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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online