Other Two–Parameter Equations of State

4. Other Two–Parameter Equations of State

Table in the Appendix section (end of this chapter) contains descriptions of other two–parameter equations of state. We comment on one of these, namely, the Dietrici equation of state that is represented by the expression

P=( R T/( v – b )) exp(– a /( R T v )).

This equation predicts a reasonable value of Z c (= 0.271) and does not yield negative values of pressures. Its performance is superior in the neighborhood of the critical point.

b. Example 2

Plot the variation of pressure with respect to v at a constant temperature of 593 K for

H 2 O using the ideal gas equation of state and the RK equation of state. Solution

Ideal gas equation. P=( R T)/ v = 0.08314 ×593/ v = 49.3/ v .

(A)

This hyperbolic behavior is illustrated by the curve QRS in Figure 4 . As v → 0, P →

Ideal Gas 200

P, bars

v, m 3 / kmole

Figure 4: The response of pressure to v at two temperatures (593 K and 900 K) using the RK equation, and of P vs. v employing the ideal gas equation at 593 K.

∞. RK equation.

a = 0.4275 R 2 T 2.5 c /P c , and b = 0.08664 R T c /P c .T c = 647.3 K and P c = 220.9 bar.

a = 142.64 bar m 6 K 0.5 kmole 2 ; b = 0.0211 m 3 kmole –1 , and P = (49.30/( V – 0.0211)) (5.86/( V + 0.0211)).

The curve BECGKNDHJMAFL in Figure 4 describes this behavior at T = 593 K. Remarks

For ideal gases, at a specified temperature and pressure P ∝ 1/ v . In general, v = v (P,T) is a single–valued function of P and T.

The real gas equation is a cubic equation in v at specified P and T. When T > T c , at given T and P the relation yields two complex roots and one real root for the volume. The pres- sure is a monotonic function of v at T>T c (e.g., illustrated by curve WXY in Figure 4 ) and hence v= v(T,P), a single valued function. A plot of the pressure with respect to vol- ume is a hyperbola at T » T c . As T →T c deviations from hyperbolic behavior occurs

around T=T c . Non–monotonic behavior occurs at T<T c inside the vapor dome (e.g., curve BECGKNDHJMAFL). This is a pressure–explicit type of equation of state, i.e., P = P(T, v ) is a single–valued function of v and T, but it is not a monotonic function of T and v , at a specified temperature T < T c , P does not increase monotonically as v decreases. As the volume decreases at constant temperature, first the pressure increases along BEC (ac- cording to the relation P ∝ R T/ v ), and as v is further reduced (along CGKN), attractive

forces become stronger so that the rate of pressure increase is reduced. As the volume is reduced further, the attractive forces become so strong that the pressure starts to decrease

(along curve NDHJM). Near the point M, the volume is so small that v ≈ b and, hence, P ≈ R T/( v – b ). The free volume available for molecular movement becomes very small, with the result that there are frequent collisions that result in an increase in the pressure

(along the curve MAFL with the attractive forces holding the matter in a liquid state). At specified pressures and temperature, multiple solutions for v can occur. Therefore, v is not a single–valued function of P and T at T < Tc for certain ranges of pressures. For in- stance, at 50<P<155 bars, three real equilibrium solutions are possible for v while at 50 and 155 bars, only two solutions exist. For P >155, and P < 50, only one solution (or v is single valued function of T and P) is possible. However, all of the solutions are not at sta- ble equilibrium states. This can be illustrated in the following manner. Consider water contained at state E in Figure 4 (i.e., 100 bar and 593 K) in a piston–cylinder–weight (PCW) assembly. Upon pushing the piston down slightly and then releasing it, the de- creased volume results in the increase of the fluid pressure and the fluid comes back to original state. On the other hand if the fluid were initially at state H (i.e., 133 bar and 593 K), the same disturbance test causes a slight decrease in volume first; but the fluid pres- sure is decreased to a value less than the external pressure. Therefore, the fluid would be compressed past state M, finally reaching an equilibrium state in the vicinity of F. State E is stable while state H is unstable. Further discussion regarding the stability of the three possible states is contained in Chapter 10. The single–valued and monotonic characteris- tics are very important while discussing the fundamental relations and thermodynamic postulates (e.g., in Chapter 5).

At large specific volumes, the vapor behaves like an ideal gas ( v » b , a / v 2 ≈0) and P ∝1/ v . Alternately, if we let a = b = 0 the ideal gas equation of state can be obtained from the real gas equation.

Liquid Like

40 Vapor Like

v, m 3 / k mole

Figure 5: The variation of pressure with specific volume for water using several state equations at 593 K.

Since the real gas state equation is cubic in terms of specific volume, explicit algebraic expressions are available for solving v (P,T). Normally there are three real solutions for v

at specified P and T for T < T c (e.g., the points F, H, and G in Figure 4 at T = 320ºC, P = 113 bar). The smallest value (point F) corresponds to a solution in the liquid phase while the largest value (point G) corresponds to a solution in the vapor phase. The middle value (point H) has no physical meaning. (These liquid–like and vapor–like solutions will be used later for determining P sat versus T or for drawing phase diagram.) Points N and M are called spinodal vapor and liquid points, and will be discussed in Chapter 10. Note that sometimes as v becomes very small, it approaches liquid–like behavior due to strong at-

tractive forces (= a / v 2 ), and the pressure becomes negative (a liquid under tension). Eqs. (26) and (37) are also known as two parameter equations of state since they involve two parameters, a and b . The variation of pressure with specific volume using several state equations is illustrated in Figure 5 . The ratio P c v c /T c = 3/8 for VW gas while Z c = 0.333 for RK gas. Some representative ex- perimentally obtained values of Z c are as follows:

0.29 ≤Z c ≤ 0.3. Hydrocarbons

Non–polar gases (e.g., Ar, He, Ne, N 2 , etc.)

0.26 ≤Z c ≤ 0.29. Polar Gases

(e.g., C n H m )

0.22 ≤Z c ≤ 0.26. Isochoric curves (i.e., v = constant or P vs. T curves) obtained from the VW equations are

(e.g., H 2 O)

linear, while those obtained from the RK equation are not necessarily so. The ideal gas equation is applicable when v » b and a /(T 1/2 v 2 )« R T/ v . In terms of re- duced variables, v R ´ » 0.4275/T 3/2

and, using the state equation, P R «T R /0.4275. At 100ºC and1 bar, the value v can be obtained by applying the RK equation (0.0264 m 3

kmole –1 ), as can the collision pressure (= R T/( v - b ) ≈5894 bar) and the pressure reduc- tion due to attractive force ≈ 5893 bar, which indicates that attractive forces cannot be ig- nored in comparison with collision forces and, hence, the ideal gas equation is not appli-

cable. On the other hand, the ideal gas equation can be used when v » b and if the pressure cable. On the other hand, the ideal gas equation can be used when v » b and if the pressure

l Avog 3 , v » a RT /( 32 / ) , i.e., l »( CkT / TP 15 . ), 13 / or l »( Cv / N T 1513 . ), / or ll / ». 3 75 / T 12 3 / B C R C 3 C ′ Avag R ′ c R . which implies that

gas behavior to apply a /(T 1/2 v 2 )« R T/ v . Since v =N

l molecular pacing at any T and P s

c molecular pacing at the ritical po s c int T R

c ´) 3 , v c =N Avog l c , and l c = 0.693 l c ´, since v c / v c ’ = 1/3. In terms of reduced variables, v

Here, v c ´=N (l

Avog

R ´ » 0.4275/T 3/2 R and, using the state equation, P R «T R /0.4275. The volume at minimum potential is related to critical volume. We will see later that v/v c » 0.7/T 0.5 R for ideal gas behavior to occur. The term a represents a measure of attractive forces between the molecules. The higher the a , higher the energy required to detach the molecules from liquid phase, the higher is the saturation temperature. Table A-1 lists boiling and melting points of several sub- stances. Figure 6 plots the boiling and melting points of several substances with respect to

a . Can you tell which substances may be more volatile: methane or water at specified T? The values of “a” are respectively 32.22 and 142.6 (bar m 6 K 0.5 )/kmole 2 for the RK equa- tion.

c. Example 3 If the number of molecules per unit volume n´ ≈≈≈ 3 ≈1/l , where l denotes the average distance

(or mean free path between molecules), determine the value of l for water vapor at 320ºC and 100 bar for the following cases: (a) when the ideal gas law is applicable, and (b) when the RK equation is applicable, and

Compare the answer from part (b) with the molecular diameter obtained from the b value. Using the LJ potential function, determine the ratio of attractive potential to minimum attractive potential. Using Tables A-4 for the saturated properties of water, determine the intermolecular spacing for the saturated liquid at 320ºC Determine the intermolecular spacing for saturated liquid and vapor at T TP , and at the critical point

Solution The temperature T = 593 K. n = 100 ÷(0.08314×593) = 2.028 kmole m –3 , and

n´ = 2.028 ×6.023×10 26 = 1.22 x 10 27 molecules m –3 ,

l = 935.44 ×10 –12 m = 9.35 Å.

P= R T/( v – b )– a /(T 1/2 v ( v + b ))

100 = 0.08314 ×593/( v – 0.0210) – 142.64/(593 1/2 v ( v + 0.0210) Solving iteratively, v = 0.375 m 3 kmole –1 , n = 2.667 kmole m –3 .

n´ = 1.606 ×10 27 molecules m –3 , l = 853.94 ×10 –12 m or 8.54 Å.

Note that the first term in P expression is 139.3 bars due to collision of molecules at 593 K with the wall while the pressure of attraction/intermolecular forces is deter- mined as 39.4 bars

d ≈ ((6/4π) 0.0210) 1/3 = 0.000255 m or 2.55 Å, l/d = 3.35, d/l = 0.299. Using the LJ function we find that (intermolecular force/maximum attractive force) ≈

a, bar K 0.5 m 6 /kmole 2

Figure 6: The boiling and melting points of several fluids with respect to a : solid diamond: MP, solid square: BP.

f = 1.4988 ×10 m kg or 0.0270 m kmole , n = 37.03 kmole m or 22.30x10 molecules m –3 , l = 355 ×10 –12 m, i.e., 0.000355 µm or 3.55 Å which is about half the spacing for vapor at 320ºC, and 100 bar. Therefore, l/d ≈ 1.39 and some space re- mains between the water molecules. The intermolecular potential is 25.8% of the

minimum potential, indicating strong attractive forces. At T

TP , v f = 0.001 m

kg

or v f = 0.0180 m kmole , n = 55.55 kmole m or

g = 206.136 m kg –1 , n = 0.000269 kmole m –3 , or 1.62 ×10 21 molecules, and l g,TP = 183 Å. At the critical point, v = 0.003155 m 3 kg c –1 , or 0.568 m 3 kmole –1 , n = 1.762 kmole m –3 or 1.06 ×10 27 mole- cules m –3 , and l f,C =l g,C = 9.8 Å. At lower pressures, for vapor states the molecules are separated at farther distances, while they are more closely packed in the liquid

33.46 ×10 27 molecules m –3 , and l f,TP = 3.1 Å, Similarly, v

state.

d. Example 4 Determine an expression for the reversible work done by a Van der Waals gas for:

A closed system. An open system. Assume the processes to be isothermal.

Solution The work done during the isothermal compression of a closed system can be written in the form

v 2 v w= 2

1 Pdv ∫ v ( RT /( vb − ) − / a v dv ) 1

= RT ln ((v 2 –b)/(v 1 –b)) + a (1/v 2 – 1/v 1 )

The work done during isothermal compression in a steady state steady flow device is: v 2 v w= 2

v vdP = − ( Pv ) 1 v 1 +

∫ Pdv

=P 1 v 1 –P 2 v 2 + RT ln ((v 2 –b)/(v 1 –b)) + a (1/v 2 – 1/v 1 )

Remarks Even though the work expression has properties at states 1 and 2, the functional form changes if we choose a different path for the same final state. Thus, w is a path de- pendent quantity.