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# A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheresn2=n1exp[−mg(h2−h1)/kBT]where n2,n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:n2=n1exp[−mgNA(ρ−ρ′)(h2−h1)/(ρRT)]where ρ is the density of the suspended particle and ρ′ that of surrounding medium. [NA is Avogadro's number, and R the universal gas constant.][Hint : Use Archimedes principle to find the apparent weight of the suspended particle.]

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Solution

## According to the law of atmospheres, we have:n2=n1exp[−mg(h2−h1)/kBT] ...(i)where,n1 is the number density at height h1, and n2 is the number density at height h2mg is the weight of the particle suspended in the gas columnDensity of the medium =ρ′Density of the suspended particle =ρMass of one suspended particle =m′Mass of the medium displaced =mVolume of a suspended particle =VAccording to Archimedes’ principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as:Weight of the medium displaced – Weight of the suspended particle =mg−m′g=mg−Vρ′g=mg−(m/ρ)ρ′g=mg(1−(ρ′/ρ)) ... (ii)Gas constant, R=kBNkB=R/N ...(iii)Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), we get:n2=n1exp[−mg(h2−h1)/kBT] =n1exp[−mg(1−ρ′/ρ))(h2−h1)(N/RT)] =n1exp[−mg(ρ−ρ′)(h2−h1)(N/RTρ)]  Suggest Corrections  0    Join BYJU'S Learning Program
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