### 6 Gases - Science

```6 Gases
A gas expand to occupy the entire volume it is placed in. Molecules in a
gas translate freely between collisions, and they all behave alike
regardless of their type.
What are some of the properties of gases?
Pressure, temperature, heat capacity, volume, density,
molar volume, color, average speed of molecules,
solubility (in water or other liquid),
absorption, compressibility,
gas-liquid equilibrium, composition, identity (compound or element),
chemical properties, combustibility, stability
Which of these properties are intensive properties, and which are
extensive properties?
6 Gases
1
Announcement
Appointments for Winter '04 enrolment are now posted on your QUEST
account – Each student has only a three day appointment time. If
missed, you will have to wait until all students have completed their
appointments. Open enrolment begins November 3rd, but courses
could be full by then
soon as possible.
6 Gases
2
6 Gases
3
Pressure
Pressure: force per area (1 Pascal = 1 N m–2)
Liquid pressure (explain these formulation in terms of physics)
F W
g*m
g* V*d
g * h *A* d
P = --- = ---- = ---------- = --------------- = ----------------- = g * h * d
A A
A
A
A
These equivalences are useful for unit conversions:
1 atm = 101.325 kPa
= 76 cm Hg
= 760 mm Hg (torr in honor of Torricelli)
= 1.01325 b = 1013.25 mb (bar & m bar)
= 14.6960 pounds / sq. inch
6 Gases
4
Torricelli’s
Barometer
Barometric pressure
Explain Torricelli’s work
(1608 - 1647)
P=ghd
1 atm
= 0.76 m Hg
13594 kg
1 m3 Hg
9.80665 N
1 kg
= 101317 N m–2 (Pascal)
= 101.32 k Pa
6 Gases
5
Torricelli Mercury Barometer
Evangelista
Torricelli invented
the Torricelli
Mercury
Barometer in
1644. He used a
long glass tube,
closed at the
upper end, open
at the lower and
filled with
mercury.
6 Gases
6
No water!!
1
Pump Water from a Well
0
The specific gravity of mercury is 13.5939. If water is used for a
.
barometer, what is the height of water corresponding to 1.00 atm?
3
Solution:
3
13.5939 g 1 cm3 H2O = 1033.14 cm H O
76 cm Hg
2
1 cm3 Hg
1g
= 10.33 m H2O
m
W
a
t
e
r
Explain water pump
and depth of well
Water water
everywhere!
6 Gases
What about a diver under water? Be sure to get that during lecture.
7
Pressure of Skater
A skater weighing 70 kg stands on one foot and
the contact between the blade and ice is 1 cm2.
What is the pressure in torr sustained by the
ice?
Solution:
70 kg
1 cm2
10000 cm2
1 m2
1 m3 Hg
13594 kg
1000 mm
= 51493 mm Hg
1m
= 51493 torr
= 67.8 atm
6 Gases
8
The ABCD of gas laws are Avogadro’s, Boyle’s, Charle’s
& Dalton’s laws of gases.
at the same temperature T and pressure P, equal volumes contain
equal amounts of gases in moles n.
at the same temperature T and pressure P, the volume V of a gas is
proportional to the number of molecules or number of moles n.
VP,T = k n
(k, a constant, 22.4 L at STP)
6 Gases
contributions to
science will be given.
9
Boyle’s Law
Robert Boyle, (1621-91)
For a certain amount (constant n) of
gas at constant temperature T, the
volume V times the pressure P is a
constant.
Mathematical aspects of
PV product will be
discussed
P
P V n, T = constant
= P1 V1 = P2 V2
• P1 V1
How do you graph the Boyles law?
State the law in another way.
T2 n1
T1 n1
• P2 V2
V
What curves are P-V plots?
6 Gases
10
Charle’s Law
(law of Charles-Gay-Lussac)
For a certain amount of gas at constant
pressure, its volume, V, is directly
proportional to its temperature T in
Kelvin.
Vn, P = b T
or Pn, V = b T
(b is a constant)
(b is a constant)
Vn, P
or Pn, V
State the Charle’s law in another way
V1 V2
----- = ----- = b,
V1T2 = V2 T1
T1 T2
Charles, Jacques-Alexandre-César (1746-1823,
top) first to ascend in a H2-balloon, developed this
law in 1787. Later Joseph Louis Gay-Lussac 6 Gases
(lower) published a paper citing Charles’s law.
T
11
General Gas Equation
Combining ABC gas laws, we have
P1 V1
T1 =
P2 V2
T2
=nR
Subscripts 1 and 2 refer to different conditions for the same
quantities of gases (n).
Experiments show that one mole of gas at STP occupies 22.4 L.
6 Gases
12
The Ideal Gas Equation
The ABC laws of gases can be combined into one and the result is an
ideal gas equation.
A+B+C  ideal
PV=nRT
1 atm * 22.4140 L
R = ---------------------------- = 0.082057 L atm mol–1 K–1
1 mol 273.15 K
101.325 kPa * 22.414 L
R = ----------------------------------- = 8.3145 L kPa mol–1 K–1
1 mol 273.15 K
Please confirm that 1 kPa L = 1 J
(1 L = 1e-3 m3)
6 Gases
13
Dalton’s law
The Partial pressure Pi is the pressure of a component in a mixture as
if others don’t exist in that system – due to the fact all gases behave as
if they are independent of each other.
Pi = ni R T
correction
V
Dalton’s law:
The total pressure Ptotal, of a mixture of gases is the sum of the partial
pressures of the components.
Ptotal = P1 + P2 + … + Pn
= (n1 + n2 + … + nn)
RT
V
6 Gases
(V is common to all)
14
Application of the Ideal gas Law
Parameters of the ideal gas law: P, V, T, n and a constant R
ideal
 A+B+C
PV=nRT
At constant n and T,
P1 V1 = P2 V2
Ideal gas law
Boyles law
At constant n and V
P= (n R / V ) T
P1 / T1 = P2 / T2
Charles law
At constant n and P
V = (n R / P ) T
V1 / T1 = V2 / T2
ditto
At constant P and T
V = (R T / P ) n
V=k n
Avogardro’s law
6 Gases
15
Gas Densities
Evaluate the density of O2 (molar mass M = 32.0) at 300 K and 2.34 atm.
Hint: Density d of a gas with mass m (= n M) and volume V
m
nM
d = ----- = -----V
V
n
d
---- = -----V
M
Thus
nRT
dRT
P = ------------- = ----------V
M
Find relationship between
density d and M.
Manipulate symbols to get a
useful formula before you
calculate the quantities.
d=PM/RT
Include and work out the units plse
d = 2.34 *32.0 / (0.08205*300) = ___
6 Gases
16
Reactions Involving Gases
How much NaN3 is required to produce 12.0 L N2 gas at 302 K
and 1.23 atm for the air bag in your designed Autotie?
Solution:
Equations: 2 NaN3 = 2 Na + 3 N2;
12.0 L
N2 1.23 atm mol K
0.08205 L atm * 302 K
n = V (P / R T )
2 mol NaN3
3 mol N2
(23+3*14) g NaN3
1 mol NaN3
= ___
Work out N2 volume for 51 g NaN3 used under the same condition.
6 Gases
17
- reaction involving gases
NO is made from oxidizing NH3 at 1123 K with platinum as a
catalyst. How many liter of O2 at 300 K and 1 atm is required for
each liter of NO measured also at 300 K and 1 atm?
Solution:
The reaction is: 4 NH3 + 5 O2  4 NO + 6 H2O
Since n = ( P/RT ) V molar relationships are the same as volumetric
relationship, providing T and P are the same.
Complicated problem may
1 L NO 5 mol O2
= 1.25 L O2
4 mol NO
have a simple solution,
Further oxidation of NO leads to NO2, which is used to
make HNO3, a valuable commodity.
How much H2O is produced?
6 Gases
18
A Mixture of Gases
What is the pressure exerted by the gas when 1.0 g of H2, 2.0 g of O2,
and 0.1 g of CO2 are all confined in a 10.0 L cylinder at 321 K?
Solution:
Dalton’s law : ntotal = i ni
(count molecules non-discriminately)
P = n R T / V;
1 mol
1.0
g
H
+ 2.0 g O2
n=
2 2gH
2
P = 0.585 mol
8.314J * 321 K
10 L mol K
1 mol
+ 0.1 g CO2
32 g O2
= 126 kPa
Calculate partial pressures of each
gas plse
6 Gases
1 mol
44 g CO2
= 0.585 mol
(note 1 J = 1 kPa L)
19
Vapor pressure
The (saturated) vapor pressure is
the partial pressure that is at
equilibrium with another phase.
Vapor pressure of ice
Vapor pressure of water
Explain
Structure of water molecule
Hydrogen bonding
Structure of water
Structure of ice
Vapor pressure of ice and water
Relative and absolute humidity
6 Gases
20
Collecting Gas Over Water
When 1.234 g of a sample containing Ag2O is heated, 40.6 mL of O2 is
collected over water at 296 K, and the atmosphere is 751 mmHg. Vapor
pressure of water at 296 K is 21.1 mmHg.
What is the percentage of Ag2O in the sample?
Solution:
Ag2O = 2 Ag + 0.5 O2
n = P V / R T;
R = 0.08205 L atm mol-1 K-1
P = (751 – 21.1) mmHg / 760 mmHg = 0.961 atm (PO2 = Ptotal – Pwater)
V = 40.6 mL = 0.0406 L
Mass of Ag2O =
0.961 atm*0.0406 L O2
0.08205 L atm mol-1 K-1*296 K
1 mol Ag2O
0.5 mol O2
231.7 g Ag2O
1 mol Ag2O
= 0.744 g Ag2O
Percentage of Ag2O = 0.744 g Ag2O / 1.234 g = 0.603 Ag2O = 60.3 % Ag2O
6 Gases
21
Assumptions of Kinetic-molecular Theory
1. Gas is composed of tiny, discrete particles (molecules or atoms).
2. Particles are small and far apart in comparison to their own size.
3. Ideal gas particles are dimensionless points occupying zero volume.
4. Particles are in rapid, random, constant straight line motion.
5. There is no attractive force between gas molecules and between
molecules and the sides of the container.
6. Molecules collide with one another and the sides of the container.
7. Energy is conserved but transferred in these collisions.
8. Energy is distributed among the molecules in a particular fashion
known as the Maxwell-Boltzmann Distribution.
6 Gases
22
Kinetic-molecular Theory of Gases
For N gas molecules, molecule mass = m, molecular mass = M
speed = u, average speed = u , Avogadro’s number N
volume = V, temperature = T, Pressure = P,
Kinetic energy = ½ m u 2
Collision frequency  u N / V
Pressure  (m v) (u) (N / V)  (N / V) m u 2
= 1/3 (N / V) m u 2
(1/3 due to 3-Dimensional space)
P V = 1/3 N m u 2 = n R T
(Meaning of T)
Thus, 3 R T = NA m u 2
correction
Furthermore, u 2 = 3 R T / M
(Temperature and speed)
6 Gasesthese formulas for sciences
Explain the significances of and apply
23
Molecular Speeds
Distributions of speed of various gases will be demonstrated using a
simulation program, and for each gas, three speeds are indicated.
In the following: m = mass of a molecule, M = molar mass,
R = gas constant, and k = R / Navogadro = Boltzmann constant.
The most probable speed
ump = (2 k T / m)1/2 = (2 R T / M)1/2
The root-mean-square speed
urms = ( 3 k T / m)1/2 = (3 RT / M)1/2
The mean or average speed
uave = ( 8 k T /π m)1/2 = (8 RT / π M)1/2
6 Gases
24
Diffusion of Gases
All gases move together because they are subjected to
Different gases diffuse at different rates
Diffusion contributes to net movement of O2 and CO2 across the
alveolar-capillary membrane (breathe).
Constant molecular motion.
Diffusion from higher to lower concentration regions.
Since uave = ( 8 k T /π m)1/2 = (8 RT / π M)1/2,
(slide 24)
1
Diffusion rate  -------Graham’s law
M
Discuss diffusion rates of H2, He, CH4, N2, O2, CO2, 235UF6, 238UF6 at lecture
6 Gases
25
Diffusion problems
Problems are usually to compare diffusion or effusion rates in the
following terms:
 3RT
urms = ( 3 k T / m)1/2 =
M
uave = ( 8 k T /π m)1/2 =  8RT
 πM
diffusion rate 
1
M
effusion time for same amount  M
distance traveled by molecules in certain period
amount of gas effused
6 Gases
26
Comparing Effusion Rates
If 1e20 N2 molecules effuse from an orifice in 1.0 min, how many H2
molecules will effuse the same orifice at the same condition (T P)? How
many minutes will be required for the same number of H2 molecules to
effuse?
H2 effusion rate = MN2 / MH2 (N2 effusion rate)
= (28 / 2 )1e20 molecules/min)
= __________ figure out value and units
Time for 1e20 H2 molecules to effuse = (2 / 28 ) 1.0 min
= __________
What is the effusion rate ratio of N62 Gases
and H2 or any two gases?
27
Effusion and Life
Breathing chemistry is complicated, and we can only scratches the surface!
O2
P2
CO2
A
A D

Vgas 
( P1  P2 )
T
P1
T
6 Gases
28
The van der Waals equation for real gases
Gases tend to behave ideally at high T and low P. Required T and P
for ideality depends on gas properties and molar mass, and van der
Waals proposed correction terms for the ideal gas equation for real
gases.
Correction for intermolecular forces
Correction for volume of molecules
2a
n
(P +
) (V – n b) = n R T
2
V
where a and b are gas-dependant constants.
Gas
He
Ne
N2
O2
CO2
Cl2
a L2 atm mol-2 b L mol-1
0.0341
0.0237
0.211
0.0171
1.39
0.0391
1.36
0.0318
3.59
0.0427
6.49
0.0562
Explain the meaning of vdW eqn
Note units for a and b
6 Gases
29
Application of van der Waal’s Equation
What is the pressure of Cl2 at 300 K occupying 20.0 L according to vdW
and ideal gas laws?
Solution: Look up data for Cl2,
a = 6.49 L2 atm mol-2, b = 0.0562 L mol-1,
n = 1 mol, R = 0.08205 L atm K-1 mol-1
P=
=
nRT
V–nb
-
n2a
V2
1 mol * 0.08205 L atm mol-1 K-1*300 K
20.0 L – 1 mol 0.0562 mol-1 L
2 mol2 * 6.49 L2 atm mol-2
1
20.02 L2
= _____________________
P from the ideal gas law.
6 Gases
30
Problems related to van der Waal’s Equation
What is the molar volume of Cl2 at 300 K and 1 atm according to vdW?
Solution: Look up data for Cl2,
a = 6.49 L2 atm mol-2, b = 0.0562 L mol-1,
n = 1 mol, R = 0.08205 L atm K-1 mol-1
V=
nRT
+ n b = 24.615 / (1+0.013) = 24.299 L
2
2
P+n a/V
try V = 22 L
= 24.625 / (1+ 0.011) = 24.434 L
= 24 L
try V = 24.434 L = ?
and calculate V
Calculate P for a definite volume is easier, and using the successive
method for V is interesting, but it’s a challenge.
6 Gases
Find out how engineers deal with real gases.
31
Volume of vdW eqn
What is the volume of occupied by 132 g CO2 gas at 12.5 atm and 300 K?
Solution:
Solve volume of van der Waals equation for V
2a
2ab
R
T
+
b
P
n
n
2
V –n(
)V + (
)V–(
)=0
P
P
P
3
a = 3.59 L2 atm mol-2, b = 0.0427 L mol-1,
n = 1 mol, R = 0.08205 L atm K-1 mol-1
This is similar to problem 6 –106, a question for
6 Gases
practicing the successive approximation
method.
32
Successive Method again
What is the volume of occupied by 132 g CO2 gas at 12.5 atm and 300 K?
Solution:
a = 3.59 L2 atm mol-2, b = 0.0427 L mol-1, T = 300 K
n = 132/44 = 3 mol, R = 0.08205 L atm K-1 mol-1
V=
nRT
+ nb
2
2
P+n a/V
= 3*0.08205*300 / (12.5 + 32* 3.59 / 12) + 3*.0427 = 1.78 L
= 73.845 / (12.5 + 32*3.59 / 32) + 0.128 = 4.7 L
= 73.845 / (12.5 + 32*3.59 / 42) + 0.128 = 5.2 L
= 73.845 / (12.5 + 32*3.59 / 52) + 0.128 = 5.5 L
= 73.845 / (12.5 + 32*3.59 / 5.62) + 0.128 = 5.58 L
6 Gases
33
Molecular Formula of Gas
Combustion of 1.110 g hydrocarbon produces 3.613 CO2 and 1.109 g
H2O. A 0.288 g sample of the same has a volume of 131 mL at 298 K
and 753 mmHg. Find the molecular formula.
Solution:
C : H = 3.613 : 1.109*2 = 0.0821 : 0.123 = 1 : 1.5 = 2 : 3
44
18
Empirical formula is C2H3
dRT
–1) * 0.08205 L atm mol–1 K –1 * 298 K
(0.288
g
/
0.131
L
M= P =
(753 / 760) atm
= 54.3_______ work out units
C4H6 has a molar mass of 54.3. Confirm and conclusion please!
6 Gases
A 2-step problem, similar to one in Advanced Exercises
34
ROOMS FOR TEST #1 (CHEM 120) – Wed., Oct. 8th
Write the test during your regular lecture time (10:30 am) on Wed., Oct. 8th.
Go to DC 1350 and ESC 146/149 according to your Surnames
Surnames
For 8:30 class
For 9:30 class
A–L
M–Z
A – Mo
Mu – Z
Room(s)
DC 1350
ESC 146 & 149
DC 1350
ESC 146 & 149
10:30 am
A – Ma
DC 1350
(For you)
Mc – Z
ESC 146 & 149 (First Year Chem Lab)
For 11:30 class
A – Ma
Mc – Z
DC 1350
ESC 146 & 149
6 Gases
35
Regarding Test 1
 Do not enter the room until directed to do so by the proctors. We
need space and time to set out the test booklets and computer answer
cards.
 Bring a calculator and a pencil for filling out the computer answer
card.
Do NOT bring your own scrap paper or periodic table. All work must
be done on the test booklet. A periodic table will be supplied.
6 Gases
36
Some concepts to review
Convert between mass and mole and vice versa
Find empirical and molecular formulas
Figure out limiting and excess reagent, calculate theoretical and percent
yields
Calculate concentrations in molarity, mass percentage, etc even when
solutions are combined (dilution)
Analyze binary mixture: extra problems B2 and B3 (handout page 8)
Figure out the net ionic reaction equations
Balance redox reaction equations (figure out oxidation states, balance
half-reaction equations and balance equations)
6 Gases
37
Concepts to review – cont.
Apply ideal gas low to various problems
Calculate stoichiometric quantities using on gas law and reaction
equations.
Apply Dalton’s partial pressure equation
Compare effusion or diffusion rates of gases
6 Gases
38
```