Saved Bookmarks
| 1. |
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 atmosphere' : n_(2) = n_(1) exp [-mg (h_(2)-h_(1))//k_(B)T] where, n_(1). n_(2) refer to number density at height h_(1) and h_(2) respectively. use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid columns: n_(2) = n_(1) exp [-mg N_(A) (roh-roh^('))(h_(2)-h_(1))//(roh RT)] where, roh is the density of the suspended particle, and roh' that of surrounding medium. [N_(A) is Avogadro's number, and R is the universal gas constant]. |
|
Answer» Solution :ACCORDING to the law of atmospheres, `n_(2) = n_(1)` exp. `[-(mg)/(k_(B)T)(h_(2)-h_(1))]` ...`(i)` Where `n_(2), n_(1)` refer to number density of particles at heights `h_(2)` and `h_(1)` respectively. If we consider the sedimention equilibriumof suspended particles in a liquid, then in place of mg, we will haveto take effective weight of the suspended particles. LET `V =` averagevolume of a suspended particle, `rho=` density of suspended particle `rho'=` density of liquid, `m=`mass of one suspendedparticle, `m' = `mass of one suspended particle, `m' =` equal volume of liquid displaced. According to Archimede'sprinciple, effective weight of one suspended particle = actual weight - weightof liquid displaced `= mg - m'G` `=mg-V rho' g = mg -(m/(rho)) rho' g = mg (1-(rho')/(rho))` Also, Boltzman constant, `k_(B)= (R)/(N_A)` where, R is gas constant and `N_(A)` is Avogadro's number. Putting , `mg (1-(rho')/(rho))` in place of mg and VALUE of `k_(B)` in `(i)` , we get `n_(2) = n_(1)` exp. `[-(mgN_(A))/(RT)(1-(rho')/(rho))(h_(2)-h_(1))]` which is the required relation. |
|