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52. (a) Using
M
= 32.0 g/mol from Table 191 and Eq. 193, we obtain
sam
12.0 g
0.375 mol.
32.0 g/mol
M
n
M
==
=
(b) This is a constant pressure process with a diatomic gas, so we use Eq. 1946 and
Table 193. We note that a change of Kelvin temperature is numerically the same as a
change of Celsius degrees.
()
3
77
0.375 mol
8.31 J/mol K 100K
1.09 10 J.
22
p
Qn
CTn R T
§·
§
·
=Δ
=
Δ
=
⋅
=
×
¨¸
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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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