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Chapter 7
BOUNDARY WORK
Frank is struck  as if for the ﬁrst time  by how much civilization
depends on not seeing certain things and pretending others never
occurred.
−Francine Prose (Amateur Voodoo)
In this chapter we will learn to evaluate Win, which is the work term
in the ﬁrst law of thermodynamics. Work can be of diﬀerent types, such
as boundary work, stirring work, electrical work, magnetic work, work of
changing the surface area, and work to overcome friction. In this chapter,
we will concentrate on the evaluation of boundary work.
110 Chapter 7
7.1 Boundary Work in Real Life
Boundary work is done when the boundary of the system moves, causing
either compression or expansion of the system. A reallife application of
boundary work, for example, is found in the diesel engine which consists of
pistons and cylinders as the prime component of the engine.
In a diesel engine, air is fed to the cylinder and is compressed by the
upward movement of the piston. The boundary work done by the piston
in compressing the air is responsible for the increase in air temperature.
Onto the hot air, diesel fuel is sprayed, which leads to spontaneous self
ignition of the fuelair mixture. The chemical energy released during this
combustion process would heat up the gases produced during combustion.
As the gases expand due to heating, they would push the piston downwards
with great force. A major portion of the boundary work done by the gases
on the piston gets converted into the energy required to rotate the shaft
of the engine.
The rotating shaft of the diesel engine is responsible for turning the
wheels of an automobile. The rotating shaft could also be used to spin a
coil between the north and south poles of an electromagnet thus generating
electricity as in the generators.
7.2 Evaluation of Boundary Work
Consider, for example, compression of the gas contained in the piston
cylinder device shown in Figure 7.1.

dx
gas
9 piston
cylinder
.....
... ...... ...
.....
.......
.........
... ........
... ........
.
.
Figure 7.1 A compression process during which boundary work is
done on the system by its surroundings.
F
..
..
..
..
..
..
..
..
..
..
..
..
.
.
.
.
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.
..
..
.
..
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..
.
..
.
..
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.
.
Boundary Work 111
Take the gas contained in the pistoncylinder device as the system. The
inner surfaces of the piston and cylinder deﬁne the boundary of the system.
Everything outside the boundary is the surroundings. During compression,
the boundary is moved by the surroundings so as to reduce the volume of
the system.
If at a certain state during the process, say, the boundary of the system
is displaced by an inﬁnitesimal (that is, a very tiny) distance dx by the
application of force Fas shown in Figure 7.1. The dashed line shown in
the ﬁgure indicates the new position of the boundary (note that the dis
placement is exaggerated.) Boundary work is evaluated by determining the
product between the displacement of a system boundary and the force caus
ing the displacement in the direction of the displacement. The inﬁnitesimal
amount of boundary work done by the surroundings to compress the gas is
therefore
dW =Fdx (7.1)
The external force Facting on one side of the piston, as shown in
Figure 7.1, tries to compress the gas. On the other side of the piston, the
gas pressure applies a force to resist the compression. If we assume that
the compression process is almost fully resisted then the forces acting
on either side of the piston become almost equal to each other. It follows
then,
F=PA (7.2)
where Pis the gas pressure that is assumed to be uniformly distributed
everywhere within the gas at any state during the compression, and Ais
the crosssectional area of the piston on which the gas pressure acts.
The change in the volume of the gas, designated by dV , is related to
dx by the following:
−dV =Adx (7.3)
since the gas has experienced a reduction in its volume, we know that dV is
a negative quantity. It is therefore a negative sign is placed in front of dV
in (7.3) to be able to equate it to the product of two positive quantities,
Aand dx.
Eliminating Fand dx from (7.1) using (7.2) and (7.3), respectively, we
get
dW =(PA)−dV
A
=−PdV (7.4)
112 Chapter 7
which is the inﬁnitesimal amount of boundary work done on the system
by its surroundings. In other words, this amount of boundary work enters
the system. Incorporating this information as a subscript on dW , (7.4) is
rewritten as
dWin =−PdV (7.5)
If the process considered were an expansion process, then the system
would have moved the boundary so as to increase the volume of the system.
The inﬁnitesimal amount of boundary work done by the system on its
surroundings (that is, the work leaving the system) would therefore be
given by
dWout =PdV (7.6)
Integrating (7.5) or (7.6) between the initial state with volume Vo,and
the ﬁnal state with volume Vf,weget
Wout =−Win =Vf
Vo
PdV (7.7)
The total boundary work is therefore expressed in terms of system properties
Pand Vby (7.7). It is important to observe that (7.7) can be applied
to evaluate the boundary work for any substance, solid, liquid, gas or any
mixture of phases, provided that the expansion or compression process
occurs under almost fully resisted condition.
Student: Teacher, one of the assumptions you made to derive at equation
(7.7) is that the system pressure Pis uniformly distributed every
where within the system at any state during the compression process. I
don’t feel comfortable with that assumption. It is because I think, dur
ing compression, the gas close to the face of the piston gets compressed
more than the gas away from the face of the piston leading to nonuniform
pressure distribution?
Teacher: Yes, if the compression process were sudden, the gas close to the
face of the piston would get compressed more than the gas away from the
face of the piston. The resulting pressure distribution within the system
would then be far from uniform. On the other hand, if the compression
or the expansion process is carried in such a manner that it is almost
fully resisted while taking place, then it will be possible to maintain an
almost uniform pressure distribution within the system as it undergoes
the compression or expansion process.
Boundary Work 113
Student: Teacher, could you please explain me how exactly a process is carried
out under almost fully resisted condition?
Teacher: Yes, I will. Take a look at Figure 7.2. Let us imagine that the lit
tle unhappy faces shown in the ﬁgure represent the gas contained in the
cylinder, trapped by the piston represented by the black solid bar shown
in the ﬁgure. The force exerted by thegaspressureonthepistonpushes
the piston upwards thus causing expansion of the gas. The total weight
of the piston, the person standing on the piston and the basket of tiny
stones carried by him causes the piston to move downwards compressing
the gas. When these two forces balance each other, the system is in equi
librium. Let us now remove a single tiny stone from the basket. Since
the mass of the tiny stone removed is extremely small, the force exerted
by the gas molecules on the piston to move it upwards becomes slightly
larger than the force exerted to prevent the expansion of gas. The piston
thus moves upwards by a very small distance allowing the gas to expand
slightly. Such an expansion process is an almost fully resisted expan
sion, since it is carried out with a very small diﬀerence (an inﬁnitesimal
diﬀerence) between the force causing the expansion and the force resist
ing the expansion. An almost fully resisted compression process
would be the reverse of the above process.
Figure 7.2 Almost fullyresisted expansion of a gas in a pistoncylinder device.
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tiny stones of extremely
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114 Chapter 7
Student: I see that the forces acting on either side of the piston almost equal
each other in an almost fully resisted process. I understand very well now
how you got equation (7.2). If we had removed the stones in the basket
that the person is carrying all at once, then the piston would have risen
rapidly causing the gas to experience sudden expansion. Or, if we had
loaded the basket with a lot more stones all at once then the piston would
have moved rapidly downwards causing sudden compression of the gas.
Am I right, Teacher?
Teacher: Yes, you are. Note that the external force Fis related to a gas
pressure Pby (7.2) which is a property of the system, only because the
compression is assumed to be almost fully resisted. If the process were
not an almost fully resisted expansion or compression then the relationship
given by (7.2), in general, would not hold.
Student: Yes, I see that. However, I still do not see how it is possible to
maintain a uniform pressure distribution during such a process.
Teacher: Since there is only an inﬁnitesimal diﬀerence between the force caus
ing the expansion (or compression) and the force resisting the expansion
(or compression) in an almost fully resisted process, the pressure under
goes only an inﬁnitesimal change at a time. Any temperature change
created by the expansion (or compression) would also be inﬁnitesimal in
its magnitude. An almost fully resisted process such as the one discussed
above could progress only very slowly so that it would allow enough time
for the inﬁnitesimal changes experienced by the properties of the system,
such as pressure, temperature, and speciﬁc volume, to be distributed al
most uniformly across the system.
Student: Thank you, Teacher. I understand now how the uniform distribution
of properties, I mean, the almost uniform distribution of properties, is
maintained in a system when the expansion or compression process is
carried out under almost fully resisted condition.
7.3 Quasistatic Process
As beginners, we will mostly be working with the special kind of pro
cess known as the quasistatic process, which is deﬁned as a process
Boundary Work 115
whose path is composed of near equilibrium states. The initial and ﬁnal
states of any quasistatic process are equilibrium states, and all the states
that a system passes through along the path of a quasistatic process are
almost equilibrium states. A quasistatic process is also known as a quasi
equilibrium process.
We have seen in Section 2.9 that an equilibrium state can be repre
sented by a point on a property diagram. Therefore, it is acceptable to
represent the path of a quasistatic process, which is made up of near equi
librium states, by a curve on a property diagram, as shown in Figure 7.3.
initial state
ﬁnal state
path of a quasistatic process
Pressure
Speciﬁc volume
r
I
rF
)
Figure 7.3 A quasistatic process with its initial state I, its ﬁnal state F,
and its path shown on a Pvdiagram.
9
Consequently, it is possible to describe the path of a quasistatic process
by an equation, such as P=f(v)on a Pvdiagram, T=f(v)on a Tv
diagram, and P=f(T)on a PTdiagram. Moreover, since each state is
almost an equilibrium state along a quasistatic process, the properties at
a state along the path of a quasistatic process can be interrelated by, for
example, the ideal gas equation of state.
The path followed by a nonquasistatic process consists of states that
may or may not be equilibrium states, and therefore the path of a non
quasistatic process cannot be represented by a curve on a property diagram.
The fully resisted expansion or compression process, which we were in
troduced to in the section above, is an example of a quasistatic process
since each state along a fully resisted expansion or compression is almost
an equilibrium state.
116 Chapter 7
7.4 Boundary Work in Real Processes
Boundary work can be evaluated using (7.7) only for processes that
approximate a quasistatic process. Strictly speaking, boundary work asso
ciated with many real systems, such as the pistoncylinder arrangements
of diesel engines described in Section 7.1, cannot be evaluated using the
work integral of (7.7). It is because the uniform nature of the properties
within the system at each and every state along the path of the process is
not maintained in such systems owing to the very high speeds at which the
piston moves.
Moreover, in the pistoncylinder device, for example, friction occurs
between the surface of contact of the piston and cylinder. The resulting
friction force on the piston acts in a direction opposite to the motion of
the piston, and therefore work is done to overcome the eﬀect of friction.
The problem here, however, is the evaluation of the exact amount of work
done against the friction force, which is not an easy task.
Besides, friction forces acting on the piston has the capacity to hold
the piston in place until the diﬀerence between the forces acting on either
side of the piston reaches a ﬁnite value. At this point, the friction could be
overcome and the piston would undergo a sudden movement, which in turn
would cause sudden expansion or compression of the gas contained in the
pistoncylinder device. The movement of the piston caused by a ﬁnite force
and the resulting sudden expansion or compression of the gas destroy the
quasistatic nature of the process. Therefore, we simplify our calculation of
boundary work in many real systems by making the assumption that the
piston is free to move within the cylinder without friction.
If the process is not quasistatic then there is no direct way of calculating
the work done by the system, or on the system. Since it is next to impossible
to do calculations with nonquasistatic processes, we assume, wherever
possible, that the given process is quasistatic. This assumption makes it
possible for us to carry out calculations with real thermodynamic processes.
Owing to the assumptions that friction forces are absent and that the
properties are almost uniformly distributed within the system, a quasistatic
process sets a limit for the best performance that is expected of a real
process. A device producing work produces the maximum amount of work
and a device requiring work requires the minimum amount of work when
they undergo quasistatic processes.
Boundary Work 117
7.5 Units for Boundary Work
Since work is a form of energy, the unit for work is the same as the unit
for energy. If Ptakes the unit Pa and Vtakes the unit m3in the work
integral of (7.7), then the unit of Win becomes
Pa×m3=(N/m
2)×m3=N·m=J.
In this textbook, we will use the unit kPa for P. And, therefore, the
unit of Win will be kJ.
When the total volume Vof (7.7) is replaced by the speciﬁc volume v,
we get
win =−vf
vo
Pdv (7.8)
in which win takes the unit kJ/kg provided vis in m3/kg. If vwere the
molar volume, taken in m3/kmol, win wouldbeinkJ/kmol. Inboththese
cases Pis in kPa.
7.6 Path Dependence of Boundary Work
Consider the gas contained in the pistoncylinder device of Figure 7.1
once again. Imagine that we could push the piston slowly to decrease the
volume occupied by the gas while maintaining the quasistatic nature of
the process. Let us say that we measured the volume occupied by the
gas and the corresponding pressure of the gas at certain times during the
compression process, and the data so gathered were plotted on a PV
diagram as shown by the dots of Figure 7.4. Connect the data points by a
smooth curve as in the ﬁgure. The curve so obtained represents the path
of the process between its initial state (o) and its ﬁnal state (f) on the PV
diagram. The equation describing the curve will provide the relationship
between Pand Valong the path of the process.
Integrating along the path of the process on the PVdiagram of Figure
7.4 in the increasing direction of V,weget
Vo
Vf
PdV
118 Chapter 7
The value of the above integral is simply the area of the shaded region
shown in Figure 7.4. Interchanging the limits in the above integral, we get
−Vf
Vo
PdV
which, according to (7.7), is Win. That is, the area under the curve rep
resenting the path of a compression process on a PVdiagram gives Win,
which is the boundary work done on the system by its surroundings.
ttttt
P
V
tttttt
(f)
(o)
VfVo
Figure 7.4 When the process takes place in the direction of decreasing
volume (that is, compression), the area of the shaded region gives Win.
Win
.....
... ...... ...
.....
.......
.........
... ........
... ........
.
.
gas is compressed
from Voto Vf
If the process considered is an expansion process, then the volume of
the system increases from Voto Vfas shown in Figure 7.5. Integrating
along the path of the process on the PVdiagram in Figure 7.5, we get
Vf
Vo
PdV
which, according to (7.7), gives Wout. That is, the area under the curve
Boundary Work 119
representing the path of an expansion process on a PVdiagram gives
Wout, which is the boundary work done by the system on its surroundings.
ttttt
P
V
tttttt
(o)
(f)
VoVf
Figure 7.5 When the process takes place in the direction of increasing
volume (that is, expansion), the area of the shaded region gives Wout.
Wout
.....
... ...... ...
.....
.......
.........
... ........
... ........
.
.gas expands
from Voto Vf
It is important to note that boundary work given by the area of the
shaded region in Figure 7.4 or in Figure 7.5 will change if the path of the
process, described by the curve on the PVdiagram, changes. This means
that even if the initial and ﬁnal states of a process are ﬁxed, the bound
ary work would be diﬀerent for the diﬀerent paths followed by a process.
Therefore, we say that boundary work is path dependent.
7.7 Other Forms of Work
Boundary work is just one mode of work transfer between the system
and its surroundings. There are also other modes of work, such as shaft
120 Chapter 7
work, surface work, electrical work, magnetic work, electrical polarization
work, etc. Shaft work is a form of work associated with a rotating shaft
which is a prime component in stirrers, pumps, compressors, turbines, diesel
and petrol engines. Shaft work per unit time is given by 2π˙nτ,where ˙n
is the number of revolutions made by the shaft per unit time and τis the
constant value of the torque applied to the shaft. When τtakes the unit
N·mand ˙ntakes the unit revolutions per second, shaft work takes the unit
J/s, which is W.
In this book we deal only with simple compressible systems (deﬁned in
Section 2.7). Since such systems do not experience electrical, magnetic and
surface tension eﬀects, we will not get into details of how the work needed
to change surface area, electrical work, magnetic work or electrical polar
ization work are evaluated. For details of the evaluation of these modes of
work, please refer to the following textbooks on thermodynamics: (i) Hol
man, J.P. 1988 Thermodynamics, 4th Edition, McGrawHill International
Editions. (ii) Wark, K. 1989 Thermodynamics, 5th Edition, McGrawHill
International Editions.
7.8 Worked Examples
Example 7.1
The following statements are incorrect. Find
the error in each statement.
(a) For a gas contained in a closed system changing from one state to
another state, f
oPdV is the same along any path as long as the
process is quasistatic.
(b) Boundary work is always given by f
oPdV.
(c) All forms of work transfer are zero for a constant volume process.
(d) A closed, adiabatic system of gas expanding from an initial pressure
Poto a ﬁnal pressure Pfagainst a constant force applied by its
surroundings is a quasistatic process.
Boundary Work 121
Solution to Example 7.1
(a) The boundary work done in a quasistatic process is evaluated using f
oPdV .
But the value of f
oPdV , and thus the amount of boundary work, change de
pending on the path of the chosen process. Therefore, even if the process is
quasistatic, the value of f
oPdV depends on the path of the quasistatic process
considered.
(b) Boundary work of a quasistatic process is given by f
oPdV . Boundary work
associated with a nonquasistatic process is nearly impossible to calculate, since
such a process could have a nonuniform distribution of the properties within the
system. Therefore, boundary work is not always given by f
oPdV .
(c) The boundary work associated with a constant volume quasistatic process is
zero, since dV is identically zero in the work integral f
oPdV used to determine
the boundary work of a quasistatic process. Even if the given constant volume
process is not a quasistatic process, there is neither boundary work nor frictional
work involved with the process, since the boundary does not move in a constant
volume process. However, a constant volume process may receive stirring work,
and therefore it is incorrect to assume that all forms of work transfer are zero
for a constant volume process.
(d) For the given adiabatic expansion process to qualify as a quasistatic process,
it should have been carried out under conditions of inﬁnitesimal diﬀerence in the
force causing the expansion and the force resisting the expansion. In the given
process, the force resisting the expansion remains constant, whereas the force
causing the expansion changes from PoAto PfA,whereAis the crosssectional
area on which the pressure acts. Therefore, the expansion is carried out under
ﬁnite diﬀerence between the force causing the expansion and the force resisting
the expansion. Therefore, the given process is not quasistatic.
Example 7.2
Determine the boundary work (a) for a qua
sistatic constantpressure process, and (b) for a quasistatic constantvolume
process. Pressures and volumes at the initial and the ﬁnal states of the pro
cess are given by (Po,Vo)and(Pf,Vf), respectively.
122 Chapter 7
Solution to Example 7.2
(a) A quasistatic constantpressure process: We employ (7.7) to eval
uate the boundary work since the given process is quasistatic. Since it is also a
constant pressure process, Po=Pf=P, say. Therefore, we get from (7.7)
Wout =−Win =Vf
Vo
PdV=PVf
Vo
dV =P(Vf−Vo)(7.9)
for a quasistatic constantpressure process.
In case we deal with ideal gases, then the ideal gas equation of state gives
PoVo=nRToand PfVf=nRTf,wherenis the amount of gas in the closed
system, Ris the universal gas constant, and Toand Tfare the temperatures at
the initial and the ﬁnal states, respectively. Therefore, (7.9) yields
Wout =−Win =nR(Tf−To)(7.10)
for a quasistatic constantpressure process of an ideal gas.
(b) A quasistatic constantvolume process: Since the given process is
quasistatic, (7.7) is employed to evaluate the boundary work. Since it is also a
constant volume process, dV = 0 always. Therefore, (7.7) gives
Wout =−Win =Vf
Vo
PdV =0 (7.11)
That is to say the boundary work for a quasistatic constantvolume process is
zero, since dV of a constant volume process is identically zero.
Example 7.3
The pistoncylinder device of Figure 7.1 con
tains hydrogen, say, which expands from an initial state given by (500 kPa,
0.10 m3) to a ﬁnal state given by (230 kPa, 0.18 m3) in two steps. In the
ﬁrst step, the gas is expanded to the volume of 0.18 m3while maintaining
the pressure constant at 500 kPa. In the second step, the gas pressure is
decreased to 230 kPa while maintaining the volume constant at 0.18 m3.
Determine the work done by hydrogen gas.
Boundary Work 123
Solution to Example 7.3
Analytical method: Let us assume that the given process takes place qua
sistatically. Boundary work done by the gas along the constantpressure qua
sistatic process is given by (7.9) as
WoutP=P(Vf−Vo)
Substituting, P= 500 kPa, Vo=0.10m
3and Vf=0.18m
3,weget
WoutP= 500 (0.18 −0.10) kPa×m3=40.0kJ
Boundary work done by the gas along the constantvolume quasistatic pro
cess is given by (7.11) as
WoutV=0
The total boundary work done by the gas in the twostep process is, therefore,
Wout =WoutP+WoutV=40.0kJ +0 =40.0kJ
Graphical method: Figure 7.6 shows the path of the process, represented by
the curve IEF on a PVdiagram. The initial state and the ﬁnal state are marked
by points I and F, respectively. The intermediate state where the process switches
from a constantpressure process to a constantvolume process is marked by point
E. The area under the curve IEF is
500 KPa×(0.180.10) m3=40kJ,
which is the work done by hydrogen (see Figure 7.5). Observe that the work
evaluated by the graphical method is the same as the work evaluated by the
analytical method, which is how it should be.
Figure 7.6 Path of the process of Example 7.3, shown on a PVdiagram.
r
r
200
300
400
500
P(in kPa)
V(in m3)
0.10 0.18
I
F
E
r
124 Chapter 7
Example 7.4
Hydrogen in the pistoncylinder device of Ex
ample 7.3, let us say, expands by pushing the piston in a way that the
pressure of the gas decreases continuously while the volume of the gas in
creases continuously. The data collected during the expansion process are
tabulated below. Determine the boundary work done by hydrogen.
Pressure (in kPa) 500 393 320.5 268.6 230
Volume (in m3)0.10 0.12 0.14 0.16 0.18
Solution to Example 7.4
The boundary work done by hydrogen could be evaluated using the work
integral of (7.7) if we know how Pchanges with Valong the path of the
process. Let us plot the pressurevolume data collected during the expansion
process on a PVdiagram. The data are represented by dots marked I, E1,E
2,
E3and F in Figure 7.7.
Figure 7.7 Path of the process of Example 7.4, shown on a PVdiagram.
sssss
200
300
400
500
P(in kPa)
V(in m3)
0.10
I
E1
E2E3F
0.18
TheareaunderthecurveIE
1E2E3F represents the boundary work done by
hydrogen (see Figure 7.5). It is therefore possible to determine this area by
plotting the curve IE1E2E3F on a graph sheet and then by counting the squares
under the curve on the graph sheet.
Alternatively, the equation of the curve that best ﬁts the data points in
Figure 7.7 would give us the equation describing the path of the process. Let us
ﬁrst try the equation
PVk=C(7.12)
Boundary Work 125
where kand Care constants to be determined from the experimental data using
the procedure outlined below. Taking the natural logarithm of (7.12) yields
ln P=−kln V+lnC
which shows that a plot of ln Pversus ln Vgives a straight line having a slope
kand an intercept ln C.
Now let us go back to the experimental data on Pand V, ﬁnd the natural
logarithms of Pand V, and plot the data on a ln Pversus ln Vdiagram, as
shown in Figure 7.8. We see in Figure 7.8 that the data points are well ﬁtted
by a straight line. Therefore, we conclude that (7.12) accurately describes the
path of the process.
Figure 7.8 ln Pversus ln Vplot of the data in Example 7.4.
s
s
s
s
5.40
6.20
ln P
ln V
1.70
I
E1
E2
F
2.30
s
E3
Substituting Pfrom (7.12) in (7.7), we get
Wout =Vf
VoC
VkdV
which can be integrated, for cases in which k=1, to yield
Wout =CV−k+1
−k+1Vf
Vo
=CV−k+1
f−V−k+1
o
−k+1 (7.13)
From the slope and the intercept of the straight line in Figure 7.8, we can
ﬁnd that k=1.32andC= 23.9 when Pis in kPa and Vin m3. Substituting
these values and the data Vo=0.10m
3and Vf=0.18m
3in (7.13), we get
Wout =23.90.18−0.32 −0.10−0.32
−0.32 =26.8kJ
126 Chapter 7
Teacher: Note that both the processes of Example 7.3 and Example 7.4
have the same initial and ﬁnal states. The path of the processes, of course,
are diﬀerent from each other as shown in Figure 7.6 and Figure 7.7. The
work done by hydrogen also diﬀers exhibiting the pathdependence of the
work integral given by (7.7).
Student: Yes, I noted that. Teacher, I have a question. Suppose that the
data points on the ln Pversus ln Vplot of the Solution to Example
7.4 could not have been ﬁtted by a straight line, How could we solve the
problem then?
Teacher: We should have then tried another form of equation in place of (7.12).
However, it is sometimes impossible to describe the path of the process
by an equation. In such cases, the work transfer is determined from
evaluating the area under the path of the process on a PVdiagram
plotted on a graph sheet.
Example 7.5
The initial volume Voof a cyclic process is
the same as the ﬁnal volume Vf. Therefore, your friend argues, the work
integral (7.7) used to evaluate the boundary work of a quasistatic process
reduces to zero owing to the upper and the lower limits of the integral of
(7.7) becoming one and the same. Consequently, your friend concludes
that the boundary work associated with a quasistatic cyclic process of a
closed system is always zero. Explain to your friend, using an appropriate
example, why he is wrong.
Solution to Example 7.5
For a constantvolume process, boundary work is zero since dV of a constant
volume process is identically zero throughout the entire process. The process
of this example is not a constant volume process having dV =0always,buta
cyclic process which has identically the same volumes at the initial and the ﬁnal
Boundary Work 127
states of the process. An example of a quasistatic cyclic process is shown on the
PVdiagram of Figure 7.9 by the path A→B→C→D→A.
Figure 7.9 An example of a quasistatic cyclic process.
s
s
P
V
A
C
B
D
EF
s
s
Since the constantpressure quasistatic process A→Boccurs in the direc
tion of increasing volume, the area under the path A→Bin Figure 7.9, enclosed
by the rectangle AB F E,gives(Wout)A→B(see Figure 7.5). Therefore,
(Wout)A→B=PA(VB−VA)
The constantpressure quasistatic process C→Doccurs in the direction of
decreasing volume, and therefore the area under the path C→Din Figure 7.9,
enclosed by the rectangle CDEF,gives(Win)C→D(see Figure 7.4). Therefore,
(Win)C→D=PC(VC−VD)
Since B→Cand D→Aare constantvolume quasistatic processes,
(Wout)B→C=0 and (Wout)D→A=0
as in (7.11). Note that the constantvolume quasistatic paths B→Cand
D→Ahave zero areas under them on the PVdiagram of Figure 7.9.
Putting all the information above together, we calculate the net work transfer
as follows:
(Wout)net =(Wout )A→B+(Wout)B→C+(Wout)C→D+(Wout)D→A
=PA(VB−VA)+0−PC(VC−VD)+0
From Figure 7.9, we see that VA=VDand VB=VC, and therefore we
reduce the above to the following:
(Wout)net =(PA−PC)(VB−VA)
128 Chapter 7
which is in fact the area enclosed by the rectangle ABC D in Figure 7.9. It is
obvious therefore that the work transfer in a process with identical initial and
ﬁnal volumes, as in a cyclic process, need not always be zero.
Teacher: The above example shows that the work exchange between the sys
tem and its surroundings in a quasistatic cyclic process is given by the area
enclosed by the path of the process on the PVdiagram. Please note that
the cyclic process shown in Figure 7.9 is in the clockwise direction, and
that the area enclosed by the path of the process represents the work done
by the system on its surroundings, i.e. Wout ,
Student: Yes, Teacher. I have noted that.
Teacher: If the cyclic process considered were in the anticlockwise direction
on the PVdiagram then the area enclosed by the path of the process
would have represented the work done on the system by its surroundings,
i.e. Win.
Student: Okay, Teacher. I will work that out myself.
Example 7.6
Determine the boundary work associated with
a quasistatic isothermal process of an ideal gas. Pressures and volumes at
the initial and the ﬁnal states of the process concerned are given by (Po,
Vo)and(Pf,Vf), respectively.
Solution to Example 7.6
In an isothermal process, the temperature remains constant. An ideal gas
satisﬁes the ideal gas equation, PV =nRT. Therefore, an isothermal process
of an ideal gas satisﬁes the following condition:
PV =nRT =C(7.14)
where Cis a constant since Tis a constant. In this process Pchanges hyper
bolically with V. Since the process is taken as a quasistatic process, we evaluate
Boundary Work 129
the work done on the system using (7.7) as follows:
Win =−Vf
Vo
PdV =−Vf
VoC
VdV =−Cln Vf
Vo
Applying PV = constant for the initial and ﬁnal states, that is PoVo=PfVf,
we get
Win =−Cln Vf
Vo=−Cln Po
Pf(7.15)
for a quasistatic isothermal process of an ideal gas.ThevalueofCin
(7.15) is determined using (7.14) as follows:
C=PoVo=PfVf=nRT (7.16)
Example 7.7
One kmol of an ideal gas originally at 300 K
and 1 bar is heated at constant pressure to a temperature of 400 K and then
compressed isothermally to a volume equal to its initial volume. Show this
twostep process on a PVdiagram, and calculate the net work exchange
of the system with its surroundings. State your assumptions clearly.
Solution to Example 7.7
During the ﬁrst step of the process, the ideal gas at 300 K and 1 bar is
heated at constant pressure to a temperature of 400 K. Since the temperature
increases at constant pressure, the volume of the gas increases as well. If we
take this constant pressure process to be quasistatic, then it can be shown on
the PVdiagram of Figure 7.10 by the path A→B.PointArepresents the
initial state having PA= 1 bar and TA= 300 K, and point Brepresents the
intermediate state having PB= 1 bar and TB= 400 K.
During the second step, the gas is compressed isothermally to a volume equal
to its initial volume. That is, the temperature remains constant at 400 K and
the volume decreases to its initial value, VA. Since the volume decreases at
constant temperature, the pressure of the gas increases. Taking this isothermal
process to be quasistatic, its path, described by the hyperbola PV =nRT =
130 Chapter 7
constant, can be shown by the curve B→Con the PVdiagram of Figure
7.10. Point Crepresents the ﬁnal state of the process having TC= 400 K and
VC=VA.
P
V
C
B
DE
Figure 7.10 The twostep process of Example 7.7 shown on a PVdiagram,
where the shaded area shows the net work done on the system.
A
Since path A→Bis in the direction of increasing volume, the area ABED
under path A→Bgives Wout (see Figure 7.5). Therefore, we get
(Wout)A→B=PA(VB−VA)
whereweknowthatPA= 1 bar, but we do not know the volumes at Aand
B. Since the temperatures at Aand Bare known, the volumes at Aand Bcan
be eliminated from the above equation using the ideal gas equation of state as
follows:
(Wout)A→B=PBVB−PAVA=nR(TB−TA)
in which we have used the fact that PA=PB.Sincen=1kmol,TA= 300
KandTB= 400 K, we get
(Wout)A→B=(1kmol)(8.314 kJ/kmol ·K) (400 K−300 K)
= 831.4kJ (7.17)
Since path B→Cis in the direction of decreasing volume, the area BCDE
under path B→Cgives Win (see Figure 7.4). Since B→Cis a quasistatic
isothermal process, we could calculate the work exchange using (7.15) as follows:
(Win)B→C=−Cln VC
VB=−Cln PB
PC(7.18)
To determine the volume ratio (VC/VB)or the pressure ratio (PB/PC)of
(7.18), let us use the fact that the mass of ideal gas contained satisﬁes the
Boundary Work 131
ideal gas equation of state at the states represented by points A,Band C.We
therefore get
PAVA
TA
=PBVB
TB
=PCVC
TC
which becomes (1 bar)VA
300 K=(1 bar)VB
400 K=PCVA
400 K
The above yields PC= 4/3 bar. Substituting the values of PCand PBin
(7.18), we get
(Win)B→C=−Cln (3/4) (7.19)
where Cis calculated using (7.16) as follows:
C=nRTB=(1kmol)(8.314 kJ/kmol ·K) (400 K) = 3326 kJ
Thus, (7.19) gives
(Win)B→C=−3326 ln (3/4) kJ = 956.8kJ (7.20)
Combining (7.17) and (7.20), we get the net work done on the system by
the surroundings as follows:
(Win)net =−(Wout)A→B+(Win)B→C=(−831.4 + 956.8) kJ = 125.4kJ
Example 7.8
Obtain an expression for the boundary work
associated with steam undergoing a quasistatic isothermal process.
Solution to Example 7.8
In the case of an ideal gas undergoing a quasistatic isothermal process, the P
Vpath of the process can be described by the simple equation PV =constant.
And, therefore the boundary work associated can be given by the simple and
straight forward expression (7.15), as proved in the Solution to Example
7.6.
132 Chapter 7
In the case of steam undergoing a quasistatic isothermal process, however,
the PVpath of the process cannot be described by a simple equation. And,
therefore the boundary work associated cannot be given by a simple and straight
forward expression. However, if the path of the process could be sketched on a
PVdiagram then it would be possible to graphically determine the area under
the path between the initial and the ﬁnal states of the process, and hence the
work required.
Example 7.9
Determine the boundary work associated with
an adiabatic process executed by a closed system.
Solution to Example 7.9
There is no heat transfer in an adiabatic process, and thus Qin =0.The
ﬁrst law of thermodynamics applied to closed system therefore gives
(Win)adiabatic =∆U(7.21)
which is one instant in which the work transfer Win, a path dependent function,
is equated to ∆U, change in a system property.
Whether the given adiabatic process is quasistatic or not, (7.21) can be used
to determine the boundary work associated with a closed system containing ideal
gas or steam or any other substance. The work given by (7.21) is known as adi
abatic work. It is important to note that the adiabatic work can be evaluated
without knowing the PVpath of the process concerned.
Example 7.10
An ideal gas of 0.1 kmol expands adiabatically
such that its temperature reduces from 400◦C to 150◦C. Calculate the
boundary work done by the gas assuming γto be 1.4.
Boundary Work 133
Solution to Example 7.10
Since the given process is adiabatic, the work transfer can be calculated
using the expression for adiabatic work given by (7.21). Since we are dealing
with ideal gas, (7.21) becomes
(Win)adiabatic =nTf
To
CvdT (7.22)
Since Cvis a constant and since Cv=R/(γ−1), (7.22) reduces to the
following:
(Win)adiabatic =nC
v(Tf−To)=n(R/γ −1) (Tf−To) (7.23)
=0.1×(8.314/0.4) ×(150 −400) kJ
=−519.6kJ
The boundary work done by the ideal gas is 519.6 kJ.
Example 7.11
Superheated steam of 0.1 kmol expands adi
abatically from 5 bar and 400◦C to 1 bar and 150◦C. Calculate the boundary
work done by the steam.
Solution to Example 7.11
Since the given process is adiabatic, the work transfer can be calculated
using the expression for adiabatic work given by (7.21). Since we are dealing
with steam, (7.21) should be expanded to
(Win)adiabatic =m(uf−uo)
where m=0.1×18 kg, and ufand uiare the speciﬁc internal energies of steam
at the ﬁnal state at 1 bar 150◦C and at the initial state at 5 bar and 400◦C,
respectively.
From a Superheated Steam Table, we can ﬁnd that uf= 2583 kJ/kg and
that ui= 2963 kJ/kg. Therefore,
(Win)adiabatic =1.8×(2583 −2963) kJ =−684.0kJ
The boundary work done by steam is 684.0 kJ.
134 Chapter 7
Example 7.12
Develop a pressurevolume relationship de
scribing the path of a quasistatic adiabatic process of an ideal gas.
Solution to Example 7.12
Consider a mass of ideal gas undergoing a quasistatic adiabatic process. Let
us suppose that this mass of gas receives dWin amount of boundary work. Since
the process is quasistatic, the boundary work shall be given by (7.5) as
dWin =−PdV (7.24)
The work entering this adiabatic system increases the internal energy of the
system. And, the diﬀerential increase in the speciﬁc or molar internal energy of
an ideal gas is given by (5.7) as du =CvdT . The diﬀerential increase in the
internal energy of the system is therefore
dU =mC
vdT (7.25)
Since no heat is transferred between the system and its surroundings in an
adiabatic process, the ﬁrst law applied to an adiabatic closed system relates
dWin to dU by
dWin =dU (7.26)
Combining (7.24), (7.25) and (7.26), we get
−PdV =mC
vdT (7.27)
for an ideal gas undergoing a quasistatic adiabatic process.
Using the ideal gas equation of state, Tin (7.27) can be replaced by
(PV/mR), as follows:
−PdV =mC
vdPV
mR
=Cv
RPdV +VdP
The above can be rearranged to give
−(R+Cv)PdV =CvVdP (7.28)
Since Cp=Cv+Rfor an ideal gas, (7.28) becomes
−CpPdV =CvVdP
Boundary Work 135
which, using γ=Cp/Cv, can be rearranged to give
dP
P=−γdV
V
Integrating the above assuming that γremains a constant, we get
ln P=−γln V+constant
which can be rearranged to give
PV
γ=C(7.29)
where Cis a constant. Equation (7.29) describes the path of a quasistatic
adiabatic process of an ideal gas on a PVdiagram.
To ﬁnd the equation describing the path of the quasistatic adiabatic process
of an ideal gas on a TVdiagram, substitute P=mRT/V in (7.29). We get
mRT
VVγ=C
which yields the required equation as follows:
TV
(γ−1) =C
mR =constant (7.30)
To ﬁnd the equation describing the path of the quasistatic adiabatic process
of an ideal gas on a PTdiagram, substitute V=mRT /P in (7.29). That
gives
PmRT
Pγ
=C
which yields the required equation as follows:
T
P(γ−1)/γ =C1/γ
(nR)=constant (7.31)
Example 7.13
Twenty moles of an ideal gas undergoes a
quasistatic process in which the gas expands adiabatically from Po=10
bar and To= 425 K to Pf= 1 bar and Tf= 220 K. Determine the
isentropic exponent γ, and evaluate the work done by the system.
136 Chapter 7
Solution to Example 7.13
Since the given process is a quasistatic adiabatic process of an ideal gas,
the process shall be described by (7.29). Since the initial and ﬁnal states of the
process are identiﬁed by Pand T,notbyPand V, we shall describe the process
by (7.31). Since the initial and the ﬁnal states of the process satisfy (7.31), we
get To
P(γ−1)/γ
o
=Tf
P(γ−1)/γ
f
Since Po=10bar,To= 425 K, Pf= 1 bar and Tf= 220 K, we get
425
10(γ−1)/γ =220
1(γ−1)/γ
from which the numerical value of γcould be evaluated as follows:
10(γ−1)/γ = 425/220
γ−1
γ=ln (425/220)
ln (10)
1
γ=1−0.286
γ=1.4
Since the process is an adiabatic process of an ideal gas with constant Cv,
(7.23) is used to determine the work transfer as follows:
Win =nR
γ−1(Tf−To)=0.02 ×8.314
1.4−1×(220 −425) kJ =−85.2kJ
Thus, the work done by the ideal gas is 85.2 kJ.
Example 7.14
A quasistatic polytropic process process
of a simple compressible gaseous system is described by PVk=constant,
where kis known as the polytropic constant. Obtain an expression for
the work done by the system undergoing a quasistatic polytropic process
in terms of the pressures and volumes at the initial and ﬁnal states of the
process.
Boundary Work 137
Solution to Example 7.14
Starting from (7.7), the work done on the system during a quasistatic poly
tropic process may be given by
Win =−Vf
VoC
VkdV
=−CV−k+1
f−V−k+1
o
−k+1 for k= 1 (7.32)
The constant Cin (7.32) is removed by the following procedure. Since
PVk=Cis applicable at both the initial and the ﬁnal states of the process,
we get
C=PoVk
o=PfVk
f(7.33)
Using (7.33) in (7.32), we get
Win =CV−k+1
f−CV−k+1
o
k−1for k=1
=PfVk
fV−k+1
f−PoVk
oV−k+1
o
k−1for k=1
=PfVf−PoVo
k−1for k= 1 (7.34)
which can be used to evaluate the boundary work in a quasistatic polytropic
process.
If the substance contained in the system were an ideal gas then the ideal gas
equation PV =nRT could be used to simplify (7.34) to
Win =nR(Tf−To)
k−1for k=1 (7.35)
which can be used to evaluate the boundary work in a quasistatic polytropic
process of an ideal gas.
138 Chapter 7
Example 7.15
A frictionless pistoncylinder device contains
0.1 kg of superheated steam at 6 bar and 450◦C. Steam is now expanded
quasistatically along a quasistatic polytropic path until the pressure and
temperature fall to 1 bar and 200◦C, respectively. Determine the work
done by the steam.
Solution to Example 7.15
Since the steam undergoes a quasistatic polytropic process, we can use (7.34)
to determine the work transfer during the process. Since 0.1 kg of steam is
expanded from 6 bar and 450◦Cto1barand200
◦C, (7.34) becomes
Win =(0.1kg)(100 kPa)vf−(600 kPa)vo
k−1(7.36)
where vfand voare the speciﬁc volumes at the ﬁnal and initial states, respec
tively, and kis the polytropic constant.
From a Superheated Steam Table, we can ﬁnd that vo= 0.5528 m3/kg and
vf= 2.173 m3/kg, and therefore (7.36) becomes
Win =−11.44
k−1kJ (7.37)
where kis an unknown.
Thereisonlyonewaytoﬁndkwhich is to use the fact that the properties
at the initial and the ﬁnal states of the process satisfy the equation describing
the path of the process, which is PVk= constant. That leads to the following:
PoVk
o=PfVk
f
Povk
o=Pfvk
f
6×(0.5528)k=1×(2.173)k
2.173
0.5528 k
=6
k=ln 6
ln (2.173/0.5528) =1.31
Using k= 1.31 in (7.37), we get Win =−38.1 kJ. The work done by steam
during the quasistatic polytropic expansion is 38.1 kJ.
Boundary Work 139
Example 7.16
Consider a quasistatic expansion process de
scribed by P=aV +b,whereaand bare constants. The pressure is 100
kPa and the volume is 0.2 m3at the initial state of the process, and the
pressure is 900 kPa and the volume is 1 m3at the ﬁnal state. Calculate
the work done by the gas during the process.
Solution to Example 7.16
Since the given process is quasistatic, the boundary work can be evaluated
using (7.7). Since the process is described by P=aV +b, (7.7) gives
Wout =Vf
Vo
(aV +b)dV =aV2
f−V2
o
2+b(Vf−Vo)
The problem statement gives Vo=0.2m
3and Vf=1m
3, and therefore
the above expression reduces to
Wout =a(1 −0.04) m6
2+b[1 −0.2] m3
=a(0.48 m6)+b(0.8m3) (7.38)
The values of the constants aand bare not given, and there is only one way
to evaluate the constants aand b, which is to use the fact that the initial state
given by (100 kPa, 0.2 m3) and the ﬁnal state given by (900 kPa, 1 m3)should
satisfy the process described by P=aV +b. Therefore, we get the following
two equations:
100 kPa =a(0.2m3)+band 900 kPa =a(1 m3)+b
solving which we ﬁnd a= 1000 kPa/m3and b=−100 kPa. Substituting the
values of aand bin (7.38), we get
Wout = 1000 kPa/m30.48 m6+(−100 kPa)(0.8) m3= 400 kJ
The work done by the gas during the process is 400 kJ.
Student: Teacher, the expansion process of Example 7.16 is described by
P=aV +b, which means pressure increases as the volume increases. I
would expect the pressure to decrease with increasing volume......
140 Chapter 7
Teacher: It is not surprising that you expect the pressure to decrease with
increasing volume. Because that’s what happens in most cases. However,
there are situations in which pressure increases with increasing volume.
OnesuchcaseisdiscussedinExample 7.17 where the expansion of a
gasisrestrainedbytheforceexertedbyaspring.
Example 7.17
A frictionless pistoncylinder device initially
contains air at 150 kPa and 0.2 m3. At this state, a linear spring (that is, a
spring for which the force Fis proportional to displacement x)istouching
the piston but exerts no force on it. The air is now heated slowly to a
ﬁnal state of 0.5 m3and 600 kPa. Plot the process on a PVdiagram.
Hence, or otherwise, determine the work done by the air stating clearly all
the assumptions made.
Of the total work done by air, ﬁnd the work done against the variable
force exerted by the spring, and the work done against the constant force
due to the weight of the piston and the atmospheric pressure. Neglect the
weight of the spring in comparison to the weight of the piston.
Solution to Example 7.17
The initial, intermediate, and the ﬁnal states of the process are shown in
Figure 7.11. At the initial state, the spring just touches the piston and exerts no
force on the piston. As the air in the pistoncylinder device is heated slowly, it
expands slowly pushing the piston against the spring to compress it. This way, the
expansion of air is almost fully resisted by the spring provided that the frictional
eﬀects are negligible, and can therefore be approximated to a quasistatic process.
The work done by air can thus be evaluated using the work integral given by
(7.7), provided we know how the pressure of air (P)changeswithvolumeofair
(V) in the given process.
Since the expansion of air is almost fully resisted by the spring, the forces
acting on the piston are almost in balance according to
PA≈PoA+Mg+kx (7.39)
where Ais the crosssectional area of the piston on which the pressure Pof air
Boundary Work 141
acts, Pois the atmospheric pressure, Mgis the weight of the piston, kx is the
force exerted by the spring when the spring is compressed by a length x,andk
is the spring constant. The above force balance is written at the intermediate
state shown in Figure 7.11. Since the compressed length of the spring is xand
thevolumeofairisVat the intermediate state, as shown in Figure 7.11, we get
volume increase =V−Vo=Ax (7.40)
air
initial state
Vo=0.2m
3
Po= 150 kPa
ﬁnal state
Vf=0.5m
3
Pf= 600 kPa
air
intermediate state
Vm3
PkPa
6
?
x
Po
Po
Po
Figure 7.11 The initial, intermediate and the ﬁnal states of the process
given in Example 7.17.
qqq qqq
qqq qqq
qq qq air
qqqqqq
qqqqq
qqqq
qqqqq
qqqqq
qqq
qqqq
qqqq
qqq
qqq
qqq
qqq
q
qq
Combining (7.39) and (7.40), we get
P=Po+Mg
A+k
AV−Vo
A(7.41)
Equation (7.41) can be readily rearranged to give
P=mV +c(7.42)
which describes a straight line on a PVdiagram since the slope mand intercept
c,givenby
m=k
A2and c=Po+Mg
A−k
A2Vo
are constants. If mand care known, the straight line of (7.42) describing the
process can be sketched on a PVdiagram. We do not know k,A,Poand M,
and therefore are unable to calculate mand c.
142 Chapter 7
We do however know the coordinates of the initial state (o) and the ﬁnal
state (f) of the process, and they are shown on the PVdiagram of Figure 7.12.
A straight line drawn through the initial and the ﬁnal states, as shown in Figure
7.12, gives the path of the process given by (7.42) on the PVdiagram.
P(kPa)
V(m3)
(o)
(f)
0.2 0.5
600
150
Figure 7.12 Process given in Example 7.17 shown on a PVdiagram.
The boundary work done by air is the area under the straight line and is
calculated as follows:
Wout =150 + 600
2(0.5−0.2) kPa ·m3= 112.5kJ (7.43)
Here, we used a graphical method to calculate the work done by air. Alterna
tively, we can directly evaluate the work integral of (7.7) using Pof (7.42) once
we know the slope mand intercept c, which of course can readily be obtained
from the PVdiagram of Figure 7.12.
Of the total work done by air, we must now ﬁnd out how much is against
the spring and how much is against the constant force exerted by the weight of
the piston and the atmospheric pressure, which is (PoA+Mg). The work done
by air to push against this constant force can be evaluated as follows:
Work done by air against the constant force
=(constant force) ·(distance moved)
=(PoA+Mg)Vf−Vo
A=Po+Mg
A(Vf−Vo)
=Po+Mg
A(0.5−0.2) m3
Boundary Work 143
We now need to know the value of (PoA+Mg). Let us get back to (7.41)
and substitute the properties at the initial state, which are P=Po= 150 kPa
and V=Vo=0.2m
3.Then,weget
Po+Mg
A= 150 kPa
Thus, the work done by air against the force exerted by the weight of the
piston and the atmospheric pressure = (150 kPa ×0.3m3)=45kJ.
The work done by air against the variable force exerted by the spring is
therefore the diﬀerence between the total work done by air and the work done
by air against the constant force. Thus, the work done by air to push the spring
is (112.5  45) kJ = 67.5 kJ.
Example 7.18
Rework Example 7.17 with a nonlinear
spring for which F∝x2. Measurements taken at an intermediate point of
the expansion process showed that the pressure of air is 500 kPa when its
volume is 0.4 m3. Determine the total work done by air.
Solution to Example 7.18
For a nonlinear spring, we get
PA≈PoA+Mg+kx
2,(7.44)
in the notation used in the Solution to Example 7.17.Usingthefactthat
thevolumeincreaseis(V−Vo)=Ax in (7.44), we get
P=Po+Mg
A+k
AV−Vo
A2
(7.45)
Equation (7.45) can be rearranged to give
P=aV2+bV +c(7.46)
which takes a parabolic shape on a PVdiagram with a,band cas constants.
Even though we do not know k,A,Poand Mto ﬁnd the constants a,b
and c, we know that the curve represented by (7.46) on the PVdiagram passes
144 Chapter 7
through the initial state (0.2 m3, 150 kPa), the intermediate state (0.4 m3, 500
kPa), and the ﬁnal state (0.5 m3, 600 kPa). Forcing (7.46) to satisfy these three
points, we get a=−2500 kPa/m6,b= 3250 kPa/m3,andc=−400 kPa.
The work done by air is evaluated by substituting (7.46) in the work integral
of (7.7) as follows:
Wout =Vf
Vo
(aV2+bV +c)dV
=aV3
f−V3
o
3+bV2
f−V2
o
2+c(Vf−Vo)
Substituting the numerical values of a,band calong with Vf=0.5m
3and
Vo=0.2m
3,wegetWout = 124 kJ.
Example 7.19
A pistoncylinder device initially contains
steam at 200 kPa, 150◦C, and 0.10 m3. At this state, a linear spring
for which F∝xis touching the piston but exerts no force on it. Heat
is now slowly added to the steam, causing the pressure and the volume to
rise to 300 kPa and 0.12 m3, respectively. Determine the work done by the
steam and the heat supplied to the steam.
Solution to Example 7.19
Whether the system contains ideal gas or steam, it could be proven using the
method discussed in the Solution to Example 7.17 that the PVrelationship
of the given process is given by
P=mV +c(7.47)
Since the initial and the ﬁnal states of the steam satisﬁes (7.47), we get the
following equations:
200 kPa =m(0.10 m3)+cand 300 kPa =m(0.12 m3)+c
which gives m= 5000 kPa/m3and c=−300 kPa. The PVrelationship
of (7.47) therefore becomes
P= 5000V−300 (7.48)
Boundary Work 145
where Pis in kPa and Vis in m3.
The work done by steam is evaluated by substituting (7.48) in the work
integral of (7.7) as follows:
Wout =0.12 m3
0.10 m3(5000V−300) dV
= 5000 0.122−0.102
2−300 (0.12 −0.10) = 5 kJ
Heat supplied to the steam, denoted by Qin, is found by applying the ﬁrst
law of thermodynamics to the steam as follows:
Qin =∆U−Win =∆U+Wout =∆U+5kJ (7.49)
of which ∆Uis not yet known.
Since we deal with steam, we could expand ∆Uas
∆U=m(uf−uo)(7.50)
where mis the mass of steam and uoand ufare the speciﬁc internal energies
at the initial and ﬁnal states.
The initial state of steam is at 200 kPa and 150◦C, and it could be found
from a Superheated Steam Table that vo= 0.9602 m3/kg and uo= 2578 kJ/kg.
Since the initial volume of steam is 0.10 m3, we can calculate the mass of steam
as follows:
m=0.10 m3
0.9602 m3/kg =0.1041 kg
The ﬁnal state of steam is at 300 kPa. Since the ﬁnal volume is 0.12 m3,
we can calculate the speciﬁc volume of steam at the ﬁnal state as follows:
vf=0.12 m3
0.1041 /kg =1.153 m3/kg
From a Superheated Steam Table, it could be found that at 300 kPa and
1.153 m3/kg, the speciﬁc internal energy uf= 3093 kJ/kg.
Substituting the numerical values of m,uoand ufin (7.50), we get
∆U=(0.1041 kg) (3093 −2578) kJ/kg =53.6kJ
Substituting the value of ∆Uin (7.49), we get Qin =53.6kJ +5kJ =
58.6kJ. Thus, the heat supplied to the steam is 58.6 kJ and the work done by
the steam is 5 kJ.
146 Chapter 7
Example 7.20
The container shown in Figure 7.13 contains
10 kg of oxygen at a temperature of 350◦C. A constant force of 20 kN is
applied on the face of the piston having a crosssectional area of 0.05 m2.
The atmospheric pressure is 101.3 kPa. Heat is added slowly to oxygen
until its temperature rises to 400◦C. Determine the work done by oxygen,
and the amount of heat delivered to oxygen. Assume that the piston is
free to move. The average value of Cvfor oxygen can be taken as 0.66
kJ/kg ·K, and the molar mass be taken as 32 kg/kmol.
6
oxygen F=20kN
Patm = 101.3 kPa
piston with A = 0.05 m2
Figure 7.13 Oxygen contained in the pistoncylinder device.
Solution to Example 7.20
Let the oxygen in the container be the closed system. We need to determine
the work done by the oxygen, and the amount of heat delivered to it. Owing to
heat addition, the temperature of oxygen increases from 350◦C to 400◦C. The
temperature increase causes the volume of oxygen to increase pushing the piston
away, and we need to evaluate the boundary work associated with the volume
change. Since heat is added slowly, let us assume that the oxygen pushes the
piston slowly enough for the process to be assumed quasistatic. The boundary
work can then be evaluated using the work integral of (7.7), provided we ﬁnd
out how the pressure Pof oxygen varies with its volume V.
Since the piston moves freely, the pressure force acting on the inner face
of the piston will almost be balanced by the applied force of 20 kN and the
atmospheric pressure acting on the outer face of the piston. Thus, a force
balance on the piston gives
P=20 kN
0.05 m2+ 101.3kPa = 501.3kPa =aconstant
This means that the process is a constant pressure process, and the work integral
Boundary Work 147
of (7.7) becomes
Wout =P(Vf−Vo)
in which P= 501.3 kPa, but Vfand Voare unknowns. Since oxygen is assumed
to behave as an ideal gas, the above expression for work can be converted to
Wout =mR (Tf−Ti)
=10 kg
32 kg/kmol (8.314 kJ/kmol ·K) (400 −350) K= 130 kJ
The amount of heat delivered to oxygen, given by Qin, is found by applying
the ﬁrst law of thermodynamics to oxygen as follows:
Qin =∆U−Win =∆U+Wout =∆U+ 130 kJ (7.51)
of which ∆Uis unknown. Assuming that the oxygen behaves like an ideal gas,
we get
∆U=mC
v(Tf−To)=(10kg)(0.66 kJ/kg ·K) (400 −350) K= 330 kJ
Substituting the value of ∆Uin (7.51), we get
Qin = 330 kJ + 130 kJ = 460 kJ
Thus, the heat supplied to oxygen in the container is 460 kJ and the work
done by oxygen is 130 kJ.
Example 7.21
The pistoncylinder device shown in Figure 7.14,
contains 0.1 kg of air (molar mass = 29 kg/kmol; Cv= 0.717 kJ/kg ·K).
Initially, air is at 100 kPa and 27◦C, and
the piston rests on stops. Air is heated
until its volume increases by 20% of the
initial volume. An air pressure of 200 kPa
is required to raise the piston. Stating the
assumptions made, mark the path of the
process on a PVdiagram, and determine
the total work done by air and the total
heat supplied to air. Figure 7.14
air
piston
X
X
X
X
X
Xy H
H
HY
/
stops
148 Chapter 7
Solution to Example 7.21
The initial pressure of air is 100 kPa, and the air pressure required to raise
the piston is 200 kPa. Thus, the piston will rest on the stops maintaining the
volume of air constant until the pressure of air becomes 200 kPa. This part of
the process is therefore a constant volume process, during which the pressure
and temperature of air increase. Assuming the process to be quasistatic, the
path of the process is shown on the PVdiagram of Figure 7.15 by the constant
volume line A→B.PointArepresents the initial state of the process with PA
= 100 kPa and TA=27
◦C = 300 K. Point B represents the state at which the
air pressure becomes 200 kPa at constant volume, and PB= 200 kPa and VB
=VA.
The air pressure required to raise the piston is 200 kPa. This means that the
weight of the piston and the atmospheric pressure acting on the piston together
exert a net force of 200 kPa which keeps the piston resting on the stops until the
air pressure reaches 200 kPa. A small increase in the air pressure at this point
would cause the piston to rise. The consequent volume increase would help to
maintain the pressure of the air to remain at 200 kPa, provided the piston rises
quasistatically. The air pressure would then remain at 200 kPa, and the volume
and temperature of air would increase as the air continues to receive heat at
constant pressure, until the volume becomes 1.2 times its initial value. The path
of this process is shown in Figure 7.15 by the constant pressure line B→C.
Point Crepresents the ﬁnal state of the process, at which PC=PB= 200 kPa
and VC=1.2VB.
The path of the combined process is given by A→B→C.
Figure 7.15 The process of Example 7.21 shown on a PVdiagram.
s
P
V
BC
A
ED
s
s
Boundary Work 149
The total work done by air on its surroundings is represented by the shaded
area BCDE since the process is in the direction of increasing volume (see Figure
7.5). Therefore, we get
(Wout)total =PB(VC−VB) = (200 kPa)(1.2VB−VB)(7.52)
where VBis unknown.
Of the states represented by A,Band C, we know the pressure and tem
perature only at A. We also know that the mass of air in the pistoncylinder
device is 0.1 kg. Therefore, volume VAis found by applying PV =mRT to
state Aas follows:
VA=mRT
A
PA
=(0.1kg)(8.314/29 kJ/kg ·K) (300 K)
100 kPa =0.086 m3
Since VB=VA,wegetVB= 0.086 m3. Thus, (7.52) gives
(Wout)total = (200 kPa)(0.2×0.086 m3)=3.44 kJ
We have yet to determine the total heat supplied to air, which can be done
by applying the ﬁrst law of thermodynamics to the entire process as follows:
(Qin)total =(∆U)total −(Win )total
=mC
v(TC−TA)+(Wout )total
=0.1×0.717 ×(TC−300) kJ +3.44 kJ (7.53)
where TCis unknown.
Temp era t ure TCis found by applying the ideal gas equation of state at states
Aand Cas follows:
PCVC
TC
=PAVA
TA
(200 kPa)(1.2VB)
TC
=(100 kPa)VB
300 K
TC= 720 K(7.54)
Combining (7.53) and (7.54), we get
(Qin)total =(0.1) (0.717) (720 −300) kJ +3.44 kJ =33.55 kJ
150 Chapter 7
Example 7.22
A rigid box made of thermally nonconducting
material is divided into two compartments of volume 0.5 m3each by a thin,
strong, thermally nonconducting diaphragm. Initially, one compartment
contains nitrogen at 5 bar and 300 K and the other is a vacuum. When the
diaphragm is raptured, the nitrogen will rush to ﬁll the entire box. Deter
mine the temperature and the pressure of nitrogen at the ﬁnal equilibrium
state, and the boundary work done by nitrogen.
Solution to Example 7.22
The initial and the ﬁnal equilibrium states are shown in Figure 7.16. At the
initial state, nitrogen exerts a pressure force equivalent to 5 bar on one side of
the diaphragm. The other side experiences no force on it since it faces a vac
uum. Once the diaphragm is raptured, nitrogen expands to ﬁll the entire box.
The expansion of nitrogen is unrestrained since there is no force acting on the
moving part of the boundary of the nitrogen. At the ﬁnal equilibrium state, the
nitrogen occupies the entire box.
initial state, ﬁnal state,
nitrogen
5bar
Figure 7.16 The initial and ﬁnal states of the system of Example 7.22.
nitrogen
diaphragm
"
"
b
b
vacuum
b
bTf=?
Pf=?
300 K
Choose the system as the entire rigid box and all of its content. Since the
rigid box is made up of nonconducting material, no heat enters or leaves the
system. Since the box is rigid, no work enters or leaves the system. Thus, by
the ﬁrst law, the internal energy content of the system remains a constant.
Boundary Work 151
The contribution of the box and the diaphragm towards the internal energy
change is negligible, and the vacuum makes no contribution, and therefore the
internal energy of nitrogen remains a constant. If we assume that the nitrogen
behaves as an ideal gas, then the temperature of nitrogen remains a constant.
Therefore, the ﬁnal equilibrium temperature of nitrogen is the same as its initial
temperature. That is, Tf= 300 K.
The ﬁnal equilibrium pressure of nitrogen can be found by the application
of the ideal gas equation of state at the initial and ﬁnal states of nitrogen as
follows:
Pf(1 m3)
300 K=(5 bar)(0.5m3)
300 K
which gives Pf=2.5bar.
To evaluate the boundary work done by nitrogen, let us take the nitrogen
as the system, and apply the ﬁrst law of thermodynamics to it. The nitrogen
is enclosed by nonconducting material and vacuum, and therefore no heat has
entered or left the nitrogen during the process. The internal energy of nitrogen
has been shown to remain constant. Therefore, the ﬁrst law of thermodynamics
applied to nitrogen gives that no work is done by nitrogen.
Student: Teacher, a part of the boundary of nitrogen in Example 7.22 has
moved all the way to the opposite wall of the box. How could then the
boundary work done by nitrogen be zero?
Teacher: We have just proved that the boundary work done by nitrogen is zero.
Haven’t we?
Student: Yes, Teacher, we have. But, please, could we evaluate the boundary
work done by nitrogen using (7.7), instead of using the ﬁrst law as we
have just done?
Teacher: Well, as stated above, nitrogen has undergone an unrestrained expan
sion. That is, the expansion did not take place under almost fullyresisted
conditions. Therefore, the expansion of nitrogen is not quasistatic, and
we cannot use (7.7) to evaluate the boundary work done by the nitrogen.
Student: Teacher, the boundary of nitrogen has moved during the expansion
of nitrogen. Therefore, nitrogen should have done some boundary work.
I am very sure of that. How did we get zero? Oh... Teacher, I am going
crazy.
152 Chapter 7
Teacher: Cool down, dear Student. I agree that the boundary of nitrogen
has moved during the expansion of nitrogen. But, the boundary has
moved with absolutely no force resisting its movement since nitrogen has
expanded into a vacuum. It is, therefore, I say that nitrogen does not do
any work when it expands into a vacuum. It is as simple as that.
Student: Yes, I see the point now. It is simple indeed, but very strange.
Teacher: It might appear strange, but very true. Remember, boundary work
is always done against a force. By the way, you know that it is more
diﬃcult to climb a mountain than to walk a ﬂat land. It is because we go
against the gravitational force when climbing the mountain. If there were
no gravitational force acting on us, we would have ﬂoated to the top of
Mount Everest with ease. Wouldn’t we?
Student: Yes, we would.
Example 7.23
The pistoncylinder device in Figure 7.17 con
tains 0.2 kmol of helium. The weight of the piston and the atmospheric
pressure acting on it exerts a constant force
equivalent to 3 bar pressure on helium. The
stops shown prevent the piston from being
pushed out by helium at 300 K and 8 bar. With
the stops removed, helium expands adiabati
cally to 3 bar. The molar mass of helium is
taken as 4 kg/kmol and its γas 1.667. Deter
mine the work transfer and the change in the
internal energy of the system.
helium
Figure 7.17
?
piston
CCCCW
stops
Solution to Example 7.23
First, let us see if the given process can be assumed to be quasistatic. Recall
that we have learned that the force causing the expansion and the force resisting
the expansion should diﬀer only by a very small amount at any given time in a
process that is quasistatic. In the given process, helium gas expands from 8 bar
Boundary Work 153
pressure to 3 bar pressure against a force that is constant at 3 bar. The large
diﬀerence between the force causing the expansion and the force resisting the ex
pansion makes the process nonquasistatic. Since the process is not quasistatic,
we cannot use (7.7) to evaluate the boundary work.
However, since the given process is adiabatic,
Win =∆U=nC
v(Tf−To)(7.55)
can be used to evaluate the work transfer for helium, which behaves as an ideal
gas. The amount of helium nis 0.2 kmol, Cvcan be evaluated according to Cv
=R/(γ−1) = 8.314/(1.667 −1) = 12.5 kJ/kmol ·K. The initial temperature
Tois 300 K, and the ﬁnal temperature Tfis to be found.
Ideal gas equation of state applied to the ﬁnal state gives
PfVf=nRT
f,
where the ﬁnal pressure Pf, the amount of helium n, and the gas constant R
are known. Had we known the ﬁnal volume Vf, we could have found Tf.Since
we do not know the ﬁnal volume, the ideal gas equation of state is of no help
to ﬁnd Tf.
We know the initial pressure Po, the initial temperature Toand the ﬁnal
pressure Pf. Therefore, it is tempting to use the relationship, derived from
(7.31),
Tf
P(γ−1)/γ
f
=To
P(γ−1)/γ
o
to determine Tf. We should NOT do that because the given process is not
quasistatic even though it is an adiabatic process of an ideal gas.
At this point it appears that the work transfer involved in this problem
cannot be found, which is however not true. Let us now try an approach that is
very diﬀerent from those discussed above. Let us neglect the friction and other
dissipative eﬀects involved in the process. Then, we can say that the work done
by the expanding helium is done only to push the piston and the atmospheric
pressure acting on the piston. Since the piston and the atmospheric pressure
acting on the piston exert a constant force equivalent to 3 bar pressure on
helium, we can say that the helium works against a constant force, given by the
3 bar pressure multiplied by the crosssectional area of the piston.
The work done by the helium can therefore be calculated using
Wout =(3 bar) ×(crosssectional area of the piston)
×(the distance moved by the piston)
=(3 bar) ×(the increase in the volume of helium)
=(3 bar) ×(Vf−Vo)
154 Chapter 7
Since we do not know the initial and the ﬁnal volumes of the above expres
sion, replace these volumes by pressures and temperatures using the ideal gas
equation of state as
Wout =(3 bar) ×nR Tf
Pf
−To
Po(7.56)
of which we once again do not know the value of Tf.
Even though work transfer cannot be evaluated using either (7.55) or (7.56),
because we do not know Tf, these two equations can be combined to ﬁnd Tf
as follows:
nC
v(Tf−To)=−(3 bar) ×nR Tf
Pf
−To
Po
Substituting all known numerical values, we get
nR
1.667 −1(Tf−300 K)=−(3 bar) ×nR Tf
3bar −300 K
8bar
which can be solved to give Tf= 225 K. Using this in (7.55), we get
Win =(0.2kmol)×8.314 kJ/kmol ·K
1.667 −1×(225 −300) K=−187 kJ
The work done by the system on its surroundings is therefore 187 kJ. Since
the system is adiabatic, the change in internal energy of the system is −187 kJ.
The internal energy of the system has decreased because the system has done
work on its surroundings under adiabatic conditions.
Example 7.24
A rigid cylinder contains a ‘ﬂoating’ piston.
Initially, it divides the cylinder in half, and on each side of the piston the
cylinder holds 0.01 kmol of the same ideal gas at 200◦C and 100 kPa. Using
an electrical heater installed on side Aof the cylinder, heat is added slowly
until the pressure in side Areaches 150 kPa. The piston and the cylinder
are perfect thermal insulators with negligible heat capacity. Determine
(a) the ﬁnal volume of gas on side B,
(b) the work done by the piston on the gas on side B,and
(c) the amount of heat supplied by the electrical heater.
Boundary Work 155
The speciﬁc heats of the gas are constant and have the values Cv= 12.56
kJ/kmol ·KandCp= 20.88 kJ/kmol ·K.
Solution to Example 7.24
The ‘ﬂoating’ piston is assumed to be free to move within the cylinder with
out friction. Therefore, the forces acting on either side of the piston equal each
other when the piston is in equilibrium. Thus, at the initial and ﬁnal states of
the system, the pressure on side Aequals the pressure on side B,asshownin
Figure 7.18, which shows all the data for the problem. Heat added to side A
causes the gas on side Ato expand pushing the piston outwards to compress the
gas on side B. Since, heat is added slowly to side A,thepistonmovesslowly.
In addition, the ‘ﬂoating’ piston is assumed to have frictionfree movement. It
is therefore safe to assume that both the expansion of gas on side Aand the
compression of gas on side Bare quasistatic processes. Since the piston and the
cylinder are perfect heat insulators, the gas on side Bshall undergo an adiabatic
process.
A
100 kPa
473 K
0.01 kmol
B
100 kPa
473 K
0.01 kmol
VoVo
A
150 kPa
TAK
B
150 kPa
TBK
VAVB
mm m m
Figure 7.18 Initial and ﬁnal states of Example 7.24.
initial state ﬁnal state
(a) Determine the ﬁnal volume of the gas on side B.
To dete r min e VB, the volume of the gas on side Bat the ﬁnal state, we
could use the fact that the gas on side Bis an ideal gas, and that it undergoes
a quasistatic adiabatic process. We then know that the pressure of the gas on
side Bchanges with its volume according to (7.29). Therefore, we get
(100 kPa)Vγ
o=(150 kPa)Vγ
B
156 Chapter 7
Using γ=Cp/Cv= 20.88/12.56 = 1.66, the above expression can be
simpliﬁed to
VB
Vo
=100
1501/1.66
=0.78 (7.57)
We can evaluate the initial volume Vowith the help of the ideal gas equation
of state as follows:
Vo=(0.01 kmol)(8.314 kJ/kmol ·K) (300 K)
(100 kPa)=0.249 m3(7.58)
Combining (7.57) and (7.58), we get
VB=(0.78) (0.249 m3)=0.194 m3(7.59)
which is the volume of the gas on side Bat the ﬁnal state.
(b) Determine the work done by the piston on the gas on side B.
To determine the work done by the piston on the gas on side B,thatis
(Win)B, let us apply the ﬁrst law of thermodynamics to the gas on side Bas
(Qin)B+(Win)B=(∆U)B
Since the gas on side Bis an ideal gas that undergoes an adiabatic process,
the above equation simpliﬁes to
(Win)B=nBCvB (TB−473 K)=0.01 ×12.56 ×(TB−473) kJ (7.60)
To dete r min e TB, let us apply the ideal gas equation of state to the gas on
side Bas follows: (100 kPa)(Vo)
473 K=(150 kPa)(VB)
TB
which gives
TB=150
100VB
Vo(473 K)
Substituting VB/Vofrom (7.57) in the above, we get
TB=150
100 (0.78) (473 K) = 553 K(7.61)
Substituting the value of TBin (7.60), we get
(Win)B=0.01 ×12.56 ×(553 −473) kJ =10kJ (7.62)
which is the work done by the piston on the gas on side B.
Boundary Work 157
(c) Determine the amount of heat supplied by the electrical heater.
Let us say that the amount of heat supplied to the gas on side Aby the
electrical heater is (Qin)A. To determine this, we may apply the ﬁrst law of
thermodynamics to the gas on side Aas
(Qin)A+(Win)A=(∆U)A
which can be rewritten as
(Qin)A=(∆U)A+(Wout )A(7.63)
Since the piston is assumed to move without friction, the work done by the
gas on side Ain pushing the piston outwards is exactly the same as the work
done by the piston on the gas on side Bto compress it. Therefore,
(Wout)A=(Win )B
which converts (7.63) to
(Qin)A=(∆U)A+(Win )B(7.64)
The gas on side Ais 0.01 kmol of an ideal gas with Cv= 12.56 kJ/kmol K
and (Win)Bis known from (7.62). When this information is used in (7.64), we
get
(Qin)A=[0.01 ×12.56 ×(TA−473) + 10] kJ (7.65)
To ﬁnd TA, let us apply the ideal gas equation of state to the gas in side A
as
TA=150
100VA
Vo(473 K)(7.66)
We can determine VAin the following manner. Since the total volume of
gases at the ﬁnal state is the same as the total volume of gases at the initial
state, we get
VA+VB=2Vo
Using VB/Vofrom (7.57), the above equation can be rearranged to give
VA/Vo=2−VB/Vo=2−0.78 = 1.22
Substituting the above in (7.66) gives TA= 866 K, and using this value of
TAin (7.65), we get
(Qin)A=0.01 ×12.56 ×(866 −473) + 10 = 59 kJ
which is the heat supplied by the electrical heater.
158 Chapter 7
In summary, we conclude that the 59 kJ of heat supplied slowly by the elec
trical heater to the gas in side Ais used to raise the temperature of the gas in
side Ato 866 K and for the gas in side Ato push the piston outwards by doing
10 kJ of work on the piston. The piston moves without friction to use the 10
kJ of work to adiabatically compress the gas in side B. The work done on the
gas on side Braises the temperature of the gas in side Bto 553 K.
7.9 Summary
•Diﬀerential boundary work is given by
dWin =−PdV (7.5)
for a quasistatic compression process, and by
dWout =PdV (7.6)
for a quasistatic expansion process.
•The total boundary work in a quasistatic process is given by
Wout =−Win =Vf
Vo
PdV (7.7)
•Equations (7.5), (7.6) or (7.7) can be applied to evaluate the boundary
work for any substance; solid, liquid, gas or any mixture of phases.
•Boundary work takes the unit J, when Pis in Pa and Vis in m3,andit
takes the unit kJ, when Pis in kPa and Vis in m3.
•Boundary work can also be evaluated using the expression,
win =−Vf
Vo
Pdv (7.8)
in which vis the speciﬁc (or molar) volume, and therefore win takes the
unit kJ/kg (or kJ/kmol), provided Pis in kPa, and vis in m3/kg (or
m3/kmol).
Boundary Work 159
•The boundary work done by a system when executing a process depends
on the path followed by the process.
•The area under the curve describing the path of a quasistatic process on a
PVdiagram provides the amount of boundary work done by the system
if the process is an expansion, and the boundary work done on the system
if the process is a compression.
•The area enclosed on a PVdiagram by the path of a quasistatic cyclic
process gives the work exchange between the system and its surroundings.
This area represents the work done by the system if the process is in a
clockwise direction, and the work done on the system if the process is
anticlockwise.
•Boundary work done by a quasistatic constantpressure process is given
by
Wout =−Win =Vf
Vo
PdV=P(Vf−Vo)(7.9)
If the system comprises an ideal gas as the working ﬂuid, (7.9) becomes
Wout =−Win =nR(Tf−To)(7.10)
If the system comprises water/steam as the working ﬂuid, (7.9) becomes
Wout =−Win =mP (vf−vo)
•A constantvolume process does not do any boundary work since the
boundary of a system executing a constantvolume process does not ex
pand or contract.
•A quasistatic isothermal process of an ideal gas is described by
PV =nRT =C(7.14)
where Cis a constant. Boundary work done on a system undergoing a
quasistatic isothermal process is given by
Win =−Cln Vf
Vo=−Cln Po
Pf.(7.15)
The constant Cmay be determined using
C=PoVo=PfVf=nRT (7.16)
160 Chapter 7
•Boundary work done on a closed system executing an adiabatic process is
given by
(Win)adiabatic =∆U(7.21)
The adiabatic work therefore does not depend on the path of the process,
but on the initial and the ﬁnal states of the process only.
If the system comprises an ideal gas, (7.21) becomes
(Win)adiabatic =nTf
To
CvdT (7.22)
If the system comprises water/steam, (7.21) becomes
(Win)adiabatic =m(uf−uo)
•The path of a quasistatic adiabatic process of an ideal gas with constant
speciﬁc heats is described by
PVγ=constant on a PVdiagram. (7.29)
TV(γ−1) =constant on a TVdiagram. (7.30)
T
P(γ−1)/γ =constant on a PTdiagram. (7.31)
•A quasistatic polytropic process is described by PVk=constant, where
kis the polytropic constant. Boundary work done on a system undergoing
a polytropic process is given by
Win =PfVf−PoVo
k−1for k=1 (7.34)
For a system comprising an ideal gas, (7.34) becomes
Win =nR(Tf−To)
k−1for k=1 (7.35)
For a system comprising water/steam, (7.34) becomes
Win =m(Pfvf−Povo)
k−1for k=1
•In order to evaluate boundary work using (7.6) to (7.8), the system has
to undergo an almost fully resisted expansion or compression. If the ex
pansion (or compression) is not almost fully resisted throughout the entire
process then the above equations cannot be used to evaluate boundary
work (see Example 7.23).
•It is important to remember that a system expanding into a vacuum does
not do any work at all (see Example 7.22).