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Chapter 5: Q. 5.44 (page 177)

Suppose that an unsaturated air mass is rising and cooling at the dry adiabatic lapse rate found in problem 1.40. If the temperature at ground level is 25 C and the relative humidity there is 50%, at what altitude will this air mass become saturated so that condensation begins and a cloud forms (see Figure 5.18)? (Refer to the vapor pressure graph drawn in Problem 5.42)

Short Answer

Expert verified

The altitude at which air mass become saturated so that condensation begins and a cloud forms is 1.37 km

Step by step solution

01

Given information

(Refer table 5.11 and graph in problem 5.42)

At 25°Cand50%relative humidity, the partial pressure of water is 0.016 bar.

The temperature at which this partial pressure is in equilibrium with vapour pressure is13.8°C.

02

Estimating the height at which clouds will start forming.

From the dry adiabatic lapse rate for unsaturated air mass is

dTdZ=9.8°C/km⇒Z=1.14km

03

The height at which saturation begins.

The relation between pressure and height is

P(z)=Pe-z8.5

Substituting P=0.016 bar and z=1.14km

P(z) = 0.0014 bar

Using the graph in problem 5.42 we find this pressure is in equilibrium with vapour pressure at T=11.8°C

For condensation to occur the temperature should reduce to 9.8°C the air should rise a height.

04

determining the height

z=11.89.8(1.14)z=1.37≅1.40km

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Most popular questions from this chapter

In Problems 3.30 and 3.31 you calculated the entropies of diamond and graphite at 500 K. Use these values to predict the slope of the graphite- diamond phase boundary at 500 K, and compare to Figure 5.17. Why is the slope almost constant at still higher temperatures? Why is the slope zero at T = 0?

Problem 5.58. In this problem you will model the mixing energy of a mixture in a relatively simple way, in order to relate the existence of a solubility gap to molecular behaviour. Consider a mixture of A and B molecules that is ideal in every way but one: The potential energy due to the interaction of neighbouring molecules depends upon whether the molecules are like or unlike. Let n be the average number of nearest neighbours of any given molecule (perhaps 6 or 8 or 10). Let n be the average potential energy associated with the interaction between neighbouring molecules that are the same (4-A or B-B), and let uAB be the potential energy associated with the interaction of a neighbouring unlike pair (4-B). There are no interactions beyond the range of the nearest neighbours; the values of μoandμABare independent of the amounts of A and B; and the entropy of mixing is the same as for an ideal solution.

(a) Show that when the system is unmixed, the total potential energy due to neighbor-neighbor interactions is 12Nnu0. (Hint: Be sure to count each neighbouring pair only once.)

(b) Find a formula for the total potential energy when the system is mixed, in terms of x, the fraction of B.

(c) Subtract the results of parts (a) and (b) to obtain the change in energy upon mixing. Simplify the result as much as possible; you should obtain an expression proportional to x(1-x). Sketch this function vs. x, for both possible signs of uAB-u0.

(d) Show that the slope of the mixing energy function is finite at both end- points, unlike the slope of the mixing entropy function.

(e) For the case uAB>u0, plot a graph of the Gibbs free energy of this system

vs. x at several temperatures. Discuss the implications.

(f) Find an expression for the maximum temperature at which this system has

a solubility gap.

(g) Make a very rough estimate of uAB-u0for a liquid mixture that has a

solubility gap below 100°C.

(h) Use a computer to plot the phase diagram (T vs. x) for this system.

Below 0.3 K the slope of the °He solid-liquid phase boundary is negative (see Figure 5.13).

(a) Which phase, solid or liquid, is more dense? Which phase has more entropy (per mole)? Explain your reasoning carefully.

(b) Use the third law of thermodynamics to argue that the slope of the phase boundary must go to zero at T = 0. (Note that the *He solid-liquid phase boundary is essentially horizontal below 1 K.)

(c) Suppose that you compress liquid *He adiabatically until it becomes a solid. If the temperature just before the phase change is 0.1 K, will the temperature after the phase change be higher or lower? Explain your reasoning carefully.

Repeat the previous problem for the diagram in Figure 5.35 (right), which has an important qualitative difference. In this phase diagram, you should find that β and liquid are in equilibrium only at temperatures below the point where the liquid is in equilibrium with infinitesimal amounts of αandβ . This point is called a peritectic point. Examples of systems with this behaviour include water + NaCl and leucite + quartz.

Suppose you cool a mixture of 50% nitrogen and 50% oxygen until it liquefies. Describe the cooling sequence in detail, including the temperatures and compositions at which liquefaction begins and ends.
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