When a liquid and its vapor are in equilibrium, the rates of evaporation of the

Question

When a liquid and its vapor are in equilibrium, the rates of evaporation of the liquid and condensation of the vapor are equal. Assume that every molecule of the vapor are equal striking the liquid surface condenses, and assume that the rate of the evaporation is the same when the vapor is rapidly pumped away from the surface, as when liquid and vapor are in equilibrium. The vapor pressure of mercury at 0 degrees celsius is 185 x10^-6 Torr and the latent heat of vaportization is about 340 J/g. Compute the rate of evaporation of mercury into a vacuum, in g/cm2s at a temparature of a) 0 degrees celsius and b) 20 degrees celsius?

Explanation / Answer

A)This difference between the bubble-point and dew-point temperatures is probably best understood from phase diagrams like those we presented for pure fluids. Unfortunately, with a binary mixture we must add another independent variable, composition of component 1. This makes the phase diagram four dimensional instead of three dimensional. We can greatly simplify the diagrams by not including volumes, which reduces the figures to 3D. Furthermore, we will pick a particular pressure in order to look at a specific T-x plane or we will pick a particular temperature and look at a specific P-x plane. The Txy graph below shows binary mixtures of n-pentane and n-heptane at a fixed pressure of 1.5 atm. Like all phase diagrams, areas outside of enclosed regions represent a single phase, and areas enclosed by coexistence lines represent two phases in equilibrium, with the boundary lines representing the two phases that are in equilibrium. The temperature is plotted as a function of x (mole fraction of n-pentane in the liquid) and y (mole fraction of n-pentane in the vapor). Everywhere above the red line we will have only vapor; everywhere below the blue line we will have only liquid. Between the blue and red lines we will have both liquid and vapor coexisting. The blue line represents the saturated liquid line, or bubble line. Suppose a liquid containing 12 mol% n-pentane is heated up at constant pressure. It would remain liquid until it reached 370 K (see the green dotted line at the left), at which point it would boil. This is the boiling point temperature of this mixture (12 mol% pentane) at this pressure (1.5 atm). The vapor that boils off will not be the same composition of the liquid. Pentane is more volatile than heptane so there will be more pentane in the vapor mixture. Remember, on these plots

B) we are using x to represent the mole fraction of n-pentane (the more volatile component) and 1-x to represent the mole fraction of n-heptane in the liquid. Likewise, we use y to represent the mole fraction of n-pentane in the vapor and 1-y to represent the mole fraction of n-heptane in the vapor. The two phases that are in equilibrium are at the same temperature and pressure, so they can be thought of as being connected through the two phase region by a horizontal line (the horizontal green dotted line on the graph). This horizontal line connecting the two equilibrium phases is called a tie line. The plot therefore shows us that the vapor that boils off at 370 K will be 40 mol% pentane (and 60 mol% hexane). This is found from the end of the tie line where it intersects the saturated vapor or dew line (red line). Notice that at the right and left ends of the graph that the bubble and dew points meet. For example, at x = 1.0 we have pure n-pentane, which is seen from the graph to boil at 322 K. The vapor given off will also be pure n-pentane and so the bubble and dew point temperatures are the same. Does it seem strange to you that x = 1 and y = 1 at this point? If it does, then you are confusing the definitions of these terms. Remember that x applies to the liquid and y to the vapor for the same component; the mole fraction of the other component is 1-x in the liquid and 1-y in the vapor. When a pure pentane boils, of course the vapor will also be pure pentane.

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