phys 369 09 hw 4 solutions

# phys 369 09 hw 4 solutions - Phys. 369, Fall 2009 Homework...

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Phys. 369, Fall 2009 Homework #4 Solutions 1. (18 points) a) Let’s first consider the position and momentum states of the atoms, and neglect the spin states. The multiplicity for position and momentum states is just that of a gas of identical atoms: () 3/ 2 , 3 2 !3 / 2 ! NN gas ident N Vm U dU Nh N U π Ω= ( 1 ) We must now consider the spins of the atoms. In writing eq. (1), we are assuming we are in the classical limit. This means that for each atom, there is a negligible probability to find a second atom within the same 6-dimensional ( ) ,,, , , x yz xyzp p p volume element (size 3 h ) of phase space. In the classical limit, for purposes of counting the multiplicity due to spin, we can therefore treat the atoms as distinguishable. That is because each atom will have a position and/or momentum coordinate which is different than all other atoms, i.e. the atoms are not identical because they are at different places in phase space. This means that for each unique N- particle position and momentum state counted by eq. (1), the gas has 2 N spin ( 2 ) different spin states, since there are N atoms, each of which can be in one of two possible spin states or . Therefore the overall multiplicity of the spin-1/2 3 He gas is 2 , 3 2 2 / 2 ! N gas ident spin N U dU N U Ω=Ω ×Ω = × (3) b) The entropy is ,, 3/2 2 2 ln ln ln ln ln 2 45 ln ln 2 32 25 ln ln 2 2 gas ident spin gas ident spin Sackur Tetrode spin N Sackur tetrode Sk k k k S S U Nk Nk h k T Nk Nk = Ω = + =+ ⎡⎤ ⎛⎞ + ⎢⎥ ⎜⎟ ⎝⎠ ⎣⎦ + (4) where we have used the relation 3 .

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## This note was uploaded on 11/06/2010 for the course PHY 369 taught by Professor Staff during the Fall '08 term at University of Texas.

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phys 369 09 hw 4 solutions - Phys. 369, Fall 2009 Homework...

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