tag:blogger.com,1999:blog-10845272528969418642014-10-04T21:46:54.991-07:00Mistakes in IPCC Global Warming CalculationsP. Wiebarhttp://www.blogger.com/profile/15987516965582899854noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-1084527252896941864.post-72613929563332651212009-08-30T13:54:00.000-07:002009-08-31T14:22:38.700-07:00Mistakes in IPCC Global Warming Calculations<strong>Abstract</strong><br />The United Nations Intergovernmental Panel on Climate Change (IPCC) released its Fourth Assessment Report (AR4) in 2007. Here we show that the accurate value for climate sensitivity is 0.277 K/(W/m^2), which is 3 times smaller than the generally accepted value of 0.8 K/(W/m^2). Thus the climate change on doubling CO2 from 300 ppm to 600 ppm will be 1.0 degree, not 3 degrees. Because the IPCC data show that doubling CO2 will not double absorption of infrared radiation, the Beer-Lambert law is not being followed, because of diminishing returns after more-than-50% absorption. Thus further doublings of CO2 to the point of suffocating levels can only result in a fraction of a degree increase. Therefore global warming by CO2 increases has been wildly overestimated. The same IPCC data show that water vapour is 1.5 times as important as CO2 as a greenhouse gas, and it still seems to follow the Beer-Lambert law (doubling the concentration doubles the absorption). Thus climate changes are more sensitive to changes in water vapour than to CO2. Since water vapour is released on the combustion of alcohols (including methanol and ethanol) and alkanes (including methane, propane, gasoline and diesel fuels), but not coal, and by transpiration in plants in forests and crops, efforts to mitigate climate change by reducing human-produced water vapour would run in exactly the opposite direction to efforts to reduce climate change by controlling CO2 alone.<br /><br /><br /><strong>Diminishing Returns on Doubling CO2 Concentration</strong><br />For reference purposes, let Fig. 1 be the graph titled "Modtran3 v1.3 upward irradiance at 20 km, U.S. Standard Atmosphere" found at en.wikipedia.org/wiki/Radiative_forcing. Using the IPCC value for the average Earth surface temperature of 288.2K (15 degrees Celsius) along with an emissivity of 0.98 in the Stefan-Boltzmann Law [see Appendix 1], the power emitted per square metre is 383 W/m^2. At 300 ppm CO2, the power/m^2 absorbed by CO2 is 383 minus the amount escaping into space = 383 - 260 = 123 W/m^2. At 600 ppm CO2, the power/m^2 absorbed by CO2 is 383 - 257 = 126 W/m^2. The increase is 126 - 123 = 3 W/m^2 (3.39 W/m^2 if we carry the two decimal places quoted in Fig. 1). 3 W/m^2 is only 2.4% of 123 W/m^2 (2.8% of 123 if we use 3.39). So <strong>doubling CO2 from 300 ppm to 600 ppm will increase the absorption by only 2.8% or so</strong>. The Beer-Lambert law says that doubling the concentration of CO2 should <strong>double</strong> the absorption (i.e. increase by 100%, not 2.8%), so clearly the Beer-Lambert law is not being followed. The reason is that CO2 is a powerful greenhouse gas, and absorption at the resonant frequencies is almost total over a path length of 20 km. Because of diminishing returns [see Appendix 2], a further doubling of CO2 from 600 ppm to 1200 ppm will result in a further increase in absorption by only 1.4%, and a correspondingly smaller climate change. Therefore calculations assuming linear increases in temperature with linear or <strong>even geometric </strong>increases in CO2 will all be wrong, with the errors increasing the longer the time of projection.<br /><br /><br /><strong>Calculation of Climate Sensitivity </strong><br />For a grey body with emissivity E and absolute temperature T, the power/m^2 emitted is<br /><br />j = EST^4 [see Appendix 1], where S is the Stefan-Boltzmann constant<br /><br />Therefore dj/dT = 4EST^3.<br /><br />Therefore dj/j = 4 dT/T, or dT/T = 1/4 dj/j [Equation 1].<br /><br />This says that the relative (or %) change in temperature is 1/4 the relative (or %) change in power/m^2 emitted. If T = 288.2 K, and j = 260 W/m^2 [see Fig. 1], then dT = 0.277 dj. Equating dj with "delta F", the radiative forcing for a climate change of "delta T" (dT), then climate sensitivity is 0.277 K/(W/m^2), and this is only 35% of the generally accepted value of 0.8 K/(W/m^2) calculated from experimental correlations over long periods of time (decades, centuries or millenia). Therefore all literature predictions of climate temperature changes will be too high by at least a factor of 3. For example, a doubling of CO2 from 300 ppm to 600 ppm has "delta F" = 3.7 W/m^2, predicting a global warming of 0.277 x 3.7 = 1.0 degree, instead of the literature value of 0.8 x 3.7 = 3 degrees. Recent measured increases of CO2 from 300 ppm to 400 ppm should therefore result in a climate change of only a fraction of a degree. Therefore larger climate changes over the last few decades cannot be explained by CO2 increases alone.<br /><br />The correct value for climate sensitivity could have been calculated from Semi-Global Cooling, the drop in temperature on the night side of the Earth when there is no direct solar heating [see Appendix 3].<br /><br /><br /><strong>Effect of Many Doublings of CO2 Concentration</strong><br />It might be argued that even a 1.0 degree rise on doubling CO2 from 300 ppm to 600 ppm is worrisome, for it is possible to more-than-double CO2. However, because of diminishing returns [see Appendix 2], the sum of an infinite series gives 2.0 degrees as the limit even at 100% CO2 atmosphere [i.e. a total climate change of 2.0 degrees when CO2 increases from 0.03% (300 ppm) to 100% (1,000,000 ppm)]! This conclusion would invalidate all concerns about excess CO2 produced by human beings (except for acidification of the oceans), if nature can withstand changes of 2.0 degrees, as shown by the historic record including volcanic eruptions and ice ages.<br /><br /><br /><strong>Other Factors that Affect Climate Change</strong><br />From Equation 1, climate sensitivity = T/(4j), which is a function of T and j. The irradiance outwards, j, is sensitive to the albedo, which varies with cloud cover and smoke/dust in the atmosphere. If 260 W/m^2 is the net loss of infrared radiation to outer space both at night and in the daytime [see Fig. 1], then 2(260) = 520 W/m^2 must be the power/m^2 distributed over the daylit hemisphere, or 2(520) = 1040 W/m^2 must be the power/m^2 with the Sun directly overhead. This is 76.1% of the total insolation of 1370 W/m^2 [see Appendix 4 or en.wikipedia.org/wiki/Insolation]. This would mean an albedo of 0.24. If the albedo changed from 0.24 to 0.25, the power/m^2 reaching the Earth's surface would be 0.75 x1370 = 1028 W/m^2, and the outward irradiance at night time would be 1028/4 = 257 W/m^2. This is a 3 x 100%/260 = 1.2% change from our present IPCC value of 260 W/m^2. Therefore the predicted global temperature change would be 1.2%/4 = 0.3%, or 0.3 x 288/100 = 0.9 degrees. Thus a change in albedo from 24% to 25% would effectively balance any global temperature rise of 1.0 degree caused by doubling CO2 from 300 ppm to 600 ppm in the atmosphere. This could be caused by an increase in water vapour forming clouds (perhaps induced by global warming initiated by an increase in CO2), or by an increase in cosmic rays which would produce ions which would act as nucleation sites for water droplets or ice crystals. Similarly, a 1.2% drop in insolation reaching the Earth's surface due to increased dust or ash from a volcano, massive forest fires, and human activities like factory smokestacks would also cool the Earth by 0.9 degrees.<br /><br /><br /><strong>Pure Rotational Spectrum & Water Vapour </strong><strong>Absorbance </strong><br />Like vibrations in molecules [see Appendix 5], rotational energy levels in molecules are also quantized, with energies increasing as J(J+1), where J is a rotational quantum number that can take on only integer values of J = 0, 1, 2, 3,... As an approximation, each widely-separated vibrational energy level acts as a base for a stack of increasingly widely separated rotational energy levels [since the difference between successive levels is (J+1)(J+2) - J(J+1) = (J+1)(J+2-J) = 2(J+1), which increases linearly with increasing J]. Thus there is a forest of different transitions possible between states when there is a change in the vibrational quantum number. This explains why the infrared spectra of molecules show bands rather than discrete widely-spaced lines of atoms in the gas phase.<br /><br />Because of symmetry, N2, O2, CO2, and CH4 (methane) do not possess permanent electric dipole moments, and therefore do not jump between rotational energy levels by absorbing a photon unless there is also a change in a vibration (or electronic state). They can, however, make quantum leaps between rotational energy levels on collisions with other molecules, and this is one way that energy is distributed until an equilibrium temperature is reached.<br /><br />Because the H2O molecule possesses a permanent electric dipole moment, it can absorb photons that cause quantum leaps between rotational energy levels within the same vibrational state. Its pure rotation spectrum includes radiation with wavenumbers around 300 cm^-1 (corresponding to wavelengths around 33 microns) [see Herzberg, p. 58]. A useful application of the water molecule's rotational energy spectrum is in microwave ovens.<br /><br />With this insight, let's re-examine Fig. 1 in this article more closely. The main absorption valley at 600-750 cm^-1 is indisputably due to the bending mode of the CO2 molecule [see Appendix 5]. Since this is the lowest energy transition possible between vibrational levels of the CO2 molecule, the jagged absorption spikes at wavenumbers lower than 600 cm^-1 cannot be due to absorption by CO2 (since transitions between rotational levels in the same v=0 vibration are not possible for molecules with no permanent electric dipole moment). The distinctive twin spikes around 1050 cm^-1 in Fig. 1 must be due to absorption by ozone, O3 (largely formed in the stratosphere when ultraviolet rays from the Sun dissociate O2 molecules, and the resulting reactive atoms combine with O2) [see "A Physicist's Desk Reference", p.273]. The broad absorption in Fig. 1 from about 1250 to the edge of the graph at 1500 cm^-1 matches the bond bending H2O absorption from about 1250 to 2000 cm^-1. Therefore it is reasonable to assume that the absorption in Fig. 1 below 600 cm^-1 is due to the pure rotational spectrum of the H2O molecule (with all vibrations in their v=0 states, and with only changes in the rotational quantum number J).<br /><br />Therefore the areas between a smooth 288 K black body radiation curve and the absorption valleys corresponding to CO2, H2O, and O3 in Fig. 1 give a direct measure of the power/m^2 absorbed by the three greenhouse gases. Because the horizontal axis in Fig. 1 is in wavenumbers, which are proportional to frequency and therefore to photon energies, equal areas on all sides of the graph correspond to equal energies or powers [This would not be true if the graph had been plotted vs wavelength]. The sides of the CO2 absorption at 600-750 cm^-1 were extended until they met at the horizontal axis, forming a "V"; this is consistent with nearly-total absorption [see Appendix 6, and previous discussion].<br /><br /><strong>[Greenhouse Gas; Power/m^2 Absorbed by Greenhouse Gas;</strong><br /><strong>% of Total 383 W/m^2 Absorbed by Greenhouse Gas;</strong><br /><strong>% of Total Greenhouse Gas Absorption]</strong><br /><br />[ CO2: 48.0; 12.5; 38.3 ]<br /><br />[ H2O: 71.2; 18.6; 57.1 ]<br /><br />[ O3: 5.9; 1.5; 4.6 ]<br /><br /><strong>[Total for 3 Gases: 125.1; 32.6; 100.0 ]</strong><br /><br />The above results show that water vapour is 1.5 times more important than CO2 as a greenhouse gas, so that the IPCC assumption at the start to ignore water vapour was flawed. Their own data show that water vapour accounts for 57% of the absorption of infrared radiation (the greenhouse effect); according to <a href="http://www.lsbu.ac.uk/water/vibrat.html">www.lsbu.ac.uk/water/vibrat.html</a>, H2O is responsible for 70%. The mechanism is the same as for CO2: infrared photons emitted by the Earth's surface are absorbed by these molecules in quantized jumps between vibrational energy levels [see Appendix 5] and for H2O, between rotational energy levels of the same vibrational state. These excited molecules can lose energy either by emitting another infrared photon or by collision with other molecules, in particular molecules of N2 or O2 in the air. At equilibrium, the energy transferred equally to all components of translation and rotation will have resulted in an increase in temperature. I.e. the air will have become warmer (the greenhouse effect).<br /><br />If 125.1 W/m^2 is the total absorbed by all three gases in the atmosphere, then 383 - 125.1 = 258 W/m^2 will escape into outer space. This is consistent with the IPCC value in Fig. 1 of 260 W/m^2, the discrepancy of less than 1% being within the expected range of calculating areas under hand-drawn curves on graph paper.<br /><br />Because the 1200 cm^-1 absorption for methane (CH4) and nitrous oxide (N2O) does not show up in Fig. 1 [see A Physicist's Desk Reference, p. 273], we can conclude that these gases at the moment are not significant for global warming.<br /><br />The vapour pressure of water increases roughly exponentially with increasing temperature [see Appendix 7]. Therefore the possibility exists that there is a positive feedback loop initiated by increasing CO2, in which increasing temperatures evaporate more water vapour, which traps more heat, which raises the water temperature, etc. Water vapour is also released into the atmosphere by the burning of ethanol, methane, other alkanes including propane, gasoline and diesel fuel (but not coal), and by transpiration in plants, which act as wicks transporting liquid water in the ground (where it does not contribute to the greenhouse effect) into the atmosphere through the stomates (pores in the leaves). Because Fig. 1 shows that water vapour absorption still follows the Beer-Lambert law (the absorption valleys have not reached the horizontal axis corresponding to 50%-or-more absorption, unlike the case for CO2), % changes in water vapour will result in greater climate changes than the same % changes in CO2. In order to reduce global warming caused by increased water vapour, we might argue for burning coal (which doesn't produce greenhouse water vapour, although it of course does produce carbon dioxide) instead of natural gas (which is mainly methane), and clear-cutting forests and crops to reduce transpiration of water vapour. Obviously, economic and social factors must then be considered in the real world where compromises must be made.<br /><br />Although water vapour is a greenhouse gas, at cooler temperatures at higher altitudes, this vapour could form clouds made of tiny liquid droplets or ice crystals which would reflect incoming sunlight back into outer space. This would result in global cooling, which would counteract the previously-explained greenhouse effect. Differing assumptions about the extent of cloud formation would lead to large uncertainties in predicted global temperatures, and this might have been a factor in the IPCC's decision to assume no change in water vapour.<br /><br /><br /><br /><strong>Prediction of Daily Temperature Swings over Desert Surfaces</strong><br />A check on our calculations, and on the relative importance of water vapour and carbon dioxide may be obtained by considering day/night temperature swings over extended desert regions and cloudless doldrum areas of the open ocean. Deserts on the Earth occur in strips about 25 degrees from the equator. Therefore when the incoming Solar radiation of 1370 W/m^2 is spread out over a semi-circle instead of a diameter at latitude 25 degrees, and after weighting by the factor cos25 to account for the angle to the Sun away from directly overhead, the perpendicular power reaching the outer Earth along a narrow strip where deserts form is cos25 x 2/pi x 1370 W/m^2 = 790 W/m^2. Let us assume that the albedo remains unchanged overall at 0.24, meaning that from outer space the cloudless desert areas reflect back the same % of visible light as the average of the cloud-covered and cloudless oceans and lands including dark forests, crops, tilled, burning and clear-cut areas. Then the outward irradiance at 20 km at night would be 0.76 x 790/2 = 300 W/m^2. This would just balance the net 600 - 300 = 300 W/m^2 gained in the daytime. At 288.2 K mean surface temperature, the Earth loses 260 W/m^2 at 20 km, so the difference of 300 - 260 = 40 W/m^2 is an increase of 15.4%. Therefore the Earth's mean surface temperature must rise by (15.4)/4 x 288/100 = 11 degrees along the strip. Therefore the mean temperature along the desert strip is 15 + 11 = 26 degrees Celsius, or 299 K. According to our previous calculation [see Table], the power/m^2 absorbed by water vapour is 71.2 W/m^2, which is 18.6% of the total 383 W/m^2 emitted from the Earth. Therefore the % temperature change due to the complete absence of water vapour is 18.5%/4 = 4.6%, and 4.6% of 299 K is 14 degrees. For extended deserts, if we assume water vapour is negligible all the way up to 20 km, the predicted temperature should therefore range on average from 12 degrees Celsius at night to 40 degrees Celsius in the daytime. If the daily temperature curve is sinusoidal, the temperature swing from nighttime minimum to daytime maximum would be 14pi = 44 degrees, corresponding to a range from 4 degrees Celsius to 48 degrees Celsius (from near-freezing to blistering hot). These numbers are consistent with desert experience.<br /><br />By contrast, if we accept the IPCC assumption that all 123 W/m^2 of the greenhouse effect is due to CO2, using their best value of climate sensitivity = 0.8 K/(W/m^2) assuming it is a constant, the temperature change due to the presence of CO2 would be 0.8 x 123 = 98 degrees on average. This absurd value is derived from a wrong assumption about the importance of only CO2 and a wrong value for climate sensitivity and a wrong assumption that it is constant. As we showed previously, climate sensitivity is T/(4j), and since j varies as the 4th power of T, climate sensitivity varies inversely as the cube of the absolute temperature. For temperatures more than a few degrees different from 15 degrees Celsius, climate sensitivity cannot be assumed to be a constant. This must be understood if one uses ice age data, or tries to calculate the total greenhouse effect for any one gas, or combination of gases. In these cases, the rule to use is that the % change in climate temperature is 1/4 the % change in irradiance outward.<br /><br /><br /><strong>Effect of Water Vapour over Areas other than Deserts</strong><br />For land areas other than deserts, water vapour in the atmosphere will provide a moderating greenhouse effect at night, even when it is cloudless. Data in the literature might possibly be used by others to connect this moderation with measured water vapour pressures (the relative humidity times the saturated vapour pressure at any temperature). Then an estimate of the radiative forcing due to change in water vapour might be possible.<br /><br />One complication arises when there are significant changes of state, for example, dewing on blades of grass. During the daytime, some of the incoming solar power evaporates water, and energy is absorbed without a temperature change (e.g. boiling water stays at 100 degrees Celsius at 760 mm Hg pressure). The energy is used to separate water vapour molecules from the surface of bulk liquid. At night, the air temperature goes down as infrared radiation is lost to outer space. When the air becomes saturated with water vapour, any further loss of heat results in the formation of liquid water droplets (dew) on solid surfaces such as blades of grass. Some of the energy of the separate water vapour molecules is transferred to the grass blade as they condense, and then to the surrounding air and ground by conduction and convection. This would make up for some of the energy lost to outer space by infrared radiation. Dew consists of hemispheres approximately 4 mm in diameter. Equating the volume of a hemisphere 2pi x r^3/3 with a cylinder of volume pi x r^2h, for r = 2mm, h = 2r/3 = 4/3 mm. After a warm summer's night, each square metre of earth might end up with a layer of liquid water approximately 1 mm high. This would correspond to a volume of (100 cm)(100 cm)(0.1 cm) = 1000 mL = 1000/18.0 moles of liquid water. Since the latent heat of vaporization of liquid water at 25 degrees Celsius is 44.0 kJ/mol [see Appendix 7], the heat released by condensing water per square metre = 56 x 44 kJ. When this is divided by the number of seconds in a 12-hour period, the power/m^2 released is 56 x 44 kJ/[12(3600)s] = 57 W/m^2. This is the same order of magnitude as the 71 W/m^2 absorbed by the water vapour in the atmosphere to a height of 20 km [see previous Table]. Therefore the heat of vaporization cannot be ignored as a mechanism for moderating temperatures over the Earth's surface where water can condense at night, either over land or over liquid water. Therefore measurements of the heat absorbed at night by water vapour must be done where dewing is insignificant. In addition, further experiments measuring the weight gain due to condensation on various flats containing grass, shrubs, decorative plants, etc. could determine effective values for power/m^2 released, for in our example the increase was 1000 g, which should be easy to measure.<br /><br />Any Canadian who has had to scrape the frost off car windshields in the morning knows that water vapour pressure even during winter is not zero. Therefore water vapour even in winter provides a greenhouse effect, complicated by the change of state called sublimation, both in daytime and night time. Flats containing plants or bare dirt as well as plate glass could be weighed before and after frost buildup overnight to determine values of power/m^2. In these cases, the heat of sublimation, which is the sum of the heat of vaporization and the heat of fusion, must be used at the appropriate temperature (all such heats depend slightly on heat capacities of solid, liquid and vapour phases, and the temperature difference from the standard temperature at which literature values of heats were measured - see Appendices 7 & 8). The slightly negative oxygen atoms displayed on the surface of windshields made of silicate glass will nucleate ice crystals more effectively than the non-polar molecules in the lacquer on the car's body panels, so different materials may end up with different amounts of frosting per m^2.<br /><br /><br /><br /><strong>Greenhouse Effect on Venus</strong><br />The idea of global warming caused by greenhouse gases, in particular CO2, may have come from the known conditions on the planet Venus, so close to the Earth in size, and only a little closer to the Sun. Its surface temperature of 460 degrees Celsius (733 K) is attributed to the fact that the mainly carbon dioxide atmosphere has a pressure 92 times that of the Earth [see 2009 RASC Handbook, p. 218]. In comparison, the mean Earth temperature is 288 K (15 degrees Celsius), and the partial pressure of CO2 is 0.0003 atmosphere, more than 300,000 times smaller than that on Venus. This fact may have led to the false assumption that linear or even geometric increases in CO2 concentration on Earth will lead to linear increases in surface temperature, until the Earth resembles Venus.<br /><br />The mean Sun-Venus distance is 0.7233 astronomical units [RASC Handbook, p. 23] = 0.7233 x 1.496 x 10^11 m = 1.982 x 10^11 m, with an albedo of 0.65.<br /><br />Since the power output of the Sun is 3.855 x 10^26 W [see Appendix 4], the insolation over the circular cross-section of Venus is 3.855 x 10^26 W/[4pi(1.082 x 10^11 m)^2] = 2620 W/m^2. This means for Venus' daytime side, the insolation is 1310 W/m^2 when the power is spread out over the daylit hemisphere instead of the circular cross-section. Assuming an albedo of 0.65, the heat absorbed by Venus in daytime is 0.35 x 1310 = 458 W/m^2. Half this power/m^2 (i.e. 229 W/m^2) must be continually emitted in all directions both in daytime and night time in order to have a net energy change of zero over time.<br /><br />But the Stefan-Boltzmann Law [see Appendix 1] for a surface at 733 K says that the power/m^2 emitted, assuming emissivity = 1, is (5.678 x 10^-8)(733)^4 = 1.64 x 10^4 W/m^2. This is 71 times our estimate of the power/m^2 that must be radiated outward at night to balance the daytime insolation. Therefore the high temperature at the surface of Venus is not due to incoming solar radiation directly, but is due to an initial even higher temperature when the planet was formed by the bombardment of incoming asteroids, comets, etc. This cooled by black body radiation to outer space, but at the moment most of the energy emitted from the planet's surface <strong>is </strong>trapped by the greenhouse gases in the atmosphere. The power/m^2 emitted into outer space on the night side of Venus is just balanced by the <strong>net </strong>input (Solar insolation minus power emitted into outer space in all directions) during the Venus daytime.<br /><br />By contrast, the incoming solar radiation at the Earth's surface in daytime is <strong>greater than </strong>that emitted from the surface on the night side, and so the greenhouse effect, though not zero, is not as important in maintaining the mean temperature as on Venus.<br /><br />Despite diminishing returns on doubling concentrations [see Appendix 2], CO2 will absorb essentially 100% of the infrared radiation at its resonant frequencies in Venus' atmosphere, but there will be gaps between the quantized vibration-rotation bands<strong>. Some other greenhouse gases </strong>must be responsible for absorbing almost all of the remaining frequencies. This conclusion has been missed by almost all researchers to date.<br /><br />It is suggested here that non-linear SO2, trigonal planar SO3, H2SO4, and dimers of H2SO4 should be the first to be investigated by others for any presence in the solar absorption spectrum as seen by spacecraft near the surface of Venus. The shapes are explained by Valence Shell Electron Pair Repulsion (VSEPR) Theory, consistent with non-zero electric dipole moments, except for SO3. This means except for SO3 they will be able to produce pure rotational spectra [see previous discussion]. As well, all of these molecules will be able to produce vibration-rotation spectra [see Appendix 5]. The SO2 molecule will have 3(3) - 6 = 3 normal modes of vibration, the SO3 molecule 3(4) - 6 = 6 vibrations, and the H2SO4 molecule 3(7) - 6 = 15 vibrations. It should be noted that these H2SO4 (sulfuric acid) molecules are in the gas phase, analogous to water vapour, while the highly reflective clouds on Venus are made of droplets of liquid sulfuric acid, analogous to Earth's clouds of liquid water droplets [see RASC Handbook, p.218]. VSEPR Theory correctly explains the tetrahedral geometry around the central sulfur atom in H2SO4 which will have polar S=O, S-O, and O-H bonds which will not all cancel out to produce a net zero electric dipole moment. Therefore this molecule with its 15 vibrations will be particularly important in absorbing infrared radiation emitted from the hot surface of Venus, and transferring this energy on collision to the major gas molecule in the atmosphere, CO2. It is possible that hydrogen bonds between the H-atoms in one H2SO4 molecule with O-atoms in another molecule will be strong enough to create a dimer, (H2SO4)2, whose complex vibrations and rotations will be capable of absorbing other resonant frequencies.<br /><br /><br /><strong>References:</strong><br />P.W. Atkins, "Physical Chemistry, Third Edition" (W.H. Freeman and Company, New York, 1978).<br /><br />Gerhard Herzberg, "Molecular Spectra and Molecular Structure. II. Infrared and Raman Spectra of Polyatomic Molecules" (D. Van Nostrand Company, 1945)<br /><br />Harold S. Johnston, "Gas Phase Reaction Rate Theory" (the Ronald Press Company, 1966)<br /><br />Walter J. Moore, "Physical Chemistry, Third Edition" (Prentice-Hall, 1962)<br /><br />Observer's Handbook 2009 (The Royal Astronomical Society of Canada)<br /><br />A Physicist's Desk Reference, The Second Edition of Physics Vade Mecum, Herbert L. Anderson, Editor in Chief (American Institute of Physics, New York, 1989)<br /><br />Handbook of Chemistry and Physics, 45th Edition (The Chemical Rubber Co., 1964)<br /><br />Handbook of Chemistry and Physics, 80th Edition (The Chemical Rubber Co., 2000)<br /><br />en.wikipedia.org/wiki/Radiative_forcing<br /><br />en.wikipedia.org/wiki/Climate_sensitivity<br /><br />en.wikipedia.org/wiki/Stefan_Boltzmann_law<br /><br />en.wikipedia.org/wiki/Insolation<br /><br />en.wikipedia.org/wiki/IPCC_Fourth_Assessment_Report<br /><br /><br /><br /><strong>Appendix 1: Stefan-Boltzmann Law </strong>[see wikipedia article]<br />For a black body, the Stefan-Boltzmann Law states that the emissive power or irradiance is<br />j = ST^4 where T is the absolute temperature in K<br />and S = 5.67 x 10^-8 J.s^-1.m^-2.K^-4 is the Stefan-Boltzmann constant.<br /><br />For a grey body (which doesn't absorb or emit the full amount), j = EST^4 where E is the emissivity. In the table below, assume E = 0.98 in keeping with the IPCC calculations.<br />The final column, 0.68j, is the irradiance escaping into space if 32% is absorbed by greenhouse gases.<br /><br /><br /><strong>[T(deg Celsius);---T(K)= T;---- j= 0.98ST^4;----0.68j--]</strong><br /><strong>[--------------------------------------(W/m^2);--- W/m^2)]</strong><br /><strong></strong><br /><strong></strong>[ -53; 220; 130; 89 ]<br /><br />[ -23; 250; 216; 147 ]<br /><br />[ -11; 262; 260; 177 ]<br /><br />[ 0; 273; 309; 210 ]<br /><br />[ 2; 275; 318; 216 ]<br /><br />[ 5; 278; 332; 226 ]<br /><br />[ 10; 283; 356; 242 ]<br /><br />[ 15; 288.2; 383; 260 ]<br /><br />[ 20; 293; 410; 278 ]<br /><br />[ 25; 298; 438; 298 ]<br /><br />For comparison, for the Sun at 5780 K and E = 1, j = 6.328 x 10^7 W/m^2.<br /><br />The data at 288.2 K correspond to the IPCC measurement of 260 W/m^2 irradiating into space at 300 ppm CO2 [see Fig. 1 of this article]. Without the greenhouse effect, the value of 260 in the third column corresponds to a surface temperature of -11 degrees Celsius. This is 26 degrees below 288.2 K = 15 degrees Celsius, with the greenhouse effect.<br /><br />Note that the 4th column shows that for a small temperature range (e.g. around 2 degrees) around the mean Earth temperature of 288.2 K, the irradiance outward is roughly constant. Around 15 degrees Celsius, a change of 5 degrees either way corresponds to a change in 0.68j of about 18 W/m^2. This means the climate sensitivity is around 5/18 = 0.28 K/(W/m^2), not 0.8 K/(W/m^2) which is the generally accepted literature value to date.<br /><br /><br /><br /><strong>Appendix 2: Diminishing Returns on Doubling CO2 Concentrations</strong><br />Let rectangle ABCD represent a <strong>100% </strong>absorption spectrum, where horizontal line segment AB represents the wavenumber range (in cm^-1) at the top of the diagram. Horizontal line segment CD below represents 100% absorption (99.9% is good enough, allowing for a tiny leakage due to a decreasing exponential transmission) from the Earth's surface to 20 km altitude for each wavenumber in the range. Let E and F be the midpoints of AB and DC respectively. Let P be any point on the line of symmetry EF. When P is at E, there is no absorption spectrum, but as P moves down towards F, the area of triangle APB increases proportionately with EP, since the base stays the same at AB. From E to F, the Beer-Lambert Law will be followed: as the concentration of the absorbing molecule (e.g. CO2 or H2O) increases, the % absorbed increases linearly. For example, if M is the center of the rectangle, triangle AMB represents <strong>25% </strong>and triangle AFB represents <strong>50% </strong>absorption. We have used <strong>bold numbers </strong>to represent areas of the absorption curve. Doubling the concentration of the absorbing molecule would double the absorption until triangle AFB is reached. Note that the slope of AF is twice the slope of AM.<br /><br />Now imagine doubling the concentration of the absorbing molecule once again, so that P is at point H, on line segment EF extended so that EF = FH. Let AH and BH intersect DC and K and L respectively. Because more than 100% of any infrared energy emitted from the Earth's surface cannot be absorbed, the area of triangle KHL cannot be part of the spectrum, which now has the shape of trapezoid AKLB, whose area by symmetry is now only <strong>75%, not 100%. </strong>So doubling the concentration will not double the absorption, but by only a factor of 50% <strong>(50% to 75%). </strong>The area <strong>not </strong>absorbed has gone from<strong> 50% to 25%, </strong>a decrease by a factor of 2. Note that the slope of AH is twice the slope of AF.<br /><br />If the concentration of absorbing molecules is again doubled, then the slope of AP becomes double that of AH, and the point of intersection of AP with CD becomes the midpoint of DK. The area <strong>not </strong>absorbed becomes half of <strong>25%, or 12.5%, </strong>with the area of the trapezoid increasing to<br /><strong>75% + 12.5% = 87.5%. </strong>The result of this further doubling of concentration is an increase in absorption by only a factor of 1/6 = 16.7% (from <strong>75% to 87.5%), </strong>not 100%. This is the effect of diminishing returns even as one DOUBLES the concentration each time (a geometric, not a linear, increase).<br /><br />Is the concentration of CO2 in the atmosphere high enough even at 0.03% (300 ppm) for diminishing returns to occur? The unusual bottom to the CO2 absorption at 600-750 cm^-1 [see Fig. 1] suggests that the authors mistakenly added the 220 K black body curve to a flat 100% absorption plateau corresponding to KL in trapezoid AKLB [see Appendix 6 for discussion]. This would mean that already more than <strong>50% </strong>of the infrared radiation at the CO2 frequencies are being absorbed, with diminishing returns already occurring.<br /><br />As previously calculated using the IPCC data, a doubling of CO2 from 300 ppm to 600 ppm would mean a change of about <strong>2.8%, </strong>from <strong>94.4% to 97.2% </strong>absorption, if all 123 W/m^2 absorbed by greenhouse gases were due to CO2 alone. This would account for a climate change of 1.0 degree, using the correct value for climate sensitivity of 0.277 K/(W/m^2). Even if we went to <strong>100% </strong>total absorption over the absorption range AB, because only another <strong>2.8% </strong>is available for absorption, this means only another 1.0 degree is possible even if we increased the CO2 concentration from 0.06% (600 ppm) to 100% (1,000,000 ppm)! This incredible (at first thought) result comes from the fact that the sum of the infinite geometric series 1/2 + 1/4 + 1/8 + 1/16 + ... = 1.0.<br /><br />It was shown previously that the power/m^2 absorbed by CO2 is only 48.0 W/m^2, not all 123 W/m^2 of the total greenhouse effect, because water vapour is the major absorber. This would mean that the IPCC result for the radiative forcing (delta F = 3.7 W/m^2) due to increasing CO2 from 300 ppm to 600 ppm would be a 7.7% change in outward irradiance, not 2.8%. Since 2 x 7.7% = 15.4%, this simply means that the doubling from 300 pm to 600 ppm occurs with a change in absorbance from <strong>84.6% to 92.3% (</strong>an increase of <strong>7.7%) </strong>which would result in a 1.0 degree climate change. However there is only another <strong>7.7% </strong>remaining until <strong>100% </strong>absorbance, so a rise to 1,000,000 ppm would result in only another 1.0 degree rise, exactly the same result as before. The % greenhouse effect due to CO2 changes the magnitude of the % absorbed, but this factor cancels out, since the % remaining changes by the same factor, leaving the ratio of 1:1 unchanged. The geometric reason is that as the slope of AK doubles, corresponding to doubling the absorbance for wavenumbers that follow the Beer-Lambert law in the non-trapezoidal part of the diagram, the point of intersection K with the CD axis moves exactly half-way toward D.<br /><br />Finally, a 100% CO2 atmosphere is not quite the sum of an infinite series. 100% is greater than 0.03% by a factor of 3333. Since 2^11 = 2048, and 2^12 = 4096, 100% CO2 results after a little more than 11 doublings, not an infinite number. The shape of the trapezoid AKLB will be hardly distinguishable from that of the rectangle ABCD, however. And by Fourier synthesis, any reasonable line shape for the absorption curve at less than 50% absorbance (e.g. a Gaussian) may be approximated by a linear combination of triangles whose frequencies are multiple of the fundamental one shown so far.<br /><br />This phenomenon of diminishing returns for doublings of CO2 concentrations may not be true for all wavenumbers at which H2O absorbs, for according to Fig. 1, the water absorption peaks have not yet formed trapezoidal absorption spectra. Therefore water vapour will continue to follow the Beer-Lambert Law (doubling the concentration will double the absorption), and therefore % changes in water vapour will be more important for climate change than equal % changes in CO2, in addition to its present factor of 1.5 greater importance. Thus efforts to control climate by controlling CO2 alone may have to be re-evaluated.<br /><br /><br /><br /><strong>Appendix 3: Calculation of Climate Sensitivity from Semi-Global Cooling</strong><br />The correct value of climate sensitivity could have been estimated from semi-global cooling, the temperature drop at night as follows: Assume a generous swing of 20 degrees over land from nighttime minimum to daytime maximum (e.g. from 5 to 25 degrees Celsius), and a swing of 1 degree over the water in the oceans and lakes.. The heat capacity of liquid water is 9.1 R per mole [in Appendix 8, a crude theoretical explanation for this number is given]. At 288 K, the molar heat capacity of metallic solids is 3R [see Appendix 8 for an explanation]. So the molar heat capacity of liquid water is about 3 times that of metallic solids. Two-thirds of the surface of the Earth is water in the oceans and lakes. Therefore the temperature change over water should be weighted 3(2) = 6 times that over land. Assuming that equal surface areas would show equal numbers of moles of liquid water and solid land particles represented by metallic atoms (obviously not exactly true, but an approximation for this calculation), the weighted average temperature swing would be [(6)(1) + 1(20)]/7 = 3.71 degrees, or a maximum drop at night from the mean of 1.85 degrees. This must be weighted by another factor of 2/pi to get a reasonable average temperature drop at night of delta T = 1.2 degrees, assuming a sinusoidal temperature curve over a 24-hour period. Using the IPCC value of 32% absorption of infrared radiation by greenhouse gases, allowing 0.68 to escape from the Earth, and emissivity = 0.98, the Stefan-Boltzmann Law [see Appendix 1] gives the outward irradiance from a 288.2 K surface as 0.68(0.98)(288.2K)^4 = 260.7 W/m^2. If the temperature dropped an average of 1.2 degrees to 287.0 K, the outward irradiance would be 0.68(0.98)(287.0)^4 = 256.4 W/m^2. The difference in outward irradiance would be 260.7 - 256.4 = 4.3 W/m^2. Then the value of climate sensitivity would be 1.2 K/(4.3 W/m^2) = 0.28 K/(W/m^2), exactly the same as our value of 0.277 K/(W/m^2) calculated from first principles. This is not an accident. The result ultimately depends only on the two temperatures and the Stefan-Boltzmann Law, assuming equal % absorption in the atmosphere and equal emissivities at the two close temperatures over a 12-hour period. All the assumptions about heat capacities, areas of water and land, and the numbers of moles of water and metallic atoms representing land were used to derive a reasonable value for the <strong>average </strong>surface temperature change at night, but do not affect the final value of climate sensitivity. The reason is that there is a direct cause-and-effect relationship between temperature change and irradiance change over a 12-hour period of night time. The same cannot be said for correlations of changes over periods of decades, centuries or millenia, during which time cloud cover, volcanic ash & dust, cosmic ray flux, solar activity, CO2 and H2O concentrations, etc. all change as well as the mean surface temperature.<br /><br />Although this sample calculation gives the correct value for climate sensitivity at 288.2 K, it fails to give the functional relationship, namely T/(4j), which shows that it is not a constant when temperature changes [j varies as the 4th power of T, so climate sensitivity varies inversely as the cube of the absolute temperature].<br /><br /><br /><br /><strong>Appendix 4: Calculation of Irradiances and Surface Temperatures</strong><br />The temperature of the surface of the Sun is 5780 K. Therefore the Stefan-Boltzmann Law says that it will emit 6.328 x 10^7 W/m^2 in all directions [see Appendix 1]. The diameter of the Sun is 1.3925 x 10^9 m [RASC Observer's Handbook 2009, p. 34], so its radius is 6.925 x 10^8 m, and its surface area is 4pi(6.925 x 10^8 m)^2 and so its total power output is 4pi(6.925 x 10^8 m)^2(6.328 x 10^7 W/m^2) = 3.855 x 10^26 W [RASC Handbook, p. 34 lists 3.85 x 10^26 W]. The mean distance from the Sun to the Earth is R = 1.496 x 10^11 m [RASC Handbook, p. 33]. Therefore a sphere with radius R has surface area 4piR^2 = 4 x pi x (1.496 x 10^11 m)^2 = 2.812 x 10^23 m^2. Each square metre of area perpendicular to the direction of the Sun at distance R will therefore receive 3.855 x 10^26 W/(2.812 x 10^23 m^2) = 1371 W/m^2. This is the value of the Solar <strong>insolation </strong>[wikipedia says 1366 W/m^2, the same to 3 significant digits if we round off both to 1370 W/m^2]. For the entire daytime Earth, the solar power/m^2 over a circular cross-section gets spread out over a hemisphere. Since the area of a circle is pi x r^2 and half the surface of a sphere is 2 x pi x r^2, the ratio is 1:2, and the average power/m^2 over the Earth's daytime hemisphere is 1370/2 = 685 W/m^2. This would be reduced by a factor taking into account the albedo. Then exactly half would be emitted as infrared radiation from the Earth's surface both at night time and during the day, in order to just balance the net heating/m^2 during the daytime.<br /><br /><br /><br /><strong>Appendix 5: Quantized Vibrations of Gas Molecules</strong><br />In the everyday macroscopic world, a vibrating object (e.g. tuning fork) can increase or decrease in energy by infinitesimal amounts of energy at a time. Individual diatomic or polyatomic molecules of gas, however, have quantized vibrational energy levels which are proportional to vibrational quantum numbers that can take on only integer values ( v = 0, 1, 2, 3, etc., but <strong>not </strong>v = 1.5, or v = 3.14159..., etc.). Transitions between a lower vibrational energy level and a higher one can occur on absorption of a photon of energy E = hf = hc/L where h = 6.626 x 10^-34 J.s is Planck's constant,<br />f is the frequency in Hz (s^-1),<br />c = 2.998 x 10^8 m/s is the vacuum speed of light, and<br />L = wavelength in m.<br /><br />It is sometimes said that at a temperature of absolute zero, all motion stops. This would be true in the limit of classical mechanics (because heat would always flow via radiation from a warm surrounding universe, the limit could never be reached even in classical mechanics). But in quantum mechanics, the energy levels are actually proportional to (v+1/2), and so even at absolute zero when v = 0 (the ground vibrational state), there is a residual zero-point vibrational energy corresponding to an uncertainty in the position and momenta of the atoms. This is consistent with the Heisenberg Uncertainty Principle (if at absolute zero the atoms had exact positions and no energy, we would know their momenta would also be exactly zero).<br /><br />Because wavelengths of light, including infrared, could sometimes be measured more accurately by spectroscopy than the speed of light historically, photon energies are expressed proportional to wavenumbers, the reciprocal of the wavelength, instead of the frequency. The wavenumber is the number of wavelengths per m; for convenience in the infrared, the wavenumbers are expressed in cm^-1, the number of wavelengths per cm.<br /><br />For a molecule containing N atoms, there are 2N-5 vibrations for linear molecules, and 3N-6 vibrations for non-linear molecules. This is so because N atoms would require 3N coordinates to be located in 3-dimensional space relative to some origin. 3 coordinates could be used to locate the center-of-mass of the system, and 3 more coordinates as angles to describe the orientation in space around the center-of-mass for non-linear molecules. This leaves 3N-6 coordinates which correspond to orthogonal vibrations of non-linear molecules. For linear molecules with almost all the mass located in the nuclei arranged in a straight line, only 2 angles are needed to describe the orientation in space around the center-of-mass, leaving 3N-5 orthogonal vibrations.<br /><br />For diatomic molecules, N=2, and it is necessarily linear, so there is only 3(2)-5 = 1 vibration, corresponding to bond stretching (and compressing). For homonuclear diatomic molecules like N2 and O2 (where the two atoms belong to the same element), there is no changing electric dipole moment during vibration (one end of the molecule is not slightly negative and the other slightly positive due to differential attraction of electrons). Therefore there is no possibility for these molecules of absorbing infrared photons by jumping from one vibrational energy level to a higher one (at least according to the electric dipole mechanism). Since the quantum gap between the ground electronic state and upper electronic states (corresponding to outer molecular orbitals for electrons) is very large for N2 and O2, these gases do not absorb visible light either. So the two main components making up 99% of our atmosphere are transparent to visible and infrared light, and sunlight in these regions of the electromagnetic spectrum can reach the Earth's surface unimpeded except for minor Rayleigh scattering (which explains why the daytime sky is blue) and absorption/scattering by dust particles (e.g. from ash thrown up from volcanoes, soot and other particulates from smokestacks or diesel exhaust). Two objects at the same temperature in the vacuum of space would exchange energy at the same rate via photons, and there would be no net energy flow (one would not get hotter, and the other colder, thereby violating the Second Law of Thermodynamics). But the Sun's surface is at 5780 K, and the Earth is at 288.2 K. So heat does flow from the hot object (the Sun) to the cold one (the Earth) by a net radiation flow, with most of the Sun's energy in the form of visible light and infrared. The Earth, on the other hand, is hotter than outer space (the cosmic background radio radiation corresponds to a temperature of about 3 K). So the Earth constantly emits infrared photons which are not absorbed by N2 and O2 in our atmosphere, and they would escape into outer space unless trapped by other gas molecules.<br /><br />It should be noted that heteronuclear diatomic molecules like HCl or CO possess permanent electric dipole moments which change during vibration, and so they do absorb infrared photons as they jump to upper vibrational levels.<br /><br /><strong>Carbon dioxide, CO2, </strong>is a linear molecule (the geometry consistent with its spectrum), so it would have 3(3)-5 = 4 vibrations called <strong>normal modes. </strong>All actual complex motions of the atoms in molecules in space can be resolved into motion of the center-of-mass, rotation around that center-of-mass, and combinations of the normal modes. One normal mode is called <strong>symmetric stretch </strong>with a frequency f1 in which the oxygen atoms move synchronously outward together, and then inwardly together toward the carbon atom at the centre. Because of symmetry, there is no changing electric dipole moment, and so it does not contribute significantly to the infrared spectrum (it does show up in the Raman spectrum). Another vibration is called <strong>asymmetric stretch </strong>in which the carbon atom moves toward one oxygen atom while the oxygen atoms move in the opposite direction; thus one C=O bond is being compressed while the other is being stretched. Its fundamental absorption frequency is f3 at 2349.3 cm^-1 [see Herzberg, p.274] and would correspond to a jump from v=0 (the lowest vibrational state) to v=1 (the next higher state). It would correspond to a wavelength of 1/2349.3 cm = 4.26 x 10^-6 m = 4.26 micrometres = 4.26 microns, where 1 micron = 10,000 Angstroms is about the limit of resolution of a light microscope. The other two vibrations of CO2 are orthogonal <strong>bond bending </strong>modes with a fundamental absorption frequency of f2 = 667.3 cm^-1, corresponding to a wavelength of 1/667.3 cm = 15.0 microns. This is the most powerful absorption band for carbon dioxide in the infrared, and explains the deep valley in Fig. 1 between 600 and 750 cm^-1. Because the oxygen atom attracts electrons more strongly than the carbon atom, in both asymmetric stretch and bond bending modes there is a changing electric dipole moment, allowing for strong absorption of electromagnetic radiation at the resonant frequencies. Unlike the spectra of atoms in the gas phase, which feature widely separated narrow <strong>lines </strong>corresponding to jumps between electronic energy levels, the infrared spectra of molecules show broader <strong>bands, </strong>because each jump between vibrational levels involves jumps between more narrowly spaced <strong>rotational </strong>energy levels stacked upon the vibrations. For rotational energies are also quantized.<br /><br /><strong>Water vapour, H2O, </strong>consists of bent molecules with a bond angle of approximately 104.5 degrees (a geometry most consistent with its spectrum and chemical properties). The number of normal modes of vibration is 3(3)-6 = 3. One normal mode is <strong>bond bending </strong>with a fundamental absorption at f2 = 1595 cm^-1 [see Herzberg, p. 281]. This powerful absorption corresponds to a wavelength of 1/1595 cm = 6.3 microns. Next in importance for water vapour absorption is <strong>asymmetric stretch </strong>with f3 = 3755.8 cm^-1 (wavelength 2.66 microns). Because it is bent, not linear, the H2O molecule possesses a permanent electric dipole moment which changes with <strong>symmetric stretch, </strong>and so it will absorb infrared photons at f1 = 3651.7 cm^-1 (wavelength 2.74 microns). Also important for the H2O molecule is the <strong>first overtone </strong>of bond bending (from v=0 to v=2) at 3151.4 cm^-1 (wavelength 3.17 microns).<br /><br /><strong>Methane, CH4, </strong>is a tetrahedral molecule (a geometry most consistent with its spectrum and chemical properties). Because N=5, the number of normal mode vibrations is 3(5)-6 = 9. Because of its symmetry, it has only 2 intense infrared absorption bands at 1306.2 cm^-1 (wavelength 7.66 microns) and 3018.4 cm^-1 (wavelength 3.31 microns) [see Herzberg, p. 306]. Because at equilibrium energy will have flowed equally into all modes of motion (the Principle of the Equipartition of Energy), methane with its many modes of vibration is a potentially powerful greenhouse gas. Because of its small concentration in the atmosphere, it does not show a massive spike at 1306 cm^-1 in Fig. 1, and because climate sensitivity is only one-third the generally accepted value, methane does not at present contribute significantly to the greenhouse effect or to global warming.<br /><br />What happens to a greenhouse gas molecule which has absorbed an infrared photon emitted from the surface of the Earth? It could subsequently emit a photon of similar energy in all directions. The result is essentially the same as scattering a photon, which would eventually escape into outer space. If the concentration of excited state molecules is high enough, another incoming infrared photon of the same frequency could <strong>stimulate </strong>the molecule to release a photon in phase with the first. This stimulated emission occurs in lasers (CO2 lasers are among the most powerful lasers), but is of little importance for global warming calculations because the population of excited states is so small. And the molecule could lose its vibrational energy (drop down to a lower vibrational state) on collision with other molecules.<br /><br />To understand collisions, first consider what happens when a rolling billiard ball strikes another non-moving one dead-on. The first comes to a dead stop, and the second moves along the original line with the same speed. This is the only way that both initial and final momenta can be equal<strong> and at the same time </strong>both initial and final kinetic energies can be equal. If, however, the initial motion of the first ball is not exactly along a line through the center of the second ball, then the two balls will move apart after the collision, and the two paths will form a "Y" with the initial one. Some translational energy will have been transferred from the first ball to the second (the first will have lost some energy, and the other will have gained some energy), but both will be moving. Similarly gas molecules on collision will transfer translational energy, but not completely. For monatomic gases like argon or neon, the most probable way of distributing a finite amount of total energy among a finite number of molecules is the Maxwell-Boltzmann distribution function at any given temperature. This function goes through (0,0), because no gas molecule is motionless for long in any coordinate system, and approaches a decreasing exponential function at high translational energies or speeds. Diatomic and polyatomic molecules can possess quantized rotational energy as well as translational kinetic energy. Experiments have shown that after a few collisions with other molecules, highly-rotating molecules will have lost some rotational energy (will be spinning more slowly) and slowly-rotating molecule may have gained some rotational energy. At equilibrium the population of molecules in the various rotational states will fit a Boltzmann distribution function (a decreasing exponential function, complicated by the possibility of degeneracy, a multiplicity of quantum states with the same energy). At equilibrium the rotational temperature will be the same as the temperature for translational kinetic energy.<br /><br />Because the gap between quantized vibrational energies might be large compared to translational or rotational energies at a given temperature, most molecules on collision will be unable to jump from one vibrational state to a higher one. For example, at 288 K, the average kinetic energy of translation of a gas molecule is 3kT/2 = 3(1.38 x 10^-23 J/K)(288 K)/2 = 5.96 x 10^-21 J, where k = Boltzmann constant. For the CO2 molecule in the first excited vibration (v=1) of the bond bending mode f2, the photon energy on dropping down to v=0 is E = hf = hc/L = (6.626 x 10^-34 J.s)(2.998 x 10^8 m/s)(667.3 cm^-1)(100 cm/m) = 13.3 x 10^-21 J, where h = Planck's constant, c = vacuum speed of light, and 1/L = 667.3 cm^-1 is the wavenumber. The average translational energy of any gas molecule at 288 K is less than half the vibrational energy gap, and so most collisions will be incapable of boosting the CO2 molecule from v=0 to v=1. However, there will be a few fast-moving molecules that can do so, and therefore the first excited vibration will be important in the heat capacity of the carbon dioxide molecule [see Appendix 8]. That is, the capacity for heat to be absorbed by boosting CO2 molecules from v=0 to v=1 in bond bending, as well as for heat to be absorbed in increasing translational and rotational kinetic energies.<br /><br />For the reverse case, an excited molecule with v=1 can drop down to v=0 without emitting a photon, but by somehow hitting other atoms hard enough to increase their translational and rotational kinetic energies. Because the vibrational energy gap is more than twice the average translational energy in this example, it might take several collisions to redistribute the energy gained by the slower molecules until equilibrium is reached. And every collsion probably does not result in de-exciting the high vibrational state. For O2 molecules at 288 K, it takes about 2 x 10^7 collisions to deactivate from v=1 to v=0 [see Harold S. Johnston, "Gas Phase Reaction Rate Theory", p.276]. This seems huge unless one knows that at room temperature and pressure, each air molecule undergoes about 2 x 10^10 collisions per second with other molecules [see Appendix 9]. This is about 1000 times per second more than that needed for deactivation. Larger molecules take much fewer collisions for vibrational deactivation, as explained qualitatively by Landau and Teller. For intermediate molecules like CO2 and H2O, therefore, we expect vibrational deactivation after tens to thousands of collisions, instead of millions like O2. This is the main mechanism by which greenhouse gases absorb infrared energy radiated from the surface of the Earth and transfer it to the much more common surrounding nitrogen and oxygen molecules, thereby keeping energy in the atmosphere and reducing the heat loss to outer space.<br /><br /><br /><br /><strong>Appendix 6: 220 K black body curve in Fig. 1</strong><br />In Fig. 1 of this article, obtained from the wikipedia article on Radiative Forcing, the jagged spectrum appears to approach the black body (Planck) curve for 22oK at wavenumbers smaller than 200 cm^-1, and the prominent absorption valley at 600-750 cm^-1 appears to stop at this curve. Perhaps the authors assumed that the tenuous gas in the atmosphere at 20 km acts like a black body at 220 K, and emits this spectrum which must be included in the upward irradiance. The temperature may indeed be at 220 K, but the assumption of black body emission at emissivity 0.98 is invalid. For the density of atoms capable of emitting is so low there compared to at the Earth's surface. Temperature is an <strong>intensive </strong>property, which doesn't increase with the number of particles, unlike the total energy which as an <strong>extensive </strong>property <strong>does </strong>increase with numbers. It is well known that the tenuous corona of the Sun has a temperature in the millions of degrees (for some unknown reason), but we do not include this in calculations using the Stefan-Boltzmann Law. Instead we simply use the temperature at the surface of the Sun where there is an abrupt change in the density of particles. When the density of electrons suddenly drops, photons can travel unimpeded outward in straight lines from the Sun, unlike the random walk involving innumerable collisions with charged particles like electrons and ions in the interior of the Sun. For our calculations in this article, we need consider only the infrared radiation emitted from the Earth's surface at 288.2 K, which would escape into outer space unless absorbed by greenhouse gases getting in the way.<br /><br />Electromagnetic radiation is the main method of transferring energy from a hot object to a cold one in outer space where conduction and convection are not possible. When they are not at the same temperature (e.g. the Sun and the Earth's surface), there will be a net flow of energy from the hot to the cold. The Sun emits mainly visible light which is not absorbed by the transparent air, and a blue solid object on the Earth absorbs low-energy red photons, but reflects blue, so the energy transfer is not 100% efficient. Similarly, a red object absorbs blue (which might have enough energy per photon to break bonds in red dye molecules, bleaching them), and reflects red, so the energy transfer again is not 100% efficient. An ideal black object would absorb 100% of the Sun's visible light, and reflect none (and that's why we see it as black in the daytime). The Earth emits photons too, but at a lower surface temperature they are in the infrared range. If it were to warm up to the Sun's temperature, the Earth would also emit visible light until the photons emitted matched those absorbed from the Sun exactly. The cosmic microwave background is black body radiation at 3 K, way below 288 K, so the Earth loses energy to outer space at night and day without getting much back. Kirchhoff noted that good absorbers are also good emitters (radio dish antennae are good for both receiving and transmitting). So the best emitters are actually black bodies, and the Stefan-Boltzmann Law gives the power emitted by such a black body. Since air is a poor approximation of a black body (it's transparent), it will also be a poor emitter (so its emissivity will be low). This is obvious at visible wavelengths, but nitrogen and oxygen are also lousy absorbers/emitters of infrared radiation (that's why even small amounts of greenhouse gases like CO2 and H2O can have such an importance). So you don't have to add (or subtract) the 220 K curve (or its area which corresponds to 89 W/m^2 [see Appendix 1]) to determine the upward irradiance at 20 km. This is the reason why the sides of the CO2 absorption curve were extended to the bottom horizontal axis to form a "V" before calculating areas and therefore % absorbance by CO2.<br /><br />Finally, a spherical shell of gas at 20 km altitude does not radiate a net energy downward, because the Stefan-Boltzmann Law is derived for outward flow from convex black bodies. The reason is that at thermal equilibrium, a photon emitted from the inner surface of a cavity will reach the other side and be absorbed. If an object like the Earth is in the middle of the cavity but is in thermal equilibrium, the photon might be absorbed by the Earth, but another of equal energy will be emitted from the other side.<br /><br /><br /><strong>Appendix 7: Water Vapour Pressure</strong><br />When liquid water evaporates, individual H2O molecules escape from the body of bulk liquid into empty space, forming a gas (vapour). The energy needed to rip water molecules from the bulk liquid is called the latent heat of vaporization, delta Hvap = 40.65 kJ/mol at 100 degrees Celsius. Assuming that the volume of a mole of liquid is negligible compared to the volume of a mole of gas, and using the Ideal Gas approximation for gases, the Clausius-Clapeyron Equation says the saturated vapour pressure P is given by:<br /><br />P = P'.exp(-delta Hvap/RT) where R = Ideal Gas Constant = 8.315 J/(K.mol),<br />T= absolute temperature in K,<br />exp = exponential function, and<br />P' = a constant.<br /><br />To calculate the vapour pressure at any temperature, we need to know or measure the actual vapour pressure at one temperature. Most people will know that the normal boiling point of water is 100 degrees Celsius = 373 K, and this occurs at 1 standard atmosphere pressure = 760 mm Hg. Therefore<br /><br />760 mm Hg = P'.exp[-40650/(8.315 x 373)] = P'.exp(-13.107) = 2.032 x 10^-6 P'.<br /><br />Therefore P' = 760/(2.032 x 10^-6) = 3.74 x 10^8 mm Hg.<br /><br />Therefore the Clausius-Clapeyron Equation for water vapour pressure is<br /><br />P = (3.74 x 10^8 mm Hg).exp[-40650/(8.315 T)].<br /><br />Let us use this equation to calculate the effect on increasing temperature from 20 to 22 degrees Celsius, a 2 degree increase.<br /><br />At 20 degrees Celsius = 293 K, P = (3.74 x 10^8 mm Hg).exp[-40650/(8.315 x 293)] = 21.2 mm Hg. This is slightly (21.1%) above the value of 17. 5 mm Hg listed in the Handbook of Chemistry and Physics, showing that extended calculations over a temperature difference of 80 degrees for a roughly exponential function will create errors (in particular, we assumed that the heat of vaporization is constant over this range, which it is not).<br /><br />At 22 degrees Celsius = 295 K, P = (3.74 x 10^8 mm Hg).exp[-40650/(8.315 x 295)] = 23.8 mm Hg. This is again slightly (20.2%) above the value of 19.8 mm Hg listed in the Handbook. Since the discrepancy in each case is about the same relative %, the 12% theoretical increase from 21.2 to 23.8 mm Hg matches (to 2 significant digits) the 13% listed increase from 17.5 to 19.8 mm Hg.<br /><br />The listed value of 17.5 mm Hg is only 2.3% of atmospheric pressure, but is 77 times the partial pressure of carbon dioxide (0.03%) in dry air. However, the vapour pressure of water assumes equilibrium (100% relative humidity). If the relative humidity is only 50% (say at ocean level), then the actual water vapour would be only 77/2 = 38.5 times that of carbon dioxide at 20 degrees Celsius.<br /><br />Because the vapour pressure of water increases roughly exponentially with increasing T (i.e. the value of -1/T increases approximately linearly over a short-enough range of T), there is the possibility of a positive feedback mechanism: if the average temperture goes up for any reason, including increasing CO2, then this increases the water vapour pressure, and since water vapour is a greenhouse gas, the amount of global warming will be amplified. Of course, this could be counterbalanced by greater cloud formation which would increase the amount of sunlight reflected back into space before it could reach the Earth's surface and the energy trapped by greenhouse gases in the troposphere. When two gas molecules approach each other, electrical attractive forces (van der Waals' forces, London dispersion forces, etc.) cause them to speed up toward each other. Then, when they collide, their kinetic energies may cause them to separate again (since they initially had enough energy to be independent). In order for them to stick together, and eventually form a condensed state like a liquid drop or solid crystal, they must lose some energy, either into internal states like higher vibrations or rotations, or more likely by transferring some energy to a third body such as another gas molecule with which they by chance simultaneously collide, or to a dust particle (a piece of solid on which they might collide and stick), or to an ionized particle created by the passage of a cosmic ray from outer space. Thus cloud formation will be very difficult to predict accurately.<br /><br />One positive effect of increased water vapour in the daytime for human comfort and habitation could be that more heat would be liberated when the vapour condenses at nighttime, thereby moderating the drop in temperature. On the other hand, this heat of condensation could drive more violent tropical storms or tornadoes which would endanger humans.<br /><br />-----------------------------------------------------------------<br /><br />In my initial rough estimate above, I assumed that the latent heat of vaporization was a constant which doesn't change with temperature. Actually, it changes from 44.016 kJ/mol at 25 degrees Celsius to 40.656 kJ/mol at 100 degrees Celsius [see P.W. Atkins, "Physical Chemistry", Table 4.7]. Why is this? And can we actually calculate the difference?<br /><br />1 calorie = 4.184 J (by definition) is the amount of heat needed to raise the temperature of 1 g of liquid water by 1 Celsius degree. Since the mass of 1 mole of H2O is 16.0 + 2(1.0) = 18.0 g, the heat needed to raise the temperature of 1 mole of liquid water is 1.00 x 4.184 x 18.0 = 75.3 J/(K.mol). This is the molar heat capacity of liquid water.<br /><br />Therefore on heating from 25 to 100 degrees Celsius, liquid water gains energy equal to 75.3 x 75 = 5647.5 J/mol. Extra digits have been carried to eliminate round-off errors.<br /><br />But water vapour also gains energy on heating from 25 to 100 degrees Celsius. From Atkins' Table 4.1, the heat capacity of water vapour is 33.58 J/(K.mol), so this energy gain is 33.58 x 75 = 2518.5 J/mol.<br /><br />Therefore the difference between these energies is 5647.5 - 2518.5 = 3129 J/mol.<br /><br />This difference must be added to the latent heat of vaporization measured at the normal boiling point of 100 degrees Celsius in order to estimate the latent heat at 25 degrees Celsius. This is so because between 4 and 100 degrees Celsius, water molecules move more violently on heating, pushing each other apart, making it easier to rip individual molecules from the bulk liquid, so the latent heat goes down as the temperature rises (and vice versa).<br /><br />Therefore the estimated latent heat of vaporization of water at 25 degrees Celsius = 40656 + 3129 = 43785 = 4.38 x 10^4 J/mol. Compare this with the presumably measured value of 4.40 x 10^4 J/mol previously quoted (our theoretical prediction is only 0.45% off).<br /><br />From p. D-92 of the 45th edition of the Handbook of Chemistry and Physics, the water vapour pressure at 25 degrees Celsius is 23.756 mm Hg. This value seems absurdly precise, with too many quoted digits. Nevertheless, if we use this in the Clausius-Clapeyron Equation,<br /><br />P = P'.exp[-delta Hvap/(RT)],<br /><br />23.756 mm Hg = P'.exp[-44016/(8.3145 x 298.15)] = P'.exp(-17.756) = P'.(1.9443 x 10^-8).<br /><br />Therefore P' = 23.756 mm Hg/(1.9443 x 10^-8) = 12.218 x 10^8 mm Hg. this is 3.27 times the value we calculated previously using data at 100 degrees Celsius. Using this new value for P':<br /><br />The vapour pressure of water at 20 degrees Celsius is (12.218 x 10^8 mm Hg).exp[-44016/(8.3145 x 293.15)] = 17.55 mm Hg. [The Handbook lists 17.54 mm Hg.]<br /><br />The vapour pressure of water at 22 degrees Celsius is (12.218 x 10^8 mm Hg).exp[-44016/(8.3145 x 295.15)] = 19.83 mm Hg. [The Handbook lists 19.83 mm Hg.]<br /><br />The exact agreement between theory and listed values makes me think that most of the tabulated values were simply calculated as above, using data measured carefully at only a few key temperatures such as the standard temperature of 25 degrees Celsius which is achievable in most Earthling labs, with interpolations at intermediate temperatures.<br /><br />----------------------------------------------------------------<br /><br />At the next level of theoretical abstraction, we try to understand WHY the heat capacities have the values listed [see Appendix 8]. Trying to calculate the latent heat of vaporization of water at the normal boiling point from first principles, i.e. the Schroedinger Equation, is too hard right now for even the mightiest computer, for it is a many-body problem involving removing one molecule from the surface of bulk liquid which contains approximately 10^24 other molecules touching each other. Even the 3-body problem, e.g. the motion of the Earth, the Moon and the Sun under their mutual forces of gravitational attraction has no exact closed-form solution.<br /><br /><br /><br /><strong>Appendix 8: Heat Capacities of Gases, Metallic Solids & Liquid Water</strong><br />In this article and in Appendix 7, the heat capacities of liquid water, metallic solids, and water vapour were used in calculations. So it would be good to understand the theoretical basis for heat capacities.<br /><br /><strong>The Heat Capacity of Argon, a Monatomic Gas</strong><br />For all monatomic gases like argon (which consists of individual Ar atoms moving in otherwise empty 3-dimensional space), the average kinetic energy is 3RT/2 per mole, where R = 8.3145 J/(K.mol) is the Ideal Gas Constant and T = absolute temperature in kelvins (K). This means that translational energy is directly proportional to the absolute temperature (and vice versa). Since heat capacity is the rate of change of energy with respect to temperature, it is (using calculus terminology) the derivative of 3RT/2 with respect to T. That is, the translational heat capacity of argon gas is 3R/2 per mole (i.e. a constant).<br /><br />Experimental data are usually measured at constant pressure in the lab, because it is convenient to leave beakers, ovens, manometers, etc. open to the atmosphere. The one important exception is in the compiling of heats of combustion (e.g. in determining the Calorie value of food samples), where a <strong>bomb calorimeter </strong>is used to keep all materials, including gases, at constant <strong>volume, </strong>not pressure. The reason is that when gases are produced and released into the atmosphere, they do work (use energy) in pushing against atmospheric air. Therefore heat changes at constant pressure equal the total of internal energy changes in the molecules + the work done in expanding gases against the atmosphere. At constant pressure P, this work of expansion is P times the change in volume, i.e. P.(delta V). For a mole of Ideal Gas, PV = RT, so V=RT/P. Therefore at constant P, delta V = R.(delta T)/P. Therefore the work of expansion is P.(delta V) = R.(delta T). Since the work of expansion comes from a change in energy, the rate of change of energy with temperature is P.(delta V)/(delta T) = R.(delta T)/(delta T) = R for a mole of Ideal Gas. That is, the heat capacity in expanding one mole of gas from essentially zero volume to its final volume V at constant pressure is R. This has to be added to the change in internal energy to get the overall heat capacity. Because a monatomic gas has no rotational or vibrational energy, the total heat capacity at constant pressure should be 3R/2 + R = 5R/2 = 5(8.3145)/2 = 20.786 J/(K.mol). This is exactly the value listed for argon in Table 4.1 of Atkins' "Physical Chemistry" as well as in the 80th edition of the Handbook of Chemistry and Physics on p.4-123 listing properties of all the elements.<br /><br /><strong>Heat Capacities of Diatomic and Polyatomic Gases (e.g. Water Vapour)</strong><br />In addition to translational energy of the center-of-mass from point-to-point in space, diatomic and polyatomic molecules can possess energy of rotation and vibration. The orientation of linear molecules can be specified by knowing 2 angles in space, and non-linear molecules by 3 angles. In classical Equipartition of Energy theory, at thermal equilibrium equal amounts of energy will have flowed into each possible mode of motion. So the rotational energy stored in linear molecules is 2RT/2 per mole, and 3RT/2 per mole in non-linear molecules. Therefore the rotational heat capacity for gases made of linear molecules (like N2, O2, HCl or CO2) is R per mole, and for non-linear molecules (like H2O or CH4) is 3R/2 per mole. The total heat capacity for translation, rotation, and expansion from zero volume is therefore 7R/2 per mole for linear molecules, and 4R per mole for non-linear molecules like H2O.<br /><br />Check: The theoretical heat capacity for a mole of <strong>water vapour </strong>so far = 4R = 4(8.3145) = 33.26 J/(K.mol). The <strong>actual </strong>heat capacity of water vapour is 33.58 J/(K.mol) [see Atkins, Table 4.1]. This is slightly higher than the theoretical value so far, and corresponds to a small amount of energy going into vibration. We used the actual value in calculating the latent heat of vaporization at 25 degrees Celsius from the known value at the normal boiling point of 100 degrees Celsius, and used this to calculate the vapour pressure of water in Appendix 7.<br /><br />For a molecule containing N atoms, the number of vibrational modes is 3N-5 for linear molecules, and 3N-6 for non-linear molecules. Therefore for water, where N=3, the number of vibrational modes is 3(3)-6 = 3. These modes are symmetric stretch, asymmetric stretch, and bond bending, and explain the infrared spectrum [see Appendix 5]. Since energy in a vibrating spring contains 2 components, one kinetic energy and one potential energy, each vibration can store R per mole, not R/2 per mole, and the theoretical vibrational heat capacity for non-linear water molecules is 3R per mole. The <strong>actual </strong>vibrational heat capacity for water vapour is only (33.58- 33.26)R/8.3145 = 0.038R per mole at 25 degrees Celsius. What's wrong?<br /><br />The answer comes from quantum mechanics. In classical mechanics, energy can increase smoothly by infinitesimal amounts as temperature increases infinitesimally, resulting in a smooth Boltzmann decreasing exponential function for the distribution of molecules with increasing energy (think of how the density of air decreases exponentially with increasing height going up a mountain, as the gravitational potential energy goes up). This occurs because the decreasing Boltzmann function is the most probable way of distributing a finite total amount of energy among a finite number of molecules. But if the gap between quantized vibrational energy levels is huge compared to average kinetic energies of translation, then all the molecules will be trapped in the lowest vibrational energy state (even at absolute zero, there is a non-zero zero-point energy due to the Heisenberg Uncertainty Principle). Even if the temperature is raised a small finite amount, increasing the kinetic energies of translation and rotation, no molecules will absorb energy on collision and end up jumping from the lowest vibrational state to the next or higher one. So the capacity to absorb heat via vibration is zero (i.e. the vibrational heat capacity goes to zero as temperature approaches absolute zero). At an intermediate temperature, <strong>some </strong>collisions will result in knocking a molecule from the lowest vibrational level to the next or higher level, but not all collisions involve sufficient energy or the proper geometry to do so. So the actual vibrational heat capacity will be smaller than the theoretical one, which can be considered the high-temperature maximum limit. In our case, apparently water vapour at 25 degrees Celsius absorbs only 1.3% of the heat capacity limit for vibrations.<br /><br />At 288 K, the average kinetic energy of translation of a gas molecule is<br /><br />3kT/2 = 3(1.389 x 10^-23 J/K)(288 K)/2 = 5.96 x 10^-21 J, where k = Boltzmann constant.<br /><br />For bond bending f2 in H2O from v=0 to v=1 at 1595 cm^-1, the photon energy is<br /><br />E= hf= hc/L = (6.626 x 10^-34 J.s)(2.988 x 10^8 m/s)(1595 cm^-1)(100 cm/m)<br />= 31.7 x 10^-21 J. This is 5.3 times the average kinetic energy of translation, so few collisions have sufficient energy to cause a quantum leap from v=0 to v=1.<br /><br />For asymmetric stretch f3 in H2O from v=0 to v=1 at 3756 cm^-1, the photon energy is 74.6 x 10-21 J. This is 12.5 times the average kinetic energy of translation.<br /><br />For symmetric stretch f1 in H2O from v=0 to v=1 at 3652 cm^-1, the photon energy is 72.5 x 10^-21 J. This is 12.1 times the average kinetic energy of translation.<br /><br />For the first overtone of bond bending, from v=0 to v=2 at 3151 cm^-1, the photon energy is 62.6 x 10^-21 J. This is 10.5 times the average kinetic energy of translation.<br /><br />So at 288 K, almost all the H2O molecules are in their ground vibrational states (v=0), with a tiny fraction being knocked by collision into v=1 for bond bending f2. ALL of these states, however, can make the quantum leap upwards on absorption of infrared photons in bands centered on the wavenumbers given, and then transfer these large amounts of energy on collision to other air molecules (e.g. N2, O2), which means atmospheric warming.<br /><br /><strong>Heat Capacity of Carbon Dioxide, CO2</strong><br />Because CO2 is a linear molecule, it has 3(3)-5 = 4 normal modes of vibration. Therefore the theoretical maximum heat capacity (when we include translation, rotation, expansion from zero volume, and vibration) is 3R/2 + R + R+ 4R = 15R/2 = 7.5R per mole. The <strong>actual </strong>heat capacity is 4.46 R per mole [using 37.11 J/(K.mol) from Table 4.1 of Atkins' "Physical Chemistry", and R = 8.314 J/(K.mol)]. Because translations and rotations rapidly reach an equilibrium temperature after a few or tens of collisions, apparently only 0.96 R of the 4R maximum per mole is used to store vibrational energy at room temperature.<br /><br /><strong>Heat Capacities of Metallic Elements </strong><br />The specific heat capacities of common metallic elements range from 0.0305 cal/(g.deg) for lead (Pb) to 0.216 cal/(g.deg) for aluminum (Al). However, if we multiply the specific heat capacities (which are <strong>per gram </strong>of substance) by the atomic weights, we get molar heat capacities (which are <strong>per mole </strong>of substance) which range from 6.32 cal/(K.mol) for Pb to 5.83 cal/(K.mol) for Al. The average for Al, Cu, Fe, Pb, Ag, Zn, and Au is 6.03 cal/(K.mol), with all within 5% of this mean. There must be a simple reason for this. Because 1 calorie = 4.184 J by definition, 6.03 calories = (6.03)(4.184) = 25.2 J. Because the Ideal Gas Constant is R = 8.3145 J/(K.mol), 25.2 R/8.3145 = 3.03 R, which is within 1% of 3R.<br /><br />Because solid metals are made up of individual atoms locked into a crystal structure, the atoms cannot translate, and since the mass of an atom is mostly in the nucleus or center-of-mass, the atoms cannot possess rotational energy. The only possible motions of atoms in metallic crystals are vibrational, and there are 3 different space directions in which to vibrate. Therefore the data are consistent with the high-temperature limit of molar heat capacity of 3R, with each mode of vibration capable of absorbing R per mole (e.g. one R per mole into each of 3 perpendicular directions in space).<br /><br />But why are these metals at the high-temperature limit for vibrational heat capacity at room temperature while H2O and CO2 are not [see previous sections]? The force experienced by a mass m attached to a spring is given by F= -kx, where k is called the force constant, and x is the extension. This is called Hooke's Law, which works for small extensions. For strong springs, k is large, to that for the same extension x, more force is needed than for weak springs with small values of k. The frequency of vibration of the mass m is proportional to the square root of k/m. So if k is small (weak forces due to large atoms, meaning greater distances, and the electric force decreases with the square of the distance) and m is large (e.g. heavy lead atom compared to hydrogen), then k/m will be really small, and the frequency of vibration will be low. This means the quantized vibrational energy levels will be small and fairly close together. If the value of the average kinetic energy = 3RT/2 for a mole of atoms is larged compared to the spacings between levels, then classical Equipartition of Energy will be achieved (equal distribution of energy into all modes of motion), and the high-temperature limit of a constant heat capacity will be reached.<br /><br />Back in 1819, <strong>Dulong and Petit </strong>discovered that it was possible to estimate atomic weights by dividing measured specific heat capacities into the number 6. For example, by boiling 100 g of copper in water at 100 degrees Celsius, and then quickly transferring the metal to a calorimeter containing 100 g of waer, and measuring the temperature changes, it is easy to calculate that 1 gram of copper loses 0.0919 calories in cooling down by 1 Celsius degree. Dividing 6 by 0.0919 gives 65 as an approximate value for the atomic weight of copper. More accurate experiments are possible by weighing things before and after a chemical reaction. For example, 31.8 g of copper might be found to react with 8.00 g of oxygen in the formation of cupric oxide. 8.00 g is the gram-equivalent weight of oxygen because the valence of oxygen is 2. Since 31.8 is about half the approximate atomic weight of copper, a more accurate value would be 2(31.8) = 63.6. Similar experiments with gaseous oxides would show that the atomic weight of oxygen is 16.00 (relative to 1.00 for hydrogen, the smallest weight). And this was discovered 78 years before J.J. Thomson discovered that the electron is a small part of an atom, and 106 years before the Schroedinger Equation explained the structure of the hydrogen atom, the simplest atom!!!<br /><br />The value of 3R for the molar heat capacity of all metallic solids is a high-temperature limit achieved at room temperature. The graph of a constant non-zero value for molar heat capacity would be a horizontal straight line above the horizontal temperature axis. When temperatures approaching absolute zero are measured, the <strong>actual </strong>molar heat capacity curve resembles the letter "S" pinned at the origin (0,0) and stretched horizontally. The molar heat capacity approaches zero as temperatures approach absolute zero because atoms get trapped in the lowest possible vibrational state, and small increases in temperature cannot result in knocking atoms into higher vibrational states, so no energy is absorbed (i.e, the capacity for absorbing heat is zero). The first person to use quantum theory to calculate the molar heat capacity curve for monatomic crystals at temperatures approaching absolute zero was Albert Einstein in 1906. The Einstein model assumed a single fundamental frequency for the vibrations of all the atoms. A better model would assume a spectrum of vibration frequencies for the crystal, but the statistical problem becomes more complicated. This problem was solved by Peter Debye (1884-1966) in the years 1911-16. Debye won the 1936 Nobel Prize for Chemistry.<br /><br /><strong>Heat Capacity of Liquid Water</strong><br />1 calorie = 4.184 J can raise the temperature of 1 gram of liquid water by 1 Celsius degree. Since the mass of 1 mole of water is 16.0 + 2(1.0) = 18.0 g, the molar heat capacity of liquid water is (4.184)(18.0) = 75.3 J/(K.mol). Since R = 8.314 J/(K.mol), the <strong>actual </strong>heat capacity of liquid water is therefore 75.3 R/8.314 = 9.06 R. I have not yet seen a clear theoretical explanation for this, so what follows is speculative on my part.<br /><br />First, the theoretical molar heat capacity must be greater than 9.5 R, since the actual value is already above 9R. Since it is a liquid, whose molar volume is more than 1000 times less than that of a gas (18.0 mL compared to 22.4 L at STP), we can ignore the heat capacity for expansion from zero volume. The translational heat capacity must be 3R/2 per mole, and the rotational heat capacity another 3R/2, for a total of 3R per mole. We have already showed that the internal vibrational states of almost all H2O molecules were trapped at v=0, so vibrational heat capacity is essentially zero at room temperature. Therefore there must be at least 6.5 R heat capacity per mole due to at least 7 other vibrations. Water molecules are held together in the liquid by so-called "hydrogen bonds", which are electrostatic forces between the slightly positive hydrogen atoms in one molecule with the slightly negative oxygen atom <strong>in another water molecule. </strong>The geometry is easier to visualize in ice-I, the normal form of ice. Valence Shell Electron Pair Repulsion (VSEPR) Theory says that the 4 pairs of valence electrons in the oxygen atom repel each other electrostatically to the vertices of a tetrahedron. Two of these pairs are involved in covalent bonds to hydrogen atoms, predicting a H-O-H bond angle of 109.5 degrees (the tetrahedral angle). This neatly explains the <strong>actual </strong>bond angle of 104.5 degrees (obtained by calculating the moments of inertia from the spectrum of the water vapour molecule). The other two valence electron pairs (lone pairs) of the oxygen atom point toward hydrogen atoms in <strong>other </strong>water molecules. The tetrahedral geometry resembles that of carbon atoms in a crystal of diamond, where each internal carbon atom is covalently bonded to 4 other carbon atoms at the vertices of a tetrahedron. The small bond distances (meaning strong electric forces and therefore strong bonds) explain why diamond is the hardest naturally occurring substance. A cube has 8 vertices (corners). If 4 atoms or molecules occupy alternate corners of a cube, and another atom or molecule is at the center of the cube, the result is tetrahedral symmetry. Thus diamond and ice-I have cubic symmetry. If one looks along the body diagonal of a clear cube from a distance, one sees a hexagon, with three diagonals joining opposite vertices. This explains the hexagonal (6-fold) symmetry of ice-I in the form of a snowflake. Most gem-quality natural diamonds are found with octahedral shapes. An octahedron has 8 faces and 6 vertices. How can this be if diamond is a cubic crystal? Imagine a transparent cube, and paint a dot on the centre of each of the 6 faces. Now join the 6 dots with straight lines in 3-dimensional space, and erase the cube. You will be left with an octahedron.<br /><br />In liquid water, the molecules are in constant motion, colliding with and rebounding off each other, and rolling around each other as well. Any one molecule, however, will at least briefly be hydrogen bonded tetrahedrally to 4 other nearest-neighbour water molecules. These attractions are not as strong as the covalent bonds within the central molecule, and so the vibrational frequencies and therefore energy levels will be smaller. Therefore room temperature may involve kinetic energies high enough to lead to populations of molecules in higher vibrational energy levels of these <strong>hydrogen bonds</strong>. Since this momentary cluster of 5 water molecules resembles the geometry of a methane (CH4) molecule, how many normal mode vibrations does the methane molecule possess? The answer is 3(5)-6=9. This would give a maximum theoretical value of 9R per mole for vibrations involving hydrogen bonds with 4 nearest-neighbour molecules. This is at least greater than the lower limit of 7R per mole necessary for a satisfactory theory. So this simplistic picture predicts a maximum theoretical heat capacity for liquid water of 3R + 9R = 12R per mole, with the actual value being 9.1 R per mole (76% of the theoretical maximum). We could then hand-wave away the difference by saying that apparently not all the vibrational levels are as equally populated as the rotational levels at room temperature.<br /><br />The very large heat capacity of liquid water makes it a useful commonly available fluid for cooling systems and firefighting, and helps explain how global temperature changes are moderated by the water in the oceans and lakes.<br /><br /><br /><br /><strong>Appendix 9: Kinetic Theory of Gases</strong><br />From "Physical Chemistry" by Walter J. Moore, pp.221-222:<br /><br />The number of collisions per second experienced by a gas molecule is:<br /><br />Z = 2^0.5 x pi x N x d^2 x v, where N = number of molecules per cm^3,<br />d = distance between centers of colliding spherical molecules in cm,<br />and v = average speed in cm/s.<br /><br />For the nitrogen (N) atom, radius = 0.15 nm (1.5 Angstroms), so the distance between N atoms in N2 is 0.30 nm, and d = 0.60 nm, assuming a rapidly tumbling molecule is equivalent to a sphere with radius 0.30 nm.<br /><br />For a molecule, the average kinetic energy = 3kT/2, where k = Boltzmann constant.<br /><br />Therefore, mv^2/2 = 3kT/2.<br /><br />v^2 = 3kT/m.<br /><br />v = [3kT/m]^0.5 = [2(1.38 x 10^-23 J/K)(293 K)/(28.0 x 10^-3 kg/(6.02 x 10^23)]^0.5<br />= 5.10 x 10^2 m/s for N2 at T= 293 K,<br />= 5.1 x 10^4 cm/s. This is an r.m.s. speed which is approximately the average speed.<br /><br />Therefore Z = 2^0.5 x pi x 6.02 x 10^23/(22.4 x 10^3 x 293/273 cm^3)(0.60 x 10^-7 cm)^2.<br />(5.1 x 10^4 cm/s)<br /><br />= 2.0 x 10^10 s^-1 at 293 K and 1 atmosphere pressure.<br /><br />Even if it takes thousands of collisions to deactivate a vibrationally excited CO2 molecule by a quantum leap,transferring this energy to translational or rotational modes of surrounding N2 or O2 molecules, each molecule near the Earth's surface experiences around 20,000,000,000 collsions a second and so is rapidly quenched. Even at altitudes of 20km where air pressure might be a tenth that at the Earth's surface (assuming that at the top of Mount Everest at 29,000 feet, air pressure is one-third that at sea level), the number of collisions per second would be about 2 billion per second (assuming similar temperatures).<br /><br /><br /><br /><br /><br />This article by Roger Taguchi, 234 Knox Crescent, Ottawa, Ontario, Canada K1G 0K8, e-mail address <a href="mailto:rtaguchi@sympatico.ca">rtaguchi@sympatico.ca</a>, was posted on Monday Aug. 31, 2009.P. Wiebarhttp://www.blogger.com/profile/15987516965582899854noreply@blogger.com1