Vapor Pressure and the Clapeyron Equation
4. Vapor Pressure and the Clapeyron Equation
Determination of saturation pressure involves inherent errors due to errors in real gas state equations and critical properties. In this section we will not use real gas state equations; rather we will use the criterion of equality of differential of Gibbs function for vapor and liq- uid phases and knowledge of saturation pressure at a single temperature (or vice versa) to de- duce saturation relations.
Inside the vapor dome g f =g g along any isotherm (or isobar, FG in Figure 19 ). As the state of a fluid is changed inside the dome, the saturation temperature and pressure change in such a manner that
Melting curve (other solids)
Figure 18: (a) Phase diagram for a substance that contracts upon melting. (b) the variation of g with respect to T.
dg f along FL = dg g along GH where (125) For example, a change of g f along FL= change of g g along GH ( Fig.18 )
(126) Using Eq. (125) in Eq. (126) we find that (dP/dT) = s fg /v fg . Since Tds + vdP = dh, at constant pressure ds = (dh/T) so that s fg =h fg /T. Therefore,
dg f = –s f dT + v f dP, and dg g = –s g dT + v g dP.
(127) Equation (127) is called the Clapeyron equation. Referring to Figure 18 , P sat = 15 at T= 111 °C
(dP/dT) = h fg /Tv fg .
while P sat = 2 bar at 120 °C. This slope is related to h
C if ∆P and ∆T are small. Thus, in bar ºC –1 , ∆P/∆T = (2–1.5)/(120–111) = 0.56. The slope (dP/dT) is represented by Eq. (127), and it varies, since h fg and v fg vary
fg , T and v fg at T= 111
along OC. It predicts the variation of saturation pressure with respect to temperature along the vaporization curve OC in the phase diagram illustrated in Figure 19 . The points O and C, re- spectively, represent the triple point and the critical point. Using the relation Pv = ZRT in Eq. (127), we obtain the relation
dP/dT = h 2
fg P/(T RZ fg ),
where Z fg =Z g –Z f . We may rewrite this expression in the form
(128) Equation (128) can be generalized to any phase transition from phase α to phase β phase, i.e.,
d ln(P)/d(1/T) = –h fg /(RZ fg )
d ln(P)/d(1/T) = – h αβ /(R(Z β –Z α )). (129) Since both h fg and Z fg both decrease with an increase in the temperature, the RHS of Eq. (129)
is a weak function of temperature and can be generally treated as a constant. In that case,
(130) Using the equalities P = sat P ref and T= T ref , to determine the constant,
ln(P) = –(1/T) h fg R/Z fg + C.
sat (P / P
sat
ref ) = exp((h fg /(RZ fg ))(1/T ref – 1/T)).
a. Remarks The slope dP/dT in Eq. (127) must be the same near the saturated liquid and the saturated vapor states. For the VW equation of state, the slope along the isochoric curve at the critical point,
( ∂P/∂T) vT c , c = R/(v c –b). Using the value b = (1/3) v c ,( ∂P/∂T) vT c , c = (3/2)(R/v c ) = (3/2) (P c /Z c T c ). In dimensionless form, the slope ( ∂P R / ∂T R ) v c = 4, since Z c = 3/8 in context of the VW state equation. Similarly, using the RK state equation ( ∂P/∂T) vT c , c = 1.861 P c /(Z c T c ), i.e., ( ∂P R / ∂T R ) v c = 5.583. Using values for water, i.e., T c = 647.3 K and P c = 220.9 bars, in the expression obtained using the RK state equation , (∂P/∂T) v c = 2.433 bar K –1 .
A first order phase transition is the one for which the property g remains continuous from a phase α to another phase β. In general, the slopes (∂g/∂T) α ≠ (∂g/∂T) β , and
( ∂g/∂P) α ≠ (∂g/∂P) β . The properties v, u, h, s, and a are discontinuous (cf. Figure 20 ). During a second order phase transition (e.g., critical point), the two phases cannot be
distinguished and the values of g and the first derivatives are continuous. However, in this case the second derivatives are discontinuous, for instance, at the critical point.
Since ∂g f / ∂T = –s f and ∂g g / ∂T = –s g , ∂g f / ∂P = v f and ∂g g / ∂P = v g , at the critical point ∂g f / ∂T = ∂g g /
f / ∂P = ∂g g / ∂P, and g f = g g . However, the value of ∂ g/ ∂P = ( ∂v/∂P) T is very large, since ∂P/∂v =0 at the critical point. Alternately, the transition from phase α to β is a first order transition if h αβ ≠ 0. The transition at the critical point is a second order transition and h αβ = 0 there.
2 ∂T, ∂g 2
A fluid containing one component exhibits a single critical point.
In context of Eq. (128), we can evaluate the value of h fg /RZ fg at low pressures, since Z g »Z f , i.e., Z fg ≈Z g . Assuming ideal gas behavior for the vapor, Z g = 1. Since v f «v g , then Z f «Z g , Eq.(131) transforms into the relation
(132) where T ref denotes the saturation temperature at the reference pressure P sat ref (T ref ).
(P sat / P sat ref ) = exp ((h fg /R)(1/T ref – 1/T)),
Equation (132) is called the Clausius Clapeyron equation and it also has the form
ln P = A – B/T, where A = ln sat P ref +h fg /RT ref and B =h fg /R. (133) For water, A = 13.082 and B = 4962 K. The slope of ln(P) with respect to T –1 is pro-
portional to heat of vaporization, and its value is nearly constant except in the vicinity of the critical point The Antoine equation is a modified form of Eq. (133), namely,
ln P = A – B/(T+C). (134) One can use experimental saturation data at three different temperatures, evaluate A,
B and C and tabulate them for computer applications. An empirical relation that de- scribes fluid behavior is of the form:
ln P sat = A +B/T + C ln T +DT. (For water, ln P sat = 12.58 +(–4692)/T + 0.0124 ln T, where the value of P sat is in bar
and that of temperature in K.) Differentiating this empirical relation and equating it with Eq. (126) we can determine
the h fg , provided v fg is known. (One solution is to consider v g ≈ RT/P at low pressures and v f ≈ 0.) Applying Eq. (127) during any phase change
(dP/dT) = h αβ /(Tv αβ ) = (h β –h α )/(T(v β –v α )). (135) Let α denote the initial solid phase s and β the final liquid phase f during the melting
of a solid so that v β –v α =v sf =v f –v s . In the case of ice, v f <v s and h sf = (h f –h s ) > 0, i.e.,
(dP/dT) < 0. As the pressure increases, the ice melting point T MP decreases, as illustrated in Figure
19(a) . In case of a sublimation process, Eq. (135) suggests that
gg. Example 33 The boiling point of water T ref at P ref = 0.1 MPa is 99.6 °C. What is the saturation (or boiling) temperature at P = 0.5 MPa? Assume that h –1
fg = 2258 kJ kg . Solution
The value for water of R = 8.315/18.02 = 0.461 kJ kg –1 K –1 . Applying Eq. (132),
÷0.1 = exp((2258 ÷ 0.461) × (372.6 –1 –T ), i.e., T = 425 K (the steam tables ( A-4 ) provide a value of 424.6 K).
0.5 –1
hh. Example 34 Obtain a relation for h fg /RT c with respect to P R for a substance using the RK state equation in terms of v R,f ´, v R,g ´, and T R .
Solution
Figure 19: (a) First order, and (b) second order phase transitions for several prop- erties.
h fg sat /RT c = (d P ref /dT R )T R (v R,g ´–v R,f ´). (A) During phase change ∫ (P(T,v))dv = P sat (v g –v f ), where P(T,v) is given by any real
gas state equation. Differentiating with temperature,
(B) (dP sat
(dP sat /dT) (v g –v f )– ∫(∂P/∂T) dv = 0,
/dT R )(v R,g ´–v R,f ´) – ∫(∂P R / ∂T R ) dv R ´ = 0, and (C) P 2
R =T R /(v R ´– 0.0864) – 0.4275/(T R v R ´ (v R ´ + 0.0864)). (D)
Therefore, using Eq. (D) in Eq. (C) for ( ∂P R / ∂T R ) and then applying Eq. (127) in re- duced form
h fg /(RT c )=T R ln ((v R,f ´ – 0.0864)/(v R,f ´ –0.0864)) + (2.4671/T 3/2 R ) ln((v R,g ´(v R,f ´ + 0.0864))/((v R,g ´ +0.0864)v R,f ´)).
(E)
If v g and v f are known, h fg can be determined.