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Chapter 2

Chapter 2.pdf

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Department
Chemistry
Course
CHEM 302
Professor
Michael Wheeler
Semester
Fall

Description
12 CHAPTER 2. ATMOSPHERIC PRESSURE 2.1 MEASURING ATMOSPHERIC PRESSURE The atmospheric pressure is the weight exerted by the overhead atmosphere on a unit area of surface. It can be measured with a mercury barometer, consisting of a long glass tube full of mercury inverted over a pool of mercury: vacuum h A B Figure 2-1 Mercury barometer When the tube is inverted over the pool, mercury flows out of the tube, creating a vacuum in the head space, and stabilizes at an equilibrium height h over the surface of the pool. This equilibrium requires that the pressure exerted on the mercury at two points on the horizontal surface of the pool, A (inside the tube) and B (outside the tube), be equal. The pressure P at point A is that of the A mercury column overhead, while the pressure P Bt point B is that of the atmosphere overhead. We obtain PAfrom measurement of h: PA = ρHg gh (2.1) where ρ Hg = 13.6 g cm is the density of mercury and g =9.8ms -2 is the acceleration of gravity. The mean value of h measured at sea level is 76.0 cm, and the corresponding atmospheric pressure is 1.013x10 kg m -1s-2in SI units. The SI pressure unit is called the -1 -2 Pascal (Pa); 1 Pa = 1 kg m s . Customary pressure units are the atmosphere (atm) (1 atm = 1.013x10 Pa), the bar (b) (1 b = 1x10 Pa), the millibar (mb) (1 mb = 100 Pa), and the torr (1 torr = 1 mm Hg = 134 Pa). The use of millibars is slowly giving way to the equivalent SI unit of hectoPascals (hPa). The mean atmospheric pressure at sea level is given equivalently as P = 1.013x10 Pa = 1013 hPa = 1013 mb = 1 atm = 760 torr. 13 2.2 MASS OF THE ATMOSPHERE The global mean pressure at the surface of the Earth is P = S84 hPa, slightly less than the mean sea-level pressure because of the elevation of land. We deduce the total mass of the atmosphere m : a 2 4πR P S 18 ma= = ------------.-x105 kg (2.2) g where R = 6400 km is the radius of the Earth. The total number of 20 moles of air in the atmosphere is Na= m aM a 1.8x10 moles. Exercise 2-1. Atmospheric 2O concentrations have increased from 280 ppmv in preindustrial times to 365 ppmv today. What is the corresponding increase in the mass of atmospheric carbon? Assume CO2to be well mixed in the atmosphere. Answer. We need to relate the mixing ratio2of CO to the corresponding mass of carbon in the atmosphere.We use the definition of the mixing ratio from equation(1.3, nCO2 N C M a m C C CO2 = =-n------N------M---⋅m------ a a C a where NC and Naare the total number of moles of carbon 2as CO ) and air in the atmosphere, and m and m are the corresponding total atmospheric masses. C a The second equality reflects the assumption tha2 the CO mixing ratio is uniform throughout the atmosphere, and the third equality reflects the relationship N= m/M. The change Δm C in the mass of carbon in the atmosphere since preindustrial times can then be related to the cCO2gin the mixing ratio of CO2. Again, always use SI units when doing numerical calculations (this is your last reminder!): M C 18 12x10 3 —6 —6 Δm C = m=a----⋅ΔC-CO2 5.2x10 ⋅ ⋅-----3--( )5x10---—280x10 M a 29x10 = 1.8x1014 kg = 180 billion tons! 14 2.3 VERTICAL PROFILES OF PRESSURE AND TEMPERATURE Figure 2-2 shows typical vertical profiles of pressure and temperature observed in the atmosphere. Pressure decreases exponentially with altitude. The fraction of total atmospheric weight located above altitude z is P(z)/P(0). At 80 km altitude the atmospheric pressure is down to 0.01 hPa, meaning that 99.999% of the atmosphere is below that altitude. You see that the atmosphere is of relatively thin vertical extent. Astronomer Fred Hoyle once said, "Outer space is not far at all; it’s only one hour away by car if your car could go straight up!" 80 80 Mesosphere 60 60 40 40 Altitude, km 20 20 Stratosphere Troposphere 0 0 0.01 0.1 1 10 100 1000 200 240 280 Pressure, hPa Temperature, K Figure 2-2 Mean pressure and temperature vs. altitude at 30 N, March Atmospheric scientists partition the atmosphere vertically into domains separated by reversals of the temperature gradient, as shown in Figure 2-2. The troposphere extends from the surface to 8-18 km altitude depending on latitude and season. It is characterized by a decrease of temperature with altitude which can be explained simply though not quite correctly by solar heating of the surface (we will come back to this issue in chapters 4 and 7). The stratosphere extends from the top of the troposphere (the tropopause) to about 50 km altitude (the stratopause) and is characterized by an increase of temperature with altitude due to absorption of solar radiation by the ozone layer (problem 1. 3). In 15 the mesosphere, above the ozone layer, the temperature decreases again with altitude. The mesosphere extends up to 80 km (mesopause) above which lies the thermosphere where temperatures increase again with altitude due to absorption of strong UV solar radiation by N and O . The troposphere and stratosphere account 2 2 together for 99.9% of total atmospheric mass and are the domains of main interest from an environmental perspective. Exercise 2-2 What fraction of total atmospheric mass at 30 Nso ihte troposphere? in the stratosphere? Use the data from Figure 2-2. Answer. The troposphere contains all of atmospheric mass except for the fraction P(tropopause)/P(surface) that lies above the tropopause. From Figure 2-2 we read P(tropopause) = 100 hPa, P(surface) = 1000 hPa. The fractiotropf total atmospheric mass in the troposphere is thus P(tr)opopause F trop = 1=—------------------------.-------- P()0 The troposphere accounts for 90% of total atmospheric mass at 30 N (85% globally). The fraction Fstratf total atmospheric mass in the stratosphere is given by the fraction above the tropopause, P(tropopause)/P(surface), minus the fraction above the stratopause, P(stratopause)/P(surface). From Figure 2-2 we read P(stratopause) = 0.9 hPa, so that Fstrat= ---=-------------------------------------0-.------------------------------- P(s)urface The stratosphere thus contains almost all the atmospheric mass above the troposphere. The mesosphere contains only about 0.1% of total atmospheric mass. 2.4 BAROMETRIC LAW We will examine the factors controlling the vertical profile of atmospheric temperature in chapters 4 and 7. We focus here on explaining the vertical profile of pressure. Consider an elementary slab of atmosphere (thickness dz, horizontal area A) at altitude z: 16 horizontal pressure-gradient force area A (P(z)-P(z+dz))A
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