P21_019 - E int = 6 p V . Since n = 1 mol, this can also be...

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19. (a) Work is done only for the ab portion of the process. This portion is at constant pressure, so the work done by the gas is W = Z 4 V 0 V 0 p 0 dV = p 0 (4 V 0 V 0 )=3 p 0 V 0 . (b) We use the Frst law: ∆ E int = Q W . Since the process is at constant volume, the work done by the gas is zero and E int = Q . The energy Q absorbed by the gas as heat is Q = nC V T ,where C V is the molar speciFc heat at constant volume and ∆ T is the change in temperature. Since the gas is a monatomic ideal gas, C V = 3 2 R . Use the ideal gas law to Fnd that the initial temperature is T b = p b V b /nR =4 p 0 V 0 /nR and that the Fnal temperature is T c = p c V c /nR =(2 p 0 )(4 V 0 ) /nR = 8 p 0 V 0 /nR .Thu s , Q = 3 2 nR µ 8 p 0 V 0 nR 4 p 0 V 0 nR =6 p 0 V 0
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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.

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