hw2_solutions_F

# hw2_solutions_F - 1. (10 points). Entropy changes during...

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1 . (10 points). Entropy changes during heat exchange . Two identical blocks of the same material with temperature independent heat capacities C p are brought together. Their initial temperatures are T 1 and T 2 , T 1 < T 2 . Heat exchange with the surroundings can be neglected. Some time later the temperature of the first block was found to be 1 T (note that 1 T is not the equilibrium temperature but a temperature measured at some point in the course of the equilibration process). A. What is the temperature of the second block at the same time? You can solve this the same way we did it in the class – by equating the heat received by one block and that lost by the other. Or you can do it by noticing that the total enthalpy of the two blocks is conserved since this is a constant P process and the combined system (the two blocks) does not exchange heat with the surroundings From H=0, we find C p T 1 + C p T 2 = C p T 1 ‘+ C p T 2 and T 2 ’ = T 1 +T 2 -T 1 B. Calculate the entropy of the system (the two blocks) at this moment. S( 1 T ) = 11 2 2 1 2 1 2 ln( / ) ln( / ) ln( / ) ln(( ) / ) pp CT T TCT T C T T T T ′′′ += + + C. Show that a process such that 1 T < T 1 would violate the 2 nd law Let TT T =+ , then we find from part A: 22 T =− and from part B: 1 2 1 2 2 12 1 2 ln(( ) / ) ln(( ) / ) ln(( )( ) /( )) ln(1 ) p p S C T C TC T T TTT C T ∆= + + = + ∆∆∆ if ∆Τ <0 and T 1 <T 2 then we find S<0, which contradicts the 2 nd law.

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D. Finally, find the maximum of the system’s entropy as a function of 1 T . Demonstrate that 1 T corresponding to this maximum is, indeed, the equilibrium temperature. Take S( 1 T ) found in part B, take its derivative with respect to 1 T and set it equal to zero, d S( 1 T )/d 1 T = 0 (this derivative has to be zero for a maximum), this will give you, after some algebra: 12 1 2 () / 2 TT T T ′′ ==+ which is the equilibrium temperature 2 . Mixing of two gases . n 1 moles of an ideal gas 1 are separated from n 2 moles of an ideal gas 2 by a partition that does not conduct heat (see Figure below). The volumes occupied by the gases are V 1 and V 2 , respectively, and their temperatures are T 1 and T 2 . There is no heat exchange between the entire system (the two gases) and the surroundings. Assume that both gases have the same, temperature independent molar heat capacity V c A. (10 points) The partition is removed and the gases attain equilibrium. What is the change of the entropy of the system as a result of this process? First, we calculate the final temperature T after equilibration by realizing that the total energy of the system must be conserved: 11 2 2 1 2 VV V ncT ncT n n cT += + 11 2 2 1 2 / ( ) Tn T n n =+ + Then use our result for the entropy of an ideal gas as a function of V and T: 1 2 22 1 2 1 1 2 2 1 1 2 2 ln ln ln ln ( )( ) n T n T SSS n R n c n R n c V T nn V Tnn +++ + ∆=∆ + ∆ = + + + ++
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## This note was uploaded on 09/06/2009 for the course CH 353M taught by Professor Lim during the Spring '08 term at University of Texas at Austin.

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hw2_solutions_F - 1. (10 points). Entropy changes during...

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