Everything in this section so far has ignored the boundary
Chapter 5, Problem 46P(choose chapter or problem)
Problem 46PEverything in this section so far has ignored the boundary between two phases, as if each molecule were unequivocally part of one phase or the other. In fact, the boundary is a kind of transition zone where molecules are in an environment that differs from both phases. Since the boundary zone is only a few molecules thick, its contribution to the total free energy of a system is very often negligible. One important exception, however, is the first tiny droplets or bubbles or grains that form as a material begins to undergo a phase transformation. The formation of these initial specks of a new phase is called nucleation. In this problem we will consider the nucleation of water droplets in a cloud.The surface forming the boundary between any two given phases generally has a fixed thickness, regardless of its area. The additional Gibbs free energy of this surface is therefore directly proportional to its area; the constant of proportionality is called the surface tension, σ:If you have a blob of liquid in equilibrium with its vapor and you wish to stretch it into a shape that has the same volume but more surface area, then σ is the minimum work that you must perform, per unit of additional area, at fixed temperature and pressure. For water at 20°C, σ = 0.073 J/m2.a) Consider a spherical droplet of water containing Nl; molecules, surrounded by N – Nl molecules of water vapour. Neglecting surface tension for the moment, write down a formula for the total Gibbs free energy of this system in terms of N, Nl, and the chemical potentials of the liquid and vapour. Rewrite Nl in terms of vt, the volume per molecule in the liquid, and r, the radius of the droplet.b) Now add to your expression for G a term to represent the surface tension, written in terms of r and σ.c) Sketch a qualitative graph of G vs. r for both signs of μg–μl, and discuss the implications. For which sign of μg–μl does there exist a nonzero equilibrium radius? Is this equilibrium stable?d) Let rc represent the critical equilibrium radius that you discussed qualitatively in part (c). Find an expression for rc in terms of μg–μl. Then rewrite the difference of chemical potentials in terms of the relative humidity (see below Problem 1), assuming that the vapuor behaves as an ideal gas. (The relative humidity is defined in terms of equilibrium of a vapour with a flat surface, or with an infinitely large droplet.) Sketch a graph of the critical radius as a function of the relative humidity, including numbers. Discuss the implications. In particular, explain why it is unlikely that the clouds in our atmosphere would form by spontaneous aggregation of water molecules into droplets. (In fact, cloud droplets form around nuclei of dust particles and other foreign material, when the relative humidity is close to 100%.)Problem 1:Ordinarily, the partial pressure of water vapour in the air is less than the equilibrium vapour pressure at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapour to the equilibrium vapour pressure is called the relative humidity. When the relative humidity is 100%, so that water vapour in the atmosphere would be in diffusive equilibrium with a cup of liquid water, we say that the air is saturated. The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapour.(a) Use the vapor pressure equation (below Problem 2) and the data in below Figure to plot a graph of the vapor pressure of water from 0°C to 40° C. Notice that the vapour pressure approximately doubles for every 10° increase in temperature.(b) The temperature on a certain summer day is 30°C. What is the dew point if the relative humidity is 90%? What if the relative humidity is 40%?Problem 2:The Clausius-Clapeyron below relation is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ΔU depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ΔU to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:P = (constant) × e–L/RT,This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.Relation:Figure: Phase diagram for H2O (not to scale). The table gives the vapor pressure and molar latent heat for the solid-gas transformation (first three entries) and the liquid-gas transformation (remaining entries). Data from Keenan et al (1978) and Lide (1994).
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