tag:blogger.com,1999:blog-16612221.post8119712795608512418..comments2024-03-19T03:14:04.172-04:00Comments on Rabett Run: EliRabetthttp://www.blogger.com/profile/07957002964638398767noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-16612221.post-85549601106561780842007-03-19T02:36:00.000-04:002007-03-19T02:36:00.000-04:00Additional Note for above argument: If one calcula...Additional Note for above argument:<BR/> <BR/>If one calculates the distance the molecule would have to move upward to experience a temperature change that is 1% of the above starting value (25 deg C), it comes out 260 million times the mean free path, or, the mean free path is 1/260,000,000 this distance. <BR/><BR/>In other words, the average distance the (Oxygen or nitrogen) molecule <I>actually</I> moves in air at 25 C at the earth's surface before it collides with another gas molecule (mean free path) is vanishingly small in comparison to the distance the molecule would <I>have</I> to move to find itself in a region of higher temperature.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-16612221.post-75806874228221978602007-03-19T02:05:00.000-04:002007-03-19T02:05:00.000-04:00The condition for local thermodynamic equilibrium ...The condition for local thermodynamic equilibrium (LTE) is that <BR/><BR/>"the mean free path of the atom is very small compared to the distance over which the temperature changes,"<BR/><BR/>From "Thermodynamic Equilibrium, Local and otherwise" <BR/>Copyright © Michael Richmond. <BR/><BR/><BR/>Under such conditions, "the atom will collide many times with other atoms, all of the same temperature, before it can possibly reach some region with a different temperature. In this case, the speed of atoms within some small region may very well be described by a Maxwell-Boltzmann distribution with a definite temperature."<BR/><BR/>http://spiff.rit.edu/classes/phys440/lectures/lte/lte.html<BR/><BR/><BR/>So, to determine whether the air at the surface of the earth is in LTE, we must determine whether the mean free path of an "air" molecule (Oxygen or nitrogen, for the earth's atmosphere) is small compared to the distance over which temperature changes significantly.<BR/><BR/>or, what is equivalent: determine how much the temperature changes for an elevation gain that is equal to the mean free path of an oxygen (or nitrogen) molecule at temperature and pressure (1 ATM) typical near the earth's surface. <BR/><BR/>If this temperature change is very small over the mean free path in question, then we can be confident that we have met LTE.<BR/><BR/>So, is this temperature change very small over the mean free path in question?<BR/><BR/>For O2 (oxygen)at one atmosphere and 25 °C, the Mean free path is only 9.7 × 10^(–6)cm <BR/><BR/>or about 10^(-5)cm<BR/><BR/>where the ^ means "raised to the power"<BR/><BR/>http://www.iupac.org/goldbook/M03778.pdf.<BR/>For nitrogen (N2) it’s of the same order<BR/><BR/>So, how much does the temperature of the atmosphere change if we move upward over a distance equal to the mean free path of Oxygen gas at 1 ATM and 25 deg C? --ie, if we move up by 10^(-5)cm<BR/><BR/>For that we need the adiabatic lapse rate, defined as the negative of the rate of change in temperature with height observed while moving upwards through an atmosphere.<BR/> <BR/>Dry adiabatic lapse rate<BR/>The DALR is a constant + 9.78 °C/km <BR/>Moist adiabatic lapse rate<BR/>MALR 4.9 °C/km <BR/><BR/>http://en.wikipedia.org/wiki/Lapse_rate<BR/><BR/>Using the larger of the two values above (dry lapse rate), gives 9.78x10^(-5) deg C per cm rise in elevation <BR/><BR/>or roughly 10^(-9) deg C per every 10^(-5)cm rise in elevation<BR/><BR/>so, in other words, for dry air at the earth's surface, the temperature changes only by 10^(-9) deg C (1 billionth of a deg C!) for a rise in elevation that is equal to the mean free path of Oxygen at the surface, when the temperature at the surface is about 25C.<BR/><BR/>Or we can look at it the other way:<BR/><BR/>Suppose we assume that we gain just enough elevation that the temperature changes by 1/10,000th of one percent (0.000001) of an assumed starting value of 25 deg C at the surface or by 0.000025 deg C.<BR/><BR/>How much elevation elevation gain will that take? <BR/><BR/>0.000025 deg/ (9.78 deg / km) = 2.6 x 10^(-6)km = 0.26cm<BR/><BR/>How many times the mean free path is this?<BR/><BR/>0.26cm/ (mean free path of O2 at 25C and 1 ATM) = 0.26/(9.7 x 10^(-6)cm) = 26000 times!<BR/><BR/>or in other words, the change in elevation that produces a temperature change of 1/10,000th of one percent is equal in length to 26,000 times the mean free path! <BR/><BR/><BR/><BR/>So, in the above case, the mean free path is a tiny fraction of the distance over which the molecule would have to move to experience a temperature change of 1/10,000th of one percent (a tiny fraction of the initial temperature!)<BR/><BR/>Recall the criterion for satisfying Local Thermodynamic equilibrium: <BR/>"the mean free path of the atom is very small compared to the distance over which the temperature changes,"<BR/><BR/>To a very high degree of accuracy, the criterion for local thermodynamic equilibrium as described above <I>is</I> met for air at the earth's surface.<BR/><BR/>Anyone who claims otherwise is either blowing smoke in your face or inhaling it himself (and it sure as hell ain't cigarette or cigar smoke!)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-16612221.post-33911458050640988382007-03-18T22:33:00.000-04:002007-03-18T22:33:00.000-04:00Eli wonders if Essex has ever calculated the devia...Eli wonders if Essex has ever calculated the deviation from a thermal distribution in a volume of air. Rabett Labs has measured it (for example the CO2 rotational distribution in an IR spectrum when you leave the sample chamber open.<BR/><BR/>Oh yes, John A will you please ask where Souda, India is? or more likely what actual data set was used.EliRabetthttps://www.blogger.com/profile/07957002964638398767noreply@blogger.comtag:blogger.com,1999:blog-16612221.post-34741065866368500672007-03-18T17:19:00.000-04:002007-03-18T17:19:00.000-04:00"In order for this to approximate to LTE in the hy..."In order for this to approximate to LTE in the hypothetical box of atmosphere would be to compress the time period under which these properties apply and assume no loss or gain of molecules during that time period."<BR/><BR/>This discussion obviously does not have much (if anything) to do with reality, does it?<BR/><BR/>It reminds me a little of the guy who sits at his desk worrying about whether he and his papers are going to fall through the gaps between the atoms in his chair and desk.<BR/><BR/>In the real world, one makes "good enough" approximations all the time. Without them, we'd all still be living in caves.<BR/><BR/>Within a parcel of air near the earth's surface, where the pressure at the top of the parcel is very nearly the same as that at the bottom, temperature still means something physical -- at least to most of us.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-16612221.post-60562207343625570452007-03-18T13:42:00.000-04:002007-03-18T13:42:00.000-04:00The simplest point of view is that thermodynamics ...<EM>The simplest point of view is that thermodynamics describes the properties of a system at equilibrium, pressure, volume, temperature, etc. if the properties of your non-equilibrium system are the same, then you have a local thermodynamic equilibrium (LTE) in your parcel of atmosphere.</EM><BR/><BR/>Except of course that this is rubbish. In order for this to approximate to LTE in the hypothetical box of atmosphere would be to compress the time period under which these properties apply and assume no loss or gain of molecules during that time period.<BR/><BR/>Under such circumstances as in the Earth's atmosphere where pressure varies continuously, as does temperature, such conditions of LTE will never happen.<BR/><BR/>That's the difference between thermodynamic and non-thermodynamic equilibrium which you've failed to grasp.John Ahttps://www.blogger.com/profile/04544387836529972922noreply@blogger.comtag:blogger.com,1999:blog-16612221.post-8811770054681526132007-03-16T15:34:00.000-04:002007-03-16T15:34:00.000-04:00My goodness, this has certainly become one of the ...My goodness, this has certainly become one of the most amusing pinata parties in some time. Odd that there's not a Pielke or two for whacking.<BR/><BR/>Mus musculus anonymousAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-16612221.post-70842280445485089182007-03-16T11:39:00.000-04:002007-03-16T11:39:00.000-04:00If all scientists had taken the view of Essex, the...If all scientists had taken the view of Essex, there would be no science (and hence no global warming -- by definition, since there would be no thermometer to measure it).<BR/><BR/>Even the ideal gas laws make assumptions that are not (quite) real.<BR/><BR/>That is not to say that there is nothing worth studying in the field of non-equilibrium thermodynamics, of course (living things, for example)Anonymousnoreply@blogger.com