Real Gases and the Virial Equation

2005-9-23 66
– The compressibility factor Z depends not only on temperature,
but also the pressure.
– In the case of nitrogen gas at100 bar, as the temperature is
reduced, the effect of intermolecular attraction increases because
the molar volume is smaller at lower temperatures and the
molecules are closer together.
– If the temperature is low enough, all gases show a minimum in
the plot of Z vs. P. Hydrogen and helium exhibit this minimum
only at temperature well below 0 ℃. Why?
1.5 Real Gases and the Virial Equation

2005-9-23 67
The Virial Equation: (Virial: derived from the Latin word for ‘force’):
• In 1901 H. Kamerlingh-Omnes proposed an equation of state for
real gases in Z as a power series in terms of 1/Vm for a pure gas:
For over wider range of P, add more terms in power series (1/Vm).
B and C: the second and third virial coefficients, and both are
temperature dependent, but not on the pressure.
L+++== 2
mm
m
V
C
V
B
1
RT
PV
Z

2005-9-23 68
Table 1.1 Second virial Coefficients, B
35014.1H2
1160-15.8Ar
1200-16.1O2
Table 1.1 Second and Third virial Coefficients at 298.15 K
1550-8.6CO
1100-4.5N2
12111.8He
C/10-12 m6 mol-2B/10-6 m3 mol-1Gas

2005-9-23 69
Fig. 1.9 Second virial
coefficient B.

2005-9-23 70
11.9-21.7Ar
Temperature
13.810.4Ne
12.9-22.0O2
Table 1.1’ Second virial Coefficients, B (10-6 m3 mol-1)
-19.6-153.7Xe
21.7-10.5N2
10.412.0He
13.7H2
600 K273 K

2005-9-23 71
Example Comparison of the ideal gas law with the virial equation
What is the molar volume of N2(g) at 600 K and 600 bar according to
(a) the ideal gas law and (b) the virial equation?
Answer: (a) The virial coefficient B of N2(g) at 600 K is 0.0217 L mol-1
[ ] ( )( ) 1-2-
1-1-2-
molL8.3145x10
bar600
K600molKbarL8.3145x10
ideal === mV
P
RT
( )
( )
1.261
molL8.3145x10
molL0.0217
11(b) 1-2-
-1
=+=+=
RT
BP
Z
( )( ) 11-2 molL0.1048molL8.3145x101.261 −− ===
P
ZRT
Vm

2005-9-23 72
The virial Equation:
It is more convenient by using P as an independent variable.
P Vm[real] = RT (1 + B’ P + C’ P 2 + ... )
or
B’ and C’: both coefficients are temperature dependent;
usually B’P ≫ C’P 2. For over wider range of P, just add more
terms in power series of P:
[ ] L+++== 2C'B'1
real
Z PP
RT
VP m

2005-9-23 73
Example 1.5 Derive the relationships between two types of virial
coefficients.
Answer: The pressures can be eliminated from equations using
the following forms:
K+++= 3
m
2
mm V
CRT
V
BRT
V
RT
P ( ) K++





= 3
mm
2
V
RT2B
V
RT
P
22
( ) KK +
+
+





+=+++= 2
m
''
m
'''
V
RTCBRTB
V
RT
BPCPBZ
2
2
11
( )2
RTCRTBBC
RTBB
''
'
+=
=
( )2
2
/
RT
BC
C
RTBB
'
'
−
=
=

2005-9-23 74
Compression Factor Z
The variation of the compression
factor, with pressure
for several gases at 0 °C. A
perfect gas has Z = 1 at all
pressures. Notice that, although
the curves approach 1 as p → 0,
they do so with different slopes.
RTPVZ m=

2005-9-23 75
Boyle Temperature TB
• A plot of Z vs. P with a slope of dZ /dP :
dZ/dP = B’ + 2CP’ + ... (The slope is B’ at P = 0).
• Although for a real gas Z ~ 1 at P ~ 0, the slope of Z against P does
not approach zero.
• The properties of real gas may not always coincide with the perfect
gas values at low P.
• There may be a temperature at which Z ~ 1 with zero slope B = 0
at P ~ 0, this temperature is called the Boyle temperature, TB .
• At the Boyle temperature TB , the properties of real gas do coincide
with the perfect gas values as P ~ 0.
• Since the second virial coeff. B vanished and the third virial coeff. C
and higher terms are negligibly small, then PVm ~ RTB over a more
extended range of P than at other temperature.
Boyle Temperature

2005-9-23 76
Boyle Temperature TB
The compression factor Z
approaches 1 at low
pressures, but does so with
different slopes. For a perfect
gas, the slope is zero, but
real gases may have either
positive or negative slopes,
and the slope may vary with
temperature. At the Boyle
temperature, the slope is zero
and the gas behaves perfectly
over a wider range of
conditions than at other
temperatures.

2005-9-23 77
Fig. 1.10
At the Boyle temperature
(B=0), a gas behaves nearly
ideally over a range of
pressures.

2005-9-23 78
9950.2520.0725113.0405.6NH3
0.2300.056220.5647.1H2O
0.274
0.269
0.275
0.288
0.290
0.306
0.301
Zc
73.8
103.0
77.0
50.5
34.0
13.0
2.27
Pc/bar TB/KTc/K
0.094
0.127
0.124
0.0734
0.0895
0.0650
0.0573
Vc/L mol-1
22.645.24He
584Br2
417Cl2
Gas
714.81304.2CO2
405.88154.6O2
327.22126.2N2
110.0433.2H2
Table 1.2 Critical Constants and Boyle Temperatures

2005-9-23 79
P, ,T Surface Plot
A region of the P, ,T
surface of a fixed amount
of perfect gas. The points
forming the surface
represent the only states
of the gas that can exist.
1.6 P- -T Surface for a one-component systemV
V
V

2005-9-23 80
Paths on Surface
Sections through the surface
at constant temperature give
the isotherms, and the
isobars shown in here too.

2005-9-23 81
Fig 1.11
P- -T surface
for a one-
component
system that
contracts on
freezing.
V

2005-9-23 82
Fig. 1.12 The Critical Point
Pressure-molar volume
relations (e.g. isotherms) in
the region of the critical point.
The dashed horizontal lines in
the two-phase region are
called tie lines.
1.7 Critical Phenomena

2005-9-23 83
Fig. 1.12 The Critical Point
The path 1-2-3-4 shows how
a liquid can be converted to a
gas without the appearance of
a meniscus. If liquid at point
4 is compressed isothermally,
the volume decreases until
the two-phase region is
reached. At this point there is
a large decrease in volume at
constant pressure (the vapor
pressure of the liquid) until
all of the gas has condensed
to liquid. As the liquid is
compressed, the pressure
rises rapidly.

2005-9-23 84
The Critical Point
Experimental isotherms of carbon
dioxide at several temperatures.
The `critical isotherm`, the
isotherm at the critical
temperature, is at 31.04 °C. The
critical point is marked with a
star(*).

2005-9-23 85
1.8 The van der Waals Equation
• For real gas, van der Waals equation is written as
where a and b are van der Waals coefficients, specific to each
gas. Term a is adjusted to represent the attractive forces of
the molecules, without giving any specific physical origin to
these forces; V-nb, not V, now represents the volume in
which molecules can move.
2
mm V
a
bV
RT
P −
−
=

2005-9-23 86
van der Waals equation
The surface of possible
states allowed by the van der
Waals equation. Compare
this surface with that shown
before.
2
mm V
a
bV
RT
P −
−
=

2005-9-23 87
a and b : van der Waals coefficients
The VDW coefficients a and b are found by fitting the
calculate curves to experimental curves. They are
characteristic of each gas but independent of the
temperature.
5.164.137Xe
2.380.0341He
4.293.610CO2
3.201.337Ar
b / 10-2 L mol-1a / atm L2 mol-2
Table 1.3 van der Waals Constants

2005-9-23 88
The van der Waals Equation (1873):
• Virial Equation are not convenient for too many constants and all
are temperature dependent.
• Van der Waals modified the ideal gas law by taking account two
effects: a). intermolecular forces, and b). molecular size.
• VDW equation has only two constants a, b. And both are
independent of temperature.
• Only “free volume” follows ideal gas law; p above ideal value by
size effect, P = RT / (Vm – b),
• Constant b as parameter for excluded volume,
b ~ 4x(molecular size);

2005-9-23 89
The van der Waals Equation (1873):
• Real P are less than ideal value by the attraction force from
interior of the gas (internal pressure), pressure on wall depends
on both collision frequency and collision force, both related to
the molar concentration (n/V = 1/Vm), so P reduced by a
correction term which is square of molar conc. a/Vm
2
P [real] = P – correction term
• Also ( ∂ P/ ∂ T)v = R/(Vm – b), constant b may be determined
from the slope of P vs. T at constant V.
• T ( ∂ P/ ∂ T)v = P + a/Vm
2, constant a may be determined from
the P-V-T data.
( ) RTbV
V
a
Por m
m
=−







+ 2
2
mm
2
V
a
bV
RT
V
n
a
nbV
nRT −
−
=−
−
= 








2005-9-23 90
Example Calculate the molar volume of Ar in 373 K and 100 bar
by treating it as a VDW gas.
(a = 1.35 L2 bar / mol2 and b = 0.0322 L / mol2 for Ar)
Ans: multiply by (Vm
2 /P ) and rearrange as
Vm
3 – (b + RT/P ) Vm
2 + (a/P ) Vm – (ab/P ) = 0
⇒ Vm = 0.298 L/mol

2005-9-23 91
Liquefaction and the VDW Equation
•The ideal gas isotherms are only at high temperature and large molar
volumes (Vm ≫ b).
•VDW loop: the calculate oscillations below the critical temperature.
•First term from kinetic energy and repulsive force, second term for the
attractive effect. VDW loop occurs when both terms have similar
magnitudes, liquids and gases coexist when cohesive and dispersing effects
are in balance.
•Maxwell construction: horizontal line drawn with equal areas above and
below the line at VDW loop.
2
mm
V
a
bV
RT
P −
−
=

2005-9-23 92
van der Waals loops
The van der Waals equation
predicts the real gas
isotherms with oscillations
below the critical temperature
which is unrealistic.
023 =−++− 























P
ab
V
P
a
V
P
RT
bV MMM

2005-9-23 93
Fig.1.13 van der Waals loops
Isotherms calculated from the
van der Waals equation. The
dashed line is the boundary
of the L + G region.
023 =−++− 























P
ab
V
P
a
V
P
RT
bV MMM

2005-9-23 94
Z for the VDW gas:
• Compressibility factor Z = P Vm/RT
• At moderate pressure, b/Vm < 1; expanding the first term using:
1/(1– b/Vm) = 1 + (b/Vm) + (b/Vm)2 + (b/Vm)3 + ...
substituted for Z, then yield the virial equation in terms of volume:
( )
L+++= 2xx1
x-1
1
mmmm
mm
V
a
VbV
a
bV
VPV
Z
RT
-
/-1
1
RT
-
-RT
===
L+





+





+=
2
1
RT
-1
mm V
b
V
a
bZ

2005-9-23 95
Example 1.6 Expansion of (1-x)-1 using the Maclaurin series
Since we will use series like (1-x)-1 = 1 + x + x2 + …
a number of times, it is important to realize that functions can
often be expressed as series by use of the Maclaurin series
Ans: In this case,
( ) ( ) ....
2!
1
0 2
2
2
+





+





+=
==
x
dx
fd
x
dx
df
fxf
0x0x
( )
( )
( ) 2andx-12
1and
x-1
1
,10
2
2
3-
2
2
2-
=





=





=





=





=
=
=
0x
0x
dx
fd
dx
fd
dx
df
dx
df
f

2005-9-23 96
• the second virial coefficient B = b – a/RT, C = b 2
the value of a is relatively more important at low T,
the value of b is relatively more important at high T.
• To obtain the virial equation in P , replace (1/Vm) = P / RTZ and
Z = 1 + B’ P + C’ P 2 + ...
= 1 + (b – a/RT) (1/ RTZ) P + (b / RTZ)2 P 2 + ...
= 1 + B’ P + C’ P 2 + ...
• In the limit of zero pressure, Z = 1 and
B’ = (1/RT) (b – a/RT) = B / RT
L+





+





+=
2
1
RT
-1
mm V
b
V
a
bZ

2005-9-23 97
Z for the VDW gas:
Z = 1 + B’(1/Z) P + (b / RTZ)2 P 2 + ...
= 1 + B’ P + C’ P 2 + ...
now B’ (1/Z – 1) P + (b / RTZ)2 P 2 = C’ P 2
or B’ [ (1–Z) /Z] (1/ P ) + (b / RTZ)2 = C’
• in the limit of zero pressure, Z = 1 and (Z–1)/ P = B’
C’ = (b / RT)2 – B’2 = (C – B2) (1 / RT)2
C’ = a (1 / RT)3 (2b – a/RT)
Z = 1 + (1 / RT) (b – a/RT) P + a (1 / RT)3 (2b – a/RT) P 2 + ...

2005-9-23 98
• let (1 / RT) = β
Z = 1 + β (b – a/RT) P + a β3 (2b – a β ) P 2 + ...
(dZ / dP ) = β (b – a β ) + 2 a β3 (2b – a β ) P + ...
• At P = 0, the initial slope
( ∂ Z/ ∂ P )p=0 = β (b – a β )
= (1 / RT) (b – a/RT)
= b (1/RT) – a (1/RT)2
• if bRT < a, attractive force dominated, (negative slope)
• if bRT > a, repulsive force dominated, (positive slope)
• at Boyle temperature, the second term vanished, B = 0;
bRT = a, then TB = a/bR
• the Boyle temperature is relate to the VDW coefficients.

2005-9-23 99
Exercise
Given the van der Waals constants for ethane gas as a = 5.562 L2 bar/mol2, b = 0.06380
L/mol, for 10.0 mol of ethane at 300 K and under 30 bar,
(a) find the second virial coefficient B at this temperature.
(b) calculate the compressibility factor Z from the first two terms.
(c) estimate the approximate molar volume from Z.
(d) what is its Boyle temperature TB?
Ans:
(a) B = b - a/RT
= 0.06380 L/mol - [5.562 L2 bar/mol2 /(0.0831451 L-bar/K-mol * 300 K)]
= -0.159 L/mol
(b) Z = 1 + B (p/RT) + ...
= 1 + (-0.159 L/mol) [30 bar / (300 K*0.0831451 L-bar/K-mol)]= 0.81
(c) V/n = ZRT/p
= (0.81) (0.0831451 L-bar/K-mol) (300 K) / 30 bar = 0.67 L/ mol
(d) TB = a/bR
= 5.562 L2 bar/mol2 / (0.06380 L/mol * 0.0831451 L-bar/K-mol)
= 1049 K

2005-9-23 100
vdw Isotherms
Van der Waals isotherms at
several values of T/Tc . The
van der Waals loops are
normally replaced by
horizontal straight lines.
The critical isotherm is the
isotherm for T/Tc = 1.
For T < TC, the calculate
isotherms oscillates from
min → max and converges
as T ⇒ TC then coincide at
T = TC with flat inflection.

2005-9-23 101
Critical Constants from vdw
Isotherms.
As T/TC → 1, the vdw
isotherm becomes the critical
isotherm with a flat inflexion
when both the first and second
derivatives are zero.
( )
0
V
a2
bV
RT
V
p
32
=+
−
−=





∂
∂
= cc
c
TT c
( )
0
V
a6
bV
RT2
V
P
=−
−
=





∂
∂
=
4
c
3
c
c
TT
2
2
c

2005-9-23 102
Critical Constants at inflexion point
At inflexion point, the critical
constants can be derived from the
VDW equation and related to the
VDW coefficients.

2005-9-23 103
Example 1.7 VDW constants expressed in terms of critical constants
The van der Waals equation may be written as
Differentiating wrt molar volume and evaluating these equations
at the critical point to obtain a and b in terms of Tc , Pc or Tc , Vc
Answer:
2
cc
c
c
V
a
bV
RT
P −
−
=
( )
0
V
6a
bV
2RT
V
p
4
C
3
C
C
T
2
2
c
=−
−
=





∂
∂
( )
0
V
2a
bV
RT
V
p
3
C
2
C
C
cT
=+
−
−=





∂
∂
2
mm V
a
bV
RTP −
−
=






==
==
⇒
3
V
P8
RT
b
VRT
8
9
P64
TR27
a
c
c
2
c
c
2
c
2
c
c

2005-9-23 104
Example 1.8 Critical constants expressed in terms of VDW constants.
Derive the expression for the molar volume (Vc) temperature (Tc)
and pressure (Pc) at the critical point in terms of the van der Waals
constants.
Answer:
38
8
9
64
27 2
22
c
c
c
cc
c
c
V
P
RT
b
VRT
P
TR
a
==
==







==
=
==
⇒
278
3
27
8
9
8
2
b
a
P
RT
P
bV
Rb
a
VR
a
T
c
c
c
c
c
c

2005-9-23 105
Critical Constants:
3bcV =
2
27b
a
cP =
27Rb
8a
cT =
8
3
cRT
cVcP
cZ ==

2005-9-23 106
Example 1.9 Calculation of the molar volume using the VDW equation.
What is the molar volume of ethane at 350 K and 70 bar according
to (a) the ideal gas law and (b) the van der Waals equation?
(a = 5.562 L2 bar / mol2 and b = 0.0638 L / mol for C2H6)
Ans:
a) Vm=RT/P=(0.083145 L bar/K mol)(350 K)(70 bar)=0.416 L/mol
b) VDW equation:
70 = (0.083145) (350 K)/( Vm - 0.0638) - 5.562/ Vm
2
solve a cubic equation using successive approximations, starting
with ideal gas molar volume:
This yields Vm = 0.23 L/mol (Comparing with the real root: 0.2297
L/ mol)
2
mm V
a
bV
RTP −
−
=
b
V
aP
RTV
2
m
m +
+
=

2005-9-23 107
Reduced Variables Plot
The compression factors of four
gases, plotted using reduced
variables. The use of reduced
variables organizes the data on
to single curves.
cP
P
rP =
cV
V
rV m=
cT
T
rT =
Thermodynamics: State of a Gas

Real Gases and the Virial Equation

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