real gases deviation from ideal gas law: -gas particles have volume - attraction exists between gas...
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Real Gases
Deviation from ideal gas Law:-Gas particles have volume
- Attraction exists between gas particles (liquefication)
Ideal gas law applies only at low gas density:
high T, low p
TRpVn
VvolumemolarV
TRn
VpTRnpV
mm
isotherm
Van der Waal’s Equation
Ideal gas law needs modification:
- Let’s consider the molecular volume:
b: size of 1 mole of gas molecules
For 1 mole of gas: Vm,total = Vm,space + b
bVbVV
VV
mtotalmidealm
spacemidealm
,,
,,
- Let’s consider the attraction forces between
molecules:
- Due to attraction: - velocity decreased, - colliding less with wall, - pressure drops
- How big is this drop in pressure?
attraction force
1f2f3f
1cFattr
1V2V
attraction force
1f2f3f
2cFattr
1V2V
- How big is this drop in pressure?
21 ccFattr
ccc 212cFattr
V
nc
2
2
V
nFattr
22222
2 111
mVnVnVV
n 2
1
mattr V
F
- How big is this drop in pressure?
drop in pressure proportional to attraction force
drop in pressure indirectly proportional to square of Vm
2mV
constppressureindrop
2m
ideal
realideal
idealreal
idealreal
V
constpppp
ppp
ppp
pp
RTVp idealmideal ,
RTbVV
ap m
m
2
n
VVm
RTnbnVV
nap
2
2
VdW Equation
of state
b: volume of 1 mole of gas molecules.
a: represents strength of attraction.
Maxwell construction
a1 = a2
Many other equations of state Mostly empirical
Virial Equation
midealmideal V
RTV
RTp
1
,
to correct, develop in a series (additional correction terms:
...1
...1
32
32
mmm
mmm
V
C
V
B
VRTp
V
CRT
V
BRT
VRTp
B,C, …: second, third, … virial coefficients: exp. determined
Compression factor z
describes deviation from ideal gas behavior.
idealm
m
V
Vz
,
p
RTVTRVp mm
RT
Vpz m
For ideal gas, z=1),,( TVpfz m
At a given temperature, z depends on the nature of gas
Why?
Let’s find for a VdW gas at p → 0
The slope of the curve z=f(p) in the previous diagram.
Tp
z
RTbVV
ap m
m
2
2mm V
a
bV
RTp
2mm V
a
bV
RTp
mm
m
m
mm
m
VTR
a
bV
Vz
RT
V
V
a
bV
RT
RT
Vpz
2
m
m
mm
m
VTR
a
VbVTR
a
bV
Vz
1
1
1...11
1 32
xforxxxx
1mV
bx
...11
132
mmm
m
V
b
V
b
V
b
Vb
mmmm VTR
a
V
b
V
b
V
bz
...1
32
TmTm
T
m
m
V
z
RTVRTz
p
z
V
TRp
TR
Vpzpat
11
,1,1,0
...1
132
mmm V
b
V
b
VTR
abz
TR
ab
RTp
z
T
1lim
0p
slope at p → 0
TR
ab
RTp
z
T
1lim
0p
> 0
= 0
< 0
bR
aT
TR
ab
TR
ab
TR
ab
RT
Boyle
BoyleBoyle
Boyle
0
01
0
0
0
patpositiveslopeTR
abif
patzeroslopeTR
abif
patpositiveslopeTR
abif
b: volume of 1 mole of gas molecules.
represents repulsive forces
a: represents strength of attraction
Size effect > attraction
Size effect = attraction
Size effect < attraction
Explain the trend!
Critical Point
Vm range in which L and G phases coexist shrink to a single point
Above Tc, L and G phases can not be distinguished from each other:
Density (L) = Density (G) No Interface
b
apthatshow
Rb
aT
b
a
b
RT
b
a
b
RT
V
a
bV
RT
bVVbVVbV
V
a
bV
RT
V
a
bV
RT
V
p
V
a
bV
RT
V
a
bV
RT
V
p
c
ccc
cmcm
c
cmcmcmcmcm
cmcm
c
cmcm
c
Tm
cmcm
c
cmcm
c
Tm
27
27
8
27
2
427
2
40
2
323332
620
62
20
2
32323,
2,
,,,,,
4,
3,
4,
3,
2
2
3,
2,
3,
2,
002
2
TmTm V
p
V
p
2mm V
a
bV
RTp
Experimental measurements of pc and Tc are more
accurate than that of Vc, a and b are calculated from the
experimental values of pc and Tc .
!
64
27
8
22
bandaofunitstheDerive
p
TRa
p
RTb
thatshow
c
c
c
c
Is it possible to find an equation of state that doesn’t
contain material specific constants?
The law of corresponding states
Reduced properties:
cm
mrm
cr
cr V
VV
T
TT
p
pp
,,
mcmrmcrcr VVVTTTppp ,,
b
apc 27
Rb
aTc 27
8 bV cm 3,
mrmrr VbVTbR
aTp
b
ap 3
27
8
27 ,2
mrmrr VbVTbR
aTp
b
ap 3
27
8
27 ,2
2mm V
a
bV
RTp
2,,
3
13
8
rmrm
rr VV
Tp
•Two gases at the same Tr and pr have the same Vr.
•These two gases are in “corresponding states’’.
•Ar (302 K and 16 atm) ↔ ethylene (381 K and 18 atm)
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