Water Vapor Pressure in the Presence of Air

The equilibrium between water and water vapor in the case of humid air is il­lustrated in Fig. 4.7.

Dry air (Pi)

Water

Vapor (ph)

Water

FIGURE 4.7 Equilibrium between water and vapor in the presence of air.

The state of equilibrium differs from the equilibrium between water and pure water vapor in that, in a gas phase, there is also inert gas (dry air} present. This means that the water pressure is equal to the total gas pressure, p = p, + ph, not to the water vapor pressure ph.

In a state of equilibrium the chemical potentials of water and water vapor are equal:

P) = Ph), (4.95)

Where the subscript v refers to water and h to water vapor. Notice that

P = Pi + Ph —

From Eq. (4.95) the partial pressure of water vapor ph can be solved for, and we see that it is dependent on temperature and the partial pressure of dry air:

Ph = Ph(T, P,)- (4,96)

Next we will show that the dependence of water vapor pressure on the partial pressure of dry air is very small, and consequently a good approximation is

Ph = Ph(Tl (4.97}

To show this, Eq. (4.95) is differentiated, and we get

On the other hand,

Dpy = = _c

DT dT, n

(4.99)

Where s„ is the specific entropy of liquid water, sh is the specific entropy of

Water vapor, vv is the specific volume of water, and vh is the specific volume

Of water vapor. Notice that vh = 1/p/,. Substituting Eqs. (4.99) in Eq. (4.98), we obtain

— s,, dT + vv dp = — si, dT + vh dp},,

And it follows that

Dph = SjLL^dT+ —dp. (4.100)

V vh r

On the other hand, while p. = h — Ts, in accordance with the balance clause (4.95),

Bv — Tsv= hh — Tsh,

And it follows that

Substituting this equation in Eq. (4.100), we obtain

Dph = h^~^dT+ Vfdp. (4.101)

 On the other hand, the differential of the total pressure is Dp = dpi + dph, which, substituted in Eq. (4.101), gives
 Hu-b„
 Dph
 -JT+-
 -dp,
 (4.102)
 T{vh-v,_) vh-vv This differs from the water pressure equation of Clapeyron, which lacks the last term. If pi = 0 or dp, = 0, then Eq. (4.102) is identical to the Clapeyron equation, as it should be. Considering that i/h » vv and vh = 1 /ph and using Eq. (4.78), an ap­proximation to Eq. (4.102) is obtained:
 Dph _ Mh(hk-h„)
 DT +
 Ph
 RT~
 (4.103)

On the other hand,

^ = D(nph), Ph

So according to Eq. (4.103) we can write

 9(ln ph) dT

 (4.104)

Mh(hh — hv)

RT2

 3(ln ph) dpi

 M. kVv RT

 (4.105)

The specific volume of water is approximately vv = 10-3 m3/kg, and an esti­mate of Eq. (4.105) at a temperature of 50 °C is

 D(lnp/7) I Dp, t

0.0180 10 j 1 r^n in9l_

8.314-323.15Pa b-/u lupa —

Integrating Eq. (4.105) with the help of this value we can examine the effect of air on the vapor pressure. When the partial pressure of air p, = 0 and when the partial pressure of air is 105 Pa, a ratio of the vapor pressures correspond­ing to these situations at the temperature of 50 °C is obtained:

In Ph(5° = 1VFa ) = 6.70 • 10 ^ • (10s — 0) Pa = 6.70 • 10~4

Ph(50 °C, pj = 0)

Or

Pfe(50 °C, pj = 10 Pa) __ 6.70 -1o“4 P*(50 °C, pi = 0) e

When pj — 0, a situation is described where water and water vapor are in an equilibrium without the presence of dry air. The corresponding vapor pressure can be found in tables for vapor:

1.0006702.

Ph(50 °C, p, = 0) = p’h(50 °C) = 0.12335 bar.

The difference between vapor pressures is pf,(50 °C, pt = 103 Pa) 0 °C,

Pi = 0) = 8 Pa.

This shows that the presence of air in the gas phase has a very small influence on the vapor pressure of water. Repeating the same calculation procedure for other temperatures, we can show that the vapor pressure of water can with good accuracy be taken from the vapor pressure tables for saturated water (water has the same pressure as water vapor when they are in equilibrium), as though there were no air in the gas phase. So the vapor pressure of water is with good accuracy also in this case just a function of temperature, and Eq. (4.97) is valid. New vapor pressure tables will not be needed for calculations with humid air.