1.5 Real Gases and the Virial Equation -...
Transcript of 1.5 Real Gases and the Virial Equation -...
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– 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
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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.
1.5 Real Gases and the Virial Equation
L+++== 2mm
mVC
VB1
RTPVZ
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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
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Fig. 1.9 Second virialcoefficient B.
1.5 Real Gases and the Virial Equation
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1.5 Real Gases and the Virial Equation
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
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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-
mol L 8.3145x10bar 600
K 600 mol K bar L8.3145x10ideal === mVP
RT
( )( ) 1.261
mol L 8.3145x10 mol L 0.02171 1 (b) 1-2-
-1=+=+=
RTBPZ
( ) ( ) 11-2 mol L 0.1048mol L8.3145x10 1.261 −− ===P
ZRTVm
1.5 Real Gases and the Virial Equation
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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:
1.5 Real Gases and the Virial Equation
[ ]L+++== 2 C' B' 1realZ PP
RTVP m
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Example 1.5 Derive the relationships between two types of virialcoefficients.
Answer: The pressures can be eliminated from equations using the following forms:
K+++= 3m
2mm V
CRTVBRT
VRTP ( )
K++
= 3
mm
2
VRT2B
VRTP
22
( )KK +++
+=+++= 2
m
''
m
'''
VRTC BRTB
VRTBPCPBZ
22 11
( )2RTCRTBBCRTBB
''
'
+=
=
( )2
2
/
RTBCC
RTBB'
'
−=
=
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Compression Factor ZThe 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.
1.5 Real Gases and the Virial Equation
RTPVZ m=
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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 = 0at 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
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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.
1.5 Real Gases and the Virial Equation
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Fig. 1.10
At the Boyle temperature (B=0), a gas behaves nearly ideally over a range of pressures.
1.5 Real Gases and the Virial Equation
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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
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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
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Paths on Surface
Sections through the surface at constant temperature give the isotherms, and the isobars shown in here too.
1.6 P- -T Surface for a one-component systemV
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Fig 1.11P- -T surface for a one-component system that contracts on freezing.
1.6 P- -T Surface for a one-component systemV
V
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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
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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.
1.7 Critical Phenomena
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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(*).
1.7 Critical Phenomena
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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.
2mm Va
bVRTP −
−=
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van der Waals equation
The surface of possible states allowed by the van der Waals equation. Compare this surface with that shown before.
2mm Va
bVRTP −
−=
1.8 The van der Waals Equation
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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
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1.8 The van der Waals Equation
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);
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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/Vm2, constant a may be determined from
the P-V-T data.
1.8 The van der Waals Equation
( ) RTbVV
aPor mm
=−
+ 2
2mm
2
Va
bVRT
VnanbV
nRT −−
=−−
=
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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 (Vm2 /P ) and rearrange as
Vm3 – (b + RT/P ) Vm
2 + (a/P ) Vm – (ab/P ) = 0
⇒ Vm = 0.298 L/mol
1.8 The van der Waals Equation
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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 similarmagnitudes, 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.
1.8 The van der Waals Equation
2mm Va
bVRTP −
−=
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van der Waals loops
The van der Waals equation predicts the real gas isotherms with oscillations below the critical temperature which is unrealistic.
023 =−++−
PabV
PaV
PRTbV MMM
1.8 The van der Waals Equation
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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 =−++−
PabV
PaV
PRTbV MMM
1.8 The van der Waals Equation
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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:
1.8 The van der Waals Equation
( ) L+++= 2xx1x-1
1
mmmm
mmVa
VbVa
bVVPVZ
RT-
/ - 11
RT-
-RT===
L+
+
+=
21RT
-1mm Vb
VabZ
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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 canoften be expressed as series by use of the Maclaurin series
Ans: In this case,
( ) ( ) .... 2!1 0 2
2
2+
+
+=
==x
dxfdx
dxdffxf
0x0x
1.8 The van der Waals Equation
( )( )
( ) 2 and x-12
1 and x-11 , 10
2
23-2
2
2-
=
=
=
=
=
=
=
0x
0x
dxfd
dxfd
dxdf
dxdff
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1.8 The van der Waals Equation
• 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+
+
+=
21RT
-1mm Vb
VabZ
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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 + ...
1.8 The van der Waals Equation
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• let (1 / RT) = βZ = 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.
1.8 The van der Waals Equation
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ExerciseGiven 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
1.8 The van der Waals Equation
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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.
1.8 The van der Waals Equation
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Critical Constants from vdwIsotherms.As T/TC → 1, the vdwisotherm becomes the critical isotherm with a flat inflexion when both the first and second derivatives are zero.
1.8 The van der Waals Equation
( ) 0V
a2bV
RTVp
32 =+−
−=
∂∂
= cc
c
TT c
( ) 0V
a6bV
RT2VP =−
−=
∂
∂
=4
c3
c
c
TT2
2
c
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Critical Constants at inflexion pointAt inflexion point, the critical constants can be derived from the VDW equation and related to the VDW coefficients.
1.8 The van der Waals Equation
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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:
1.8 The van der Waals Equation
2cc
cc V
abV
RTP −−=
( ) 0V6a
bV2RT
Vp
4C
3C
C
T2
2
c
=−−
=
∂∂
( )
0V2a
bVRT
Vp
3C
2C
C
cT
=+−
−=
∂∂
2mm Va
bVRTP −
−=
==
==⇒
3V
P8RTb
VRT89
P64TR27a
c
c2c
c
2c
2
c
c
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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:
1.8 The van der Waals Equation
38
89
6427 2
22
c
c
c
ccc
c
VP
RTb
VRTPTRa
==
==
==
=
==
⇒
278
3
27
89
8
2ba
PRTP
bVRba
VRaT
c
cc
c
cc
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Critical Constants:
3bcV =
227ba
cP =
27Rb8a
cT =
83
cRTcVcPcZ ==
1.8 The van der Waals Equation
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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 accordingto (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/molb) VDW equation:
70 = (0.083145) (350 K)/( Vm - 0.0638) - 5.562/ Vm2
solve a cubic equation using successive approximations, startingwith ideal gas molar volume:
This yields Vm = 0.23 L/mol (Comparing with the real root: 0.2297 L/ mol)
1.8 The van der Waals Equation
2mm Va
bVRTP −
−=
bVaP
RTV2
m
m ++
=
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Reduced Variables PlotThe compression factors of four gases, plotted using reduced variables. The use of reduced variables organizes the data on to single curves.
cPP
rP =cVV
rV m=cTT
rT =
Thermodynamics: State of a Gas