P21_020

# P21_020 - 20. (a) The ﬁnal pressure is pf = (5.00...

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Unformatted text preview: 20. (a) The ﬁnal pressure is pf = (5.00 kPa)e(Vi −Vf )/a = (5.00 kPa)e(1.00 m 3 −2.00 m3 )/1.00 m3 = 1.84 kPa . (b) We use the ratio form of the gas law (see Sample Problem 20-1) to ﬁnd the ﬁnal temperature of the gas: p f Vf (1.84 kPa)(2.00 m3 ) Tf = Ti = 441 K . = (600 K) p i Vi (5.00 kPa)(1.00 m3 ) For later purposes, we note that this result can be written “exactly” as Tf = Ti 2e−1 . In our solution, we are avoiding using the “one mole” datum since it is not clear how precise it is. (c) The work done by the gas is f W Vf p dV = = (5.00 kPa)e(Vi −V )/a dV Vi i Vf = (5.00 kPa)eVi /a · − ae−V /a = = (5.00 kPa)e1.00 (1.00 m3 ) e−1.00 − e−2.00 3.16 kJ . Vi (d) Consideration of a two-stage process as suggested in the hint, brings us simply to Eq. 21-4. Consequently, with CV = 3 R (see Eq. 20-43), we ﬁnd 2 ∆S Vf Vi = nR ln = nR ln 2 + = = = +n Tf 3 R ln 2 Ti 3 ln 2e−1 2 3 3 p i Vi ln 2 + ln 2 + ln e−1 Ti 2 2 3 (5000 Pa)(1.00 m ) 5 3 ln 2 − 600 K 2 2 1.94 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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