HW14_Sol_F10

HW14_Sol_F10 - Dec. 3, 2010 PHYS 2101 HW#14 WileyPlus...

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Dec. 3, 2010 PHYS 2101 HW#14 WileyPlus Problem Solutions 1 HW #14 Solutions – or at least most of them… (USING BOOK VALUES 9 th Edition in problems; problem # indicated by []) 1) [20-CQ4] From the information, “ An ideal monatomic gas at initial temperature T 0 (in kelvins) expands from initial volume V 0 to volume 2V 0 by each of the five processes indicated in the T-V diagram” , we know that C V = 3 2 R , C P = C V + R = 5 2 R , and γ = C P V = 5 3 [NOTE: this is a T - V diagram, not p - V ] (a) In an isothermal expansion, Δ T = T f T 0 = 0 . The only process in which the temperature remains constant is AE. (b) In an isobaric expansion, the pressure ( p ) remains constant. From the ideal gas law, pV = nRT , this means that T 0 V 0 = T f V f T f = T 0 V 0 2 V 0 ( ) = 2 T 0 . The process in which the temperature increases by a factor of 2 is AC. (c) In an adiabatic expansion, Q = 0. In Section 19.11, we know for an adiabatic process, p 0 V 0 = p f V f [Eqn. 19–53] and T 0 V 0 1 = T f V f 1 [Eqn. 19–55]. Here T 0 V 0 1 = T f 2 V 0 ( ) 1 , or T f = T 0 1 2 ( ) 1 = T 0 1 2 ( ) 5 3 1 0.63 T 0 . The process in which the temperature decrease by this factor is AF. (d) In general, the entropy change of a reversible gas process is given by [see Eqn. 20–4]: Δ S gas , reversible = nR ln V f V 0 + nC V ln T f T 0 . In this case, Δ S gas , reversible = nR ln V f V 0 + 3 2 ln T f T 0 = nR ln 2 V 0 V 0 T f T 0 3 2 = nR ln 2 T f T 0 3 2 is never negative (the entropy never decreases; it either remains the same AF , or increases AE , AD , AC , AB ).
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HW14_Sol_F10 - Dec. 3, 2010 PHYS 2101 HW#14 WileyPlus...

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