P21_061 - Q = W , which further implies (regarding Eq....

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61. (a) It is a reversible set of processes returning the system to its initial state; clearly, ∆ S net =0. (b) Process 1 is adiabatic and reversible (as opposed to, say, a free expansion) so that Eq. 21-1 applies with dQ =0andy ie lds∆ S 1
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Unformatted text preview: Q = W , which further implies (regarding Eq. 21-1) dQ = pdV . Therefore, Z dQ T = Z pdV ³ pV nR ´ = nR Z dV V which leads to ∆ S 3 = nR ln 1 2 = − 23 . 0 J/K. (d) By part (a), ∆ S 1 + ∆ S 2 + ∆ S 3 = 0. Then, part (b) implies ∆ S 2 = − ∆ S 3 . Therefore, ∆ S 2 = 23 . 0 J/K....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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