Transcript Chs 10 & 11

Chs 10 & 11
Gases and Kinetic Molecular Theory
1
Comparison of Solids, Liquids,
and Gases
• The density of gases is much less than that of
solids or liquids.
Densities
Solid
Liquid
Gas
(g/mL)
H2O
0.917
0.998
0.000588
CCl4
1.70
1.59
0.00503
Gas molecules must be very far apart
compared to liquids and solids.

2
Composition of the Atmosphere and Some
Common Properties of Gases
Composition of Dry Air
Gas
N2
O2
Ar
CO2
He, Ne, Kr, Xe
CH4
H2
% by Volume
78.09
20.94
0.93
0.03
0.002
0.00015
0.00005
3
Pressure
• Pressure is force per unit area.
– lb/in2
– N/m2
• Gas pressure as most people think of it.
4
Pressure
• Atmospheric pressure is measured using a
barometer.
• Definitions of standard pressure
–
–
–
–
–
76 cm Hg
760 mm Hg
760 torr
1 atmosphere
101.3 kPa
Hg density = 13.6 g/mL
5
Boyle’s Law:
The Volume-Pressure Relationship
V  1/P or
V= k (1/P) or PV = k
P1V1 = k1 for one sample of a gas.
P2V2 = k2 for a second sample of a gas.
k1 = k2 for the same sample of a gas at the
same T.
• Thus we can write Boyle’s Law
mathematically as P1V1 = P2V2
•
•
•
•
•
6
Boyle’s Law:
The Volume-Pressure Relationship
• Example 12-1: At 25oC a sample of He has a
volume of 4.00 x 102 mL under a pressure of 7.60
x 102 torr. What volume would it occupy under a
pressure of 2.00 atm at the same T?
V
PPP
V111 PP22 V
V222
111V
PP11 V
V11
V

V22 
PP2
2



760
torr
400
mL



760
torr
400
mL


1520
1520torr
torr
 2.00 10 2 mL
7
Boyle’s Law:
The Volume-Pressure Relationship
• Notice that in Boyle’s law we can use any pressure
or volume units as long as we consistently use the
same units for both P1 and P2 or V1 and V2.
• Use your intuition to help you decide if the volume
will go up or down as the pressure is changed and
vice versa.
8
Charles’ Law:
The Volume-Temperature Relationship
35
30
25
Volume (L)
vs.
Temperature (K)
20
15
Gases liquefy
before reaching 0K
10
5
0
0
50
100
150
200
250
300
350
400
absolute zero = -273.15 0C
9
Charles’ Law:
The Volume-Temperature Relationship
• Charles’s law states that the volume of a gas is
directly proportional to the absolute
temperature at constant pressure.
– Gas laws must use the Kelvin scale to be correct.
• Relationship between Kelvin and centigrade.
K = o C + 273
10
Charles’ Law:
The Volume-Temperature Relationship
• Mathematical form of Charles’ law.
V V
V
= kT
or or k  k
V
TTororVV
= kT
T T
V1
V2
 k and
 k however th e k' s are equal so
T1
T2
V1 V2

in the most useful form
T1 T2
11
Charles’ Law:
The Volume-Temperature Relationship
• Example 12-2: A sample of hydrogen, H2,
occupies 1.00 x 102 mL at 25.0oC and 1.00 atm.
What volume would it occupy at 50.0oC under
the same pressure?
T1 = 25 + 273 = 298
T2 = 50 + 273 = 323
V1 V2
V1T2

 V2 =
T1 T2
T1
1.00 10 2 mL  323 K
V2 =
298 K
 108 mL
12
Standard Temperature and Pressure
• Standard temperature and pressure is given
the symbol STP.
– It is a reference point for some gas calculations.
• Standard P  1.00000 atm or 101.3 kPa
• Standard T  273.15 K or 0.00oC
13
The Combined Gas Law Equation
• Boyle’s and Charles’ Laws combined into one
statement is called the combined gas law equation.
– Useful when the V, T, and P of a gas are changing.
Boyle's sLaw
Law
Boyle'
Charles'
Charles'Law
Law
V1P V
P2 V2
P1VP1 1
2 2
V2
VV11  V
2

T2
TT11 T
2
For a given sample of gas : The combined gas law is :
PV
P1 V1 P2 V2
k

T
T1
T2
14
The Combined Gas Law Equation
• Example 12-3: A sample of nitrogen gas, N2, occupies
7.50 x 102 mL at 75.00C under a pressure of 8.10 x
102 torr. What volume would it occupy at STP?
V1 =V
750
mL V2 = ? V2 = ?
1 = 750 mL
T1 =K348 K
T2 K= 273 K
T1 = 348
T2 = 273
P1 =torr
810 torr
P2 torr
= 760 torr
P1 = 810
P2 = 760
P1 V1 T2
P
V
T
1 2 1= 2
SolveSolve
for V2for
= V
P2 T1
P2 T1

810 torr 750 mL 273 K 

760 torr 348 K 
 627 mL
15
The Combined Gas Law Equation
• Example 12-4 : A sample of methane, CH4,
occupies 2.60 x 102 mL at 32oC under a
pressure of 0.500 atm. At what temperature
would it occupy 5.00 x 102 mL under a
pressure of 1.20 x 103 torr?
You do it!
16
The Combined Gas Law Equation
V1 = 260 mL
V2 = 500 mL
P1 = 0.500 atm
P2 = 1200 torr
= 380 torr
T1 = 305 K
T2 = ?
T1 P2 V2 305 K 1200 torr 500 mL 
T2 =

380 torr 260 mL 
P1 V1
= 1852 K  1580 C
o
17
Avogadro’s Law and the
Standard Molar Volume
18
Avogadro’s Law and the
Standard Molar Volume
• Avogadro’s Law states that at the same temperature and
pressure, equal volumes of two gases contain the same
number of molecules (or moles) of gas.
• If we set the temperature and pressure for any gas to be STP,
then one mole of that gas has a volume called the standard
molar volume.
• The standard molar volume is 22.4 L at STP.
– This is another way to measure moles.
– For gases, the volume is proportional to the number of moles.
• 11.2 L of a gas at STP = 0.500 mole
– 44.8 L = ? moles
19
Avogadro’s Law and the
Standard Molar Volume
• Example 12-5: One mole of a gas occupies 36.5 L and
its density is 1.36 g/L at a given temperature and
pressure. (a) What is its molar mass? (b) What is its
density at STP?
? g 36.5 L 136
. g


 49.6 g / mol
mol mol
L
?g
49.6 g 1 mol


 2.21 g/L
LSTP
mol
22.4 L
20
Summary of Gas Laws:
The Ideal Gas Law
Boyle’s Law - V  1/P (at constant T & n)
Charles’ Law – V  T (at constant P & n)
Avogadro’s Law – V  n (at constant T & P)
Combine these three laws into one statement
V  nT/P
• Convert the proportionality into an equality.
V = nRT/P
• This provides the Ideal Gas Law.
PV = nRT
•
•
•
•
• R is a proportionality constant called the universal gas constant.
21
Summary of Gas Laws:
The Ideal Gas Law
• We must determine the value of R.
– Recognize that for one mole of a gas at 1.00 atm, and 273
K (STP), the volume is 22.4 L.
– Use these values in the ideal gas law.
PV 1.00 atm  22.4 L
R =

nT 1.00 mol 273 K 
L atm
 0.0821
mol K
22
Summary of Gas Laws:
The Ideal Gas Law
• R has other values if the units are changed.
• R = 8.314 J/mol K
– Use this value in thermodynamics.
• R = 8.314 kg m2/s2 K mol
– Use this later in this chapter for gas velocities.
• R = 8.314 dm3 kPa/K mol
– This is R in all metric units.
• R = 1.987 cal/K mol
– This the value of R in calories rather than J.
23
Summary of Gas Laws:
The Ideal Gas Law
•
Example 12-6: What volume would 50.0 g of ethane, C2H6,
occupy at 1.40 x 102 oC under a pressure of 1.82 x 103 torr?
–
1.
2.
3.
To use the ideal gas law correctly, it is very important that all of
your values be in the correct units!
T = 140 + 273 = 413 K
P = 1820 torr (1 atm/760 torr) = 2.39 atm
50 g (1 mol/30 g) = 1.67 mol
24
Summary of Gas Laws:
The Ideal Gas Law
nR
T
T
n
R
V
=
V= P
P
LLatm


atm
11..67


mol
0
.
0821

67 mol  0.0821mol K 413
413KK

mol K


22..39
39atm
atm
 23.6 L
25
Summary of Gas Laws:
The Ideal Gas Law
• Example 12-7: Calculate the number of moles
in, and the mass of, an 8.96 L sample of
methane, CH4, measured at standard
conditions.
You do it!
26
Summary of Gas Laws:
The Ideal Gas Law
1.00 atm 8.96 L
PV
n =

 0.400 mol CH 4
L atm 
RT 
 0.0821
  273 K

mol K
16.0 g
? g CH 4  0.400 mol 
 6.40 g
mol
27
Summary of Gas Laws:
The Ideal Gas Law
• Example 12-8: Calculate the pressure exerted by 50.0
g of ethane, C2H6, in a 25.0 L container at 25.0oC.
You do it!
n = 1.67 mol and T = 298 K
nRT
P=
V
L atm 

1.67 mol  0.0821
298 K 
mol K 

P
25.0 L
P  1.63 atm
28
Determination of Molecular Weights and
Molecular Formulas of Gaseous Substances
• Example 12-9: A compound that contains only
carbon and hydrogen is 80.0% carbon and
20.0% hydrogen by mass. At STP, 546 mL of
the gas has a mass of 0.732 g . What is the
molecular (true) formula for the compound?
• 100 g of compound contains 80 g of C and 20
g of H.
29
Determination of Molecular Weights and
Molecular Formulas of Gaseous Substances
CC
11mol
C
1mol
mol
? mol
C
atoms===
80.0gggCC
 6..67
mol
CC C
? mol
C
80.0

67
mol
mol
Catoms
atoms
80.0
C
6.67
mol
12.0
CC
12.0
gg C
12.0
g
1
mol
H
1
mol
H
? mol
HHatoms
.19
.8 mol
? mol
atoms == 20.0
20.0 ggHH
 19
8 mol
H H
1.01
gH
1.01
gH
Determine the smallest w hole number ratio.
19.8
 3  the empirical formula is CH 3 with mass = 15
6.67
30
Determination of Molecular Weights and
Molecular Formulas of Gaseous Substances
Remember, the molar mass is the
mass divided by the number of moles.
0.732 g
g
n=
 30.0
0.0244 mol
mol
actual mass
30.0

2
empirical mass 15.0
Thus the molecular formula is
the empirical formula doubled.
CH3 2  C 2 H 6
31
Determination of Molecular Weights and
Molecular Formulas of Gaseous Substances
• Example 12-10: A 1.74 g sample of a
compound that contains only carbon and
hydrogen contains 1.44 g of carbon and
0.300 g of hydrogen. At STP 101 mL of the
gas has a mass of 0.262 gram. What is its
molecular formula?
You do it!
32
Determination of Molecular Weights and
Molecular Formulas of Gaseous Substances
1 mol C
? mol C atoms = 1.44 g C 
 0.120 mol C
12.0 g C
1 mol H
? mol H atoms = 0.300 g H 
 0.297 mol H
1.01 g H
0.297
 2.5  C 2 H 5 with mass = 29
0.120

PV
1.00 atm 0.101 L 
n=

 0.00451 mol
L atm 
RT 
 0.0821
273 K 
mol K 

33
Determination of Molecular Weights and
Molecular Formulas of Gaseous Substances
?g
0.262 g

 58.1 g/mol
mol 0.00451 mol
58.1
 2  C 2 H 5 2  C 4 H10
29
34
Dalton’s Law of Partial Pressures
• Dalton’s law states that the pressure exerted
by a mixture of gases is the sum of the partial
pressures of the individual gases.
Ptotal = PA + PB + PC + .....
35
Dalton’s Law of Partial Pressures
• Example 12-11: If 1.00 x 102 mL of hydrogen,
measured at 25.0 oC and 3.00 atm pressure, and 1.00
x 102 mL of oxygen, measured at 25.0 oC and 2.00
atm pressure, were forced into one of the containers
at 25.0 oC, what would be the pressure of the
mixture of gases?
PTotal  PH 2  PO 2
 3.00 atm + 2.00 atm
= 5.00 atm
36
Dalton’s Law of Partial Pressures
• Vapor Pressure is the pressure exerted by a
substance’s vapor over the substance’s liquid
at equilibrium.
37
Dalton’s Law of Partial Pressures
• Example 12-12: A sample of hydrogen was collected
by displacement of water at 25.0 oC. The
atmospheric pressure was 748 torr. What pressure
would the dry hydrogen exert in the same container?
PTotal  PH 2  PH 2O  PH 2  PTotal  PH 2O
PH 2  748  24 torr = 724 torr
PH 2O from table
38
Dalton’s Law of Partial Pressures
• Example 12-13: A sample of oxygen was
collected by displacement of water. The
oxygen occupied 742 mL at 27.0 oC. The
barometric pressure was 753 torr. What
volume would the dry oxygen occupy at
STP?
You do it!
39
Dalton’s Law of Partial Pressures
V1  742 mL
V2  ?
T1  300 K
T2  273 K
P1  753  27 = 726 torr P2  760 torr
273 K 726 torr
V2  742 mL 

 645 mL @ STP
300 K 760 torr
40
Mass-Volume Relationships
in Reactions Involving Gases
•In this section we are looking at
reaction stoichiometry, like in Chapter 3,
just including gases in the calculations.
MnO 2 &
2 KClO 3(s) 
 2 KCl(s) + 3 O2 (g)
2 mol KClO3 yields 2 mol KCl and 3 mol O2
2(122.6g) yields
2 (74.6g) and 3 (32.0g)
Those 3 moles of O2 can also be thought of as:
3(22.4L)
or 67.2 L at STP
41
Mass-Volume Relationships in
Reactions Involving Gases
• Example 12-14: What volume of oxygen
measured at STP, can be produced by the
thermal decomposition of 120.0 g of KClO3?
You do it!
42
Mass-Volume Relationships in
Reactions Involving Gases
? LSTP O 2  120.0 g KClO 3 
1 mol KClO 3
22.4 LSTP O 2
3 mol O 2


122.6 g KClO 3 2 mol KClO 3
1 mol O 2
? LSTP O 2  32.9 LSTP O 2
43
The Kinetic-Molecular Theory
• The basic assumptions of kinetic-molecular
theory are:
• Postulate 1
– Gases consist of discrete molecules that are
relatively far apart.
– Gases have few intermolecular attractions.
– The volume of individual molecules is very small
compared to the gas’s volume.
• Proof - Gases are easily
compressible.
44
The Kinetic-Molecular Theory
• Postulate 2
– Gas molecules are in constant, random, straight
line motion with varying velocities.
• Proof - Brownian motion displays molecular
motion.
45
The Kinetic-Molecular Theory
• Postulate 3
– Gas molecules have elastic collisions with
themselves and the container.
– Total energy is conserved during a collision.
• Proof - A sealed, confined gas exhibits no
pressure drop over time.
46
The Kinetic-Molecular Theory
• Postulate 4
– The kinetic energy of the molecules is
proportional to the absolute temperature.
– The average kinetic energies of molecules of
different gases are equal at a given
temperature.
• Proof - Brownian motion increases as
temperature increases.
47
The Kinetic-Molecular Theory
• The kinetic energy of the molecules is
proportional to the absolute temperature. The
kinetic energy of the molecules is proportional to
the absolute temperature.
• Displayed in a Maxwellian distribution.
48
The Kinetic-Molecular Theory
• The gas laws that we have looked at earlier in this chapter are proofs that
kinetic-molecular theory is the basis of gaseous behavior.
• Boyle’s Law
– P  1/V
– As the V increases the molecular collisions with container walls
decrease and the P decreases.
• Dalton’s Law
– Ptotal = PA + PB + PC + .....
– Because gases have few intermolecular attractions, their pressures
are independent of other gases in the container.
• Charles’ Law
– VT
– An increase in temperature raises the molecular velocities, thus the V
increases to keep the P constant.
49
Real Gases:
Deviations from Ideality
•
•
•
Real gases behave ideally at ordinary
temperatures and pressures.
At low temperatures and high pressures real
gases do not behave ideally.
The reasons for the deviations from ideality are:
1. The molecules are very close to one another, thus
their volume is important.
2. The molecular interactions also become important.
50
Real Gases:
Deviations from Ideality
• van der Waals’ equation accounts for the
behavior of real gases at low temperatures and
high pressures.

n 2a 
V  nb  nRT
P +
2 
V 

• The van der Waals constants a and b take into
account two things:
1. a accounts for intermolecular attraction
2. b accounts for volume of gas molecules
• At large volumes a and b are relatively small
and van der Waal’s equation reduces to ideal
gas law at high temperatures and low
pressures.
51
Real Gases:
Deviations from Ideality
•
What are the intermolecular forces in
gases that cause them to deviate from
ideality?
1. For nonpolar gases the attractive forces
are London Forces.
2. For polar gases the attractive forces are
dipole-dipole attractions or hydrogen
bonds.
52