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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 201) 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 twostage process as suggested in the hint, brings us simply to Eq. 214. Consequently, with CV = 3 R (see Eq. 2043), 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.
 Fall '08
 SPRUNGER
 Physics

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