Transcript Chapter 10: Gases - John A. Ferguson Senior High School

CHM 1045: General Chemistry and
Qualitative Analysis
Unit 6:
Gases & The Kinetic
Molecular Theory
Dr. Jorge L. Alonso
Miami-Dade College –
Kendall Campus
Miami, FL
Textbook Reference:
•Module # 8
Gases
Characteristics of Gases
Condensed phases
• Unlike liquids and solids, gases . . . .
 Are highly compressible.
 Expand to fill their containers.
 Have extremely low densities.
Gases
Characteristics of Gases
• Variables affecting the behavior of gases




Amount = number of moles ()
Pressure (P)
Volume (V)
Temperature (T in Kelvin)
{PropGases*}
Gases
Pressure
• Pressure is the
amount of force
applied to an area.
F
P=
A
=
105
Newtons
meter2
• Atmospheric
pressure is the
weight of air per unit
of area.
Force = mass x acceleration
Newton = 1kg . m/sec2
105 Newtons = (104kg)(10 m/sec2)
Approx. 12 miles
= 101.325 kPa
Gases
Units of Pressure
Torricelli’s
Atmosphere
1.00 atm
= 760 mm Hg (torr)
= 101.325 kPa
760 mm Hg =
weight of
equal surface
area of the
atmosphere
(Normal atmospheric
pressure at sea level).
Gases
Barometer
33 ft H2O =
weight of
equal surface
area of the
atmosphere
Gases
Manometer
instrument used to measure the difference in pressure
between atmospheric pressure and that of a gas in a vessel.
{Manometer}
Gases
Manometer
Used to measure the difference in pressure between
atmospheric pressure and that of a gas in a vessel.
Pgas = 760 torr
Pgas = 760 + 6 torrs
Gases
Pgas = 760 - 6 torrs
Gas Laws
Variables affecting gases: moles (η), pressure (P), volume (V) and
Temperature (T)
1. Boyle’s Law
 Compared: P versus V ( & T are held constant).
2. Charles’s Law
 Compared: V versus T ( & P are held constant).
3. Avogadro’s Law
 Compared: V versus η (P & T are held constant).
4. Combined Gas Law
 Compared: P vs V vs. T ( is held constant).
5. Ideal Gas Law
 Compared: P vs V vs. η vs T (no variable held constant).
6. Dalton’s Law of Partial Pressure
 Compared: individual pressures of gases in a mixture
Gases
Boyle’s Law: Pressure-Volume Relationship
( & T are held constant).
2x
V
1
P
V?
Vx½
The volume of a fixed quantity of gas at constant
temperature is inversely proportional to the pressure.
{Boyle’s Law}
Gases
P & V: inversely
proportional
V
1
P
Also,
OR
V=
k
P
P↑V ↓ = k
This means a plot of V
versus 1/P will be a
straight line.
{PV.Graphs}
Gases
Boyle’s Law

Gases
Charles’s Law: Temp. – Volume Relationship
( & P are held constant).
V T
V?
V2x
T2x
• The volume of a fixed amount of gas at constant
pressure is directly proportional to its absolute
Gases
temperature.
{*Charles’s Law Liq N2}
Charles’s Law
• The volume of a gas is directly proportional to its
absolute temperature.
V T
V=kT
or
V =k
T
A plot of V versus T will be a straight line.
Gases
Charles’s Law
Gases
{AvogLaw}
Avogadro’s Law: Moles-Volume Relationship
(P & T are held constant).
Vn
V?
V2x
2x
• The volume is directly proportional to the
number of moles of the gas.
Gases
{Avogadro’s Law}
Avogadro’s Law
• Mathematically, this means
or,
Vn
V=kn
Gases
{*Avogadro’s Law in Reactions}
Standard Temperature & Pressure
(STP) and Molar Volume
• Standard Temperature: 00C or 273 K
• Standard Pressure: 760 torr (1 atm)
At STP the Molar Volume of any gas is 22.4 L
(11.1 in)3 or
(28.2cm)3
1 mole = 6.022 x 1023 part. = gMM = 22.4 L
Gases
Standard Temperature & Pressure
(STP) and Molar Volume
At STP the Molar Volume of any gas is 22.4 L
1 mole = 6.023 x 1023 part. = gMM = 22.4 L
H2 = 2.0g
O2 = 32.0g
CO2 = 44.0 g
Problem: At STP, what volume in mL would 75g of CO2 occupy?
 1  CO 2   22.4 L   1000 mL 
4

3.8
x
10
mL
 
 

? mL  75 g CO2 
 44.0 g   1  CO 2   1 L 
Gases
Ideal-Gas Equation
The Gas Laws:
V  1/P (Boyle’s law)
V  T (Charles’s law)
V  n (Avogadro’s law)
Combining these, we get
nT
V
P
or
knT
V=
P
or
nT
k=
PV
Gases
Ideal-Gas Equation
The relationship
nT
1 =
k=
PV
R
PV
R=
nT
then becomes
PV = nRT
Gases
Ideal-Gas Equation:
Useful for pure gas under one set of conditions.
PV = nRT
Units: (torr) (L) = (mol)
{PV= nRT RapVideo}
RapVideoLinkYouTube
R
 L  torr 


 mol  K 
(K)
Gases
Ideal Gas Law
Problems
What volume (in mL) would a 2.20 g
sample of hydrogen gas (H2) at 50.00C
occupying at 443 torr?
PV = nRT
V=
V = nRT
P
L  torr 
 1  
2.20g 
  62.36 R
50.0  273 K
2.0
g
 K 


443 torr
1000 mL 
= 50.0 L 

 1L 
Gases
Ideal-Gas Equation: Densities of Gases
For Ideal Gas Equation:
PV = nRT
 1 mole 

Since mole s(n) (# g) 
 g - MW 
Then
 #g 
PV  
 RT
 MW 
and
# g RT
MW 
PV
Dividing both sides of the equation on the left by V we get
( )
#g R T
P
V MW
RT
P d
MW
Where d = Density of Gas
If we solve the equation for density, we get……..
Gases
Ideal-Gas Equation: Densities
and Molecular Weigh of Gases
Problem: What is the density of the oxygen in a tank in
an AC room (25°C) and whose pressure gauge reads
25.0 atm
P (MW)
d
RT

(25.0 atm) (32.0g/  )
 32.7 g
L
 0.0821 L  atm
 (298 K)
K  

Problem: A gas whose density is 0.0131 g/mL and is in
a container at room temperature and whose
pressure gauge reads 1.9 x 104 mmHg. What is its
MW?
 1000 mL 
g
 0.0131
 
mL  
d RT 
MW 

P
 62.36 L  torr   K  ( 298 K)

L
Gases

1.9 x 10 4 torrs
 12.8 g

Ideal-Gas Equation: Densities &
Molecular Weigh of Gases Problems
What is the density (in g/mL) of SO2 at STP?
 #g 
PV  
 RT
 MW 
PV = nRT
()
RT
P d
MW
#g R T
P
V MW
P (MW)
d
RT
=
760 torr (64.0 g  )


L  torr 
0
273
K
 62.36

0

 K 


 1L 

= 2.62 g/L 
 1000 mL 
 0.00262 g
Gases
mL
Ideal-Gas Equation: Densities &
Molecular Weigh of Gases Problems
What is the molecular weight of a gas whose density
@ STP is 7.78 g/L?
 #g 
PV  
 RT
 MW 
PV = nRT
()
RT
P d
MW
#g R T
P
V MW
(7.78 g L)  62.36 0  2730K 
 K 

 174 g 
=

d RT
MW 
P
L  torr 
760 torr
Gases
2006 A
Gases
Gases
Combined Gas Law Equation
The Gas Laws:
V  1/P (Boyle’s law)
V  T (Charles’s law)
Combining, we can get
The Combined Gas Law
P1V1
P
V
2
2
k
T1 =
T2
Useful for a constant amount of a pure gas under
Gases
two different conditions.
Combined Gas Law Equation
Constant
P1V1 =
T1
P2V2
T2
Gases
Combined Gas Law Problem
A scuba diver takes a gas filled 1.0 L balloon from the surface
where the temperature is 34 0C down to a depth of 66 ft (33 ft H2O
= 1 atm). What volume will the gas balloon have at that depth if the
temperature is 15 0C?
P1V1
T1
=
V2
P2V2
T2

1 atm1 L

3 atm ( V2)
34  273 K  15  273 K 
0
1 atm1 L 288 0 K 
307 0 K  3 atm
0
 .313L
Gases
Dalton’s Law of
Partial Pressures
• The total pressure of a mixture of gases
equals the sum of the pressures that
each would exert if it were present
alone.
• In other words,
Ptotal = P1 + P2 + P3 + …
Pair = P N2 + PO2 + PH2O + …
Gases
Partial Pressures
• When one collects a gas over water, there is water vapor mixed
in with the gas.
P of gas
P of atm
• To find only the pressure of the desired gas, one must subtract
Gases
the vapor pressure of water from the total pressure.
Vapor Pressure
of Water
• Daltons Law:
Ptotal = Pgas + PH2O
• To find only the pressure
of the desired gas, one
must subtract the vapor
pressure of water from the
total pressure.
Pgas = Ptotal - PH2O
{Press on can}
Vapor
Press
(torr)
Vapor
Press
(torr)
-10
2.15
40
55.3
0
4.58
60
149.4
5
6.54
80
355.1
10
9.21
95
634
11
9.84
96
658
12
10.52
97
682
13
11.23
98
707
14
11.99
99
733
15
12.79
100
760
20
17.54
101
788
25
23.76
110
1074.6
30
31.8
120
1489
37
47.07
200
11659
Gases
Evaporation vs Boiling in terms of
Vapor Pressure
Patm
Patm
+
+
Pvap
Patm
=
Pvap
Pvap
Vapor Pressure (v.p. or Pvap)
• Caused by the tendency
of solids & liquids to
evaporate to gaseous
form. It is temperatureGases
(K.E.) dependent.
Stoichiometry with Gases
Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)
Problem: If 2.0 g of Mg are
reacted with excess HCl, what
volume of H2 will be produced
at 250C and 775 torr? At STP?
 1 Mg   1 H 2 
 
  0.082 H 2
? η H2  2.0 g Mg 
 24.3g Mg   1 Mg 
PV = nRT
V
0.082η H 2  62.36 L o torr 298 0K 
 mol  K 
(775 - 24 torr)
 2.0 L
Gases
Kinetic-Molecular Theory
A model that aids in our understanding of what happens to gas
particles as environmental variables change.
Main Tenets:
1. Gases consist of large
numbers of molecules
that are in continuous,
random motion.
2. Collisions between gas molecules and between
gas molecules and the walls of the container must
be completely elastic (energy may be transferred
Gases
between molecules, but none is lost).
Kinetic-Molecular Theory
Main Tenets:
3. Attractive and repulsive
forces between gas
molecules are
negligible.
4. The combined volume of all the molecules of
the gas is negligible (excluded volume)
relative to the total volume in which the gas is
contained.
Gases
Kinetic-Molecular Theory
Main Tenets:
@ 100 0C
5. Energy can be transferred
between molecules during
collisions, but the average
kinetic energy of the
molecules does not change
with time, as long as the
temperature of the gas
remains constant.
6. The average kinetic energy (KE=½mv2) of the
molecules is proportional to the absolute
{KE  T(K)}
temperature.
Gases
Diffusion
Movement of
molecules from an area
of high concentration to
an area of low
concentration until
equilibrium is reached
(homogeneity).
Effusion
The escape (diffusion) of gas
molecules through a tiny hole
into an evacuated space.
Gases
Effect of Molecular Mass on
Rate of Effusion and Diffusion
Thomas Graham (1846): rate of diffusion is
inversely proportional to the square root of its
molar mass
Kinetic Energy per individual molecule:
RateA 
1
MWA
1 2
KE  m v
2
Gases
Rate of Diffusion & Effusion
Dropper
with Br (l)
Thomas Graham (1846): rate of
diffusion is inversely proportional
to the square root of its molar mass
RateA 
1
MWA
{BrDiffusion}
Comparing the rates of two gases:
Graham’s Law of Diffusion and Effusion of Gases
RateA

RateB
MWB
MWA
Rate NH 3

Rate HCl
36 g/mol
 2.1Gases
 1.4
17 g/mol
{GasDiff}
Effusion and Diffusion
• This is the most widespread uranium
enrichment method. Uranium is
reacted with fluorine to make uranium
hexafluoride gas: 235UF6 & 238UF6
• The physical principle is that the
diffusion speed of a gas molecule
depends on the mass of the
molecule: the lighter ones diffuse
faster and get through a porous
material easier.
• In gas diffusion units, uraniumhexafluoride gas diffuses through an
etched foil made of either an aluminum
alloy or teflon, due to artificially
maintained difference in pressure. The
lighter molecules (i.e. those
containing 235U) get through easier to
the other side, therefore the gas
accumulating there will be richer in
235U.
Gases
Gas Centrifugation
The gas centrifuge is essentially a bowl, in which there is a rotor spinning round at a
very high speed. The gas (UF6) directed to the centrifuge is forced to spin by the rotor.
Due to the centrifugal force the heavier molecules (those which contain 238U) will
accumulate near the wall of the bowl, while the lighter molecules containing 235UGases
will
stay closer to the center of the centrifuge.
Boltzmann Distributions
The Maxwell–Boltzmann distribution is the statistical distribution of
molecular speeds in a gas. It corresponds to the most probable speed
distribution in a collisionally-dominated system consisting of a large
number of non-interacting particles.
{Boltzman Plot}
Gases
Kinetic Energy of Gas Molecules
Kinetic Energy per individual molecule:
Kinetic Energy per mole:
1
KE  m v 2
2
3
KE  RT
2
☺
☺
Combining above equations and solving for velocity we get:
urms
3RT

Mm
☺
•
The root-mean square velocity of gases
is a very close approximation to the
average gas velocity.
• To calculate this correctly:
 The value of R = 8.314 kg m2/s2 K mol
 Mm = molar mass, and it must be in kg/mol.
Gases
The Kinetic-Molecular Theory
• Example: What is the root mean square velocity of N2
molecules at room T, 25.0oC?
urms
3RT

Mm
u rms

kg m 2 
298 K 
3 8.314 2
sec K mol


0.028 kg/mol
 515 m/s = 1159 mi/hr
• To calculate this correctly:
 The value of R = 8.314 kg m2/s2 K mol
 And M must be in kg/mol.
Gases
The Kinetic-Molecular Theory
Problem: What is the root mean square velocity of He
atoms at room T, 25.0oC?
You do it!
urms
3RT

Mm
urms
kg m 2 

3 8.314
298 K 
2
sec K mol 


0.004 kg/mol
 1363 m/s = 3067 mi/hr
• To calculate this correctly:
 The value of R = 8.314 kg m2/s2 K mol
 And M must be in kg/mol.
N2 was  515 m/s = 1159mi/hr
• Can you think of a physical situation that proves He molecules
have a velocity that is so much greater than N2 molecules?
• What happens to your voice when you breathe He(g) or SF6 (g)?
Gas MW
HeGases 4
N2
28
SF6 146
Ideal
vs.
Real
Gases
In the real world, the behavior of gases only
conforms to the ideal-gas equation at relatively high
Gases
temperature and low pressure.
Deviations from Ideal Behavior
Two particular assumptions made in the kineticmolecular model break down at high pressure
and/or low temperature:
Gases
(1) attractive forces and (2) excluded volume.
2003 A
Gases
Corrections for Non-ideal
Gas Behavior
• The ideal-gas equation can be adjusted
to take these deviations from ideal
behavior into account.
• The corrected ideal-gas equation is
known as the van der Waals equation.
Gases
The van der Waals Equation
n2a
P + 2 (V − nb) = nRT
V
(
)
Gases
Real Gases:
Deviations from Ideality
• van der Waals’ equation accounts for the
behavior of real gases at low temperatures
and high pressures.
n 2 a

 P + 2 V  n b  nRT
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
Gases
gas law at high temperatures and low
pressures.
Real Gases:
Deviations from Ideality
• Example: Calculate the pressure exerted by 84.0 g of
ammonia, NH3, in a 5.00 L container at 200. oC using
the ideal gas law.
You do it!
n = 84.0 g NH3 
nRT
P=
V

1 mol
 4.94 mol
17.0 g
4.94 mol  0.0821 L atm 473 K 

mol K 
5.00 L
 38.4 atm
Gases
Real Gases:
Deviations from Ideality
• Example: Calculate the pressure exerted by 84.0 g of
ammonia, NH3, in a 5.00 L container at 200. oC using
the van der Waal’s equation
You do it!
n = 4.94 mol
L2 atm
a = 4.17
b = 0.0371 L
2
mol
mol
nRT
n 2a
P=
 2
V - nb V
n a

P
+

V - nb   nRT
2
V 

2
L atm



4
.
94
mol
4.17 Lmolatm 

4.94 mol 0.0821 mol K 473K 
P

L
5.00 L  ( 4.94 mol )( 0.0371 mol )
5.00 L 2
2
2
2
P
191.8 L atm
 4.07 atm  (39.8 atm  4.1 atm )  35.7 atm
4.817 L
Gases
Comparedto38.4atm,thisis a 7.6%differencefromideal behavior