1.5 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

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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.


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.


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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


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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:


[ ]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.


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.


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Fig. 1.10

At the Boyle temperature (B=0), a gas behaves nearly ideally over a range of pressures.


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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.


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Fig 1.11P- -T surface for a one-component system that contracts on freezing.


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.


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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(*).


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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 −

−=


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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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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.


( ) 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


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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.


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


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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


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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:


( ) 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


( )( )

( ) 2 and x-12

1 and x-11 , 10

2

23-2

2

2-

=

=

=

=

=

=

=

0x

0x

dxfd

dxfd

dxdf

dxdff

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• 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 + ...


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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.


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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


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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.


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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.


( ) 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.


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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:


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:


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 ==


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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)


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

1.5 Real Gases and the Virial Equation -...

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