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Unformatted text preview: E int = 6 p V . Since n = 1 mol, this can also be written Q = 6 RT . Since the process is at constant volume, use dQ = nC V dT to obtain S = Z dQ T = nC V Z T c T b dT T = nC V ln T c T b . Substituting C V = 3 2 R and using the ideal gas law, we write T c T b = p c V c p b V b = (2 p )(4 V ) p (4 V ) = 2 . Thus, S = 3 2 nR ln 2. Since n = 1, this is S = 3 2 R ln 2. (c) or a complete cycle, E int = 0 and S = 0....
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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.
 Fall '08
 SPRUNGER
 Physics, Work

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