Quasi-equilibrium transitions

Transcription

1 Quasi-equilibrium transitions We have defined a two important equilibrium conditions. he first is one in which there is no heating, or the system is adiabatic, and dh/ =0, where h is the total enthalpy associated with all potentials. = dh d p =0=c d p + g dz +... (1) Assuming equilibrium, we were able to derive, for example the dry adiabatic lapse rate (@/@z) p = g/c p. he second sense is that there is heating but that dh/ =0because heating is balanced by working through a change in pressure at constant temperature. his leads to the following key result = dw = (2) Now what we ask how this second equilibrium solution changes as temperature is changed. Specifically, what will happen to the relationship between heating and working if the temperature is changed a little bit for the constant temperature condition (Eq. 2). Effectively, we are asking what is the sensitivity of the equilibrium condition to small changes. We are still looking for an equilibrium condition. However, we now want to know how the equilibrium solution is different at different temperatures. Effectively what we are seeking is an expression for quasi-equilibrium transitions. emperatures are changing slowly enough that something very close to equilibrium is maintained. One place where this question is particularly important arises where there are phase changes and we are looking for how the vapor pressure above a surface of water, at equilibrium, changes as the temperature changes. o proceed, at constant temperature, we have working equals heating and = dw = (3) so the equilibrium solution is that the change in volume for a given heating is equal 1

2 to inverse of the current When we look at phase changes, we are concerned with the amount of heat that must be applied to a liquid to cause it go from a dense liquid phase to a gas that occupies much more space per molecules. Because we are looking at a jump, since this is a phase change and the difference between a gas and a liquid is very large, we are interested in differences. hus, the relationship is no longer a differential, but rather q We are interested in vapor so p e = R v / v. Also, the corresponding amount of heat = 1 p = 1 p that must be applied to enable a phase change is q = which is about J/kg a huge number. hus = v R v his is our equilibrium expression for how the specific volume changes in response to a phase change. How does this equilibrium expression change if the temperature changes a small amount? aking the derivative with respect to temperature d d = v R v 2 Here, we make a simplification. For specific volume, the difference between vapor and liquid is = v l = 1 v 1 l But the density of vapor is tiny compared to that of liquid, by about a factor of 1000, 2

3 so ' 1 v = v hus, the expression for the quasi-equilibrium transition to another temperature becomes or d v d = R v 2 d ln v d ln = R v (4) Where heating is balanced by working, and we are dealing with phase changes. Normally, this is expressed a little differently. Recognizing that d ln v = d ln e, we get d ln e d ln = R v where e p is the vapor pressure and R R v. his is the famous Clausius Clapeyron equation that gives the sensitivity of the saturation (equilibrium) vapor pressure over a plane surface of pure water e ( ). One would be tempted to derive a solution for this equation as Lv 1 1 e ( )=e 0 ( )exp R v 0 At freezing the latent heat is J kg 1 and R v = 461 J/K/kg. his is fine, but only if the difference between and 0 is small. he expression d ln p/d ln is extraordinarily powerful, but it must not be solved without care, because it is a sensitivity and other things that we have assumed to be constant might be changing with temperature also. he latent heat does vary somewhat by a few percent over the normal range of atmospheric temperatures, and this does bad things to the accuracy of our nice exponential solution for e ( ). A more accurate solution takes into account a solution for ( ). e s ( )=611.2 exp( ) where, is in C. he reason for the slightly different form of the equation is that itself is a weak function of temperature. Often you will hear the expression that at warm temperatures the air can hold more water vapor. Certainly from the C-C equation this seems to be correct, but semantically it is misleading. Nowhere in our discussion of the derivation have we needed to mention air at all. Rather, the C-C equation would hold even in the absence 3

4 e s (mb) (C) Figure 1: Clausius-Clapeyron Equation for water vapor over liquid water 4

5 of dry air. hat said, in our atmosphere at least, having air present is what enables the temperature to be as high as it is, since it is molecular collisions that enable radiative absorption to show up as a measurable temperature. 5