Transcript CH 5 CHEM1A GASES - Santa Rosa Junior College

Chemistry 1A
General
Chemistry
Instructor:
Dr. Orlando E. Raola
Chapter
Santa Rosa Junior College
5:
Properties of
Gases
Gases and the
Kinetic Molecular Theory
5.1 An Overview of the Physical States of
Matter
5.2 Gas Pressure and Its Measurement
5.3 The Gas Laws and Their Experimental
Foundations
5.4 Applications of the Ideal Gas Law
The distinction of gases from liquids
and solids
1. Gas volume changes greatly with pressure.
2. Gas volume changes greatly with
temperature.
3. Gases have relatively low viscosity.
4. Most gases have relatively low densities
under normal conditions.
5. Gases are miscible.
The three states of matter
A mercury barometer
Units of pressure
Sample Problem 5.1
PROBLEM:
Converting Units of Pressure
A geochemist heats a limestone (CaCO3) sample and collects
the CO2 released in an evacuated flask attached to a closedend manometer. After the system comes to room temperature,
Dh = 291.4 mm Hg. Express the CO2 pressure in torr,
atmosphere, and kilopascal.
PLAN: Construct conversion factors to find the other units of pressure.
SOLUTION:
291.4 mmHg
1torr
= 291.4 torr
1 mmHg
291.4 torr
1 atm
= 0.3834 atm
760 torr
0.3834 atm 101.325 kPa
1 atm
= 38.85 kPa
The relationship between the volume
and pressure of a gas.
Boyle’s Law
The relationship between the
volume and temperature of a
gas.
Charles’s Law
V a
Boyle’s Law
VxP
V a T
V
= constant
T
Amontons’s Law
T
combined gas law
P
= constant
Charles’s Law
P
1
P a T
= constant
V a
n and T are fixed
V = constant / P
P and n are fixed
V = constant x T
V and n are fixed
P = constant x T
T
P
V = constant x
T
PV
P
T
= constant
Avogadro’s Law
For a gas at constant temperature and pressure, the
volume is directly proportional to the number of moles of
gas
V = an
a = proportionality constant
V = volume of the gas (m3)
n = chemical amount of gas (mol)
Ideal Gas Law
An equation of state for a gas.
“state” is the condition of the gas at a given time.
PV = nRT
-1
-1
R = 8.314 J∙mol ·K
(in common units 0.082 atm∙L∙mol-1·K-1)
Standard molar volume.
THE IDEAL GAS LAW
PV = nRT
R is the universal gas constant
IDEAL GAS LAW
nRT
PV = nRT or V =
fixed n and T
Boyle’s Law
V=
constant
P
P
fixed n and P
fixed P and T
Charles’s Law
Avogadro’s Law
V=
constant X T
V=
constant X n
Sample Problem 5.2
PROBLEM:
Applying the Volume-Pressure Relationship
Boyle’s apprentice finds that the air trapped in a J tube occupies
24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases
the pressure on the trapped air to 2.64 atm. Assuming constant
temperature, what is the new volume of air (in L)?
PLAN:
n and T are constant
SOLUTION:
V1 in cm3
1cm3=1mL
V1 in mL
unit
conversion
103 mL=1L
V1 in L
xP1/P2
gas law
calculation
P1 = 1.12 atm
P2 = 2.64 atm
V1 = 24.8 cm3
V2 = unknown
L
24.8 cm3 1 mL
1 cm3 103 mL
P1V1
V2 in L
n1T1
V2 =
P1V1
P2
=
P2V2
= 0.0248 L
P1V1 = P2V2
n2T2
1.12 atm
= 0.0248 L
2.46 atm
= 0.0105 L
Sample Problem 5.3
PROBLEM:
Applying the Temperature-Pressure Relationship
A steel tank used for fuel delivery is fitted with a safety valve that
opens when the internal pressure exceeds 1.00x103 torr. It is
filled with methane at 230C and 0.991 atm and placed in boiling
water at exactly 1000C. Will the safety valve open?
PLAN:
SOLUTION:
T1 and T2(0C)
P1(atm)
1atm=760torr
P1(torr)
K=0C+273.15
T1 and T2(K)
x T2/T1
P2(torr)
P1 = 0.991atm
P2 = unknown
T1 = 230C
T2 = 1000C
P1V1
=
n1T1
P2V2
n2T2
0.991 atm 760 torr = 753 torr
1 atm
P2 = P1
T2
T1
= 753 torr
373K
296K
= 949 torr
P1
T1
=
P2
T2
Sample Problem 5.4
PROBLEM:
Applying the Volume-Amount Relationship
A scale model of a blimp rises when it is filled with helium to a
volume of 55 dm3. When 1.10 mol of He is added to the blimp, the
volume is 26.2 dm3. How many more grams of He must be added
to make it rise? Assume constant T and P.
PLAN: We are given initial n1 and V1 as well as the final V2. We have to find
n2 and convert it from moles to grams.
n1(mol) of He
P and T are constant
SOLUTION:
x V2/V1
n1 = 1.10 mol
n2 = unknown
n2(mol) of He
V1 = 26.2 dm3
V2 = 55.0 dm3
subtract n1
mol to be added
V1
n1
=
V2
n2
n2 = n1
xM
g to be added
55.0 dm3
P1V1
n1T1
=
P2V2
n2T2
V2
V1
4.003 g He
= 9.24 g He
n2 = 1.10 mol
= 2.31 mol
3
26.2 dm
mol He
Sample Problem 5.5
PROBLEM:
PLAN:
Solving for an Unknown Gas Variable at Fixed
Conditions
A steel tank has a volume of 438 L and is filled with 0.885 kg of
O2. Calculate the pressure of O2 at 210C.
V, T and mass are given. From mass, find amount (n), and use the
ideal gas law to find P.
V  438 L 
SOLUTION:
1m
3
3
 0.438 m
T  21C  273.15  294.15 K
3
10 L
m = 0.885 kg
n =
m
M
P 
P 
0.885 kg 

1000 g
1kg
32.00 g  m ol
 27.7 m ol
-1
n RT
V
n RT
V
27.7 m ol  8.314 kg  m  s
2

0.438 m
3
2
 294.15 K
 1.55  10 P a
5
Relationship between density and
molar mass for gases
S in ce n 
m
a n d P V  n R T , th e n
M
PV 
mRT
M
M 
The density of a gas is
directly proportional to its
molar mass.
mRT
PV
a n d b e ca u se d 
m
V
fin a lly M 
dRT
P
The density of a gas is
inversely proportional to the
temperature.
Sample Problem 5.7
Calculating Gas Density
To apply a green chemistry approach, a chemical engineer uses
waste CO2 from a manufacturing process, instead of
chlorofluorocarbons, as a “blowing agent” in the production of
polystyrene containers. Find the density (in g/L) of CO2 and the
number of molecules (a) at STP (00C and 1 atm) and (b) at
room conditions (20.0C and 1.00 atm).
PROBLEM:
PLAN: Density is mass/unit volume; substitute for volume in the ideal gas
equation. Since the identity of the gas is known, we can find the molar
mass. Convert mass/L to molecules/L with Avogadro’s number.
MxP
d = mass/volume
PV = nRT
V = nRT/P
d =
RT
SOLUTION:
44.01 g/mol
(a)
x 1atm
d=
0.0821
1.96 g
mol CO2
L
44.01 g CO2
atm*L
mol*K
6.022x1023 molecules
mol
= 1.96 g/L
x 273.15K
= 2.68x1022 molecules CO2/L
Sample Problem 5.6
Calculating Gas Density
continued
(b)
44.01 g/mol x 1 atm
d=
0.0821
1.83g
mol CO2
L
44.01g CO2
= 1.83 g/L
atm*L x 293K
mol*K
6.022x1023 molecules
mol
= 2.50x1022 molecules CO2/L
Determining the molar
mass of an unknown
volatile liquid.
based on the method of
J.B.A. Dumas (1800-1884)
Sample Problem 5.8
PROBLEM:
Finding the Molar Mass of a Volatile Liquid
An organic chemist isolates a colorless liquid from a petroleum
sample. She uses the Dumas method and obtains the following
data:
Volume of flask = 213 mL
T = 100.00C
P = 754 torr
Mass of flask + gas = 78.416 g Mass of flask = 77.834 g
Calculate the molar mass of the liquid.
PLAN: Use unit conversions, mass of gas, and density-M relationship.
SOLUTION:
M=
m = (78.416 - 77.834) g = 0.582 g
m RT
VP
=
0.582 g x 0.0821
atm*L
mol*K
0.213 L x 0.992 atm
x 373K
= 84.4 g/mol
Relationship between density and
molar mass for gases
S in ce n 
m
a n d P V  n R T , th e n
M
PV 
mRT
M
M 
The density of a gas is
directly proportional to its
molar mass.
mRT
PV
a n d b e ca u se d 
m
V
fin a lly M 
dRT
P
The density of a gas is
inversely proportional to the
temperature.
Sample Problem 5.7
Calculating Gas Density
To apply a green chemistry approach, a chemical engineer uses
waste CO2 from a manufacturing process, instead of
chlorofluorocarbons, as a “blowing agent” in the production of
polystyrene containers. Find the density (in g/L) of CO2 and the
number of molecules (a) at STP (273.15 K and 1 bar) and (b) at
room conditions (20.0C and 1.00 atm).
PROBLEM:
PLAN: Density is mass/unit volume; substitute for volume in the ideal gas
equation. Since the identity of the gas is known, we can find the molar
mass. Convert mass/L to molecules/L with Avogadro’s number.
MxP
d = mass/volume
PV = nRT
V = nRT/P
d =
RT
SOLUTION:
 4 4 .0 1 g ·m o l  1 b a r 
d=
= 1 .9 4
 0 .0 8 3 1 4 b a r·L ·m o l ·K   2 7 5 .1 5 K 
-1
(a)
-1
1.94 g
mol CO2
L
44.01 g CO2
6.022x1023 molecules
mol
-1
g ·L
= 2.65x1022 molecules CO2/L
-1
Sample Problem 5.6
Calculating Gas Density
continued
(b)
44.01 g/mol x 1 atm
d=
0.0821
1.83g
mol CO2
L
44.01g CO2
= 1.83 g/L
atm*L x 293K
mol*K
6.022x1023 molecules
mol
= 2.50x1022 molecules CO2/L
Determining the molar
mass of an unknown
volatile liquid.
based on the method of
J.B.A. Dumas (1800-1884)
Sample Problem 5.8
PROBLEM:
Finding the Molar Mass of a Volatile Liquid
An organic chemist isolates a colorless liquid from a petroleum
sample. She uses the Dumas method and obtains the following
data:
Volume of flask = 213 mL
T = 100.00C
P = 754 torr
Mass of flask + gas = 78.416 g Mass of flask = 77.834 g
Calculate the molar mass of the liquid.
PLAN: Use unit conversions, mass of gas, and density-M relationship.
SOLUTION:
M=
m = (78.416 - 77.834) g = 0.582 g
m RT
VP
=
0.582 g x 0.0821
atm*L
mol*K
0.213 L x 0.992 atm
x 373K
= 84.4 g/mol
Sample Problem 5.5
PROBLEM:
PLAN:
Solving for an Unknown Gas Variable at Fixed
Conditions
A steel tank has a volume of 438 L and is filled with 0.885 kg of
O2. Calculate the pressure of O2 at 210C.
V, T and mass are given. From mass, find amount (n), and use the
ideal gas law to find P.
V  438 L 
SOLUTION:
1m
3
3
 0.438 m
T  21C  273.15  294.15 K
3
10 L
m = 0.885 kg
n =
m
M
P 
P 
0.885 kg 

1000 g
1kg
32.00 g  m ol
 27.7 m ol
-1
n RT
V
n RT
V
27.7 m ol  8.314 kg  m  s
2

0.438 m
3
2
 294.15 K
 1.55  10 P a
5
Mixtures of gases
•Gases mix homogeneously in any proportions.
•Each gas in a mixture behaves as if it were the
only gas present.
Dalton’s Law of Partial Pressures
Ptotal = P1 + P2 + P3 + ...
P1= c1 x Ptotal
where c1 is the mole fraction
c1 =
n1
n1 + n2 + n3 +...
=
n1
ntotal
Mole fraction and partial pressure
For each component we define the mole
fraction xB
xB 
nB
n A  n B  n C  ...

nB
 na
a
and because of Dalton’s law, we can
calculate the partial pressure of each
component as
p B  x B Ptot
Sample Problem 5.9
PROBLEM:
Applying Dalton’s Law of Partial Pressures
In a study of O2 uptake by muscle at high altitude, a physiologist
prepares an atmosphere consisting of 79 mol% N2, 17 mol%
16O and 4.0 mol% 18O . (The isotope 18O will be measured to
2,
2
determine the O2 uptake.) The pressure of the mixture is
0.75atm to simulate high altitude. Calculate the mole fraction
and partial pressure of 18O2 in the mixture.
18
PLAN: Find the c 18Oand P18O from Ptotal and mol% O2.
2
mol%
18O
2
SOLUTION:
2
divide by 100
c
18O
P18
2
multiply by Ptotal
partial pressure P
18O
2
O2
c
18O
=
2
4.0 mol% 18O2
= 0.040
100
= c 18 x Ptotal = 0.040 x 0.75 atm = 0.030 atm
O2
Table 5.3 Vapor Pressure of Water (P
H2O
T(0C)
0
5
10
11
12
13
14
15
16
18
20
22
24
26
28
) at Different T
P (torr)
T(0C)
P (torr)
4.6
6.5
9.2
9.8
10.5
11.2
12.0
12.8
13.6
15.5
17.5
19.8
22.4
25.2
28.3
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
31.8
42.2
55.3
71.9
92.5
118.0
149.4
187.5
233.7
289.1
355.1
433.6
525.8
633.9
760.0
Collecting a water-insoluble gaseous reaction
product and determining its pressure.
Sample Problem 5.10 Calculating the Amount of Gas Collected Over Water
PROBLEM:
Acetylene (C2H2), an important fuel in welding, is produced in
the laboratory when calcium carbide (CaC2) reaction with water:
CaC2(s) + 2H2O(l)
C2H2(g) + Ca(OH)2(aq)
For a sample of acetylene that is collected over water, the total
gas pressure (adjusted to barometric pressure) is 98.32 kPa and
the volume is 523 mL. At the temperature of the gas (230C), the
vapor pressure of water is 2.798 kPa. How many grams of
acetylene are collected?
PLAN: The difference in pressures will give us the P for the C2H2. The ideal
gas law will allow us to find n. Converting n to grams requires the
molar mass, M.
Ptotal
P
H2O
P
C2H2
n=
SOLUTION:
PV
RT
n
g
C2H2
C2H2
xM
P
C2H2 = (98.32-2.798) kPa =95.52 kPa
Sample Problem 5.9
Calculating the Amount of Gas Collected Over Water
continued
9.552  10 P a 5.23  10 m 

8.314 J  m ol  K 296 K 
4
4
nC
2
H2

0.0203mol 
1
26.04 g C2H2
mol C2H2
1
3
0.0203 m ol
= 0.529 g C2H2
Summary of the stoichiometric relationships among the
amount (mol,n) of gaseous reactant or product and the gas
variables pressure (P), volume (V), and temperature (T).
P,V,T
of gas A
ideal
gas
law
amount
(mol)
amount
(mol)
P,V,T
of gas A
of gas B
of gas B
molar ratio from
balanced equation
ideal
gas
law
Sample Problem 5.11
Using Gas Variables to Find Amount of
Reactants and Products
PROBLEM: Dispersed copper in absorbent beds is used to react with
oxygen impurities in the ethylene used for producing
polyethylene. The beds are regenerated when hot H2 reduces
the metal oxide, forming the pure metal and H2O. On a
laboratory scale, what volume of H2 at 765 torr and 2250C is
needed to reduce 35.5 g of copper(II) oxide?
PLAN: Since this problem requires stoichiometry and the gas laws, we have
to write a balanced equation, use the moles of Cu to calculate mols
and then volume of H2 gas.
mass (g) of Cu
SOLUTION:
divide by M
mol of Cu
35.5 g Cu
CuO(s) + H2(g)
mol Cu
0.559 mol H2 x 0.0821
use known P and T to find V
L of H2
1 mol H2
63.55 g Cu 1 mol Cu
molar ratio
mol of H2
Cu(s) + H2O(g)
atm*L
x
mol*K
1.01 atm
= 0.559 mol H2
498K
= 22.6 L
Sample Problem 5.12
Using the Ideal Gas Law in a Limiting-Reactant
Problem
PROBLEM: The alkali metals (Group 1) react with the halogens (Group 17) to form
ionic metal halides. What mass of potassium chloride forms when
5.25 L of chlorine gas at 0.950 atm and 293K reacts with 17.0 g of
potassium?
PLAN: After writing the balanced equation, we use the ideal gas law to find the
number of moles of reactants, the limiting reactant and moles of product.
SOLUTION:
2K(s) + Cl2(g)
2KCl(s)
P = 0.950 atm
V = 5.25 L
PV
0.950 atm x 5.25L
T = 293K
n = unknown
n =
=
= 0.207 mol
Cl2
RT
atm*L
0.0821
x 293K
mol*K
2 mol KCl
mol
K
0.207
mol
Cl
= 0.414 mol
2
17.0g
= 0.435 mol K
1 mol Cl2
KCl formed
39.10 g K
2 mol KCl
Cl2 is the limiting reactant.
0.435 mol K
= 0.435 mol
2 mol K
KCl formed
74.55 g KCl
0.414 mol KCl
= 30.9 g KCl
mol KCl
Problem
Gaseous iodine pentafluoride can be prepared by
the reaction between solid iodine and gaseous
fluorine. A 5.00-L flask containing 10.0 g of I2 is
charged with 10.0 g of F2 and the reaction proceeds
until one of the reactants is completely consumed.
After the reaction is complete, the temperature in
the flask is 125 ºC.
a) What is the partial pressure of IF5 in the flask?
b) What is the mole fraction of IF5 in the flask?
Postulates of the Kinetic-Molecular Theory
Postulate 1: Particle Volume
Because the volume of an individual gas particle is so
small compared to the volume of its container, the gas
particles are considered to have mass, but no volume.
Postulate 2: Particle Motion
Gas particles are in constant, random, straight-line
motion except when they collide with each other or with
the container walls.
Postulate 3: Particle Collisions
Collisions are elastic therefore the total kinetic
energy(Ek) of the particles is constant.
Distribution of molecular speeds as a function of temperature
Distribution of molecular speeds as a
function of temperature
Ek a T
E k  c onstant  T
Molecular description of Boyle’s Law
Molecular description of Dalton’s law of partial pressures
Molecular description of Charles’s law
Molecular description of Avogadro’s
law
Relationship between molecular
speed and mass
Ek 
1
2
mu
2
u rm s 
3R T
M
The meaning of temperature
3  R 
 T
Ek 


2  N A 
Absolute temperature is a
measure of the average kinetic
energy of the molecular
random motion
Effusion and difussion
Effusion: describes the passage of
gas through a small orifice into an
evacuated chamber.
Diffusion: describes the mixing of
gases. The rate of diffusion is the
rate of gas mixing.
Effusion:
R a te o f e ffu sio n fo r g a s 1
=
R a te o f e ffu sio n fo r g a s 2
M2
M1
Diffusion:
D istance trav eled by gas 1
D istance trav eled by gas 2
=
M2
M1
Effusion and KMT
Diffusion through space
distribution of molecular speeds
mean free path
collision frequency
Sample Problem 5.13
Applying Graham’s Law of Effusion
PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH4).
PLAN: The effusion rate is inversely proportional to the square root of the
molar mass for each gas. Find the molar mass of both gases and find
the inverse square root of their masses.
SOLUTION:
M of CH4 = 16.04g/mol
rate
He
rate
CH4
=
√
16.04
4.003
= 2.002
M of He = 4.003g/mol
Table 5.4 Molar Volume of Some Common Gases at STP
(00C and 1 atm)
Gas
He
H2
Ne
Ideal gas
Ar
N2
O2
CO
Cl2
NH3
Molar Volume
(L/mol)
22.435
22.432
22.422
22.414
22.397
22.396
22.390
22.388
22.184
22.079
Condensation Point
(0C)
-268.9
-252.8
-246.1
---185.9
-195.8
-183.0
-191.5
-34.0
-33.4
The behavior of
several real gases
with increasing
external pressure
Effect of molecular attractions on pressure
Effect of molecular volume on measured volume
van der Waals equation

 n  
 P +a     V -n b  =n R T
 V  

2
Van der Waals
equation for n
moles of a real gas
a (atm·L2·mol-2)
Gas
He
Ne
Ar
Kr
Xe
H2
N2
O2
Cl2
CO2
CH4
NH3
H2O
0.034
0.211
1.35
2.32
4.19
0.244
1.39
1.36
6.49
3.59
2.25
4.17
5.46
b (L·mol-1)
0.0237
0.0171
0.0322
0.0398
0.0511
0.0266
0.0391
0.0318
0.0562
0.0427
0.0428
0.0371
0.0305