phys 369 09 hw 8 solutions

phys 369 09 hw 8 solutions - Phys. 369, Fall 2009 Homework...

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1 Phys. 369, Fall 2009 Homework #8 Solutions 1. (14 points) a) The energy of the ball is () () ( () ) osc th Ut U t U Tt = + , where 2 1 2 osc Utm g h t m v t =+ is the energy of the ball’s oscillation in the valley, and (() ) th UT t is the thermal energy of the ball, with m the mass of the ball, h its elevation, v its velocity, and T ( t ) its temperature. An expression for U th can be derived from the thermodynamic relation. With P fixed ( dP = 0), it can be re-written as ( ) th P dU C T dT T PV dT β =− ( 1 ) where ( ) T is the volume thermal expansion coefficient and V the volume of the ball. When I integrate this equation I will have one constant of integration that I can choose however I like. I will choose the integration constant such that eq. (1) integrates to give ( ) 0 ( ) valley Tt th P T t CTd T P V Td T ′′ ≈− ∫∫ ( 2 ) For your solution, I will give full credit for anything of the form 0 T th UC T d T . Here, I am just showing one way to handle the details correctly so you can see how to do that. Since the temperature of the ball begins and ends at value T valley , the initial and final values of the thermal energy are both 0 valley T th valley P T = . ( 3 ) We’ll set the arbitrary zero for the elevation h of the ball to be the elevation at the bottom of the valley. Then, the initial and final values of U osc are (0) mgh and 0, respectively. So, the initial and final values of U are ( ) ( ) ( ) 00 th valley Ut m g h U T == + and ( ) ( ) th valley →∞ = ( 4 a , b ) The entropy from thermal excitations of the ball is within two orders of magnitude of Nk , where N is the number of atoms in the ball (something like 10 24 ). Thermodynamically, the oscillation mode of the ball is equivalent to the excitation of N = 1 “atom.” Therefore the entropy arising from the oscillation motion of the ball in the valley is completely negligible. We can define the entropy of the ball from the relation P CdT dQ dS TT , where in the second equality we are using the fact that the pressure is constant. Therefore the entropy of the ball is just 0 ) P T STt T = ( 5 )
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2 The initial and final values of the entropy are the same, () 0 (0 ) ( ) valley T P valley CTd T St ST T == = = ( 6 ) The free energy is F UT S =− . From eqs. (4a,b) and (6), the initial and final values of F are ( ) () ( ) 00 th valley valley valley Ft m g h U T T ST + ( 7 a ) () () th valley valley valley →∞ = ( 7 b ) where th valley and valley are given in eqs. (3) and (5).
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phys 369 09 hw 8 solutions - Phys. 369, Fall 2009 Homework...

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