Chapter 19
–
First Law of Thermodynamics Hints

10/2
7
/10
1. The Work

Energy Principle in mechanics was written as follows:
W
non

cons
= ∆U + ∆K
. The 1
st
Law of
Thermodynamics is written as
Q
–
W= ∆U
.
a)
Discuss the similarities and differences between these tw
o versions of what is essentially the same
principle.
Check your notes…and the answers below.
b)
What does +Q vs.
–
Q mean in the 1
st
Law? What does +W vs.
–
W mean? What does +U vs.
–
U mean?
Q is heat added or subtracted from the system; W is the work done by
the system (+W when the gas
expands,

W when it contracts); U is the internal energy of a gas (for ideal gases it consists only of
random kinetic energy, but real gases have additional energies).
c)
Explain why there is a (
–
) sign in front of “W” in the 1
st
Law, but not in the mechanics “W”.
The
–
sign in front of “W” is because this work is
done by
the gas on the surroundings, instead of the
other way around as is the “W” in the work

energy principle.
d)
What is the difference between the ∆U’s in the two princi
ples? Why isn’t there a ∆K in the 1
st
Law?
“
U” in the 1
st
law includes the kinetic and potential energies of the gas. For an ideal gas potential
energy is negligible and “U” is all kinetic.
e)
When applied to a gas W=∫PdV. If the gas expands, is W positive or
negative?
Expanding gases do
+W.
f)
Most people think of “heat” as a form of energy, but it is more like a form of “microscopic work”.
Explain what is meant by “microscopic work”.
Heat generally “flows” between two objects at different temperatures in therma
l contact. What
actually happens is that the particles at the thermal contact boundary are colliding and doing work of
each other. The faster particles (higher T) do more work on average on the slower (lower T) particles
and that is how energy is transfere
d through the heat process.
g)
In the following examples, is the work done by the system on the environment + or

? (i) expansion of
the burned gasoline

air mix
ture in the cylinder of a car engine; (ii) opening a bottle of champagne; (iii)
filling a scuba t
ank with compressed air; (iv) crumpling of a sealed, empty water bottle, as you drive
down a mountains. (The system is defined as the expanding or contracting gas).
(i) +W; (ii) +W; (iii)
–
W; (iv)
–
W.
2. What is meant by the “state of a system”?
The col
lection of physical properties of the system.
a)
What is an “equation of state”?
An equation relating “state variables”.
b)
Why are P, V, T, and U called
state variables? Because they define the state of a system.
c)
Why is ∆U a state function, but Q and W are not?
∆U depends only on the initial an final states of the
system, but Q and W depends on the path taken form the initial to final states.
d)
If you are told
only
the initial and final states of a system, can you determine whether the internal
energy change was d
ue to work or to heat transfer? Explain.
No, a system can change from one state to
another through an infinite number of processes.
3. Complete the following table summarizing the basic differences between the basic thermal processes:
Arrow directions a
re arbitrary.
Isochoric
Isobaric
Isothermal
Adiabatic
Cyclic
Constant
Ratio
∆V=0
P/T= constant
∆P=0
V/T= constant
∆T=0
PV= constant
Q=0
PV
= constant
x
P vs. V
Graph
Q
nC
V
∆T
nC
P
∆
T
Q=W
Q=0
Q=W
W
0
P∆V=nR∆T
nRT ln(V/V
o)

nC
V
∆T
=

∆U
Area inside figure
∆U
=
nC
V
∆T=Q
nC
V
∆T=Q

W
0
nC
V
∆T=

W
0
4. Do you understand these
concepts!
a)
In a constan
t

volume process, the change in internal energy
∆U=nC
V
∆T,
but in a constant

pressure
process
∆U≠nC
P
∆T; ∆U=nC
V
∆T
still. In fact,
∆U=nC
V
∆T,
in
all
processes! Explain why.
P
V
P
V
Iso

therms
P
V
Iso

therms
P
V
Iso

therms
P
V
Iso

therms
∆U depends only on the temperature difference, not the path,
so if ∆U=nC
V
∆T in a constant

volume process is has to be same in all processes.
b)
When a gas expands
adiabatically, it does work on its sur
roundings. But if there is no heat input
into the gas, where does the gas get the energy to do work? Conversely, whe
n a gas contracts
adiabatically it heats up, but there is no heat entering the system, so how does the temperature of
the gas increase?
It comes from the internal energy of the gas, and that is why gases cool when they
expand adiabatically. In an adiabatic
compression the gas heats up because work is done on it by
an external agent
5. In the following scenarios identify the most likely process (assuming an ideal gas) and explain what
happens.
a)
A sealed inflated balloon is held over a hot air vent and it slo
wly expands. After it’s removed from
the vent it cools back to room temperature. During the expansion, which was larger: the heat added
to the balloon or the work done by the air inside it?
Approx. isobaric…the heat added is more than
the work in an expans
ion since the internal energy increases and Q
=∆U + W.
b)
You kick a soccer ball, compressing it suddenly into a smaller volume. Does the temperature
inside the ball increase, decrease, or stay the same?
Approx. adiabatic compression…temperature
increases since the work is done on the gas by the kick.
c)
W
eights are added slowly to a frictionless piston pushing it down into a container of gas that is
surrounded by a constant temperature bath. What happens to the temperature of the gas as the
volume decreases?
Isothermal…the temperature of the gas remains th
e same. The heat transferred
to the gas is used by the gas to do work.
6. a) In practical terms, an isochoric process is easy to carry out, and (approximate) adiabatic processes are
fairly common. Explain why.
Discussed in class…it is easy to seal a conta
iner so V doesn’t change…any
sudden change in volume doesn’t allow time for heat to flow and Q~0…
b) Isobaric and isothermal processes are more difficult to carry out. Why? How do you approximate an
isobaric or an isothermal process in the real world?
Disc
ussed in class…to keep a gas at constant pressure
would take a frictionless, movable piston that allows to the gas to keep changing volumes to maintain the
same pressure…to keep a constant temperature a gas must be able to freely exchange heat with its
sur
rounding and, since it takes time for heat to flow, the process must be slowed down as much as possible.
7. It is common to confuse temperature, heat
(Q)
and internal energy. Clearly distinguish among the three.
Consider the following scenarios:
a)
A sealed
Thermos bottle full of hot coffee is shaken, what are the changes,
if
any, in the
temperature and the internal energy of the coffee?
Very little to none…there is no net work done by
the coffee and no heat transfer.
b)
A sealed insulated tube full of oil has
several
small pebble
s inside. The tube is turned upside down
and right

side up repeatedly. Does its temperature and internal energy change? Explain.
Yes, the
potential energy lost by the falling
pebbles
would increase the internal kinetic energy and
temper
ature of the liquid.
8A. In a
constant

volume
process, 200
J
of heat is transferred to 1 g of Helium (a monatomic gas) initially
at 300 K at 1 atm. Find (a) the initial and final internal energies of the gas, (b) the work done by the gas,
and (c) its fina
l temperature and pressure. d) Draw a P

V diagram for this process.
a) U
o
=3nRT/2=935 J and U= 935 + 200= 1135 J. b) W=0
c) Q=nC
V
∆T
∆T=200/(.25)(3R/2)=64 Cº
T= 364 ºC; P=P
o
(364/300)=1.2 atm
d) Look at Problem 3.
8B. In a
constant

pressure
process, 200
J
of
heat is transferred to 1 g of Helium gas initially at 300 K at 1
atm. Find (a) the initial and final internal energies of the gas, (b) the work done by the gas, and (c) its final
temperature and volume.
d) Draw a P

V diagram for this process.
a) U
o
=3nRT/2
=935 J ; to find U you need to find T first; using Q=nC
P
∆T
∆T=200/(.25)(5R/2)=38.5
Cº
T= 338.5 ºC
U=3nRT/2=1055 J
b) W=nR
∆T
=(.25)(8.314)(
38.5
)=80 J, or you can use W=Q
–
∆U=200
–
120=80 J
c) From (a
) T =338.5 ºC; V =V
o
(338.5/300) =1.13
V
o
. To find
V
o
remember 1 mole occupies 22.4 liters at
STP, here we have ¼ m
ole at 300 K
V
o
=(22.4/4)
(300/273)=6.2 and finally V=1.13
(6.2)=7.0 liters
d) Look at Problem 3.
9. The two previous problems should make it clear that any change in temperature in a gas depends not
only on the heat transferred to it, but also on the work
th
e gas does. Since the internal energy (U) is a state
function, the ∆U will
only
depend on the
initial and final temperatures
, but the heat (Q) and the work (W)
required for the change will depend on the
path
.
a)
Show that in an
isobaric
process (P is constant
), (∆U/Q) = 1/
and (W/Q)= 1

(1/
).
Use the expressions ∆U=nC
V
∆T and Q=nC
P
∆T, and W=Q

∆U…
b)
For an isobaric process in a
monatomic
gas, what percentage of the heat transferred goes into
expanding the gas and what percentage goes into increasing the internal
energy of the gas?
Using
derived formulas
,
(∆U/Q)
60% and (W/Q)
40%.
c)
Repeat (b) for
diatomic
gas.
Using derived formulas
,
(∆U/Q)
71% and (W/Q)
29%.
d)
What if the process is
isothermal
? …
isochoric
?...
adiabatic
?
Isothermal:
(∆U/Q)
0% and (W/Q)
100%.
Isochoric
:
(∆U/Q)
100% and (W/Q)
0%.
Adiabatic
:
(∆U/Q) and (W/Q)
infinity, since (Q=0)
10A. If a molecule has
f
degrees of freedom
, show that a gas consisting of such molecules has the
following properties:
(i)
its total thermal energy is
fnRT/2;
(ii) its molar s
pecific heat at constant volume is
fR/2;
(iii) its molar specific heat at constant pressure is
(f+
2)R/2;
and (iv) the ratio
~
C
P
/C
V
=
(f +
2)/f.
The basic rule is that “each degree of freedom contributes, on average, kT/2 of energy per
molecule”…and it all f
ollows from there.
10B. Same as 18

18b
10C. Gases often behave like polyatomic gases (f=7) at high temperatures, diatomic (f=5) at lower
temperatures, and monatomic (f=3) at even lower temperatures. How does the simple theory of
degrees of
freedom
for g
ases developed in class explain this phenomenon?
You can read about this in the
textbook…the basic idea is that rotational and vibrational energy modes require higher energy levels and
if the temperature is low these higher energy modes are “frozen out” of
the energy sharing process.
10D. At high temperatures many crystalline solids exhibit a molar heat capacity value of 25 J/mK (3R),
corresponding to 6 degrees of freedom. How does our simplified kinetic molecular theory account for this?
Again, the textbo
ok explains this quite well…
11. A cylinder containing
n
moles of an ideal gas undergoes a reversible
isothermal
process.
a)
Show that the work done by the gas is
W=nRT[ln(
V
/V
o
)].
Done in class…∫PdV=nRT∫dV/V=nRT ln(V/V
o
). Note that final values are denoted as P&V and
initial values as P
o
and V
o
.
b)
Show that the work formula can also be written as
W=
PV
[ln(
P
o
/P)].
According to the gas law, nRT=PV=P
o
V
o
for an isobaric process.
c)
Compare
to the work formula for an adiabatic process described in the problem below.
12. A cylinder containing
n
mol of an ideal gas undergoes a reversible
adiabatic
process.
a)
Starting with the
1st

law
equation prove that the work done is
W
=
nC
V
(T
o
–
T
).
Using
the 1
st
law: W=

∆U=

nC
V
∆T
for an adiabatic process…
b)
Using the
ideal gas law
show that the formula in (a) is equivalent to
W= [1/(

1)](P
o
V
o
–
PV ).
W
=
nC
V
(T
o
–
T
)=
nC
V
(P
o
V
o
–
PV )/nR
(C
V
/R)(P
o
V
o
–
PV )
[C
V
/
(
C
P

C
V
)](P
o
V
o
–
PV )…
c)
You can also derive the eq
uation above from the work definition:
W
=
∫
PdV
, using the fact that
PV
=
constant
in adiabatic processes
.
Since PV
=constant
∫PdV=C∫dV/
V
=C/(1

)[(V

+1
)

(V
o

+1
)] and C=
P
o
V
o
PV
13. Earlier in the course we derived a formula for the speed of s
ound that depended on the
bulk modulus
,
B=

∆P/(∆V/V), and density of the medium: v
sound
=√(
B/
).
a)
Show that if the compressions in a gas are
adiabatic
(Q=0) then B=
P.
For adiabatic
PV
=constant
P=C/V
Taking derivatives d
P/dV=

P/
V
~ ∆P/∆V, substituting
i
nto the definition for B we get that B=
P.
b)
Use the result above to derive a new formula for the speed of sound in a gas: v
sound
= √(
RT/M).
After you substitute for B recall that the Gas Law can be written as PM=
RT…
c)
Since the speed of sound depends on th
e temperature and molar mass, one can also derive the
following formula:
v
sound
~ (
R273/M)
1/2
[1+ T/546], where T here is in Celsius degrees. When
applied to air this formula gives v
air
~ [331.4+ 0.6T], which was the given formula used in the
Speed of Soun
d lab. Hint: This proof requires the binomial approximation: (1+x
n
) ~ (1 + nx), which
is satisfactory for x<<1.
Change the units of “T” from Kelvin to Celsius, T
K
= (273+T
ºC
), then you can factor out all the
constant terms to get the formula
v
sound
=(
R27
3/M)
1/2
[1+T
ºC
/273]
1/2
. Since the term (T/273)<<1,
we can apply the binomial approximation and simplify. The formula for the velocity of sound in air
is obtained by substituting the values for air.
14. Air (
=
1.4) at 27°C and atmospheric pressure is i
nitially drawn into a bicycle pump that has a cylinder
with an inner diameter of 2.5 cm and length 50 cm. Then a quick down stroke on the handle adiabatically
compresses the air, increasing the gauge pressure to 850 kPa inside the pump before the air enter
s the tire.
Assume that friction between the piston and the cylinder wall is negligible.
a)
When the air is first drawn into the pump, the temperature of the air decreases, but when the air is
compressed quickly the air increases in temperature. Explain.
a
) The initial intake of the air into an empty container would allow the air to expand quickly with
little chance for air to flow, so the air would cool. When the air is later compressed quickly the
work done by the pump on the air would heat it. Both of th
ese are approximate adiabatic processes.
b)
Show that the column of air will be reduced from 50 cm to about 10 cm by the quick compression.
b) Calling the cross

section of the pump “A” the initial volume of the air is “A50”. Since process is
adiabatic, P
a
(A50
)
=(P
a
+850x10
3
)(Ah)
h ~10.1 cm
c)
Find the final temperature of the compressed air above.
c) Here we need T(V
)=constant
300(A50
)=
T(A10
)
T~570 K
d)
The pump is made of steel and has an inner wall that is 2.0 mm thick. Assume that the 10 cm
colu
mn of compressed air is allowed to come to thermal equilibrium with the10 cm of the
cylinder's length in contact with the air. What will be the increase
in the temperature of the cylinder
wall
? The specific heat of steel is 470 J/kgCº and its density is 7.
8 g/cm
3.
d) The volume of the pump that will co
me to equilibrium can be found to be V=2π(2.5/2)(10.1)(0.2)
cm
3
which gives a mass of steel= 124 g. You also need to find the # of moles of air from the gas
law: n=PV/RT=0.00994 mole.
Now set up an equilibrium calorimetry problem:
/
(mc∆T)
/
steel
=
/
(nC
v
∆T)
/
a
ir
I got an equilibrium
temperature T
e
=301 K, which means the temperature of the pump is hardly affected by the
adiabatic heating. You may notice that the bicycle pump used in the “Ratio of Cs” lab gets quite
warm, this problem tells us that the heating
was due to the friction between the piston and wall of
the pump and not the adiabatic compression.
e)
How would the situation change if you had done this very slowly?
e) If done slowly the process would not be adiabatic, but closer to isothermal, and heat wou
ld flow
through the pump walls and keep the temperature inside close to room temp. To reach the same
final gauge pressure in the compression, the final volume would be less than before since
PV=constant in this process
P
a
(A50)=(
P
a
+850 x10
3
)(Ah)
h~5.3 cm
.
50
cm
10
cm
15. During the power stroke in a four

stroke automobile engine, the air

fuel mixture is approximately
undergoing an adiabatic expansion. We can model the air

fuel mixture in the piston as an ideal diatomic
gas. We estimate that the
gauge
pressure right
before the expansion is 20 atm and it
decreases to zero as the volume of the mixture expands from 50 cm
3
and
400
cm
3
.
Find the average power generated during the expansion, assuming the engine is
running at
2500
rpm (250/6 rps) and that the time involve
d in the expansion is ¼
that of the total cycle period.
Recall that power =work/time, the time given for the expansion is ¼ of the cycle period
∆t=(6/250)/4 =6
msec. The work in an adiabatic process can be calculated from the formula derived in 12b, that is
W=
[1/(

1)](P
o
V
o
–
PV)
I got W=164 J. Finally, Power=164/0.006= 27.4 kwatts.
16. One can also apply the 1
st
law of thermodynamics to liquid
s and solids. Since their volumes change
very little when heated or cooled, processes with liquids and solids are nearly always considered
isochoric
.
For example:
a)
Consider a 10 g cube of aluminum with a density of 2.7 g/cm
3
. The specific heat of aluminum
is 0.24
cal/g∙Cº, and its linear expansion coefficient is 24 x 10

6
/Cº. The temperature is raised 100 Cº at 1
atm, determine: (i) the heat added, (ii) the work done by the expanding cube, and (iii) the change in
the internal energy of the cube. Is it reas
onable to ignore the volume change here?
a) (i) Q=cm∆T= 10(.24)(100)= 240 cal= 1003 J.
(ii)
Now
W=P∆V= (P
a
)(V3
∆T)=
(101x10
3
)(3.7
x 10

6
)(
3x24 x 10

6
)(100)=2.69
x 10

3
J.
Comparing the two: W<<<Q and it’s negligible. So, (iii) ∆U=Q=1003 J
b)
When ice me
lts at
0ºC
,
its volume decreases (density changes from 0.92 to 1.0
g/cm
3
). Is the
change in internal energy more than, less than, or equal to the heat added? Why did we ignore this
in previous calorimetry problems?
Here W=P(

∆V) is negative, so ∆U=Q+P∆V. S
o the internal energy increase is larger than the
heat added but the work done would be negligible and is not worth considering.
c)
There are a few materials that contract when their temper
ature is increased, such as water between
0ºC
and 4°C. Would you expe
ct
C
p
for such materials to be greater or less than
C
v
?
Would there be
a significant difference?
Explain.
Since ∆U=nC
V
∆T and Q=
nC
P
∆T, and W is negative the 1
st
law can be written: nC
V
∆T =
nC
P
∆T
–
P(

∆V)
So C
V
>C
P
, unlike ideal gases. But since the work is negligible C
V
~C
P.
d)
An extreme form of matter that we don’t usually talk about is a “plasma”, which is an i
onized gas
that exists at very high temperatures.
Imagine a gas made up entirely of negatively charged elec

trons. Like charges repel, so the electrons exert repulsive forces on each other. Would you expect
that the temperature of such a gas would rise, fa
ll, or stay the same in a free expansion? Why?
As the electrons push each other apart they would accelerate, their potential energies are
decreasing as their kinetic energies are increasing, hence the temperature would be going up, but
the gas would also b
e getting less dense.
Note: It would be nearly impossible to assemble such a gas since the forces required would be
extraordinarily high. Common “plasmas” are made up of equal no. of + and
–
ions so that they are
electrically neutral.
Below are more exa
mples of “1
st
Law problems”. All these problems feature one or a combination of
thermodynamics processes. Information is given in written or graphic form, and you are asked to draw a
P

V graph (if none is provided) and to solve for unknow
ns such as Q, W, ∆U, and/or initial and final
values of n, P, V, T, & U. The challenge is to understand the different processes, molar heat capacities,
and work expressions, as you apply the 1
st
Law
and
the Gas Law.
17. In problem 21 in the “Gases” proble
m set, you were asked to find the
work done in taking the gas from I to F via the three different paths illustrated.
a) Now determine the heat transferred and internal energy change during these
processes, first assuming a
monatomic
gas, then assuming a
diatomic
gas.
P(atm)
4
I
A
2
B
F
2
4 V(lit)
Answers in the chart below:
W=∫PdV=Area
畮摥爠灡瑨
=
r
mona
=(3/2)nRT=(3/2)PV
∆U=(3/2)(PV
J
=
m
o
V
o
)
Q=∆U+W
=
r
dia
=(5/2)nRT=(3/2)PV
∆U=(5/2)(PV
J
=
m
o
V
o
)
Q=∆U+W
=
fAc
=
㠰㠠U
=
⠳⼲E⠴〴
J
㠰㠩U==
J
㘰㘠6
=
=
㈰㈠2
=
J
⁊
=
J
㈰㈠2
=
fBc
=
㈰㈠2
=
J
㘰㘠6
=
J
㐰㐠Q
=
J
⁊
=
J
㠰㠠U
=
䥆
=
㔰㔠5
=
J
㘰㘠6
=
J
ㄠ1
=
J
⁊
=
J
㔰㔠5
=
=
戩bf猠楴⁰潳獩扬攠瑯楮搠潵i⁴桥畭扥爠潦潬e猠楮⁴桩猠ha浰me?⁗潵汤lyo畲湳睥牳渠ra⤠c桡nge=楦⁴桥=
ga猠睥re楡瑯浩c?=
There’s not enough information to find the no. of moles. You would need the
temperature at one of the states. Then n=PV/RT…But th
e U’s and therefore
the ∆U’s and Q’s would be
different for a diatomic gas since C
V
=5R/2, as shown in the chart.
c) Now assume you return to I in a
closed cycle
process. Determine the net work done and the net heat
transferred in the following cyclical processes: IAFBI, IBFI,
IAFI.
c) In a closed path, ∆U=0 so W=Q always…W
IAFBI
=808

202=606 J;
W
IBFI
=202

505=

303 J;W
IAFI
=505

202=+303 J.
d) Can you verify that for
clockwise
cycles the net work done and heat transferred are positive and for
counterclockwise
cycles the net work a
nd heat transferred is negative.
If the cycle is clockwise the
expansion part of the cycle (+W) will be bigger that the compression part of the cycle (

W). The opposite is
true for a counter

clockwise cycle.
18. A system undergoes the cyclic process illus
trated. The cycle consists of two closed loops: I and ll.
a) Over one complete cycle, does the system do posi
tive or negative work?
(+W)
b) In each of the loops I and II, is the net work done by the system + or

?
W
I
is + and W
II
is

. You should real
ize that cycle I expands more than contracts
but the reverse is true for cycle II. The net work is equal to the area enclosed by the
closed path.
c) Over one complete cycle, does heat flow into or out of the system?
(into; +Q)
d) In each of loops I and
II does heat flow into or out of the system?
Q
I
is + and Q
II
is
–
. Heat flows into the system when it does +W and vice

versa.
19.
A well

insulated cylinder contains ¼ mole of carbon dioxide (CO
2
) gas at a temperature of 27 °C. The
cylinder is provided
with a movable, fric
tionless piston that moves as required to keep the gas at a
constant
pressure
of 1.00 atm (in equilibrium with atmospheric pressure outside). The gas is heated until its
temperature increases to 127 °C. Assume that the CO
2
can be treat
ed as an ideal polyatomic gas (
f=7
).
a) Draw a pV

diagram for this process.
See problem 3.
b) How much work is done by the gas in this process, and on what is this work done?
W=P∆V=nR∆T=(.25)(8.31)(100)= 208 J, the work is done on the air outside.
c) Wha
t is the change in internal energy of the gas?
∆U=nC
V
∆T and since CO
2
is polyatomic C
V
=7R/2
∆U=(.25)(7/2)(8.31)(100)=727 J
d) How much heat was supplied to the gas?
Q =
nC
P
∆T or
Q =∆U + W=935 J
e) How much work would have been done if the pressure had been
0.50 atm but ∆T was the same?
The same amount of work, the volume change would have been twice as large, though.
20. If you keep raising the pressure on a CO
2
sample the gas will display
non

ideal behavior and it will
eventually solidify without going
through the
liquid state in a process called
deposition
(the opposite process is called
sublimation
). The graph shown illustrates an isothermal curve of this process.
Explain what is happening at various stages in this thermodynamic process.
What is th
e phase of the CO
2
in different sections of the graph?
The graph suggests that at low pressures and large volumes the gas behaves like a typical ideal gas, but
when the volume and pressure combination reaches a particular value the gas begins to change ph
ase
(from gas to solid in the case of CO
2
). This is the flat part of the graph, and the volume decreases
P
V
I
II
P
V
solid
solid

gas
gas
dramatically while the pressure hardly changes. After the gas has all turned to solid, the pressure
increases dramatically while the volume hardly chan
ges, since solids are almost incompressible, and the
curve rises steeply.
21.
On a warm day, a large mass of air (atmospheric pressure 101
kPa
) is heated by the ground to a
temperature of 26 ºC and then begins to rise through the cooler surrounding air.
(This can be treated
approximately as an adiabatic process (why?). Calculate the temperature of the air mass when it reaches a
level where atmospheric pressure is only 85
kPa (0.8415 atm)
. Assume that air is an ideal gas (
=1.4)
.
(Note of interest: This co
ol
ing rate for dry, rising air, is roughly 1°C per 100 m of altitude, and is called the
dry adiabatic lapse rate.
The reverse process, when air descends rapidly from the mountains, is called
adiabatic heating
and is what generally leads to the dry, hot, w
indy conditions we call the
Santa Anas
).
You should convince yourself that PV
=constant is equivalent to (P
1

)T
=constant for adiabatic
processes.
Set up a ratio using atm units for pressure and K for temperature
(1
1

)299
=(.8415
1

)T
T=285 K.
2
2. Consider the following two graphs that apply to the same sample of an ideal gas. In both cases the
initial internal energy and temperature of the gas is 1000 J and 300 K at point
a
. The initial pressure and
volume is 1 atm and 4 liters. From
a
to
b
the
gas is compressed
to half its original volume at constant pressure. But in
the first graph the path from
a
to
c
is
isothermal
and
in the second graph the path from
a
to
d
is
adiabatic
.
From
b
to the final state (
c
or
d
), the path is
isobaric
.
a) Determ
ine the number of moles of the gas
and whether it’s
monatomic or diatomic.
a) n=PV/RT=404/(8.31)(300)=0.162 mole. Recall that
U=n(f/2)RT, where “f” is the degrees

of

freedom
f=2U/nRT=2000/404=4.95, which is close to the no. for a diatomic gas (5).
b) Fi
ll in the charts below:
My answers…
P(kPa)
V(m
3
)
T(K)
U(J)
P(kPa)
V(m
3
)
T(K)
U(J)
a
101
4x10

3
300
1000
a
101
4x10

3
300
1000
b
101
2x10

3
150
500
b
101
2x10

3
150
500
c
202
2x10

3
300
1000
d
264
2x10

3
396
1320
Q(J)
W(J)
∆U(J)
=
=
=
儨g)
=
t⡊
)
=
∆U(J)
=
慢
=

702

202

500
ab

702

202

500
bc
+500
0
+500
bd
+820
0
+820
ac

280

280
0
ad
0

320
+320
abca*
+78
+78
0
abda*
+118
+118
0
*Note the change in direction of processes ac and ad in the cycle.
Some solutions: W
ac
=PVln((1/2)=404(

0.693)
=

280 J; P
d
V
d
=
P
a
V
a
P
d
(2)
=
(1)(4)
P
d
=264 kPa…
23. A certain ideal gas has molar heat capacity at constant vol
ume
C
v
.
A sample of this gas initially
occupies a volume
V
o
at pres
sure
P
o
and absolute temperature
T
o
.
The gas expands isobarically to
a volume
2V
o
and then expands further adiabatically to a final vol
ume of
4V
o
.
a) Draw a PV

diagram for this sequence of processes.
b) Compute the total work done by the gas for this sequence of processes.
b) It can be shown that (
–
1)
=(R/
C
v
), and P
o
(2V
o
)
=P(4V
o
)
for
the adiabatic process. Then the expressions can all be written
in terms of the given quantities: W
total
=W
isobaric
+W
adiabatic
=
P
o
V
o
+(
)

1
(2P
o
V
o
–
4PV
o
)=P
o
V
o
+(
C
v
/R)(2P
o
V
o
)[1
–
2
R/Cv
]
c) Find the final temperature of the gas.
c) Use
(2
T
o
)(2V
o
)
=T(4V
o
)
T=4T
o
/
2
.
P (atm)
V (lit)
d
1
b
a
2
4
P (atm)
V (lit)
c
1
b
a
2
4
Po
2T
o
P, T
(not given)
V
o
2
V
o
4V
o
d) Find the absolute value Q of the total heat flow into or out of the gas for this sequence of
processes,
and state the direction of heat flow.
d) Heat flows only during the isobaric process and since the gas expan
ds heat flows into the gas
(+Q) from the surroundings: Q=nC
P
∆T= (
C
v
+R)n∆T=(
C
v
+R)P
o
∆V/R=[(
C
v
/R) +1]P
o
V
o
Challenge problem:
24. Engine Turbochargers and Intercoolers. The power output of an automobile engine is directly
proportional to the mass of air tha
t can be forced into the volume of the engine's cylinders to react
chemically with gasoline. Many cars have a
turbocharger,
which compresses the air before it enters the
engine, giving a greater mass of air per volume. This rapid, essentially adiabatic com
pression also heats the
air. To compress it further, the air then passes through an
intercooler
in which the air exchanges heat with
its surroundings at essentially constant pressure. The air is then drawn into the cylinders. In a typical
installation, air
is taken into the turbocharger at atmospheric pressure (101 kPa), density
= 1.23
kg/m
3
,
and temperature 15.0°C. It is compressed adia
batically to 145 kPa. In the intercooler, the air is cooled to
the original temperature of 15.0°C at a constant pressur
e of 145 kPa.
a)
Draw a pV

diagram for this sequence of processes.
a) The PV graph looks the same as the one in the above problem but the process is occurring in the
reverse order. The initial and final pressures are given (1 atm and 145/101 atms); the init
ial and final
temps are the same (288 K); and the initial density.
b)
If the volume of one of the engine's cylinders is 575 cm
3
, what mass of air exiting from the intercooler
will fill the cylinder at 145 kPa? Compared to the power output of an engine that ta
kes in air at 101 kPa
at 15.0°C, what percent
age increase in power is obtained by using the turbocharger and intercooler?
b) The given volume (
575 cm
3
) is the final volume in the process (call it V
3
), you need to determine the
volumes at the beginning and
end of the adiabatic process (call them V
1
and V
2
). To do this you must
first find the temperature at the end of the adiabatic process, T. Using the appropriate constant
adiabatic ratios and the constant isobaric ratio, I got T= 319 K, V
2
=638
cm
3
, and V
1
=825
cm
3
.
Now you can determine the initial mass taken into the turbocharger,
m =
V=(1.23)(
825 x10

6
) = 1.015
x10

3
kg. Compare to the mass without the turbocharger and intercooler,
m’=
V= (1.23) (
575 x10

6
)=0.707 x10

3
kg
1.015/.707=1.44
~44% more eff
icient.
c)
If the intercooler is not used, what mass of air exit
ing from the turbocharger will fill the cylinder at 145
kPa? Compared to the power output of an engine that takes in air at 101 kPa at 15.0 ºC, what
percentage increase in power is obtained by u
sing the turbocharger alone?
c) With no intercooler, the volume into the turbocharger would have been less (I got 744
cm
3
) and the
corresponding mass is 0.915 g. Compared to 0.707 g
.915/.707=1.29
~29 % more efficient.
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