Transcript Document 7356074

Chapter 5
Gases
5.1 Pressure
Properties of a gas
-
Uniformly fills any container.
Mixes completely with any other gas
Exerts pressure on its surroundings.
Compressible
Gas pressure varies with altitudes and
storms
Measuring atmospheric pressure
Torricellian barometer



Torricelli (1608-1647) studied the problem using
mercury rather than H2O.
Mercury is denser than water, so the column wasn’t
quite so high.
Gas Pressure
Pascal (SI units)
Force (N)
P (Pa) =
Area (m2)

Liquid Pressure = g ·h ·d
Pascal and Torricelli
Blaise Pascal
(1623-1662)
Evangelista Torricelli
(1608-1647)
Barometer
760 mmHg
atmospheric
pressure
P = d·g·h
d - density
g - acc. of
gravity
h
atmospheric
pressure
Units of Pressure
One atmosphere (1 atm)
Is the average pressure of the atmosphere
at sea level
Is the standard atmospheric pressure
Standard Atmospheric Pressure:
1 atm = 76 cm Hg = 760 mm Hg = 760 Torr =
101,325 Pa
Very small unit, thus it is not
commonly used
Example
A. What is 475 mm Hg expressed in atm?
485 mm Hg
x
1 atm
= 0.625 atm
760 mm Hg
B. The pressure of a tire is measured as 29.4 psi.
What is this pressure in mm Hg?
29.4 psi x 1.00 atm x 760 mmHg = 1.52 x 103 mmHg
14.7 psi 1.00 atm
Manometer
Device for Measuring the Pressure of a Gas in a Container
5.2 The Gas Laws of Boyle, Charles
and Avogadro
 Boyle’s
Law:
PV = const
 at constant n, T
 Charles’ Law:
V/T = const
 at constant n, P
 Avogadro’s Law:
V/n = const
 at constant P, T
Boyle’s Law
1
V
p
Slope=
1/k
Boyle's Law
0.012
Pressure
(Torr)
1/Pressure
(1/Torr)
0.01
PV =k
(at constant
T and n)
Boyle's Law
800
0.008
700
0.006
600
500
0.004
400
0.002
300
0
200
P1V1 = P2V2
0
50
100
100
Volume (L)
150
200
0
0
50
100
Volume (L)
150
200
A Plot of PV Versus P for Several Gases at
Pressure Below 1 atm
Boyle’s holds
Only at very
Low pressures
A gas strictly
obeys Boyle’s
law is called
Ideal gas
Charles’s Law
•V/T = b
•V = bT
(constant P & n)
•V1/T1 = V2/T2
Charles' Law
35
Volume (L)
30
25
20
15
10
5
0
0
100
200
300
Temperature (K)
400
500
Plots of V Versus T(Celsius) for Several Gases
Volume of a gas
Changes by 1
273
When the temp.
Changes by
1oC.
I.e., at -273oC ,
V=0 ???
All gases will solidify or liquefy before reaching zero volume.
• Vn
• V = an
Avogadro’s Law
(constant P& T)
V V

n n
Avogadro's Law
120
Volume (L)
100
80
60
40
20
0
0
1
2 moles 3
4
5
1
2
1
2
5.3 The Ideal Gas Law
k
V 
p
V  bT
V  an
Tn
V  R( )
P
Boyle’s law
Charles's law
Universal gas
constant
Avogadro’s law
PV  nRT
The Ideal Gas Law
PV = nRT
R = 0.0821 atm L mol-1 K-1
V nR P nR


T
P T V
V RT

n
P
The Ideal Gas Law can be used to derive
the gas laws as needed!
Molar Volume
At STP
4.0 g He
1 mole
(STP)
V = 22.4 L
16.0 g CH4
1 mole
(STP)
44.0 g CO2
1mole
(STP)
V = 22.4 L
V = 22.4 L
The value of R
 What
is R for 1.00 mol of an ideal gas at
STP (25 oC and 1.00 atm)?Given that
V of 1 mol of gas at STOP= 22.4L
PV
PV  nRT R 
nT
1.00 atmatm
22.4.L
L
R

R  01..0821
00 mol (273K )
mol.K
Example
A
reaction produces enough CO2(g)
to fill a 500 mL flask to a pressure of
1.45 atm at a temperature of 22 oC.
How many moles of CO2(g) are
produced?
PV
n
PV = nRT
RT
mol K
0.500 L
1.45 atm 

0.0821atm L 295 K
The Ideal Gas Law: Final and initial state problems
PV
 nR  Constant
T
PV
PV

T
T
1
1
1
2
2
2
(Ideal gas equation)
Ideal Gas Law
 The
ideal gas law is an equation of state.
 Independent of how you end up where
you are at. Does not depend on the path.
 The state of the gas is described by: P, V,
T and n
 Given 3 you can determine the fourth.
 Ideal gas equation is an empirical
equation - based on experimental
evidence.
Example
A sample of helium gas has a volume of
0.180 L, a pressure of 0.800 atm and a
temperature of 29°C. What is the new
temperature(°C) of the gas at a volume of
90.0 mL and a pressure of 3.20 atm?
Data Table
Set up Data Table
P1 = 0.800 atm
V1 = 0.180 L
T1 = 302 K
P2 = 3.20 atm
V2= 90.0 mL
T2 = ??
??
Solution
Solve for T2
Enter data
T2 = 302 K x
T2 =
atm x
atm
K - 273 =
mL =
mL
°C
K
Calculation
Solve for T2
T2 = 302 K x 3.20 atm x 90.0 mL = 604 K
0.800 atm 180.0 mL
T2 = 604 K - 273 = 331 °C
Example
A gas has a volume of 675 mL at 35°C and
0.850 atm pressure. What is the
temperature in °C when the gas has a
volume of 0.315 L and a pressure of 802
mm Hg?
Solution
T1 = 308 K
T2 = ?
V1 = 675 mL
V2 = 0.315 L = 315 mL
P1 = 0.850 atm
= 646 mm Hg
P2 = 802 mm Hg
T2 = 308 K x 802 mm Hg x
646 mm Hg
= 178 K - 273 = - 95°C
315 mL
675 mL
5.4 Gas Stoichiometry
 Reactions
happen in moles
 At Standard Temperature and Pressure
(STP, 0ºC and 1 atm) 1 mole of gas
occuppies 22.4 L.
 If not at STP, use the ideal gas law to
calculate moles of reactant or volume
of product.
Example
A.What is the volume at STP of 4.00 g of CH4?
4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L
16.0 g CH4
1 mole CH4
B. How many grams of He are present in 8.0 L of gas
at STP?
8.00 L x
1 mole He x
22.4 He
4.00 g He = 1.43 g He
1 mole He
Example
A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC.
Determine the pressure in the cylinder.
P = nRT
V
PV = nRT
= (3.74 mol)(0.082L•atm)(297.5K)
(12.25 L)
mol•K
P= ?
V = 12.25 L
n=
75.5 g
mol
= 3.74 mol
20.18 g
R = 0.082 L•atm
mol•K
T = 24.5 + 273 = 297.5 K
= 1.009 atm
= 5670 torr
Example
 30.2
mL of 1.00 M HCl are reacted
with excess FeS. What volume of
gas is generated at STP?
STP means standard temperature and
pressure . . . 0 oC and 1 atm.
2 HCl
HCl
+ FeS
+ FeS

FeCl
FeCl
+ 2H
S2S
2 +2 H
Now . . . Go for moles,
2 HCl + FeS  FeCl2 + H2S

1.00 mol HCl mol H 2S  0.0821 atm L 
mol
HCl
 0.0302 L 111..00
.00
00mol
molHCl
HCl  mol
molH
H222SS  0.0821 atm L 273.15 K 
 2 mol HCl  mol K 273.15 K 
000..0302
.0302
0302LLL
L
22mol
molHCl
HCl  mol K
LLL


1.00 atm
 0.339 L
PV = nRT
V = nRT/P
Example
The decomposition of sodium azide, NaN3, at
high temperatures produces N2(g). What
volume of N2(g), measured at 735 mm Hg and
26°C, is produced when 70.0 g NaN3 is
decomposed.
2 NaN3(s) → 2 Na(l) + 3 N2(g)
2 NaN3(s) → 2 Na(l) + 3 N2(g)
Determine moles of N2:
nN2 = 70 g NaN3 X
1 mol NaN3
X
3 mol N2
= 1.62 mol N2
65.01 g N3/mol N3 2 mol NaN3
Determine volume of N2:
nRT
(1.62 mol)(0.08206 L atm mol-1 K-1)(299 K)
V=
=
P
1.00 atm
(735 mm Hg)
760 mm Hg
= 41.1 L
Molar mass of a gas
PxV=
P x
V=
nxRxT
m
n
M
mxRxT
M
 m = mass, in grams
 M = molar mass, in g/mol
 Molar mass = m R T
PV
Density
 Density
(d) is mass divided by volume
P x V =
mxRxT
M
P
=
d
mxRxT
V xM
=
P
=
m
V
dxRxT
M

PM
d
RT
Example
A glass vessel weighs 40.1305 g when clean, dry and
evacuated; it weighs 138.2410 when filled with water
at 25°C (d=0.9970 g cm-3) and 40.2959 g when filled
with propylene gas at 740.3 mm Hg and 24.0°C.
What is the molar mass of polypropylene?
PV  nRT
mRT
M
PV

m
PV 
RT
M
Volume of the vessel
Vflask
m
d
H 2O
=
138.2410g – 40.1305g
(0.9970 g cm )
-3
H 2O
= 98.41
cm3 = 0.09841 L
mgas = mfilled - mempty = (40.2959 g – 40.1305 g)
= 0.1654 g
PV = nRT
M=
m
RT
PV =
M
m RT
M=
PV
(0.6145 g)(0.08206 L atm mol-1 K-1)(297.2 K)
(0.9741 atm)(0.09841 L)
M = 42.08 g/mol
Example
Calculate the density in g/L of O2 gas at STP.
From STP, we know the P and T.
P = 1.00 atm
T = 273 K
Rearrange the ideal gas equation for moles/L
d=
PXM
RxT
PXM
d
RXT
g
(1.00 atm ) X (32.0
)
mol
d

L.atm
(0.0821
)X (273K)
mol.K
1.43g / L
The density of O2 gas at STP is
1.43 grams per liter
Example

2.00 g sample of SX6(g) has a volume of
329.5 Cm3 at 1.00 atm and 20oC. Identify the
element X. Name the compound
 P= 1.00 atm
1L
V  392.5 Cm3X
 0.3295L
1000Cm3
T
= 273+20 = 293K
m RT
M=
PV
L.atm
(2.00 g )(0.0821
)(293K )
mol.K
MM 
(1.00atm)(0.3295L)
= 146 g SX6 /mol
f
Molar mass of (X6 )= 146- 32 = 114 g/mol
Molar mass of X = (114 g/mol X6) /6 = 19
X = with a molar mass of 19 = F
The compound is SF6
5.5 Dalton’s Law of Partial Pressures
 For
a mixture of gases in a container,
the total pressure is the sum of the
pressure each gas would exert if it were
alone in the container.
 The total pressure is the sum of the
partial pressures.
 PTotal = P1 + P2 + P3 + P4 + P5 ...
 For
each gas:
nRT
P
V
•Partial pressure
–Each component of a gas mixture exerts a pressure
that it would exert if it were in the container alone
 PTotal
= n1RT + n2RT + n3RT +...
V
V
V
 In
the same container R, T and V are
the same.
 PTotal
Thus,
= (n1+ n2 + n3+...)RT
V
P
Total
RT
n ( )
V
Total
A 250.0 mL flask contains 1.00 mg of He and 2.00 mg
of H2 at 25.0oC. Calculate the total gas pressure in the
flask in atmospheres.
The total pressure is due to the partial pressures of each
of these gases.
so:
Ptotal  PHe  PH 2
 RT 
 (n He  n H 2 )

 V 
For He:
1.00
x 10-3 g He mol = 2.50 x 10-4 mol He
_____________________
4.00 g
For H2:
2.00
x 10-3 g H2
mol
______________________
= 9.92 x 10-4 mol H2
2.016 g
A 250.0 mL flask contains 1.00 mg of He and and 2.00
mg of H2 at 25.0oC. Calculate the total gas pressure in
the flask in atmospheres.
RT 

so: Ptotal  PHe  PH  (n He  n H )

2
For He:
For H2:
2
 V 
1.00
x 10-3 g He mol = 2.50 x 10-4 mol He
_____________________
4.00 g
2.00
x 10-3 g H2
mol
______________________
= 9.92 x 10-4 mol H2
2.016 g
And: Ptotal = (2.50 x 10-4 + 9.92 x 10-4)(RT/V)
= (0.001242 mol)(0.0821 L•atm)(25 + 273)K
mol•K (0.2500 L)
Ptotal= 0.1216 atm
A 250.0 mL flask contains 1.00 mg of He and and 2.00
mg of H2 at 25.0oC. Calculate the total gas pressure in
the flask in atmospheres.
RT 

so: Ptotal  PHe  PH  (n He  n H )

2
For He:
For H2:
2
 V 
1.00
x 10-3 g He
mol = 2.50 x 10-4 mol He
_____________________
4.00 g
2.00
x 10-3 g H2
mol = 9.92 x 10-4 mol H2
______________________
2.016 g
Calculate the pressure due just to He (???):
n He RT = 0.0245 atm
PHe 
V
and Phydrogen= ?
0.1216 - 0.0245 = 0.0971 atm
Magnesium is an active metal that replaces hydrogen
from an acid by the following reaction:
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
How many g of Mg are needed to produce 5.0 L of H2 at a
temperature of 25 oC and a pressure of 745 mmHg?
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
?g
5.0 L
Hint: find moles of H2 using PV = nRT then work as a
stoichiometry problem.
n = PV =____________________________________
745 mmHg 5.0 L
mol•K
RT
62.4 L•mmHg 298 K
n = 0.20 mol
The mole fraction
 Mole
fraction: number of moles of one
component in a mixture relative to the
total number of moles in the mixture
 symbol
is Greek letter chi
n
c 
n
1
1
Total
c
n
=
n  n  n  ....
1
1
2
3
Mole fraction expressed in pressures
n
n
c

n
n  n  n  ....
V
V
n  P ( ); n  P ( ); ..........
RT
RT
1
1
Total
1
1
1
2
2
3
2
P(
V
)
RT
n
c 

V
V
V
n
P ( )  P ( )  P ( )  .....
RT
RT
RT
1
1
1
Total
1
2
3
P
P
c 

( P  P  P  ....) P
1
1
1
1
2
3
Total
P cP
1
1
Total
c 
1
V
P( )
RT
1
V
( )( P  P  P  ....)
RT
1
2
3
n
P
c 

n
P
1
1
Total
Total
1
Example

A 1.00 L sample of dry air at 786 Torr and
25 oC contains 0.925 g N2 plus other
gasses (such as O2, Ar and CO2.) a) What
is the partial pressure of N2? b) What is
the mole fraction of N2?
mol
0.925 g 
 0.0330 mol
28.0 g
0.0330 mol 0.0821atm L mol 1 K 1 298 K   0.807 atm
1.00 L
613Torr
760 Torr
 0.780
0.807 atm 
 613 Torr X N 2 
786 Torr
atm
Collecting gas over water
An insoluble gas is passed into a container of water, the
gas rises because its density is much less than that of
water and the water must be displaced
KClO3
O2 gas
Collection of Gases over Water
Assuming the gas is saturated with water vapor, the
partial pressure of the water vapor is the vapor
pressure of the water.
Ptotal = Pgas + PH2O(g)
Pgas = Ptotal – PH2O(g)
Example
Oxygen gas generated was collected over water. If the
volume of the gas is 245 mL and the barometric pressure
is 758 torr at 25oC, what is the volume of the “dry” oxygen
gas at STP? (Pwater = 23.8 torr at 25oC)
PO2 = PTotal - Pwater = (758 - 23.8) torr = 734 torr
P1V1T2
P1V1 P2V2
P1= PO2 = 734 torr;
V2 


P2= SP = 760. torr
P2T1
T1
T2
V1= 245mL
T1= 298K;
T2= 273K;
V2 = (245mL)(734torr)(273K)
V2= ?
(298K)(760.torr)
=217 mL
5.6 The Kinetic Molecular Theory of Gases


It explains why ideal gases behave the way they do.
Assumptions are made to simplify the theory, but
don’t work in real gases.
 Postulates
of the kinetic Theory:
 Gas particles (atoms or molecules) are
so small compared with distances
between them, thus we can ignore their
volume.
 The particles are in constant motion
and their collisions with walls cause
pressure exerted by the gas
Kinetic Molecular Theory
 The particles do not affect each other,
neither attracting or repelling
 The average kinetic energy (KE ) is
proportional to the Kelvin temperature.
 KE = 1/2 mu2
m=mass of gas particle
 v=average velocity of particles

KMT explains ideal gas laws
• P&V: P = (nRT) . (1/V)  P 1/V
– # collisions increases when V decreases
• P & T: P = (nR/V).T  P T
– When T increases hits with walls become stronger and more frequent
• V & T: V=(nR/P).T  V  T
– When T increases hits with walls become stronger and more frequent.
To keep P constant, V must increase to compensate for particles speeds
• V & n: V= (RT/P). N  V  n
– When n increase P would increase if the volume to be kept constant. V
must increase to return P to its original value
• Dalton’s law: Individual particles are independent of each
other and their volumes are negligible. Thus identities of gas
particles do not matter
Driving the ideal gas law from KMT
• The following expression was derived for pressure of an
ideal gas:
2 nN A (1 / 2mu 2 )
P [
]
3
V
N  Avogadro' s number
A
m  mass of each particle
u  average of square velocities of particles
2
(KE)  N (1/2mu )
2
avg
A
2  n(KE) 
P  [
]
3  V

avg
PV 2
 ( KE )  T
n
3
avg
PV
T
n
PV
RT
n
The meaning of temperature
PV
RT
n
( KE)
avg
=
2
( KE )
3
3
RT
=
2
avg
Root mean square velocity
Combine
these two equations
= NA(1/2 mu 2 )
 (KE)avg = 3/2 RT
 (KE)avg
 Where
M is the molar mass in kg/mole,
and R has the units 8.3145 J/Kmol.
 The velocity will be in m/s
Molecular speed for same gas at two different temperatures
3RT
u (
)
M
1
1/ 2
T1
3RT
u (
)
M
2
1/ 2
T2
u
T
( )
u
T
T1
1
T2
2
1/ 2
u
TM
(
)
u
TM
T1
1
2
T2
2
1
1/ 2
Molecular speed for two different gases at two different
temperatures
3RT
u (
)
M
3RT
u (
)
M
1/ 2
1
T1
1
2
1/ 2
T2
2
u
TM
(
)
u
TM
T1
1
2
T2
2
1
1/ 2
Effects of Molar Mass on urms
At constant T, 273 K, the
most probable speed for
O2 > CH4 > H2
urms  M–1/2
so smaller molar masses
result in higher molecular
speeds
EOS
Range of velocities
 The
average distance a molecule
travels before colliding with another is
called the mean free path and is small
(near 10-7)
 Temperature is an average. There are
molecules of many speeds in the
average.
Effects of Temperature on urms
At constant mass the most
probable speeds for O2
increase with temperature
urms  T1/2 so higher
temperatures result in
higher molecular speeds
Summary of Behaviors
EOS
number of particles
273 K
Molecular Velocity
number of particles
273 K
1273 K
Molecular Velocity
number of particles
273 K
1273 K
1273 K
Molecular Velocity
Molecular Velocity
 Average
increases as temperature
increases.
 Spread increases as temperature
increases.
 Smaller molar masses result in higher
molecular speeds
 Passage
of gas through a small hole,
into a vacuum.
 The effusion rate measures how fast
this happens.
 Graham’s Law the rate of effusion is
inversely proportional to the square
root of the mass of its particles.
5.7 Effusion and Diffusion
 Passage
of gas through a small hole,
into a vacuum.
 The effusion rate measures how fast
this happens.
 Graham’s Law : the rate of effusion is
inversely proportional to the square
root of the mass of its particles.
Rate of effusion for gas 1
M2

Rate of effusion for gas 2
M1
Deriving
 The
rate of effusion should be
proportional to urms

Effusion Rate 1
=
Effusion Rate 2
effusion rate 1

effusion rate 2
u rms 1

u rms 2
urms for gas 1
urms for gas 2
3RT
M1

3RT
M2
M2
M1
Diffusion
 The
spreading of a gas through a room.
 Slow considering molecules move at
100’s of meters per second.
 Collisions with other molecules slow
down diffusions.
 Best estimate is Graham’s Law.
Diffusion
 The
spreading of a gas through a room.
 Slow considering molecules move at
100’s of meters per second.
 Collisions with other molecules slow
down diffusions.
 Best estimate is Graham’s Law.
5.8 Real Gases
 Real
molecules do take up space and
they do interact with each other
(especially polar molecules).
 Need to add correction factors to the
ideal gas law to account for these.
Volume Correction
 The
actual volume free to move in is
less because of particle size.
 More molecules will have more effect.
 Corrected volume V’ = V - nb
 b is a constant that differs for each gas.
nRT
P' 
V  nb
Pressure correction
 Because
the molecules are attracted to
each other, the pressure on the
container will be less than ideal
 depends on the number of molecules
per liter.
 since two molecules interact, the effect
must be squared.
Pressure correction
Because the molecules are attracted
to each other, the pressure on the
container will be less than ideal
 depends on the number of molecules
per liter.
 since two molecules interact, the
effect must be squared

Pobserved
n 2
 P ' a ( )
V
Van der Wall’s equation
2

n

 Pobs + a    x  V - nb  nRT
V 

Corrected
Pressure
Corrected
Volume
a
and b are determined by experiment.
 Different for each gas.
 Bigger molecules have larger b.
 a depends on both size and polarity.
Example
 Calculate
the pressure exerted by
0.5000 mol Cl2 in a 1.000 L container at
25.0ºC
 Using the ideal gas law.
 Van der Waal’s equation


a = 6.49 atm L2 /mol2
b = 0.0562 L/mol