Chapter 5 Gases. 2 Early Experiments Torricelli performed experiments that showed that air in the...
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Transcript of Chapter 5 Gases. 2 Early Experiments Torricelli performed experiments that showed that air in the...
Chapter 5
Gases
2
Early Experiments
Torricelli performed experiments that showed that air in the atmosphere exert pressure.
Torricelli designed the first barometer.
3
A torricellian barometer
4
Simple manometer
5
Unit of Pressure
1 standard atmosphere
=1 atm
=760 mm Hg
=760 torr
=101325 Pa
2
force F NPressure pascal
area A m
6
Boyle’s Experiment
7
Boyle’s Law : PV=k A gas that obeys Boyle’s law is called an ideal
gas
8
Charles’s Law: V=bT
The volume of a gas at constant pressure increases linearly with the temperature of the gas.
9
Plot V vs. T using kelvin scale
10
Avogadro’s Law : V=an A gas at constant temperature and pressure
the volume is directly proportional to the number of moles of gas.
11
Ideal Gas Law
Equation of state for a gas PV=nRT P: atm V: Liter n: moles R: 0.082 L atm K-1
T: K
12
Laws for Gas experiments
Boyle’s Law : PV=k Charles’s Law : V=bT Avogadro’s Law : V=an The ideal Gas Law : PV=nRT
(so called equation of state for idea gas)
13
Ideal Gas The volume of the individual particles can
be assumed to be negligible. The particles are assumed to exert no
force on each other. It expresses behavior that real gases
approach at low pressure and high temperature.
14
Gas StoichiometryStandard Temperature and Pressure (STP) T=0oC P=1 atm V=22.4 L
Natural Temperature and Pressure (NTP) T=25oC P=1 atm V=24.5L
15
Plot of PV versus P for 1 mol of ammonia.
16
17
2NaN3(s)→2Na(s)+3N2(g)
18
Dalton’s Law of Partial Pressures
total 1 2 3
31 2
1 2 3
total
P =P +P +P +......
n RTn RT n RT = + + +....
V V VRT
=(n +n +n ...)( )V
RT =n ( )
V
19
Mole Fraction and Partial Pressure
toraltotal
total
total
PxPP
P
PPP
Px
RTVPRTVPRTVP
RTVP
n
nx
RT
VPn
RT
VPn
RT
VPn
nnn
n
n
nx
111
321
11
321
111
2211
321
111
...
...)/()/()/(
)/(
..... ),( ),()(
...
20
The Model of Ideal Gas in Kinetic-Molecular Theory
The volume of the individual particles can be assumed to be negligible.
The particles are assumed to exert no force on each other.
The particles are in constant motion. The average kinetic energy of a collection of
gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.
21
An ideal gas particle in a cube wholse sides are of length L (in meters).
22
The velocity u can be broken down into three perpendicular components, ux, uy, and u2.
23
Pressure of an Ideal Gas
Let the container be a rectangular box with
sides of length Lx, Ly and Lz.
Let v be the velocity of a given molecule.
2 2 2 2
2 2 21 1 1
2 2 2: translational energy
x y z
tr x y z
tr
v v v v
mv mv mv
24
In the xy plane,vx
2 + vy2 = vyx
2 by the Pythagorean theorem.
25
26
A molecule collide with wall W where W is parallel to the xz plane. Let i have the velocity components vx, vy, vz
2
1
2
1
2
1
2
2
1221
t
t xxxw
t
t xx
t
t xxx
xx
xxxx
dtFmv so Fives Fhird law gNewton's t
dtFmv
dtF)(tP)(t P to tg from tIntegratin
dtFPmomentum ddt
dP
dt
dvmmaF
27
The integration can be extended over the whole time interval t1 to t2.
x
xw
xx
x
xw
x
xxx
w
t
twx
l
vmNF
N
vv
l
mvF
v
ltttvl
F
dttFtt
FttFmv
222
2
12
1212
2 ,
Won wall exerted force average theis where
])(1
[ )(22
1
速度平方之平均值
28
The translational energy Etr= 1/2mv2
2 2 21
22
3
2
3
trtr
tr
tr tr
tr
m v vm
PV N
N is the total translation kinetic energy E of gas
PV E
V
vTherefore
vvvv
V
vmN
lll
vmN
ll
lv
mN
ll
F
A
FP
zyx
x
xyz
x
yz
x
x
yz
ww
3
mNP
2
2222
22
2
single particle
29
Temperature dependence with translation kinetic energy
0
0
2 3
3 23 3
2 2
3
2
3 ( )
2'
tr tr
tr tr
tr
PV E nRT E nRT
NE N nRT RT
N
RT
N
kT translational kinetic energy of a molecule
k Boltzmann s constant
single particle
one mole of particles
30
2
2
12 2
1 3 =
2 23
- -
3 3
:
:
tr
rms
rms
m v kT
kTv
m
v is called root mean square speed v
kT RTv
m Mm mass of one gas molecule
M molar mass of the gas
31
Distribution of Molecular Speeds in an Ideal Gas
Root mean square speed is assumed that all
molecules move at the same speed.
The motions of gas molecules should have
distribution of molecular speeds in
equilibrium.
32
( )
( )
( ) :
:
v
v
Use the distribution function G v
dNG v dv
NG v the fraction of molecules with speed
in the range v to v dv
dNthe probability that a molecule will have
Nits speed between v and v dv
33
2
2
3
222
-
( ) 42
.
mvv kT
cv
dN mG v dv e v dv
N kT
Maxwell distribution laws
The function with Ae form is called
normal or gaussian distribution
指數函數→機率
34
Plot of O2 molecules
35
Plot of N2 molecules
36
Application of The Maxwell Distribution
23
232
0 0
3 1
2 2
2
( ) 42
1 8 8 4 or
2 2( )2
8
mv
kT
average speed
mv vG v dv e v dv
kT
m kT kTv
mkT m mkT
RTv
M
37
Application of The Maxwell Distribution
1.225:1.128:1382
20
M
RT:
πM
RT:
M
RT:vv:v
M
RTs v one find
dv
dG(v)when
peedprobable smost
rmsmp
mp
38
Velocity distribution for nitrogen
39
Collisions with a Wall-2 -11
(collisions cm s )wdN
A dt,
,
,
0
0
( ) ( )
:
( ) ( )
( )
1( )
v yy y v y y y
yy
y
y yv y y y
y y
wy y y
y
wy y y
dNg v dv dN Ng v dv
Nv dt
the fraction of molecules within distance v dt of Wl
v dt vdN Ng v dv dt
l l
dN Ng v v dv dt
A l A
dN Ng v v dv
A dt V
vydt
ly
dNw:粒子撞擊牆壁的數目
40
21 1
2 22
0 0
1
2
0
0
1
20
1( )
2 2 4
1
2
1 1 1 8
4 4
ymvy kT
y y y yy
w
w
v m RTg v dv e v dv v
l kT M
dN N RT
A dt V M
PNN NPV nRT RT
N V RT
dN PNN RTv
A dt V RT M
41
3 3 8.314 2981928( / )
0.002
8 8 8.314 2981776( / )
0.002
2 2 8.314 2981574( / )
0.002
rms
mp
RTv m s
M
RTv m s
M
RTv m s
M
H2 at 25oC and 1atm
分子量單位 :Kg
42
1
20
23 11
3 1 -1
24 2 1
1 1 1 8
4 4
1 (1 )(6.02 10 )1776 100
4 (82.06 )(298 )
4.4 10
wdN PNN RTv
A dt V RT M
atm molcm s
cm atm mol K K
cm s
43
Definition of Pressure The pressure of a gas results from collisions
between the gas particles and the walls of the container.
Each time a gas particle hits the wall, it exerts a force on the wall.
An increase in the number of gas particles in the container increases the frequency of collisions with the walls and therefore the pressure of the gas.
44
The effusion of a gas into an evacuated chamber.
45
Suppose there is a tiny hole of area A in the wall
and that outside the container is a vacuum.
Escape of a gas through a tiny hole is called
effusion.
01
2
2
1
(2 )
'
1
2
A PN AdN
dtMRT
Graham law of effusion
Rate of effusion for gas M
Rate of effusion for gas M
collisions × area
46
Diffusion
Relative diffusion rates of NH3 and HCl molecules
47
Molecules Collisions and Mean Free Path
Intermolecular collisions are important in reaction kinetics.
Assume a molecule as a hard sphere. No intermolecular forces exist except at the
moment of collision.zAA: the number of collisions per unit time
that one particular A molecule makes with
other A molecule [collisions s-1]
48
Cylinder swept by gas particles
1( )
2 A B A Bd d r r
49
Calculate zAA and zAB
2
:
,
( ) .
AB
AB
cyl AB A B
v average speed of A relative to B
In time dt the moving molecule will travel a
distance v dt and will sweep out a cylinder
of volume V v dt r r
(rA+rB=d)
50A B
51
Since the stationary B molecules are uniformly distributed throughout the container volume V, the number of B molecules with centers in the cylinder equals (Vcyl/V)NB.
2
2
( ) ( ) ( )
( )
cyl B BAB AB A B AB
cyl AB A B
V N Nz z r r v
V dt V
V v dt r r
52
2 2
2
12 2 2 2
12 2
1 1 12 2 02 2 2
8 8
( ) ( )
( ) [ ] ( )
8 1 1( ) [ ( )] ( )
82 2 ( )
AB A BA B
BAB A B AB
BAB A B A B
BAB A B
A B
AAAA A A A
A
RT RTv v v
M M
Nz r r v
V
Nz r r v v
V
RT Nz r r
M M V
If B A
P NN RTz d v d
V M RT
53
Mean Free Path The average distance of a molecule travels
between collisions. In a mixture of gases A and B, A differs from
B.
:
:
: ( )
A
A
AA AB
average speed of A v
distance v t
number of collisions z z t
54
average distance traveled by an A molecule
between collisions
( )A
AAA AB
v
z z
In pure gas A, there are no A-B collisions, zAB=0
1 12 2 02 2
1 1
2 ( ) 2
A
AA
v RT
z PNNd d
V
55
Diameter d of H2 in the hard-sphere is 1.48 Å
56
1
9 1
5
1776 100
4.3 10
4.1 10 4100A
AAA
AA
v cm s
z collisions s
cm
57
Real Gas No gas exactly follows the ideal gas law. Compression factor (壓縮因子 )
Z(P,T)=PV/nRT
Z<1, P<Pid, V<Vid strong intermolecular attraction
Z=1, P→0 and T→∞
Z>1, P>Pid, V>Vid strong intermolecular repulsion
58
59
60
Van der Waals equation
2( )( ) ( )
a: the effect of intermolecular attractive forces on the gas pressure
b: nonzero volume of molecules and volume
exclusive by intermolecular repulsive force
a VP V b RT V
nV
61
62
63
64
Calculate pressure for 1 mole of CO2 at 0oC in containers with 22.4 L
P = 0.995 atm
Use idea gas equation
Use van der Waals equation
65
Calculate pressure for 1 mole of CO2 at 0oC in containers with 0.2 L
Use idea gas equation
Use van der Waals equation
P = 52.6 atm
66
Calculate pressure for 1 mole of CO2 at 0oC in containers with 0.05 L
Use idea gas equation
Use van der Waals equation
P = 1620 atm
67
Analysis of the van der Waals Constants~a constant
The a constant corrects for the force of attraction between gas particles.
attraction between particles↑ a ↑ As the force of attraction between gas particles
becomes stronger, we have to go to higher temperatures for the molecules in the liquid to form a gas.
Gases with very small values of a, such as H2 and He, must be cooled to almost absolute zero before they condense to form a liquid.
68
Analysis of the van der Waals Constants~b constant
a rough measure of the size of a gas particle the volume of a mole of Ar atoms is 0.03219 L r = 2.3 x 10-8 cm
69
Chemistry in the Atmosphere
70
The variation of temperature and pressure with altitude.
71
Concentration (in molecules per million molecules of “air”) of some smog components versus time of day.
72
Diagram of the process for scrubbing sulfur dioxide from stack gases in power plants.