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Unformatted text preview: PHYS 333 Winter 2009 Assignment 8 1. A question on Osmosis. For this problem you need two additions to the theory of osmosis found in the textbook (page 203) and discussed in class. They are that: • the chemical potential of a solvent containing two dilute solutes is μ = μ k B T ( x 1 + x 2 ) where x 1 and x 2 are the respective concentrations. • in the presence of an electrostatic potential Φ , the chemical potential of the positive (negative) ions is increased (decreased) by Φ e , where e is the magnitude of the charge per ion. Note also that the concentration x is defined as the ratio of the number of molecules of solute to the number of water molecules (solvent). Likewise, the concentration of ions is defined as the ratio of the number of ions to the number of water molecules. Two containers A and B, of equal volume, are full of water. Both contain N molecules of water. They are connected by a semipermeable membrane. In the first part of an experiment, a dilute concentration x of NaF is introduced into the container A. The NaF completely disocciates into Na + and F ions, and both Na + and F ions can pass through the membrane. (a) When thermodynamic equilibrium is established, with both containers at the same pres sure and temperature, the chemical potentials in the two containers must be equal. Use the chemical potentials of the solute ions to show that the equilibrium concentrations will therefore be x A Na = x B Na = x / 2 and x A F = x B F = x / 2, so that the condition x A F x A Na = x B F x B Na is satisfied. The notation x A Y means “the concentration of Y in A”. Note, first, that in class, we obtained the condition μ A = μ B for conditions of constant V and N . This is not at all the case here. Rather, under conditions of constant P and T , we have that the Gibbs function G is constant, so that dG = 0 . However, the Gibbs function for the Na ions is G Na = G A Na + G B Na – remember that the Gibbs function is extensive, like the internal energy U – so that for these ions, dG = 0 becomes dG Na = 0 = μ A Na dN A Na + μ B Na dN B Na . Since no Na ions are lost, we also have that dN A Na = dN B Na . Necessarily, therefore, μ A Na = μ B Na . The same argument holds for the F ions, so that μ A F = μ B F . To answer the present question, proceed as follows: The expressions for μ Na and μ F are respectively μ Na = f Na + k B T log x Na and μ F = f F + k B T log x F 1 Therefore, x A Na = x B Na and x A F = x B F . This is true no matter whether the volume of the containers is the same, or not. However, suppose that the numbers N of water molecules in containers A and B are the same. The total number of Na ions is then x N . But, since, in equilibrium, we also have G A Na = μ A Na N A Na = μ B Na N B Na = G B Na , it follows that N A Na = N B Na = x N/ 2 so that x A Na = x B Na = x / 2 ....
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 Winter '09
 Harris
 Physics, Thermodynamics, Gibbs, xa, chemical potential, XA Xa

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