Chapter 3: Atmospheric pressure and temperature

Transcription

1 Chapter 3: Atmospheric pressure and temperature 3.1 Distribution of pressure with altitude The barometric law Atmospheric pressure declines with altitude, a fact familiar to everyone who has flown in an airplane and felt pressure changes in their ears, or climbed a mountain and struggled to breathe. The decrease of pressure with altitude is a simple consequence of the relationship between weight of overlying fluid and pressure: each increment in altitude leaves below a slab of atmosphere with a certain weight due to its mass and the force of gravity. As we go up, pressure must decline by exactly the weight of the slab with unit area, we leave behind. An equivalent relationship holds descending into the ocean, apart from a change in the convention of measuring depth vs. altitude (see Figure 3.1). Figure 3.1. Changes in pressure with altitude in the atmosphere (left) and depth in the ocean (right). Pressure always increases as the observer moves downward because the weight of the overlying column of fluid (air or water) increases. Altitude is conventionally measured increasing upwards from the surface of the earth, and depth increasing downwards. Therefore pressure decreases with increasing altitude in the atmosphere and pressure increases with increasing depth in the ocean. Air and water are both fluids, but they differ with respect to compressibility. We know about compressibility from everyday experience. For example, if we take a bicycle pump and seal the outlet, the plunger can be pressed in, decreasing the volume of the air trapped inside. The trapped air exerts a spring-like force on the handle, the smaller the volume into which the air is squeezed, the stronger the force. This is Boyle's Law. If we place water in the pump, we cannot significantly compress the volume. This simple thought experiment illustrates that air is compressible, density depends on pressure, following the Perfect Gas Law. Water is almost incompressible. This difference makes the change in pressure with height very different between the atmosphere and the ocean. The relationship between density, pressure and altitude is illustrated in Figures 3.2 and 3.3. In Figure 3.2, the weight of the slab of fluid between Z1 and Z2 is given by the density, ρ, multiplied by (volume of the slab) g = ρ (area x height) g. If we let the area be 1 m 2, the weight is therefore ρ g (Z2 Z1). If the atmosphere is not being accelerated, there must be a difference in pressure (P2 - P1) across the slab that balances the force of gravity. Therefore,

2 (P2 P1) = weight = ρ g (Z2 Z1). (3.1) (Note the minus sign, pressure is lower at P2.) We can insert the Perfect Gas Law into Eq. 3.1 (Chapter 2, use the form P = ρr' T) to calculate ρ, ρ =P av /(R'T), (3.2) where P av = (P1+P2)/2. If we put together (3.1) and (3.2) and use the value for the constant R' (=k/m) from Chapter 2, we obtain the barometric law, P=-P av mg / (kt) Z. (3.3a) Here the notation PrepresentsP2 P1, Z isz2 Z1, and P is the pressure in the middle of the slab (approximating P av ). P 2 P 2 Z 2 ============================ Z 2 ρg (Z 2 Z 1 ) =P 1 P 2 weight of slab = pressure difference P 1 P 1 Z 1 ============================ Z 1 P 1 P 1 Figure 3.2. The balance of pressure forces across a slab of fluid. The pressure P1 is greater P2 by P, corresponding to the weight of the slab over unit area. This weight is given by the mass (mass density (ρ) times the volume, 1m 2 Z g. Because air is compressible the change in pressure across the slab (Eq. 3.3a) is proportional to both the pressure itself and to the thickness of the slab. Therefore, for a given altitude change, the atmospheric pressure changes by a fixed percentage that depends on temperature but not on altitude or pressure. The scale height, H, and the simple form of the barometric law We make reference to Eq. 3.3 to define the scale height H kt/(mg). Hrepresentsa key length in the atmosphere, as it is the only parameter in the barometric law, P=-P Z/H. (3.3b) The value of the scale height, H, at room temperature (298 K) is approximately 8.7 km. Temperatures decline with altitude (as discussed later), and an approximate mean scale height is about 7km(corresponding to 240 K) for the whole atmosphere. If we had an atmosphere where the temperature did not change with altitude, the barometric law would have a very simple form in terms of the exponential function exp(), which appears on most hand calculators, exp(x) e x, where e= , P(Z) = P o e -Z/H. (3.4) Here P o is the pressure at the ground (1000 mb, see Chapter 2). You may readily check that the exponential function in (3.4) has the property defined in the barometric law, i.e.

3 that the pressure declines by a fixed fraction or percentage for a fixed increase in altitude (14% per km in the example). Although temperatures do vary with altitude, the changes are modest (H varies from about 6 to 8.7 km), thus Eq. 3.4 is useful as an approximation to the barometric law giving variation of pressure with altitude Depth (km) 2 Altitude (km) Pressure (bar) oceans Pressure (bar) atmosphere Figure 3.3 The distribution of pressure with depth in the oceans (left) and with altitude in the atmosphere (right, scale height H = 7km). Pressure increases linearly with depth in the ocean because water is incompressible, and therefore density (ρ) is constant. Pressure decreases exponentially with altitude in the atmosphere because density not constant, but it proportional to pressure according to the Perfect Gas Law (Chapter 2). Eq. (3.4), with H=7 km, is illustrated in Figure 3.3 (right panel), which shows the dependence of pressure on altitude. We have placed pressure on the x-axis and altitude on the y-axis, a convention often used in atmospheric science because it illustrates altitude as the vertical dimension. 3.2 Pressure vs depth in the ocean The weight of a slab of ocean with unit area (1 m 2 )isthe[massoftheslab] g= ρ g(d1- D2) which gives the pressure difference between the top and bottom of the slab,

4 P=ρ g D. (3.5) Since ρ is essentially constant for water (water is incompressible), the change in pressure across the slab is proportional to the thickness of the slab but not proportional to pressure itself (contrast to atmosphere). The pressure changes by the same increment for a given depth change, and pressure increases linearly, not exponentially, with depth in the ocean, P=P o + ρ w gd, (3.6) where ρ w is the mean mass density of seawater (density of seawater changes slightly with salinity and temperature) and P o is the pressure at the surface (1 atm). Since the mass density of liquid water is about 1000 times greater than the density of air, the pressure becomes very large in the deep ocean is quite large (see Figure 3.3). 3.3 Buoyancy Buoyancy is the tendency for less dense fluids to be forced upwards by more dense fluids under the influence of gravity. Buoyancy is extremely significant as a driving force for motions in the atmosphere and oceans, and hence we will examine the concept very carefully here. The mass density of air ρ is given by mn, wherem is the mean mass of an air molecule ( kg molecule -1 for dry air), and n isthenumberdensityofair(n = molecules m -3 at T=0 o C, or K). Therefore the density of dry air at 0 C is ρ =1.29 kg m -3. Ifweraisethetemperatureto10 C ( K), the density is about 4% less, or 1.24 kg m -3. This seemingly small difference in density would cause air to move in the atmosphere, i.e. to cause winds. Example 1a: Buoyant block in a tank of water To understand buoyancy, let's first consider the familiar example of a solid object with density ρ b held below the surface in a tank of water with density ρ w (see Figure 3.4a). Figure 3.4a. Buoyancy forece Forces on a solid body immersed in a tank of water. The solid is assumed less dense than water and to area A (m 2 ) on all sides. P1 is the fluid pressure at level 1, and P1x is the downward pressure exerted by the weight of overlying atmosphere, plus fluid between the top of the tank and level 2, plus the object. The buoyancy force is P1 P1x (up ) per unit area of the submerged block. The pressure at depth D1in the water is the weight of the overlying atmosphere plus the weight of a column of water, per unit area. At depth D1, the pressure P1 is uniform everywhere at that depth in the water, and the force pushing up on the block is the pressure area of the block = ρ w g A D1, since pressure P1 = ρ w g D1. P1A=A ρ w gd1a (3.7a)

5 where 10 5 is the pressure of the overlying atmosphere. However, the pressure P1x in the downward direction, and thus the force on the block, is different that the upward force at the base of the object. We already have the tools to calculate the magnitude of this difference by computing the weight of (fluid + object): P1x = ρ w gd2+ρ b g (D1-D2), (3.7b) where the height of the block is D1 D2. Thus there is a net force on the object, the difference in pressure A, that will accelerate the object, from Eq. 3.7 F net =A (P1x P1) = (ρ b ρ w )ga(d1-d2)=(ρ b ρ w )gv block. (3.8) The difference in pressure at the bottom of the object creates an upward (negative) buoyant force given by Equation 3.8, if ρ w > ρ b, as in the familiar case of wood (ρ b = 700 kg m -3 )andwater(ρ w = 1000 kg m -3 ). In Eq. 3.8, we identify (ρ b gv block )asthe weight of the block, and ρ w gv block is the buoyancy force, an upward force on the object that we see is equal to the weight of the water displaced by the block. Eq. 3.8 is therefore a form of Archimedes' principle: the buoyancy force is equal to the weight of fluid displaced by the object. Note that the total pressure of overlying water and atmosphere above level 2 is irrelevant, since only pressure differences lead to unbalanced forces, and thus to acceleration of fluids or objects. Example 1b: Floating block in a tank of water, add mineral oil Figure 3.4b shows the behavior of a block of Teflon immersed in water and in cooking oil, and addresses the question as to what will happen if the block is first placed into a bucket of water, then oil place on top.? A B C Fig. 3.4b. Buoyancy experiment: A block of Teflon has slightly lower density than water, and therefore it floats [A] in water. Teflon is more dense than oil, and therefore it sinks [B] in oil. What will happen if we have the block floating in water, and then add oil on top [C] (will it move up, down, or stay put)?? In panel A, about 1/8 of the volume of the block is surrounded by air, and 7/8 by water. The block is floating, so the net force on it is zero (no acceleration, it stays put), F=0=(ρ b - ρ air )fv b g+(ρ b - ρ w )(1 f)v b g (ρ B =880; ρ AIR =1.3; ρ W 1000; ρ OIL =830 kg/m 3 ) 3.8'

6 where f =1/8. The weight of the block is ρ b V b g, the sum of the terms in f and(1 f ), and buoyancy force, which just balances the weight, is the sum of the other two terms, (-1) (ρ air f + ρ w (1 f )). The density of air is about 800 times smaller than the density of water, so almost all of the buoyancy comes from the water. When oil is added it forms a separate layer that floats on the surface of the water, surrounding the top of the block. The magnitude of the buoyancy force on the upper part of the block, initially ρ air f, increases to ρ oil f, since the density of oil is about 700 times that of air. There is a larger upward force and the block therefore rises into the oil. Eventually it reaches a new position where only a small bit is supported by water, and most of the block is immersed in the oil, and none is in the air. Example 2. Buoyancy due to density variations in a fluid The next example illustrates how buoyancy forces tend to push less dense fluids (e.g. warm air) upwards in the presence of more dense fluids (Fig. 3.5). Figure 3.5. Tank with connected reservoirs, filled with dense fluid (light blue, e.g. cold air, density ρ d) and less-dense fluid (dark red, e.g. warm air, density ρ l )). We assume that the fluids cannot mix. If fluid heights are equal (left panel, height = h), the pressure at the bottom of the left reservoir (ρ d h) will be higher than the right reservoir (ρ l h) because ρ d > ρ l. The denser fluid will flow under the less-dense fluid (right panel), until the pressures are equalized. The tendency for the less-dense fluid to be displaced upward by the more-dense fluid is a consequence of gravity acting on density difference to produce pressure gradients and buoyancy. As discussed above, the pressure in each tank of a pair is equal to the weight of the column of liquid per unit area, i.e. at the bottom of the light blue tank, P blue = ρ blue (height of liquid), and in the dark red tank, P red = ρ red x (height of liquid). Since ρ blue >ρ red, itfollowsthatp blue >P red. This puts a higher pressure on the left side of the tube

7 connecting the two columns, and the liquid will flow from left to right until the pressure is the same on each side of the pipe connecting the tanks. The less dense liquid is buoyant relative to the blue liquid; the difference in density causes the less dense fluid to float on top of the more dense liquid, or or to rise if it is inserted into the denser liquid. Example 3. Buoyancy: warm and cold air The Perfect Gas Law tells us that, if a gas is kept at a constant pressure, increasing the temperature will cause it to expand, reducing its number density and its mass density (Charles' Law). The liquids in the previous example could be cold air (more dense, light blue) and warm air (less dense, dark red). Therefore, if an air parcel is heated it will become less dense than the surrounding air and it will rise, as illustrated in Fig BUOYANCY AND MASS DENSITY Hot and cold gas, both at P = 1 {n, P, T} -->> add heat => molecules move faster hit the piston harder PRESSURE STAYS FIXED the number density is lower for higher T the mass density is lower for higher T -->> the dense (cold) air displaces the less dense (warmer) air under the force of gravity {n/2,p,tx2} Figure 3.6. (upper) If two air parcels are at the same pressure, the warmer parcel is less dense than the colder air, following the Perfect Gas Law: P=ρ 1 R'T 1 =ρ 2 R'T 2 the product ρt=constant. (lower) If a parcel of warm air is surrounded by a cooler volume of air, the pressures are the same in both parcels, thus the density of the warmer parcel is lower than the surroundings and it experiences a buoyancy force just as the fluids in Figure 3.5. cold warm warm air rises above the cold air We now understand the real meaning of the well-known idea that warm air rises. Air rises if it is warmer than surrounding air at the same pressure, because it is less dense, and therefore buoyant. (Hence the phrase is not accurate: it's temperature differences, not "warmth", that counts.) Curiously, the force that pushes buoyant air parcels upwards is the force of gravity, acting in a fluid as illustrated in Figures 3.5 and 3.6. Buoyancy is responsible for the over-turning of the troposphere, as we discuss in the next chapter. The sun heats the earth s surface and air near the surface is warmed, becoming buoyant. As they rise, they lose their heat to the surrounding air or by radiating heat to space, and then eventually sink. The buoyant rising motion is called convection and the over-turning moves heat from the equator to higher latitudes, representing the driving force for climate.

8 3.4 Water vapor in the atmosphere It takes energy to evaporate liquid water, and thus when water vapor condenses, energy is released, warming the air. Thus wet air carries latent energy, the potential for condensation of water vapor to heat the air, a key factor in storms and rain. In addition, water molecules have a lower mass than air molecules, and wet air at a given pressure has lower density than dry air. If water vapor is added to dry air, buoyancy will be created, causing wet air at the same temperature and pressure to be lifted over dry air. In the following section we study evaporation and condensation of water vapor in air, the effect of water vapor on atmospheric density, and the energy content associated with condensation. An air parcel is a useful hypothetical construct, defined as an unconfined volume of air that may be followed around as a unit for some period of time. It must therefore be small enough that it has a single temperature, pressure, and composition, but large enough that it can be treated as a fluid rather than individual molecules. This concept is used to aid the discussion in this and following chapters. Vapor pressure of water and the Clausius-Clapeyron Equation If we place liquid water or ice in a closed vessel with dry air and maintain a constant temperature, we will find that, over time, the amount of water vapor in the air becomes steady. This amount depends only on the temperature of the liquid or solid. Chemists summarize the results of this experiment by plotting the partial pressure P s of water vapor as a function of absolute temperature T of the condensed phase. Partial pressure is defined as the pressure due to water molecules hitting the walls of the vessel, a concept readily visualized in terms of the ping-pong-ball atmosphere (paint the water molecules a distinctive color). The vapor pressure (partial pressure) of water over liquid and solid water increases steeply with temperature (Fig. 3.7). This relationship (the Clausius-Clapeyron Equation, named for the scientists who measured it in the 19 th century), has the approximate form P s = A exp(b(1/ /t), (Eq. 3.9) with A= 6.11 mbar (the vapor pressure at 0 C) and B=5308 ( K). Dew point, frost point, and relative humidity The Clausius-Clapeyron Equation (3.9)(Figure 3.7) describes the maximum content of H 2 O in air at a given temperature. An air parcel can have less if the air is dry, but no more: if we try to add more than this amount of water, liquid or solid particles will condense, removing the excess water until the partial pressure is given by Eq For example, suppose we have an air parcel at 16 C ( K) with 10.2 mbar of water vapor. If I inject a drop of liquid water into this air, the vapor pressure over the liquid is higher than in the air, and the droplet will evaporate, moistening the parcel. If we cool that air to 7.48 C and try again, the vapor pressure over the drop is equal to that in the air, and the drop will remain as it was. The liquid and vapor are said to be in equilibrium, or equivalently, the air at 7.48 C with 10.2 mbar of water vapor is said to be saturated with water vapor (see Figure 3.8). If we cool this air parcel further, water will condense onto

9 the droplet. There are always particles around to help water to condense, so we never observe significant excess vapor pressure over the saturation vapor pressure. T (Celcius) T (Celcius) Vapor pressure of water (mbar) Vapor pressure of water (mbar) Vapor pressure of water (mbar) T (Kelvin) T (Celcius) liquid ice fraction, mole/mole T (Kelvin) T (Kelvin) Figure 3.7a. Vapor pressure of water (mbar) over liquid, plotted as a function of temperature (K). Figure 3.7b (upper). Vapor pressure of water on an expanded scale. The scale on the right shows the fraction of air molecules that are H 2 Oat1atm. pressure for air in equilibrium with water at the given temperature. (lower). Vapor pressures of water ( ) and ice (---) at temperatures below 0 C ( K). Figures 3.7b and 3.8b shows what happens when liquid water is cooled below 0 C: The vapor pressure over ice becomes lower than that over liquid water. This makes ice more stable than liquid, as may be visualized in the following thought experiment. Imagine ice and water both present in a closed vessel at a temperature below 0. The liquid will tend to saturate the air with vapor, but the corresponding partial pressure of water is higher than the saturation pressure over ice (see Fig. 3.7b, lower panel), so the excess vapor will condense on the ice. The process will continue until all the liquid has disappeared and only ice remains.

10 Relative Humidity and Dew Point Relative Humidity and Frost Point Saturated Vapor Pressure (mb) Dew Point 7.48 C Air Parcel Saturated Vapor Pressure (mb) Frost Point -8.3 C Air Parcel Temperature (C) RH = 10.2 mb / 17.9 mb x 100% = 57% Temperature (C) RH = 3 mb / 12.1 mb x 100% = 25% Figure 3.8a Diagram illustrating condensation of water vapor from an air parcel initially at 16 C with 10.2 mb water. Liquid will not condense from this parcel, because the vapor pressure over liquid water at 16 C (18 mb, see diagram) is higher than in the air. As we cool the air (follow horizontal dotted line), condensation will start at 7.5 C, where the vapor pressure over liquid water equals that in the air. If the parcel is cooled further, condensation will occur and the partial pressure of water will fall following the Clausius-Clapeyron equation (solid line). Figure 3.8b Same as for Figure 3.8a, except the parcel is initially at 10 C and contains 3 mb of H 2 O vapor. Condensation starts at -8.2 C, and ice may form rather than liquid water. Since the water vapor content of air is a key property for weather and climate, it is often reported as part of weather forecasts. However, most people are unfamiliar with partial pressures, and therefore alternative measures are used, as illustrated in Fig. 3.8: partial pressure of water, also called absolute humidity P w, is the pressure in Newtons m -2 or mbar due to water molecules only, i.e. equal to n water kt. relative humidity is the ratio of actual water vapor partial pressure P w to the saturation pressure P s. The air parcel in our first example had relative humidity of 57%. specific humidity is the number of grams of water vapor per kilogram of dry air. dew point or frost point refer to the temperature to which an air parcel must be cooled to first make condensation occur.

11 -->> add water keep P & T FIXED (P = n kt, both sides) n in dry piston (unit vol) = n in wet unit vol *BUT < mass air molecule > mass water molecule mass in dry piston > mass in wet piston =>> mass density dry air > mass density wet air ****************************************** Fig 3.9. Density of wet and dry air at fixed pressure P and temperature T. Diagram showing why dry air has higher mass density than wet air when both parcels at the same temperature and pressure. Molecular weights: air=29, water=18 (grams/mole). dry air {n,p,t} wet air {n,p,t} Adding water vapor to air reduces its density. We know from the Perfect Gas Law, P= nkt, that pressure depends on the number density n (number of molecules per cubic meter) and temperature, but not on the mass of the molecules. (Don t be confused by the form of the Perfect Gas Law P=ρR T, the dependence of ρ and R on molecular mass cancel out!) Therefore if we have two air parcels at the same P and T, n must also be the same. However the average molecular mass, and hence the density ρ, is lower in the wet parcel (see Fig. 3.9). Thus addition of moisture to dry air may create buoyancy. Discussion. The rapid rise of P w with increasing temperature is very important for climate. Powerful storms occur in the tropics, and at midlatitudes in summer, such as huge thunderstorms or hurricanes; these storms draw their energy from release of latent heat. The Clausius- Clapeyron equation tells us that warm air can contain much more water vapor than cooler air, and thus potentially much more latent heat can be released to produce buoyancy, convection, and rain or snow. Thus convective storms involving warm, moist air are often more intense than storms that develop in cooler regions. Water vapor is also an important contributor to warming of the surface of the earth through the greenhouse effect, the absorption and re-emission of radiant heat by the atmosphere. Through the Clausius-Clapeyron equation, warming the atmosphere allows higher amounts of water vapor, increasing the efficiency of the greenhouse effect. This phenomenon will be discussed in detail in a later chapter. 3.5: Demonstrations for this chapter These laboratory demonstrations illustrate phenomena discussed in this chapter. 1) Mercury barometer: The height of mercury in the column is supported by air pressure pushing down on the surface of the fluid reservoir. Atmospheric

12 pressure changes produce changes in the height of the column. We use mercury instead of water because the density of mercury is 13.6 the density of water: Height of a mercury barometer = 0.76m (760 mm Hg). Height of a water barometer = 10 m (33 feet!) Difficult to use. 2) Buoyancy of fluids and density. Colored water vs. paint thinner: The two fluids are placed in a glass U-tube. The paint thinner is less dense than the water, and the column of the less dense fluid is push upwards as in Figure 3.5. The pressure at the bottom of the tube with water (neglecting the pressure of the atmosphere, which is the same for both tubes) is P water = ρ water gh, where h = height of the water column. Initially h was the same in the tube with paint thinner, and since ρ thinner < ρ water,p thinner = ρ thinner gh was less than P water. After the valve is opened the less dense liquid rises higher in the tube, equalizing the pressure in both tubes. This demo shows that the driving force for buoyancy, and for stratification (light fluid floating on top of denser fluid) is the difference in mass density (ρ thinner <ρ water) under the influence of gravity. 3) Latent heat of vaporization. We showed that water can be made to boil at room temperature by lowering the pressure to the point at which atmospheric pressure over the water is less than its vapor pressure. As soon as the water begins to boil, its temperature begins to drop--the latent heat of vaporization is supplied by cooling the water itself (and by heat transferred from air around the flask). We can actually make the water freeze if we have a good vacuum line, a phenomenon exploited in freezing of foods and proteins. 3.6 Summary of main points of this chapter Barometric law The pressure in the atmosphere declines with height according to the barometric law, which is given approximately by P(Z) = P o e (-Z/H) where P o is the pressure at the ground, Z is altitude, H is the scale height [H kt/(mg) 7kmonaveragenthe atmosphere, where the mean T ~ 250 K]. Atmospheric pressure declines by factor 1/e 1/2.7 for each scale height or a factor of about 10 for two scale heights. The exact form of the barometric law is P/ P = - Z/ H, where P is the change in pressure that occurs for vertical displacement Z. The barometric law results from hydrostatic balance combined with the Perfect Gas Law: the pressure of the atmosphere at each altitude supports the weight of the overlying atmosphere. Buoyancy Buoyancy is the force of gravity acting on fluids with different density (commonly due to a difference in temperatures); buoyancy may drive atmospheric motions such as convection("warm air rises", more accurately, air warmer than its surroundings becomes buoyant).

13 Water vapor The vapor pressure of water increases steeply (exponentially) as temperature increases. The latent heat of vaporization of water is large (due to hydrogen bonding between water molecules, to be discussed in a later chapter). Transport of water vapor in the earth's atmosphere by winds provide an efficient way to move energy from one location to another. We defined specific humidity (grams of water vapor per gram of air) and relative humidity (fraction or % of saturation vapor pressure) and showed how to calculate these quantities using information about the thermodynamic properties of water, and atmospheric temperature and pressure. Here are some useful numbers about water: molecular weight of water = 18 g/mole (compared to 29 gm/mole = mean molecular weight of dry air). ρ w, density of liquid water = 1 g/cm 3 = 1000 kg/m 3 (compare to air, ρ air = 1.3 kg/m 3 ). 1 kg = mass of 1 liter (1 liter = 1000 cm 3 =10-3 m 3 ) of water. latent heat of vaporization = J/kg to evaporate water. latent heat of sublimation of ice = J/kg (to evaporate ice); latent heat of freezing = J/kg [=( ) 10 6 ] specific heat of water =4.2J/g(or /kg), energy required to raise the temperature of 1 g of liquid water (or 1 kg) by 1 C. The definition of the traditional measure of heat, the calorie, is the heat required to raise the temperature of 1 cm 3 of liquid water by 1 C thus 4.2 J 1 calorie. 3.7 Exercises Problem 1: Compute the temperature rise from condensation of a particular amount of water vapor. Compare to solar input. Problem 2: Compute the density difference between wet and dry air at 305K. Show the effects of this small density difference in the dry line in New Mexico in summer. Problem 3: (a)find the heat required to raise the temperature of 1 mole of water from 0 to 100 C (the boiling point). (b)by what factor is the latent heat of vaporization larger than the amount needed to raise the temperature by 1 C? Problem 4: Hurricane Mitch dropped.7 m of rain over a large area in 2 days. Compute the amount of latent heat release from the storm and compare to total US energy use for a year.

14 Chapter 3: Atmospheric pressure and temperature Distribution of pressure with altitude... 1 The barometric law... 1 The scale height, H, and the simple form of the barometric law Pressure vs depth in the ocean Buoyancy... 4 Example 1: Buoyant block in a tank of water... 4 Example 2. Buoyancy due to density variations in a fluid... 6 Example 3. Buoyancy: warm and cold air Water vapor in the atmosphere... 8 Vapor pressure of water and the Clausius-Clapeyron Equation... 8 Dew point, frost point, and relative humidity : Demonstrations for this chapter Summary of main points of this chapter Barometric law Buoyancy Water vapor... 13