Van der Waals (VW) Equation of State

2. Van der Waals (VW) Equation of State

We now develop a rational approach to develop a real gas equation of state. Later we will use various equations of state in determining the stability characteristics (Chapter 10). Prior to the presentation of VW equation, Clausius I equation of state will be described because of its relevance to VW equation.

a. Clausius–I Equation of State Consider N moles of a gas that occupy a volume V in a container at some pressure and temperature. The gas molecules undergo random motion and pressure forces exist due to their impact on the walls of the container. For instance, at room temperature the force due to impact of air molecules on any surface is 10 N cm –2 . This is the average pressure experienced by the surface due to the impact of molecules that travel at velocities of approximately 350 m s –1 . If b´ denotes the volume of each molecule and N´ the number of molecules in the volume

V, then the total body volume of the molecules is given by the product N´b´. If the volume N´b´ is insignificant compared to geometrical volume V, the molecules can be assumed to be point masses, an approximation that is valid at low pressures. The volume V then denotes the free volume in which these point mass molecules can move. In case intermolecular forces are negligible (except upon impact) one can derive the ideal gas or perfect gas equation of state (cf. Chapter 1) in the form

PV = N R T, where N = N´/N Avog .

In case the geometrical volume is reduced while keeping the number of molecules unchanged so that the volume N´b´ is comparable to V, the ideal gas equation of state must be modified. In this case V–N´b´ denotes the free space available for equivalent point mass mole- cules to move randomly.

Assume, for sake of illustration, that N´ = 8 and consider a cube of side 2 σ (where σ denotes the molecular diameter) and volume V. The volume of each molecule (assumed to be

spherical) b´ = πσ 3 /6 and, if the eight molecules are tightly packed inside the cube, the free volume available to them to move around is V– 8( πσ 3 /6). However, since the intermolecular distance in the cube is 2 σ and adjacent molecules touch each other, the empty space between any two molecules is unavailable for movement. Therefore, knowing the molecular diameter

alone is insufficient information regarding the free volume. The shortest possible distance be- tween any two molecules at which there is contact is σ (= σ/2 + σ/2). No other molecule can

be included within the volume defined by the radius σ (or diameter 2σ), and the “forbidden volume” per pair of molecules is π(2σ) 3 /6. For a single molecule the forbidden volume is π(2σ) 3 /12, and for N ′ molecules it is

b´ = N´ π(2σ) 3 /12 = 4 N´ πσ 3 /6 = N´ (collisional volume ÷ 2), (18)

where the collisional volume is defined as 4 πσ 3 /3.

The ideal gas equation can be corrected by subtracting the product N´b´ from the geometrical volume V. The free volume V–N´b´ is the volume occupied by equivalent point masses in an ideal gas and N´b´ is considered to be the apparent body volume of the molecules contained in the volume V. The reduced free volume increases the number of molecules per unit free volume, which, in turn, increases the number of collisions per unit time. This results in a higher pressure. Based on the reduced free volume, the ideal gas equation may be modi- fied into the form

(19a) where

P=N R T/(V – N´b´) = R T/( v – b ),

(19b) The product πσ 3 is the forbidden volume per kmole of the gas, and is four times the body vol-

b = N´b´/N = (2/3)N Avog πσ 3 .

ume. The number of molecules and moles are related by the expression N´ = N Avog N. Rewriting Eq. (19a) (which is known as the Clausius–I equation of state),

(20) Note that v 0 would have been the volume had the gas been ideal.

v = R T/P + b = v 0 + b .

The compressibility factor

(21) Z–1= b (P/ R T).

Z= v (T,P)/ v 0 (T,P) = P v / R T=1+P b / R T, or

(22) Equation (22) is also called the deviation function for the compressibility factor. The deviation

function tends to zero as P →0 at a specified temperature. According to the Clausius–I equation of state, the geometrical volume at a specified state (i.e., fixed T and P) is equal to ideal gas

volume (predicted by ideal gas equation) plus a correction for the molecular body volume. Upon comparing Eq. (21) with Eq. (14), it is clear that volume (predicted by ideal gas equation) plus a correction for the molecular body volume. Upon comparing Eq. (21) with Eq. (14), it is clear that

intermediate pressures.

b. VW Equation The Clausius–I equation of state (Eq. (19)) does not account for the intermolecular

attraction forces that are significant when the molecular spacing is relatively close (e.g., at high pressures). Therefore, Eq. (19) must be appropriately modified.

Let us denote the pressure that would exist in absence of attractive forces as P´. In or- der to understand the effects of attractive forces on the pressure P´ we use the analogy of the gas molecules represented by groups of five particles. Each analogous group consists of a sin- gle particle labeled as “M” surrounded by four other particles that are geometrically located 90º apart. Assume that a group travels at a velocity of 350 m s –1 . There is no net attractive force on the particle contained in the interior of the group, since the attractive forces due to the four exterior particles cancel each other out. On the other hand when the particle “M’ (and similar particles) impinges on a surface so as to create a pressure, the particle that was origi- nally in the interior is now surrounded by only three particles (since one particle has already struck the surface). Therefore, at this time, the particle group will exert a lower pressure than

10 5 Nm –2 , since there is now a net attraction force exerted on the interior particle “M”. In order to determine the reduction in pressure, we need a functional form for the at- tractive force exerted between a pair of molecules separated by an intermolecular distance l.

Such a relation can either be represented by empirical relations (e.g., by using the Len- nard–Jones (LJ) empirical intermolecular potential energy Φ(l) between a pair of molecules) that was discussed in Chapter 1, or it can be deduced through a phenomenological approach.

Applying the LJ approach for a like molecular pair, the intermolecular force function (cf. Chapter 1)

(24) where ε denotes the characteristic energy of interaction between the molecules (which corre-

F( ) l

=− 13 ( 4 εσ 7 / )( ( / ) 12 σ l − 6 (/) σ l ,

sponds to the maximum attraction energy ≈ 0.77 k B T c, k B is the Boltzmann constant), σ the characteristic or collision diameter of the molecule (= 2 σ´), and l the intermolecular distance. The first term on the RHS of Eq. (24) represents the repulsion force between a molecular pair,

and the second term arises due to the intermolecular attraction between the two molecules. The repulsive force is only of interest if the substance is a solid or a liquid. The reduction in pres- sure ∆P attr due to attractive forces is derived in the Appendix, and is

(25) where n´ denotes the number of molecules per unit volume= N Avog n, and n the number of

∆P = 2.667π εσ 2 3 n´ 2 ≈3π εσ 3 n´

moles per unit volume. By using Eqs. (19a) and (25) we obtain the Van der Waals (VW) equa- tion of state (named after Johannes Diderik Van der Waals, 1837–1923) in the form

P= R T/( v – b )–( a / v 2 ),

(26) where a = 2.667 πεσ 3 N 2 Avog . The units for b and v are identical (in m 3 kmole –1 ), while those

for a are atm m 6 kmole –2 . According to a more rigorous derivation based on the potential

a = 2.667 πεσ 3 N Avog (27) The first term in Eq.(26) is the pressure exerted due to collision and bouncing off an

imaginary plane and is proportional to thermal part of energy (i.e te+ve+re etc) of all the molecules within unit volume of free space. The second term is the reduction in force due to attractive force exerted on those molecules by neighboring molecules. A typical experimen- imaginary plane and is proportional to thermal part of energy (i.e te+ve+re etc) of all the molecules within unit volume of free space. The second term is the reduction in force due to attractive force exerted on those molecules by neighboring molecules. A typical experimen-

T c ,P=P c , and v = v c ). Thus, one must select values for “a” and “b” such that the following inflexion condition is satisfied, i.e.,

= 0. (28) Therefore, in context of Eq. (26)

c TT = c

TT = c =− RT c /( v c − b ) + 2 av / c = 0 , and

TT = c = 2 RT c /( v − b ) + 6 av / c = 0 .

b = v c / 3. (32) Finally, combining Eqs. (31) and (32) we obtain

a =(/) 98 RT v cc .

(33) If critical data on T c and v c are available then the constants a and b can be determined from

Eqs. (32) and (33) and the critical pressure can then be determined from the state equation (27) at the critical point, i.e., by using the result for a and b from Eqs. (32) and (33)

(34) Therefore,

P c =( RT v c / c )(/) 38.

Z c = Pv cc / RT c = / 3 8.

(35) The critical temperature and volume for water are, respectively, T c = 647.3K and v c =

0.0558 m 3 kmole –1 ( Table A-1 ) Thereafter, using Eq. (32), b = 0.0186 m 3 kmole –1 , and from Eq. (33) a = 60.54 bar m 6 kmole –2 . (The calculated value of P c from Eq. (34) is 362 bar, but the measured value is 220.9 bars, i.e., the experimental data for the critical pressure of water at specified T c and v c deviates far from values obtained from the VW equation of state when Z c ≠ 3/8. For the moment we will presume that this equation of state is accurate at the critical

point and proceed to express a and b in terms of P c and T c using Eqs. (32)–(34). Therefore,

a = (27/64) R 2 T 2 2 c 2 /P c =c 1 R T c /P c , and b = R T c /(8P c )=c 2 R T c /P c , (36) where the constants c 1 =27/64= 0.4219 and c 2 =1/8= 0.125. If measured data for all three criti-

cal properties P C ,T C ,  v C are available, one has a choice of either Eqs. (32) and (33) or Eqs. (36). Use of eqs. (36) is recommended for better comparison with experimental data. One may

tabulate “a” and “b” values for a few substances using Eqs. (36) as is done in many texts. Ta-

ble A-1 lists the VW values of a and b for many substances.

A possible experiment to measure the critical properties of a substance (say, water) can be constructed in the following manner. First, pour the water into a quartz made pis- ton–cylinder assembly that is maintained at a constant pressure. Then, slowly heat the water until a small bubble appears. Determine the specific volume of the water just as bubble begins

AP

Figure 3: A typical experimentally determined P–v diagram. to appear (i.e, v f , saturated liquid). Thereafter, determine the gas–phase volume when the entire

mass of the water has been evaporated (i.e., v g for saturated vapor). Also measure the boiling temperature of water under the specified conditions. Repeat the experiment several times after incrementally increasing the system pressure and plot the response of v f to the pressure (i.e., curve FAC in Figure 3 ), and of v g to P (curve GBC in the Figure 3 ). Two distinct phases will

be observed until the critical point is reached (where the curves for both v f and v g vs. P inter- sect). The pressure at which we cannot observe a clear demarcation between the liquid and

vapor (i.e., v f being equal to v g ) is the critical pressure.

i. Comments Equation (26) can be rewritten in the form

v 3 + v 2 (– b P– R T)/P + v ( a /P) (– ab /P) = 0.

which is a cubic equation in terms of v . Therefore, for a specified pressure and tempera- ture, there are one real and two imaginary solutions, and/or three real positive solutions for v . The constants a and b are related to the critical properties. The appendix presents explicit solutions for cubic equations.

Recall that b = (2/3)N Avog πσ 3 , so that the collision diameter and, hence, the molecular size

can be determined once we know b from critical properties.

Instead of using the inflection approach for evaluation of a and b , the constants c 1 and

c 2 (in Eq. (36)) can be selected to minimize the difference between the experimental data and theoretical results obtained from Eq. (26). If we assume that v » b (i.e., a point mass approximation is used), P ≈ R T/ v – a / v 2 ,

which is a quadratic equation. In this case, the compressibility factor Z = P v /( R T) = 1 – which is a quadratic equation. In this case, the compressibility factor Z = P v /( R T) = 1 –

Thus when both effects are included Z>1 or Z<1 For a closed system the reversible work w = ΙP(T,v) dv. If a real gas is involved, one

must use the real gas equation of state Eq. (26) for P (T, v). Similarly, for an open system the reversible specific work w = –Ιv(T,P) dP. Once v is specified, ∂P/∂T = R/(v–b), i.e., isochoric curves are linear in P-T plane for a fluid following VW equation of state.

Since a = (27/64) R 2 T 2 c /P c and b = R T c /(8P c ), then using Eqs. (19b) and (27), we can obtain the following relations:

σ = 0.3908k 1/3 B (T c /P c ) 1/3 and ε = (27/64 R 2 T 2 c /P c )/(3 N 2 Avag πσ 3 ). Simplifying

ε = 0.75 k B T c ,

which is close the value of 0.77 k B T c cited in the literature.

a. Example 1 Determine v for H 2 O(g) at P = 140 bars and T = 673 K using the ideal gas equation, the compressibility chart, the steam tables and the VW equation of state. What is the size of a single molecule?

Solution Ideal gas. v = R T/P = 0.08314 bar m 3 kmole –1 K –1 × 673 K ÷ 140 bar = 0.4 m 3 kmole –1 .

Compressibility chart. P R = P/P c = 140 bar ÷ 220.9 bar = 0.634,

T R = T/T c = 673 K ÷ 647.3 K = 1.04. From the compressibility chart, Z = 0.78. Since P v =Z R T, v = 0.78 × 0.08314 bar m 3 kmole –1 K –1 × 673 K÷140 bar = 0.312 m 3 kmole –1 . Tables.

18.02 kg kmole –1 = 0.31 m 3 kmole –1 . Van der Waals equation.

v = 17.22 cm 3 g –1 × 10 –6 m 3 cm –3 × 10 3 g kg –1

a = 27 R 2 T 2 /64 P = 27 ×0.08314 2 bar 2 m 6 kmole –2 K c –2 ×647.3 2 K c 2 ÷(64 × 220.9 bar), i.e.,

a = 5.531 bar m 6 kmole –2 .

Likewise, b = R T c /(8P c ) = (0.08314 bar m 3 kmole –1 K –1 × 647.3 K÷(8 × 220.9 bar), i.e.,

b = 0.0305 m 3 kmole –1 . However, using Eq. (27), at 140 bar, T= 673 K, v = 0.31 m 3 kmole –1 , according to VW equation. (Note that use of this equation produces a closer prediction to the tabulated value than was obtained using the ideal gas law.)

Molecular diameter Recall that b represents the finite volume or 4 × (body volume) of the molecules. A kmole contains 6.023 × 10 26 molecules. Therefore, the body volume of a single mole- cule ≈ 0.0305 m 3 kmole –1 ÷ (4 × 6.023 × 10 26 molecules) = 1.264 × 10 –28 m 3 . The molecular volume

πd 3 /6 = 1.264 ×10 –29 m 3 , i.e., d = 2.89 ×10 –10 m = 2.89 Å.

Remarks The term – a / v 2 in the VW equation corresponds to the reduction in pressure due to in- termolecular forces. When v = 0.31 m 3 kmole –1 ,– a / v 2 = –57.6 bar, while the first term

R T/( v – b ) = 199.5 bar. This implies that the frequent molecular collisions create a pres- sure of 199.5 bars, but due to the strong attraction forces in the dense gas phase the mole- cules are pulled back together with a pressure equivalent of 57.6 bar. Therefore, the net pressure is 141.9 bar.

Since σ = ((6/π)( b /4)/N Avog ) 1/3 , and b = (1/8) R T c /P c using the VW equation of state, σ= ((6/32 π) R T /(P N )) 1/3 = 0.391(k T

B c /P c ) . Using the value k B = 1.3804 × 10 kJ K molecule –1 , σ = = 2.02(T c (K)/P 1/3 c ) Å with P c expressed in bar. For instance, if T c = 647 K and P c = 221 bar, σ = 2.89 Å.

–26 –1 c c Avog