| The drier air (City T) will have the higher surface pressure since each city should have essentially the same number of molecules in its column and water molecules have a smaller mass than nitrogen or oxygen (air) molecules. |
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The gas law states that pressure is proportional to temperature times density. Use the gas law to explain why a balloon will deflate when placed inside a refrigator.9-QT2.
First consider the air outside the balloon. Inside a refrigerator, the pressure is the same as outside the refrigerator. (otherwise air would rush in or out whenever the door was opened.) Thus, since pressure is proportional to temperature times density, the density of the colder air in the refrigerator must be greater than the density outside the refrigerator, to compensate for the lower temperature. When a balloon is placed inside a refrigerator its air temperature decreases and its density increases just as that of air outside the balloon. Thus, in order for the density to increase, since the mass of air in the balloon is fixed, the volume of the balloon must undergo a commensurate decrease, i.e., the balloon shrinks or deflates even though the mass of the enclosed air does not change.
9-QT3. Suppose the air column above city Q is completely saturated with water vapor, and the air column above city T is completely dry. If the temperature of the air in both columns is the same, which column will have the higher atmospheric pressure at the surface? Explain. (Hint: Refer back to the Focus section on p.125 in Chapter 5, "Is Humid Air or Dry Air More Dense?")
The drier air (City T) will have the higher surface pressure since each city should have essentially the same number of molecules in its column and water molecules have a smaller mass than nitrogen or oxygen (air) molecules.
9-QT12. If you live in the Northern Hemisphere and a region of surface low pressure is directly west of you, what would probably be the surface win direction at your home? If an upper-level low is also directly west of your location, describe the probable wind direction alort and the direction in which middle-type clouds would move. How would the wind direction and speed change from the surface to where the middle clouds are located?
At such a location, the winds aloft (at the middle-type cloud level) would be parallel to the isobars or from the south. At the surface, with a frictional deflection of α = 30° the winds would be from 30° E of S or, roughly, from the SSE. Between the surface and the middle-level clouds, the wind direction would gradually shift from 30° E of S to directly S. 9-PE2. Figure 9.34 is a sea-level pressure chart (Northern Hemisphere), with isobars drawn for every 4 millibars. Answer the following questions, which refer to this map. (a) What is the lowest possible pressure in whole millibars that there can be in the canter of the closed low? What is the highest pressure possible? (b) Place a dashed line through the ridge and a dotted line through the trough. (c) What would be the wind direction at point A and at point B? (d) Where would the stronger wind be blowing, at point A or B? Explain. (e) Compute the pressure gradient between points 1 and 2, and between points 3 and 4. How do the computed pressure gradients relate to the pressure gradient force? (f) If point A and point B are located at 30EN, and if the air density is 1.2kg/m3, use geostrophic wind equation in the Focus section on p.235 to compute the geostrophic wind at point A and point B. Hint: Be sure to convert km to m and mb to Newtons/m2, where 1 mb = 100 N/m2.) (g) Would the actual winds at point A and point B be greater than, less than, or equal to the wind speeds computed in problem f? Explain.
9-PE3. (a) Suppose the atmospheric pressure at the bottom of a deep air column 5.6 km thick is 1000 mb. If the average air density of the column is 0.91 kg/m3 , and the acceleration of gravity is 9.8 m/s2 , use the hydrostatic equation on p. 243 to determine the atmospheric pressure at the top of the column. (Hint: Be sure to convert km to m and mb to Newtons/m2 , where 1 mb = 100 N/m2.) (b) If the air in the column of problem (a) becomes much colder than average, would the atmospheric pressure at the top of the new column be greater than, less than, or equal to the pressure computed in problem (a)? Explain. (c) Determine the atmospheric pressure at the top of the air column in problem (a) if the air in the column is quite cold and has an average density of 0.97 kg/m3.
(See Figure 9.34 on p.246.) (a) Since contours are drawn at 4-mb intervals, the lowest possible pressure in the closed low is 997 mb and the highest possible pressure is 1000- mb. (b) See drawing. (c) The wind direction at point A would be approximately SE. The wind direction at point B would be approximately SW or W. (d) The wind would be stronger at B than at A because of the greater pressure gradient. (e) grad (P12) = 8 mb/1050 km = 0.0076 mb/km. grad (P34) = 12 mb/1025 km = 0.0117 mb/km. The pressure gradient force is proportional to the pressure gradient. (f) The speed of the geostrophic wind is given by Vg = (grad P)/(2W sinf r) where, from part (e) above, for point A, grad (P) = grad(P12) = 0.0076 mb/km = 0.00076 N/m3, and, for point B, P = grad(P34) = 0.0117 mb/km = 0.00117 N/m3. Also W = 7.29×10-5 rad/s, sinf = sin 30°=0.500, and r =1.2 kg/m3. Hence Vg,A=8.7m/s, and Vg,B=13.4m/s. (g) Because the actual winds are surface winds where there is friction while the calculated winds are geostrophic, in the absence of friction, the actual wind speeds would be smaller. (a) Since ΔP = ρgΔz = (0.91 kg/m3)(9.8 m/s2)(5600 m)(0.01 mb/[N/m2]) = 499.4 mb, it follows that the pressure at the top is 1000 mb - 499 mb = 501 mb. (b) If the air were colder, the air density in the column would be greater so there would be a greater pressure drop over the thickness of the column and the pressure at the top would be less. (c) For a density of 0.97 kg/m3 the new pressure change would be ΔP = ρgΔz = (0.97 kg/m3)(9.8 m/s2)(5600 m)(0.01 mb/[N/m2]) = 532.3 mb and the new pressure at the top of the column would be 1000 mb - 532 mb = 468 mb.
9-PE5. Suppose air in a closed container has a pressure of 1000 mb and a temperature of 20°C. (a) Use the gas law to determine the air density in the container. (b) If the density in the container remains constant, but the pressure doubles, what would be the new temperature?
(a) The gas law (see p.220-221) is P[mb] = Cρ[kg/m3]T [K], where C ≈ 2.87. Hence ρ[kg/m3] ≈ P[mb] / CT [K] ≈
1000 / [2.87 ´ (20 + 273)] ≈ 1.19 kg/m3. (b) If the pressure were to double with no change in density, the entire change would be attributable to temperature change, hence the new temperature would be 2 ´ (293 K) = 586 K = 313°C.
