Jahnavi said:
OK . Thanks !
@Charles Link ,
@Chestermiller Kindly help me understand another point given in the book especially the second and third lines .
View attachment 225038
It would appear they are referring to a result that follows from Statistical and Thermal Physics, that the density of a gas as a function of energy will be of the form ## \rho(E)=\rho_o \, e^{-E/(k_B T)} ## where ## E=+mgh +(\frac{1}{2})mv^2 ##. This calculation is somewhat advanced. In addition, the average of ## \frac{1}{2} mv^2=\frac{3}{2}k_B T ##, so that you can compute the effect that the ## mgh ## term will have on the density. ## \\ ## If ## mgh<<k_B T ## , the height ## h ## will have very little effect on the density. ## \\ ## Edit: The result of the above is basically that the density as a function of height ## h ## will be ## \rho(h)=\rho(0) \, e^{-mgh/(k_B T)} ##. If ## mgh <<k_BT ## for all ## h ##, then the density is nearly uniform. And if ## mgh <<k_B T ##, you can also say that ## mgh << \frac{1}{2}mv^2 ##.## \\ ## The book's author apparently is familiar with this type of calculation, but it is really a somewhat advanced upper level undergraduate topic. ## \\ ## ................................## \\ ## And it may be worthwhile to relate this to Archimedes' result: force per unit volume from the pressure gradient must offset the gravitational force per unit volume, so that ## -\nabla P=\rho g \, \hat{z} ##. ## \\ ## Now ## PV=nRT ## and ## \rho=\frac{n \, M.W. }{V} ##, so that ## P=\frac{\rho RT }{ M.W.} ##. ## \\ ## The result is ## (\frac{RT}{M.W.}) \frac{d \rho}{dz}=-\rho g ##. ## \\ ## Solving this differential equation, we get ## \rho=\rho_o \, e^{-(M.W.) (g z)/(RT)} ## . Here the gas constant ## R=N_o \, k_B ##, where ## N_o ## is Avagadro's number (particles/mole). Thereby, ##\frac{ M.W. (in \, grams/mole)}{N_o}=m ##, where ## m ## is the mass of the gas molecule. ## \\ ## This gives ## \rho=\rho_o \, e^{-mg z/(k_B T)} ##. ## \\ ## These results from Archimedes and the ideal gas law are consistent with our formula above ## \rho(h)=\rho(0) \, e^{-mgh/k_B T} ##, which can also be written in the form ## P(h)=P_o \, e^{-mgh/(k_BT)} ##. Note that although the pressure has a somewhat complicated form as a function of height ## h ##, (more complicated than for a liquid), Archimedes principle still works with a gas as well: The buoyant force is equal to the weight of the gas that is displaced. ## \\ ## It may also be of interest, that for an uncompressible liquid, the pressure varies with depth ##d ## as ## P(d)=P(0) +\rho_o \, g \, d ##. Gases are compressible, so that the equation describing pressure as a function of height is much different. ## \\ ## .............................. ## \\ ## And let's try some gas molecules at room temperature: ## \\ ## ## k_B T=1.381 \cdot 10^{-23} (300) \approx 4.1 \cdot 10^{-21} \, joules ##, ## \\ ## and ## mgh ## for a height of ##h= 1.0 \, meters## , (## M.W.=30 ##), is ## \\ ## ## mgh \approx 30 (1.67 \cdot 10^{-27}) (9.8)(1.0)=4.9 \cdot 10^{-25} \, joules ##. ## \\ ## It looks like ## k_BT>> mgh ##. ## \\ ## The equation also gives the result that the atmospheric pressure should be ## P(h) \approx 0.5 \, atm ## at ## h=5800 \, meters ## altitude. ## \\ ## Edit: Note that the atmosphere is 78% ## N_2 ## with M.W.=28, and 21% ## O_2 ## with M.W.=32, so that using a number of M.W. ## \approx ## 30 for the atmosphere should give us reasonably accurate results for the pressure as a function of altitude. (Temperature is also known to vary somewhat with altitude=we're just looking for approximate results for the pressure vs. altitude calculation).
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