Answer, Key – Homework 6 – David McIntyre
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Chapters 20 and 21 problems.
001 (part 1 of 1) 0 points
A 0.0505 kg ingot of metal is heated to 173◦ C
and then is dropped into a beaker containing
0.426 kg of water initially at 23◦ C.
If the final equilibrium state of the mixed
system is 25.4 ◦ C, find the specific heat of
the metal. The specific heat of water is
4186 J/kg · ◦ C.
Correct answer: 574.172 J/kg · ◦ C.
Explanation:
Given :
mw
mx
cw
Tw
Ti
Tf
= 0.426 kg ,
= 0.0505 kg ,
= 4186 J/kg · ◦ C ,
= 23◦ C ,
= 173◦ C , and
= 25.4◦ C .
Because the thermal energy lost by the ingot equals the thermal energy gained by the
water,
mx cx (Ti − Tf ) = mw cw (Tf − Tw ) ,
mw cw (Tf − Tw )
mx (Ti − Tf )
(0.426 kg) (4186 J/kg · ◦ C)
=
0.0505 kg
25.4◦ C − 23◦ C
×
173◦ C − 25.4◦ C
= 574.172 J/kg · ◦ C .
cx =
002 (part 1 of 1) 0 points
337 g of water at 70◦ C is poured into a 25 g
aluminum cup containing 66 g of water at
29◦ C.
What is the equilibrium temperature of the
system? Assume the specific heat of aluminum is 0.215 cal/g ·◦ C.
Correct answer: 62.8341 ◦C.
1
Explanation:
Given :
mh
mc
mAl
cw
cAl
Tc
Th
= 337 g ,
= 66 g ,
= 25 g ,
= 1 cal/g ·◦ C ,
= 0.215 cal/g ·◦ C ,
= 29◦ C , and
= 70◦ C .
The heat lost by hot water equals the heat
gained by cold metal plus the water.
mh cw (Th − Tf )
= mAl cAl (Tf − Tc ) + mc cw (Tf − Tc )
(mh cw + mAl cAl + mc cw ) Tf
= mh cw Tf + mAl cAl Tc + mc cw Tc
Tf =
mh cw Th + mAl cAl Tc + mc cw Tc
mh cw + mAl cAl + mc cw
Since
mh cw Th + mAl cAl Tc + mc cw Tc
= (337 g) (1 cal/g ·◦ C) (70◦ C)
+ (25 g) (0.215 cal/g ·◦ C) (29◦ C)
+ (66 g) (1 cal/g ·◦ C) (29◦ C)
= 25659.9 cal
and
mh cw + mAl cAl + mc cw
= (337 g) (1 cal/g ·◦ C)
+ (25 g) (0.215 cal/g ·◦ C)
+ (66 g) (1 cal/g ·◦ C)
= 408.375 cal/◦C ,
then the equilibrium temperature will be
25659.9 cal
408.375 cal/◦C
= 62.8341◦C ,
Tf =
003 (part 1 of 1) 10 points
A calorimeter contains 410 mL of water at
48 ◦ C and 43 g of ice at 0 ◦ C.
Find the final temperature of the system.
Correct answer: 35.8923 ◦ C.
Explanation:
Answer, Key – Homework 6 – David McIntyre
Given : V
mw
mice
cw
Lf
Tw
Ti
∆T
= 410 mL ,
= 410 g ,
= 0.043 kg ,
= 1 cal/g ·◦ C ,
= 333000 J/kg ,
= 48 ◦ C ,
= 0 ◦ C , and
= 48 ◦ C .
The 410 mL = mv of water cools to 0 ◦ C,
giving off
Qw = mw cw ∆T
= (410 g) (1 cal/g ·◦ C) (48 ◦ C)
= 19680 cal
heat.
If all of the ice melts, it absorbs
Qi = mice Lf
= (0.043 kg) (333000 J/kg) (0.2389 cal/J)
= 3420.81 cal
heat.
Since Qw > Qi , all of the ice melts and also
warms up to a temperature above 0 ◦ C: Thus
Qlostwater = Qmeltice + Qgained
mw cw ( Tw − Tf ) = Qi + mice cw ( Tf − Tice )
mw cw Tw − mw cw Tf = Qi + mice cw ( Tf − 0)
mw cw Tw − Qi = ( mice cw + mw cw )Tf
2
Calculate the work done by the gas during
this process.
Correct answer: 367.364 J.
Explanation:
Given : P = 101300 Pa ,
V1 = 4 cm3 ,
V2 = 3630.5 cm3 ,
m = 0.004 kg , and
Lv = 2.26 × 106 J/kg .
W = P ∆V
= P (V2 − V1 )
= (101300 Pa)(3630.5 cm3 − 4 cm3 )
µ
¶3
1m
·
100 cm
= 367.364 J.
005 (part 2 of 3) 0 points
Find the amount of heat added to the water
to accomplish this process.
Correct answer: 9040 J.
Explanation:
Q = m Lv
= (0.004 kg)(2.26 × 106 J/kg)
= 9040 J.
006 (part 3 of 3) 0 points
m w c w Tw − Q i
Find the change in internal energy.
Tf =
mice cw + mw cw
Correct answer: 8672.63 J.
(410 g)(1 cal/g ·◦ C)(48 ◦ C) − 3420.81 cal
Explanation:
=
(0.043 kg)(1 cal/g ·◦ C) + (410 g)(1 cal/g ·◦ C)
19680 cal − 3420.81 cal
=
∆U = Q − W = 9040 J − 367.364 J
43 cal/◦ C + 410 cal/◦ C
= 8672.63 J.
= 35.8923 ◦ C .
004 (part 1 of 3) 0 points
4 cm of water is boiled at atmospheric pressure to become 3630.5 cm3 of steam, also at
atmospheric pressure.
3
007 (part 1 of 3) 4 points
Two moles of helium gas initially at 252 K
and 0.33 atm are compressed isothermally to
1.57 atm.
Answer, Key – Homework 6 – David McIntyre
Find the final volume of the gas. Assume
the helium to behave as an ideal gas.
Correct answer: 0.0263486 m3 .
Explanation:
Given : n = 2 mol ,
R = 8.31451 J/K · mol ,
Tf = 252 K , and
Pf = 1.57 atm .
From the ideal gas law,
Pf V f = n R T
nRT
Vf =
Pf
(2 mol)(8.31451 J/K · mol)(252 K)
=
(1.57 atm)(101300 Pa/atm)
= 0.0263486 m3 .
3
Explanation:
From the first law of thermodynamics, since
∆U = 0,
Q=W
= −6.5361 kJ .
010 (part 1 of 2) 5 points
A gas is taken through the cyclic process described by the figure. Let a = 9, b = 9.
P (kPa)
14
B
12
10
8
6
A
4
C
2
008 (part 2 of 3) 3 points
Find the work done by the gas.
Correct answer: −6.5361 kJ.
Explanation:
Given : Pi = 0.33 atm and
Ti = 252 K .
The initial volume was
nRT
Pi
(2 mol)(8.31451 J/K · mol)(252 K)
=
(0.33 atm)(101300 Pa/atm)
= 0.125356 m3
Vi =
and the work done was
µ ¶
Z
Vf
W = P dV = n R T ln
Vi
= (2 mol)(8.31451 J/K · mol)(252 K)
µ
¶
0.0263486 m3
1 kJ
ln
3
0.125356 m
1000 J
= −6.5361 kJ .
009 (part 3 of 3) 3 points
Find the thermal energy transferred.
Correct answer: −6.5361 kJ.
2
4
6
8 10 12 14
V (m3 )
Find the net energy transferred to the system by heat during one complete cycle.
Correct answer: 40.5 kJ.
Explanation:
Given :
∆P = 9000 Pa and
∆V = 9 m3 .
The change in internal energy is
∆Ucycle = Qcycle + Wcycle = 0
Qcycle = −WABCA
= (area enclosed in P V diagram)
¢
1¡
=
9 m3 (9000 Pa)
2
= 40500 J
= 40.5 kJ .
Alternate Solution:
Answer, Key – Homework 6 – David McIntyre
For each step the work is the negative of
the area under the curve on the PV diagram:
¡
¢
WAB = −[(3.6 kPa) 10.5 m3 − 6.5 m3
1
+ (9.6 kPa − 3.6 kPa)
2 ¡
¢
· 10.5 m3 − 6.5 m3 ]
= −26.4 kJ
WBC = 0 kJ
¢
¡
WCA = −(3.6 kPa) 6.5 m3 − 10.5 m3
= 14.4 kJ
Qcycle = −WABC
= −(−26.4 kJ + 0 kJ + 14.4 kJ)
= 12 kJ .
011 (part 2 of 2) 5 points
If the cycle is reversed (that is, the process
follows the path ACBA), what is the net energy transferred by heat per cycle?
Correct answer: −40.5 kJ.
Explanation:
If the cycle is reversed, then
Qcycle = WACBA = −WABCA = −40.5 kJ .
012 (part 1 of 2) 0 points
An ideal gas initially at 333 K undergoes an
isobaric expansion at 2 kPa.
If the volume increases from 1.6 m3 to 4 m3
and 15.9 kJ of thermal energy is transferred to
the gas, find the change in its internal energy.
Correct answer: 11.1 kJ.
Explanation:
Given :
P = 2 kPa ,
Vf = 4 m 3 ,
Vi = 1.6 m3 ,
Ti = 333 K .
and
From the first law of thermodynamics
∆U = Q − W = Q − P (Vf − Vi )
= 15.9 kJ − (2 kPa)(4 m3 − 1.6 m3 )
= 11.1 kJ .
4
013 (part 2 of 2) 0 points
Find the final temperature of the gas.
Correct answer: 832.5 K.
Explanation:
From the ideal gas law
Vf
Vi
=
Ti
Tf
V f Ti
Tf =
Vi
(4 m3 )(333 K)
=
(1.6 m3 )
= 832.5 K .
014 (part 1 of 2) 0 points
One mole of an ideal gas does 2519 J of work
on the surroundings as it expands isothermally to a final pressure of 0.8 atm and volume of 33 L.
Determine the initial volume.
(R =
8.31451 J/K · mol.)
Correct answer: 12.866 L.
Explanation:
Given : n = 1 mol ,
W = 2519 J ,
R = 8.31451 J/K · mol ,
Pf = 0.8 atm , and
Vf = 33 L .
By the ideal gas law, P V = n R T .
The work done in an isothermal process is
µ ¶
Vf
W = n R T ln
Vi
µ ¶
Vf
= P V ln
Vi
µ ¶
Vf
W
ln
=
Vi
PV
·
¸
Vf
W
= exp
Vi
PV
·
¸
−W
Vi = Vf exp
PV
Since
Answer, Key – Homework 6 – David McIntyre
W
2519 J
=
PV
(0.8 atm)(101300)(33 L)(1e − 3)
= 0.941922
Vi = 33 L exp [−0.941922]
= 12.866 L .
5
Find the final temperature under the same
assumptions as above.
Correct answer: 443.865 ◦ C.
Explanation:
Given : Ti = 21.8◦ C = 294.8 K .
015 (part 2 of 2) 0 points
Determine the temperature of the gas.
Correct answer: 321.645 K.
Explanation:
According to the equation of state for an ideal
gas, the temperature is
Pf V f
Tf =
nR
(0.8 atm)(101300)(33 L)(1e − 3)
=
(1 mol)(8.31451 J/K · mol)
= 321.645 K .
016 (part 1 of 2) 0 points
Air in the cylinder of a diesel engine at
21.8 ◦ C is compressed from an initial pressure of 0.821 atm and of volume of 793 cm3 to
a volume of 86 cm3 .
Assuming that air behaves as an ideal gas
(γ = 1.40) and that the compression is adiabatic and reversible, find the final pressure.
Correct answer: 18.4089 atm.
Explanation:
Given : Pi = 0.821 atm ,
Vi = 793 cm3 , and
Vf = 86 cm3 .
Using the relation
Pi Viγ = Pf Vfγ ,
we find that
Pf = P i
µ
Vi
Vf
P V = nRT
is always valid during the process and since
no gas escapes from the cylinder,
Pf V f
Pi V i
= nR.
=
Ti
Tf
Therefore,
Pf V f
Ti
Pi V i
(18.4089 atm) (86 cm3 )
=
(294.8 K)
(0.821 atm) (793 cm3 )
= 716.865 K
= 443.865◦ C .
Tf =
018 (part 1 of 4) 3 points
4.2 L of diatomic gas (γ = 1.4) confined to a
cylinder are put through a closed cycle. The
gas is initially at 1 atm and at 330 K. First,
its pressure is tripled under constant volume.
Then it expands adiabatically to its original
pressure and finally is compressed isobarically
to its original volume.
Determine the volume at the end of the
adiabatic expansion.
Correct answer: 9.20556 L.
Explanation:
Given : V0 = 4.2 L = 0.0042 m3
Pf = 3 P i .
¶γ
= (0.821 atm )
Since
µ
793 cm3
86 cm3
¶1.40
= 18.4089 atm .
Basic Concepts
P V = nRT
∆U = Q − Wgas
017 (part 2 of 2) 0 points
U = n Cv T
and
Answer, Key – Homework 6 – David McIntyre
6
Q = n C ∆T
Solution: For an adiabatic process we have
PB VBγ = PC VCγ ,
so
3 P0 V0γ = P0 VCγ .
It follows from this that
VC = V 0
µ
3 P0
P0
¶1
γ
021 (part 4 of 4) 2 points
What is the net work done for this cycle?
Correct answer: 352.579 J.
Explanation:
Given : P0 = 1 atm = 101300 Pa .
Since for a closed cycle ∆U = 0, net work
done for this cycle equals net heat transferred
to the gas. In AB this heat is
1
= (4.2 L) (3) 1.4
= 9.20556 L .
019 (part 2 of 4) 3 points
Find the temperature of the gas at the start
of the adiabatic expansion.
Correct answer: 990 K.
Explanation:
QAB = n Cv ∆T
5
= n R (TB − TA )
2
5
= n R (3 T0 − T0 )
2
= 5 n R T0
= 5 P 0 V0
= 5 (101300 Pa) (0.0042 m3 )
= 2127.3 J .
Given : T0 = 330 K .
From the equation of state for an ideal gas we
obtain
PB V B = n R T B = 3 P 0 V 0 = 3 n R T 0
Therefore
TB = 3 T 0
= 3 (330 K)
= 990 K .
020 (part 3 of 4) 2 points
Find the temperature at the end of the adiabatic expansion.
Correct answer: 723.294 K.
Explanation:
TB VBγ−1 = TC VCγ−1
µ ¶γ−1
VB
TC = T B
VC
µ
¶1.4−1
4.2 L
= (990 K)
9.20556 L
= 723.294 K .
In BC
QBC = 0,
since the process is adiabatic, and with the
help of the equation of the state for an ideal
gas, we have
QCA = n Cp ∆T
7
= n R (TA − TC )
2
µ
¶
7
VC
= n R T0 − T 0
2
V0
µ
¶
VC
7
= P0 V 0 1 −
2
V0
7
= (101300 Pa) (0.0042 m3 )
2 µ
¶
9.20556 L
× 1−
4.2 L
= −1774.72 J .
For the whole cycle
QABCA = QAB + QCA
= (2127.3 J) + (−1774.72 J)
= 352.579 J .
Answer, Key – Homework 6 – David McIntyre
022 (part 1 of 2) 0 points
A cylinder contains 2.46 mol of helium gas at
a temperature of 285 K.
How much heat must be transferred to the
gas to increase its temperature to 602 K if
it is heated at constant volume? The molar
specific heat at constant volume, cv , of helium
is 12.5 J/mol/K.
Correct answer: 9747.75 J.
Explanation:
Given :
n = 2.46 mol ,
Ti = 285 K , and
Tf = 602 K .
For the constant-volume process, the work
done is zero. Therefore for the heat Q1 transferred to the gas during the process we have
Q1 = ∆U
3
= n R ∆T
2
= n cv (Tf − Ti ) ,
7
slowly and adiabatically from a pressure of
4 atm and a volume of 13 L to a final volume
of 33 L.
What is the final pressure?
Correct answer: 1.08558 atm.
Explanation:
Given : P0 = 4 atm = 405200 Pa ,
V0 = 13 L = 0.013 m3 , and
V1 = 33 L = 0.033 m3 .
Basic Concepts
P V γ = const
P V = nRT
Solution: Since an adiabatic process for an
ideal gas is described by P V γ = const, we
have
P0 V0γ = P1 V1γ ,
so
P1 = P 0
µ
V0
V1
¶γ
¶1.4
Q1 = (2.46 mol) (12.5 J/mol/K)
· (602 K − 285 K)
= 9747.75 J .
13 L
= (4 atm)
33 L
= 1.08558 atm .
023 (part 2 of 2) 0 points
How much thermal energy must be transferred to the gas at constant pressure to raise
the temperature to 602 K? The molar specific heat at constant presure, cp , of helium is
20.8 J/mol/K.
Correct answer: 16220.3 J.
Explanation:
For the thermal energy Q2 transferred to
the gas at constant pressure,
025 (part 2 of 3) 0 points
What is the initial temperature?
Correct answer: 316.772 K.
Explanation:
From the equation of state for an ideal gas
we have
Q2 = n cp ∆T
= (2.46 mol) (20.8 J/mol/K) (317 K)
= 16220.3 J .
024 (part 1 of 3) 0 points
Two moles of an idea gas (γ = 1.4) expands
µ
P0 V 0
nR
(405200 Pa) (0.013 m3 )
=
(2 mol) (8.31451 J/K · mol)
= 316.772 K .
T0 =
026 (part 3 of 3) 0 points
What is the final temperature?
Correct answer: 218.232 K.
Explanation:
Answer, Key – Homework 6 – David McIntyre
From the equation of state for an ideal gas
we have
P1 V 1
nR
(109969 Pa) (0.033 m3 )
=
(2 mol) (8.31451 J/K · mol)
= 218.232 K .
T1 =
027 (part 1 of 4) 0 points
Assume that a molecule has f degrees of
freedom. Consider a gas consisting of such
molecules.
Determine its total thermal energy.
3nRT
2
nRT
2. U = f
3
2nRT
3. U = f
3
nRT
4. U = f
correct
2
1. U = f
5. U = f n R T
Explanation:
µ
dU
dT
¶
According to the equipartition theorem, the
amount of energy per each degree of freedom
kB T
is
, so that
2
U =Nf
nRT
kB T
=f
.
2
2
028 (part 2 of 4) 0 points
Find its molar specific heat at constant volume.
1. Cv = f R
1
f R correct
2
2
3. Cv = f R
3
1
4. Cv = f R
4
2. Cv =
1
fR
3
Explanation:
The specific heat is given by
µ
¶
1
1 dU
= f R.
Cv =
n dT
2
5. Cv =
029 (part 3 of 4) 0 points
Find its molar specific heat at constant pressure.
1
1. Cp = (f + 3)R
2
3
2. Cp = (f + 2) R
2
1
3. Cp = (f + 1) R
2
1
4. Cp = (f + 2) R
3
1
5. Cp = (f + 2) R correct
2
Explanation:
Since Cp = Cv + R, we have
Cp = C v + R =
1
Cv =
n
8
1
(f + 2) R .
2
030 (part 4 of 4) 0 points
Cp
Determine the ratio γ =
Cv
f +2
1. γ =
2f
f +2
2. γ =
3f
f +2
3. γ =
correct
f
f +3
4. γ =
f
f +1
5. γ =
f
Explanation:
Taking the magnitudes of Cp and Cv from
the previous sections, we obtain
γ=
Cp
f +2
=
.
Cv
f