6.10
If the heater operates as rated, then the total amount of heat transferred to
the
cylinder will be
1
1
4
(100 J s )(10 min)(60 s min )
6.0 10 J or 60 kJ
−
−
⋅
⋅
=
×
Work will be given by
ext
= −
Δ
w
P
V
in this case because it is an expansion
against
a constant opposing pressure:
(0.975 atm)(10.00 L
2.00 L)
7.80 L atm
= −
−
= −
⋅
w
In order to combine the two terms to get the internal energy change, we
must first convert the units on the energy terms to the same values. A
handy trick to get the conversion factor for L atm
⋅
to J or
vice versa
is to
make use of the equivalence of the ideal gas constant
R
in terms of
L atm
⋅
and J:
1
1
1
1
8.314 J K
mol
7.80 L atm
790 J or 0.790 kJ
0.082 06 L atm K
mol
−
−
−
−
⋅
⋅
= −
⋅
= −
−
⋅
⋅
⋅
w
The internal energy change is then the sum of these two numbers:
60 kJ
( 0.790 kJ)
59 kJ
U
q
w
Δ
=
+
=
+ −
≅
The energy change due to the work term turns out to be negligible in this
problem.
6.14
(a) true only if
q
= 0;
(b) always true, volume is fixed and no work can be
done;
(c) never true;
(d) always true;
(e) true if
q
= 0
6.24
(a) Because the process is isothermal,
0
Δ
=
U
and
.
= −
q
w
For a
reversible
process,
2
1
ln
= −
V
w
nRT
V
n
is obtained from the ideal gas law:
1
1
1
1
(2.57 atm)(3.42 L)
0.359 mol
(0.082 06 L atm K
mol )(298 K)
7.39
(0.359 mol)(8.314 J K
mol )(298 K) ln
685 J
3.42
685 J
−
−
−
−
=
=
=
⋅
⋅
⋅
= −
⋅
⋅
= −
= +
PV
n
RT
w
q
(b) For step 1, because the volume is constant,
0 and
.
=
Δ
=
w
U
q
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentIn step 2, there is an irreversible expansion against a constant opposing
pressure, which is calculated from
ex
= −
Δ
w
P
V
The constant opposing pressure is given, and
Δ
V
can be obtained from
final
initial
7.39 L
3.42 L
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 Spring '08
 Crowell
 Thermodynamics, Enthalpy, Energy

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