romasko (qrr58) – Homework 9 – sutcliffe – (50985)
1
This
printout
should
have
22
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
What’s the chemical formula of water?
HIJKLMNO (a little geek humor for you)
NOTE: Read the questions carefully. If it
asks for a CHANGE (Delta) in a quantity or
the VALUE of that quantity, then you must
be sure to include the correct sign. Remem
ber that the quantities are for the SYSTEM,
unless stated differently, and sometimes we
switch our viewpoint of what the system is
once we have finished a calculation. This cov
ers Ch. 9. The rest will be on HW 10, the last
HW.
001
10.0 points
When a hydrocarbon is burned in the pres
ence of oxygen, the amount of energy in the
universe
1.
increases.
2.
decreases.
3.
stays the same.
correct
Explanation:
002
10.0 points
A sample of an ideal gas having a volume of
0.220 L at 298 K and 10.0 atm pressure is al
lowed to expand against a constant opposing
pressure of 0.825 atm until it has a volume of
1.000 L at 200 K and the pressure equals the
opposing pressure. What is the work for the
system?
Correct answer:

65
.
1865 J.
Explanation:
P
1
= 10.0 atm
P
2
= 0.825 atm
V
1
= 0.220 L
V
2
= 1.000 L
T
1
= 298 K
T
2
= 200 K
In this case, the
P
is the constant opposing
pressure 0.825 atm.
Δ
V
is the change in
volume you observe, from 0.220 L to 1.000 L.
Plugging these values into the equation, you
get
w
=

P
Δ
V
w
=

(0
.
825 atm)(1
.
000 L

0
.
220 L)
=

(0
.
825 atm)(0
.
780 L)
However, some conversion factors will have
to be applied.
To get to Joules, which is
the same as N
·
m, two conversions will be
needed: L to m
3
and atm to unit of pressure
which includes newtons (which is Pa, defined
as N/m
2
). Recalling that 1 mL = 1 cm
3
, and
100 cm = 1 m, the equation should be
w
=

(0
.
825 atm)(0
.
780 L)
×
parenleftbigg
1000 cm
3
1 L
parenrightbigg parenleftbigg
101325 N
/
m
2
1 atm
parenrightbigg
×
parenleftbigg
1 m
3
10
6
cm
3
parenrightbigg
=

65
.
2 J
003
10.0 points
A CD player and its battery together do 500
kJ of work, and the battery also releases 250
kJ of energy as heat and the CD player re
leases 50 kJ as heat due to friction from spin
ning.
What is the change in internal energy
of the system, with the system regarded as
the CD player alone?
Assume that the bat
tery does 500 kJ of work on the CD player,
which then does the same amount of work on
the surroundings.
1.
+450 kJ
2.

50 kJ
correct
3.

550 kJ
4.

950 kJ
5.

800 kJ
Explanation:
Heat
from
the CD player is

50 kJ.
Battery (part of the
surroundings
) does

500 kJ work on the CD player.
CD player does

500 kJ work on some other
part of the
surroundings
.
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romasko (qrr58) – Homework 9 – sutcliffe – (50985)
2
This question is testing your ability to see
what the system is, and then look at ONLY
the energy flow for the system. Here the sys
tem is just the CD player. What the battery
player does is irrelevant, unless it involves the
CD player:
Δ
U
=
q
+
w
=

50 kJ + [500 kJ + (

500 kJ)]
=

50 kJ
004
10.0 points
Calculate the energy required to heat 436 g
of mercury from 37
◦
C to 42
◦
C. The specific
heat capacity of Hg(
ℓ
) is 0
.
138 J
/
g
·
◦
C.
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 Fall '07
 Fakhreddine/Lyon
 Chemistry, Enthalpy, Energy, Hrxn

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