PS6Sol_f10

# PS6Sol_f10 - Physics 112 Fall 2010 Professor William...

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Unformatted text preview: Physics 112 Fall 2010 Professor William Holzapfel Homework 6 Solutions Problem 1, Kittel 5.1: Centrifuge When the gas is in equilibrium, we know that the total chemical potential will be constant through- out. The total chemical potential is μ = μ int + μ ext , and we know that the internal chemical potential (of an ideal gas) is related to the number density by μ = τ ln n ( r ) n Q , so we see that a spatial variation in the external chemical potential will create a spatially varying number density. A rotating gas feels a centrifugal force F = Mv 2 /r = Mrω 2 . We can consider this force to be caused by an effective potential such that F =- dV/dr : μ ext ( r ) = V ( r ) =- Z drF ( r ) =- M 2 r 2 ω 2 . The total chemical potential μ ( r ) = τ ln n ( r ) n Q- M 2 r 2 ω 2 will be constant throughout the centrifuge in equilibrium, so we can equate the chemical potential at a radius r with the chemical potential at the axis of rotation: μ (0) = τ ln n (0) n Q = τ ln n ( r ) n Q- M 2 r 2 ω 2 = μ ( r ) Solving for n ( r ) we have: n ( r ) = n (0) exp Mr 2 ω 2 2 τ (1) So we see that we get an exponentially higher concentration of particles at larger r (as expected). Problem 2, Kittel 5.4: Active Transport We know that the ratio of the concentration of K + is n sap /n water = 10 4 , so we can find the difference in the chemical potentials of K + of the sap and the water by treating the ions as an ideal gas: 4 μ = μ sap- μ water = τ ln n sap n Q- τ ln n water n Q = τ ln n sap n water = τ ln(10 4 ) = (300 K ) k b * 4 ln(10) = (1200 K) ( 1 . 38 ×...
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## This note was uploaded on 11/18/2010 for the course PHYSICS 112 taught by Professor Steveng.louie during the Fall '06 term at Berkeley.

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PS6Sol_f10 - Physics 112 Fall 2010 Professor William...

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