This preview shows pages 1–3. Sign up to view the full content.
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
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Crowell

Click to edit the document details