hw2_sol - 1. Two identical blocks of the same material with...

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1. 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 . A. What is the temperature of the second block at the same time? 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. 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: 1 2 () / 2 T T ′′ ==+ which is the equilibrium temperature
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2 . 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. 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 12 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 +++ + ∆=∆ + ∆ = + + + ++ B* Answer question A for the case where both gases consist of the same molecules.
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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.

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hw2_sol - 1. Two identical blocks of the same material with...

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