15. (a) At point
a
, we know enough information to compute
n
:
()
(
)
3
2500Pa 1.0m
1.5mol.
8.31 J/mol K
200K
pV
n
RT
==
=
⋅
(b) We can use the answer to part (a) with the new values of pressure and volume, and
solve the ideal gas law for the new temperature, or we could set up the gas law as in
Sample Problem 191 in terms of ratios (note:
n
a
=
n
b
and cancels out):
3
3
7.5kPa
3.0m
2.5kPa
1.0m
bb
b
b
aa
a
pV
T
T
T
§·
=
¡
=
¨¸
©¹
which yields an absolute temperature at
b
of
T
b
= 1.8×10
3
K.
(c) As in the previous part, we choose to approach this using the gas law in ratio form
(see Sample Problem 191):
3
3
2.5kPa
2.5kPa
cc
c
c
a
T
T
T
=
¡
=
which yields an absolute temperature at
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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