =  
 T1 
 T2 
Chemistry-140
Lecture 30
Applications of the Ideal Gas Equation
Answer:
 
 
Step 3: Rearrange  P1  =  P2  and calculate P2
 T1 
 T2 
 P1T2 

P2 = 
 T2 
 1.5 atm 723 K  
= 

298 K 


=
3.6 atm
Chemistry-140
Lecture 31
Chapter 12: Gases
 Chapter Highlights

pressure measurements

concept of STP

gas laws (Charle’s, Boyle’s, Avogadro’s)

ideal gas law
 applications of ideal gas law
 partial pressures

kinetic theory of gases

diffusion & effusion (Graham’s law)

van der Waal’s equation
Chemistry-140
Lecture 31
Molar Mass & Gas Density
 Recall that:
d (density) = m(mass)/V(volume)
n(moles) = m(mass)/M(molar mass)
 n=  P 
 

 V   RT 
So if we write the ideal-gas equation as 
 m/ M  P 
 m   MP 
substitute for n: 
or   = 
= 

 V   RT 
 V   RT 
thus:
 MP 
d= 

 RT 
and
 dRT 
M =

 P 
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?
Chemistry-140
Lecture 31
Molar Mass & Gas Density
Answer:
 MP 
Step 1: Recall that d = 

 RT 
We need the molar mass of CCl4
M(CCl4) = (12.0 + 4(35.5)) = 154 g/mol
 154 g / mol714 / 760 atm 

d = 
 0.0821 L atm / mol K  398 K  
= 4.43 g/L
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)
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.
 1067.9 g  134.6 g 
 = 936 cm3 = 0.936 L
V = 
3
0.997 g / cm  

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 =
 2.889 g 

=
 0.936 L 
2.889 g
3.09 g/L
  3.09 g / L0.0821 L atm / mol K  304 K  
 = 79.7 g/mol
M = 
735 / 760 atm


Chemistry-140
Lecture 31
Gas Mixtures & Partial Pressure
 Partial pressure: The pressure of each component gas in a
mixture. The total pressure is the sum of the partial pressure.
Pt = P1 + P2 + P3 ...
 ni RT 
Each partial pressure is then given by Pi = 

 V 
 RT 
and Pt = 
 (n1 + n2 + n3 ...)
 V 
Chemistry-140
Lecture 31
Gas Mixtures & Partial Pressure
 The partial pressures of the gases present in a mixture are
given by the mole fraction, Xi of the gases in the mixture;
 Pi   ni RT / V   ni 
 =   = Xi
 =
 Pt   nt RT / V   nt 
Pi = XiPt
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?
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
 1 mol O 2

=
n(O2) = (6.00 g O2) 
 32.0 g / mol O 2 
0.188 mol O2
 1 mol CH4 
 = 0.563 mol CH4
n(CH4) = (9.00 g CH4) 
 16.0 g / mol CH4 
Chemistry-140
Answer:
Lecture 31
Partial Pressures
Step 2: Use the ideal-gas equation to calculate the partial
pressure of each gas.
 n(O 2 ) RT 
P(O2) = 



V
 0.188 mol0.0821 L atm / mol K  273 K  
=
 = 0.281 atm
15.0 L 


 n(CH 4 ) RT 
P(CH4) = 



V
 0.563 mol0.0821 L atm / mol K  273 K  
=
 = 0.841 atm
15.0 L 


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
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.
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?
Chemistry-140
Lecture 31
Partial Pressures
Answer:
Step 1: Use the ideal-gas equation to determine the moles
of NO2.
 PV 
n= 

 RT 
5.00 atm450 L 



= 
 0.0821 L atm / mol K  295 K  
= 92.9 mol
Chemistry-140
Lecture 31
Partial Pressures
Answer:
Step 2: Use the stoichiometry of the equation to calculate
the moles of nitric acid produced.
 2 mol HNO 3 
(92.9 mol NO2) 
 = 61.9 mol HNO3
 3 mol NO 2 
Chemistry-140
Lecture 31
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
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.
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) = 740 Torr
 P (O 2 )V 
n(O2) = 

 RT 
740 / 760 Torr 0.250 L  

 = 9.92 x 10-3 mol O2
=
 0.0821 L atm / mol K  299 K  
Chemistry-140
Lecture 31
Trapped Gases
Answer:
Step 2: We can use the stoichiometry of the equation to
calculate the moles of KClO3 and then convert to grams.
(9.92 x
10-3
 122.6 g KClO 3   2 mol KClO 3 

mol O2) 

 1 mol KClO 3   3 mol O 2 
= 0.811 g KClO3
Chemistry-140
Lecture 32
Chapter 12: Gases
 Chapter Highlights

pressure measurements

concept of STP

gas laws (Charle’s, Boyle’s, Avogadro’s)

ideal gas law

applications of ideal gas law

partial pressures
 kinetic theory of gases
 diffusion & effusion (Graham’s law)
 van der Waals equation
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.
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.
Chemistry-140
Lecture 32
Kinetic Molecular Theory
 The collisions are elastic. Energy is transferred between
molecules during collisions.
 The average kinetic energy is proportional to the absolute
temperature.
Chemistry-140
Lecture 32
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.
KE = 1/2mu2
KE = the average kinetic energy
u = root mean squared speed
m = mass of the molecule
Chemistry-140
Lecture 32
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.
Chemistry-140
Lecture 32
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
(@ constant n, V)
Chemistry-140
Lecture 32
Kinetic Molecular Theory Explains...
 Boyle’s 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 (@ constant n, T)
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)
Chemistry-140
Lecture 32
Kinetic Molecular Theory Explains...
 nmu 2 
P is proportional to 

 V 
 Since:
KE = 1/2mu2 and
KE is proportional to T
 Then:
mu2 is proportional to T
 and:
P is proportional to  nT 
Ideal Gas Equation
V 
 RnT 
P=

 V 
Chemistry-140
Lecture 32
RMS Speed
 The root-mean-square speed (rms speed) of a gas, u2 , is
given by Maxwell’s Equation
2
u =
3RT
M
As temperature increases, the rms speed of the gas increases.
As molecular mass increases, the rms speed of the gas decreases.
Chemistry-140
Lecture 32
Molecular Speed (Boltzmann Distribution)
Chemistry-140
Lecture 32
Molecular Speed (Boltzmann Distribution)
Chemistry-140
Lecture 32
Molecular Speeds
Exercise 12.13:
Calculate the rms speed of a N2 molecule at 25oC.
Chemistry-140
Lecture 32
Molecular Speeds
Answer:
Collect known quantities in SI units.
T = (25 + 273) = 298 K
M = 28.0 g/mol = 28.0 x 10-3 kg/mol
R = 8.314 J/mol K = 8.314 kg m2/s2 mol K
u =
2
3RT
M
=
3(8.314 kg m 2 / s 2 mol K)(298 K)
(28.0 x 10 -3 kg / mol)
=
5.15 x 102 m/s
Chemistry-140
Lecture 32
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.
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)
r1
=
r2
M2
M1
Chemistry-140
Lecture 32
Graham’s Law of Molecular Effusion
r( H 2 )
=
r( N 2 )
M (N2 )
M (H2 )
Chemistry-140
Lecture 32
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?
Chemistry-140
Answer:
Lecture 32
Molecular Effusion
Use Graham's Law. Let r(X2) and M(X2) be the rate of
effusion and molar mass of the unknown diatomic gas.
r( X 2 )
=
r( O 2 )
0.355 =
M(X2) =
M (O2 )
M ( X2 )
32.0 g / mol
M ( X2 )
 32.0g / mol
(0.355)
2
= 254 g/mol
Since molecule is diatomic AW(X) = 254/2 = 127. Molecule is I2.
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) < P(ideal)
Chemistry-140
Lecture 32
The van der Waals Equation
2

 n 
P  a   V  bn  nRT
 V  

correction for
intermolecular forces
correction for
molecular volume
a and b = van der Waals constants for the particular gas
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)
Chemistry-140
Lecture 32
The van der Waals Equation
Answer:
Ideal gas equation: P 
nRT
V
(8.00 mol)(0.0821 L atm / mol K)(300 K)
P
( 4.00 L )
=
49.2 atm
Chemistry-140
Lecture 32
The van der Waals Equation
Answer:
2

 n 
van der Waals equation:  P  a   V  bn  nRT
 V  

2

 nRT 
 n 
P 
  a   
 V  bn    V  
 (8.00 mol)(0.0821 L atm / mol K)(300 K) 


 4.00 L  (0.0562 L / mol)(8.00 mol) 
2


8.00
mol


2
2
 
(6.49 atm L / mol )

 
4.00
L

=
29.5 atm
Chemistry-140
Lecture 32
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
Chemistry-140
Lecture 32
Final Exam
Monday December 10th, 2001
12:00 Noon
St. Denis Centre Field House