Download 253 Chapter 12 Thermodynamics GOALS When you have mastered

Physics Including Human Applications
253
Chapter 12
Thermodynamics
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
PV diagram
isochoric process
isobaric process
isothermal process
adiabatic process
efficiency of a heat engine
Carnot cycle
refrigerator
coefficient of performance of a refrigerator
heat engine
Laws of Thermodynamics
State three laws of thermodynamics, and explain the operation of a physical system in
terms of these laws.
Thermodynamics Problems
Solve problems consistent with the laws of thermodynamics.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 5, Energy,
and Chapter 10, Temperature and Heat.
Physics Including Human Applications
254
Chapter 12
Thermodynamics
12.1 Introduction
Much of your travel is made possible by a device that we call a heat engine. We can
think of a heat engine as being any device which operates by heat energy input, does
work, and has a heat energy output. The human body is a heat engine. The refrigerator
in your home is a type of heat engine. In this chapter you will be introduced to the
fundamental principles of these and other forms of heat engines.
In Chapter 10, it was indicated that when the temperature of an object is changed
many different effects may take place - such as change in length or volume, change in
resistance, change in pressure, change in electromotive force, change in color, and so on.
In such processes there is usually a transfer of heat energy and also a performance of
work, a force acting through a displacement. The study of the phenomena that result
from energy changes produced by a transfer of heat and performance of work is called
thermodynamics. These phenomena involve heat energy, temperature, and work.
Thermodynamics is the study of the laws that govern thermal processes. When you
have completed this study of thermodynamics, you will be able to explain the
significance of these laws as they apply to both living and nonliving systems.
Thermodynamics concerns itself with a well-defined system, which interacts directly
with its surroundings, and by this interaction performs some useful function.
Thermodynamics is not concerned with internal effects by themselves.
12.2 The Zeroth Law of Thermodynamics
You know that when two equal amounts of ice are added to two identical containers of
hot tea to make iced tea, after the ice has melted both containers of hot tea will be at the
same final temperature. This is an example of the zeroth law of thermodynamics.
The zeroth law of thermodynamics can be stated as follows:
If system A is in thermal equilibrium with system B, and system B is in thermal equilibrium
with system C, then A, B, and C are in equilibrium with each other, and they are at the same
temperature.
Thermal equilibrium is defined as the condition in which there is no net energy
exchange between the systems in equilibrium. The zeroth law is seemingly trivial, but it
is the basis of thermometers since it defines operationally temperature as the property
of a system that determines its thermal equilibrium with another system. The zeroth
law of thermodynamics clearly points out that temperature, as measured by
thermometers, involves systems in equilibrium; that is, the temperature being measured
by the thermometer in the system must be constant in time, indicating equilibrium,
before a measurement is made. In this connection it is important to point out that the
thermal inertia of the thermometer determines the time required for the thermometer to
reach equilibrium with the system, as well as the effect of the thermometer on the
system being monitored. The larger the thermal inertia (mass x specific heat of the
thermometer), the longer the response time, and the more energy transferred to or from
Physics Including Human Applications
255
the thermometer in attaining equilibrium. In this chapter we will use the symbol T to
refer to absolute temperature in degrees Kelvin (0øC = 273øK).
12.3 External Work
Consider as a system a cylinder containing a
working substance (a gas) and closed by a movable
piston (Figure 12.1). If the gas pushes the piston out,
work is done by the system. If an outside agent
pushes in the piston, work is done on the system. In
either case the work is called external work.
Work that is done by one part of the system on
another part of the same system is called internal
work. The study of internal work as such is not part
of thermodynamics.
Returning to our example system, the cylinder containing gas enclosed by a movable
piston, assume the piston has an area A, and a pressure P is acting on the piston. The
total force acting on the piston is given by the product of the pressure times area, PA. If
the piston is displaced a small distance Δx, then work is done, and the amount of work
ΔW is given by the product of the force times the displacement,
ΔW = (force)(displacement) = (PA) (Δx)
(12.1)
We notice that the product of area times displacement, A Δx, is equal to the change in
the volume ΔV of the system as the piston moves. Then we can express the amount of
work done by the system when it pushes the piston out a small distance Δx by the
following equation
ΔW = P ΔV
(12.2)
If the pressure remains constant over a whole series of small displacements, then the
total amount of work done is given by the pressure times the change in volume.
W = P (V2 -V1)
(12.3)
where the difference between the final and initial volumes (V2 -V1) is the change in
volume. If the pressure is not constant, we have to use calculus methods, as we discuss
in the Enrichment section of this chapter, Section 12.15.
EXAMPLE
During normal breathing the volume of the lungs increases by about 500 ml in each
inspiration. What is the amount of work done on the lungs during inspiration?
The standard pressure of the atmosphere is about 1.00 x 10 5 N/m2.
W = (1.00 x 105 N/m2) (500 ml x 10-6 m3/ 1 ml)
W = 50 J
We can interpret the meaning of Equation 12.3 by studying a PV diagram, the graph
of the pressure versus the volume of a confined gas (Figure 12.2). If we allow the system
to expand, the pressure of the gas will decrease as the volume increased from Vi to Vf.
Physics Including Human Applications
256
The amount of work done by the gas for a small displacement Δx is given by Equation
12.2 and is shown as the shaded area in the figure since PΔV is equal to the area of the
shaded portion of the graph. This area on the graph has the units of joules.
If we add the work done by many of these small displacements as the system expands
from V1 to V2, we find that the work done by the gas is represented by the area under
the curve from C to D between the vertical lines at V1 and V2 (shown as the large shaded
portion of the graph). If we allow the gas to expand, the system goes from C to D, the
gas pushes the piston out, and we say the system does work. By convention we call this
positive work. If we compress the gas so that the volume of the system goes from V2 to
V1 (the pressure changes from D to C), we say work is done on the system. We
designate this as negative work.
Let us consider a cyclic system as shown in Figure 12.3. First, the working substance
expands from V1 to V2 in accordance with line CD. The work done is positive and
represented by the area between the curve CD and the volume axis. Then the working
substance is compressed from V2 to V1 in a process represented by the curve DEC. The
work of compression is represented by the area under DEC, DV2V1CE, and is negative.
For this cycle, the area under CD is greater than the area under DEC. The total amount
of work done in one cycle is equal to the enclosed area between CD and DEC. This area
represents the net work that the system does in one cycle. It is evident that the amount
of net work done per cycle depends upon the path that is followed from C to D and
back again. That is, it depends upon the processes used during the expansion and
compression portions of the cycle.
12.4 The First Law of Thermodynamics
You have probably had the experience of doing mechanical work on a system and
thereby changing the temperature of the system. When rapidly pumping up the tires of
a bicycle with a hand pump, you may have noticed that both the piston of the pump
and the tire become hot. The quantitative relationship between the work done of a
system and its thermodynamic properties is expressed by the first law of thermodynamics.
Physics Including Human Applications
257
The first law of thermodynamics which is a formulation of the conservation of
energy states that
the change in the internal energy of a system is equal to the heat added to the system minus the
work done by the system.
In equation form this becomes
ΔU = ΔQ - ΔW
(12.4)
or
ΔQ = ΔU + ΔW
or
ΔW = ΔQ - ΔU
where ΔU is the change in internal energy, ΔQ is the change in heat energy, and ΔW is
the positive work done by the system. Let us look at this equation carefully. The
internal energy of a system depends only on the state of the system. For this reason it is
called a state function. A state function is dependent only on the variables defining the
state of the system such as the pressure, temperature, and volume for an ideal gas. A
state function is independent of the process involved in getting the system to a given
state. When a system undergoes a change and moves from one thermodynamic state to
another, the change in the internal energy of the system will depend only on the final
and initial states. The change in internal energy will be the same regardless of the
processes involved in changing states. In Chapter 5 we had an analogous case in which
the work, or energy change, was independent of path and depended only upon the
beginning and ending points. In that situation we defined the system as conservative,
and we said we were dealing only with conservative forces. In the first law of
thermodynamics the change in internal energy depends only upon the change in
absolute temperature.
The heat and work terms in Equation 12.4 are not state functions; they are dependent
on the process involved in the change of state. For example, heat added at constant
pressure produces a different change than the same amount of heat added at a constant
volume as we shall see later in this chapter. Likewise, the work done on the system
depends on the process involved. For example, work at constant temperature produces
a different effect than the same amount of work applied at constant pressure. These
terms are parallel to work against nonconservative force as discussed in Chapter 5.
The importance of the first law of thermodynamics is that the sum of the work done
and the heat added in moving between two different thermodynamic states always
produces the same change in the internal energy of the system. The work done on the
system may be mechanical, electrical, or chemical. All three of these forms of work
occur in living systems.
EXAMPLE
Consider a diet problem. A person undertakes a diet program that provides 2000 kcal
per day, and the person expends energy in all forms to a total of 3000 kcal per day. In
order to do this much work on this diet, the person must obtain energy stored as
internal energy in the body. Loss of body fat will occur as this energy is used. Using the
first law of thermodynamics, we see that the heat energy input is represented by the
Physics Including Human Applications
258
2000 kcal (ΔQ = 2000 kcal) from the food when burned in the body. The body does work
amounting to 3000 kcal (ΔW = + 3000 kcal). This work takes many forms, mechanical
work in terms of body motion, chemical changes in the muscles, and electrical energy in
nerve activity. From the first law it follows that the internal energy must decrease by
1000 kcal per day, ΔU = ΔQ -ΔW = 2000 kcal - 3000 kcal = - 1000 kcal. If this energy were
stored as adipose (fat) tissue (7500 kcal/kg), then the person would lose about 1 kg in a
week.
12.5 Isochoric Processes
An isochoric process is one that takes place at a constant volume. An isochoric process
is represented on a PV diagram by a vertical line (Figure 12.4). You see that the area
under the curve is 0. We know that the pressure times the change in volume is the work
done. The volume does not change, so no work is done:
ΔW = PΔV
where ΔV = 0.
Therefore ΔW = 0. For an isochoric process the first law of thermodynamics reduces to
an equality between the change in internal energy and the change in heat energy.
ΔU = ΔQ
(12.5)
If the heat is added to the system, the internal energy is increased, and the temperature
rises. If heat is given up by the system, there is a decrease in internal energy and a
decrease in the temperature.
For a confined ideal gas the heat added to the gas during an isochoric process is
given by the product of the number of moles of gas, the specific heat of the gas at
constant volume, and the change in temperature,
ΔU = ΔQ = ncv ΔT
(12.6)
where n is the number of moles, (See footnote 1), cv, the molar specific heat at constant
volume, is the amount of heat required to raise the temperature of one mole of gas one
degree during an isochoric process see Table 12.1 , and ΔT is the change in temperature.
Physics Including Human Applications
259
EXAMPLE
Find the amount of heat to be added to raise the pressure of 1.50 moles of helium from
75.0 cm Hg to 100 cm Hg at constant volume. How much is the temperature raised? The
original temperature is 27.0øC.
For a gas P1V1/T1 = P2V2/T2
For this case, V1 = V2, so
P1/T1 =P2/T2
75.0 cm hg/300øK = 100 cm Hg/T2
T2 = 400øK
temperature increase = 100øK
ΔQ = ncv ΔT
cv = 3.00 cal /mole-øK
ΔQ = (1.50 moles)(3.00 cal/mole-øK)(100øK) = 450 cal
12.6 Isobaric Processes
An isobaric process is one that takes place at a constant pressure and can be represented
by a horizontal line in a PV diagram (Figure 12.5). The area under the curve CD is equal
to the product of the pressure times the change in volume, PΔV, and represents the
work that is done during the isobaric process.
Let us consider a system that is a confined ideal gas. In Section 10.9 on the gas laws, we
learned that if the pressure is constant and the volume changes, there is also a change in
temperature. A change in temperature means that there has been a change in internal
energy. Hence if heat which is added to the system causes work to be done and an
increase in internal energy, the first law of thermodynamics for an isobaric process is
given by,
ΔQ = ΔU + ΔW = ΔU + PΔV
(12.7)
As an example, consider the case of changing one gram of water at 100øC into one
gram of steam at 100øC. The heat added goes into external work (change in volume) and
internal energy.
The heat which is added during an isobaric process is given by
Physics Including Human Applications
260
ΔQ = ncpΔT
(12.8)
where n is the number of moles, cp, the molar specific heat at constant pressure, is the
amount of heat required to raise the temperature of one mole one degree during an
isobaric process, and ΔT is the change in temperature. The change in internal energy,
which is independent of the process, is the same as would have occurred by an
isochoric process,
ΔU = ncv ΔT
(12.6)
Now we can rewrite the first law of thermodynamics for an isobaric process,
ncpΔT = ncv ΔT +PΔV
(12.9)
This equation shows that the molar specific heat of a gas at constant pressure is always
greater than the molar specific heat of a gas at constant volume. By rearranging the
above equation, and using the relationship that PΔV = (constant) ΔT for an ideal gas, we
can show that the difference between the two different molar specific heats is a constant
number, independent of the kind of gas that is used in the system (see Table 12.1),
cp - cv = constant = R
(12.10)
ø
where R is the universal gas constant and has a value of 8.31 J/mole- K.
EXAMPLE
Helium (3.00 moles) originally at 273øK and a volume of 0.067 m 3 expands at a constant
pressure of 1.01 x 105 N/m2 to a volume of 0.134 m3. Find the change in temperature, the
change in internal energy, the work done, and the heat added to the system.
From the general gas law P1V1/T1 = P2V2/T2. Because P1 =P2
V1/T1 =V2/T2
0.0067 m3/273øK = 0.134 m3/ T2
T2 = 546øK
T2 - T1 = 273øK
The change in internal energy ΔU,
ΔU = ncvΔT = (3.00 moles) (3.00 cal/mole-øK) (273øK) = 2460 cal
The work done ΔW,
ΔW = PΔV
= (1.01 x 105 N/m2) (0.134 -0.067)m3 = 6.77 x 103 J.
The heat added ΔQ,
ΔQ = ncpΔT = (3.00 moles) (4.98 cal/mole-øK) (273øK) = 4080 calories
In order to verify the first law of thermodynamics, you must express all energy
quantities in the same units. The work done in calories is work done in joules divided
by 4.19 since the mechanical energy equivalent of 1 calorie of heat energy is 4.19 joules
(see Chapter 10).
Physics Including Human Applications
261
From the first law of thermodynamics, ΔW =PΔV = ΔQ - ΔU. Now expressing each term
in joules, we have
ΔQ - ΔU = (4080 - 2460) cal = 6.79 x 103 joules
ΔW = PΔV = 6.77 x 103 J
We find that we have a discrepancy of less than one percent, which is not surprising.
We may introduce a discrepancy of that size by using only three significant figures.
12.7 Isothermal Processes
An isothermal process is one that takes place at a constant temperature. For a system of
a confined ideal gas, we find that at constant temperature the pressure and volume of
the system are given by Boyle's law (Section 10.9), PV = constant. On the PV diagram an
isothermal process for an ideal gas is represented by a hyperbolic curve whose equation
is
PV = constant
See Figure 12.6.
Since the internal energy of a system is a state variable
dependent only upon the temperature of the system, for an isothermal process the
internal energy of the system remains constant, ΔU = 0. The first law of
thermodynamics for an isothermal process states the equality between the work done
and the change in heat energy:
ΔQ = ΔW =PΔV
(12.11)
For an isothermal process the heat added equals the work done by the system and is
represented by the shaded area under the curve in Figure 12.6. If the system consists of
an ideal gas that expands from a volume of V1 at a pressure P1 to a volume of V2 at a
constant temperature, the work done by the gas is given by
W = P1V1 lnV2/V1
(12.12)
(A derivation of this result is given in Section 12.16)
Physics Including Human Applications
262
EXAMPLE
If 50 cal of heat are added to an ideal gas in a cylinder under the condition of a constant
temperature, what is the work done by the expanding gas?
The internal energy of an ideal gas depends only on its temperature. In this problem
the temperature is constant (i.e., ΔU = 0). Therefore, ΔQ = +ΔW, the heat input is equal
to the work done by the gas. Note that ΔW is positive for work done by the gas. Under
the constant temperature conditions the work done by the gas equal to 50 cal, the same
as the heat input energy.
12.8 Adiabatic Processes
Any process that takes place very rapidly is essentially adiabatic. An adiabatic process
occurs if no heat is added to or taken from the system – that is, ΔQ = 0. The curve GH in
Figure 12.7 represents an adiabatic process for an ideal gas. The equation for the curve
γ
GH is PV = constant, where γ is the ratio of the molar specific heats cp/cv (Table 12.1).
The constant γ has a maximum value of 5/3 for a monoatomic ideal gas, and is always
greater than 1. For the adiabatic process in Figure 12.7 , the first law of thermodynamics
can be expressed by the following equation,
ΔU = -ΔW = -PΔV = ncvΔT = (P1V1 - P2V2)/(1 - γ)
(12.13)
A derivation of this result is given in Section 12.16.
For an adiabatic process if the system does work, there is a decrease in internal energy;
if work is done on the system, there is an increase in internal energy - the heating that
occurs when you rapidly compress the air in a tire pump, for example.
Physics Including Human Applications
263
EXAMPLE
The air in an automobile tire is released. What happens to the temperature of the air as
it expands into the atmosphere? Calculate the values of the changes assuming that the
original air pressure in the tire was 3.00 atmospheres (atm) and the temperature was
20.0øC.
This expansion is a rapid process and can be approximated by an adiabatic process.
The expanding gas does work on the atmosphere so ΔW is positive, and thus ΔU is
negative. The internal energy of the gas is reduced; that is, its temperature is lowered.
For a unit volume of air (1.00 m3) in the tire, we can use the expression for an adiabatic
γ
γ
process 3(1) = 1(Vf) where the γ for air is 1.40.
Vf = 3(1/1.4) = 2.19 m3
Now we can use the ideal gas law to find the change in temperature,
PoVo/To =PfVf/Tf
(3.00)(1.00)/293 = (1.00)(2.19)/Tf
Tf = 214øK, or - 59øC
12.9 The Second Law of Thermodynamics
The first law of thermodynamics is necessary, but not sufficient, to explain the thermal
phenomena that you observe every day. When you mix two glasses of water of different
temperatures, what is the final state of the water? Is this final state the only possible
state consistent with the first law of thermodynamics? There are an infinite number of
ways in which energy could be conserved for this system, but the situation actually
occurring is the one in which the final temperature of the water is a definite
temperature falling somewhere between the initial temperatures of the two glasses of
water. This final state is result of the second law of thermodynamics.
There are several equivalent ways of stating the second law of thermodynamics. The
Clausius statement of the second law is:
Heat energy cannot pass spontaneously from a system at a lower temperature to one at a higher
temperature.
Your experiences with hot and cold objects may make this statement of the second law
seem trivial. You have never seen an isolated object at room temperature become hot.
Why not? We know the total heat energy available in all the objects in the room is
enough warm up one object. But nature does not behave in that fashion. To transfer
heat energy from a lower temperature object to a higher temperature we must do work
on the system. If the Clausius statement were not true, we would be able to transfer
heat energy from inside our homes to the hotter outside air in the summer to aircondition our homes without using any other energy.
The Kelvin-Planck statement of the second law of thermodynamics is that
it is not possible to perform a process which only extracts heat energy from a constant
temperature object and performs an equivalent amount of work.
You may think that an isothermal process would violate this statement of the second
law because the change in heat energy is equal to the work done by the system,
Physics Including Human Applications
264
Equation 12.11. However, an isothermal expansion of the system leaves the final system
at a different volume than at the beginning of the process. To test this statement of the
second law we must return the system to its original volume since the only process that
can occur is the change of heat energy into work. But it is just this return process that is
impossible without changing the way nature performs.
The operational equivalence of the Kelvin-Planck and Clausius statements of the
second law is shown by the following example. Suppose the Kelvin-Planck statement
can be violated. Then we could construct a heat engine that only takes heat energy from
a reservoir of heat energy at constant temperature and converts the heat energy into
work. We can then use this engine to operate a refrigerator that extracts heat from the
inside of the refrigerator and delivers it to a hotter object. The net result of the heat
engine-refrigerator combination is the transfer of heat energy from a cold object to a hot
object without doing any net work on the total system, a violation of the Clausius
statement.
12.10 Heat Engines
The internal combustion gasoline engine and
the steam engine are examples of engines that
makes use of a combustion process to produce
motion and to do work. These are both
examples of heat engines. A heat engine absorbs
a quantity of heat energy Qh from a hot
reservoir of heat at temperature Th, does work
W, rejects a quantity of heat energy Qc to a cold
heat reservoir at a temperature Tc, and returns
to its original state. For example, your
automobile engine burns gasoline at a high
temperature Th from which the engine extracts
heat energy Qh to turn the crank shaft and
wheels of the automobile to do work W. The
rejected heat energy Qc is returned, via the
exhaust gases and other thermal transport
processes, to the atmosphere at a temperature
Tc. Part of the work is used to return the engine
to its original state, ready to begin the cycle
again. Heat engines operate in cycles, using the
heat absorbed during each cycle to do work. A
schematic diagram of a heat engine is shown in
Figure 12.8 .
We can combine this cyclic nature of a heat engine with the first law of
thermodynamics to develop an expression for the efficiency of a heat engine. Since the
initial and final states of a cycle of the heat engine are the same, the internal energy of
the engine must be the same at the beginning and end of each cycle, ΔU = 0; then ΔW =
ΔQ, or for the whole cycle
W = Qh - Qc
(12.14)
Physics Including Human Applications
265
The efficiency of a heat engine is defined as the useful work output divided by the
energy input. For the heat engine the energy input is given by the heat energy absorbed
from the high temperature reservoir, so
efficiency = work output/energy input
= W/Qh = (Qh - Qc)/Qh
= 1 - Qc/Qh
(12.15)
EXAMPLE
What is the efficiency of a person whose daily diet is equivalent to 4.0 kg of milk and
who does useful work at the rate of 50 watts for an eight hour day? Heat of combustion
of milk = 650 cal/g.
Qh = 650 x 4.0 x 103 x 4.19 = 11 x 106 J
W = (50 J/sec)(8 hr x 60 min/1 hr x 60 sec/1 min) = 1.4 x 106 J
efficiency = 1.4 x 106 J/11 x 106 = .13 = 13 percent.
Where does all the energy go?
12.11 Carnot's Engine
Both the first and second law of thermodynamics are basic to the study and
understanding of heat engines. How do you make the heat engine more efficient? Is
there a limit to the efficiency of a heat engine? In 1824 Sadi Carnot, a French engineer,
was the first to approach the problem of heat engines from fundamental considerations.
Carnot's approach was an entirely theoretical one. He disregarded the mechanical
operation and details and considered the fundamentals. In fact, our discussion in the
previous section followed Carnot's approach. Carnot also introduced the concept of a
reversible cycle which we now know as Carnot's cycle. The theoretical Carnot cycle is for
an ideal engine that has a cylinder and piston made of perfect insulating materials and
has three separate parts - one part is a perfect heat conductor and a high temperature
heat reservoir at temperature Th, one part is a perfect insulator, and the third part is a
perfect conductor and cold heat reservoir at temperature Tc (Figure 12.9). The cylinder
was filled with an ideal working substance, such as an ideal gas.
Physics Including Human Applications
266
The steps in the Carnot cycle are as follows:
1. The high temperature part Th is connected to the cylinder and the piston moves,
giving the isothermal expansion ΔQ = ΔW along curve AB
2. The insulating part replaces the heat reservoir Th, and the gas is further expanded as
an adiabatic process, ΔQ = 0. As work is done, the temperature of the gas is lowered to
temperature Tc along curve BC.
3. The lower temperature Tc reservoir replaces the insulating part. The gas is then
compressed at a constant temperature Tc. As the gas is compressed, the part at Tc must
absorb some heat, and the amount absorbed is equal to the work of compression (curve
CD).
4. Part Tc is replaced by the insulating head, and the gas is adiabatically compressed to
the starting point. The work done in compressing the gas increases the internal energy
of the gas until it reaches temperature Th (curve DA).
A summary of the cycle shows that heat Qh is put into the system in step 1 and that
heat Qc is taken from the system in step 3. The difference between Qh and Qc represents
the amount of work done during the cycle, W = Qh - Qc.
Follow the Carnot cycle around the curves on the PV diagram of Figure 12.9.
Starting at point A (P1V1Th), step 1 is represented by the curve AB, an isothermal
expansion at temperature Th to P2V2. The heat input is Qh, work is equal to area under
the curve AB and bounded by the corners ABV2V1. Step 2 is represented by BC an
adiabatic expansion from P2V2 Th to P3V3Tc. The work done is equal to the area under the
curve BC and is bounded by the corners BCU3V2. There is a decrease in the internal
energy so the temperature drops from Th to Tc. Step 3, represented by curve CD, is an
isothermal compression at temperature Tc, heat Qc is exhausted from the system. Work
is done on the system in compressing the ideal gas. It is negative work equal in
magnitude to the area under the curve CD and bounded by the corners CV3V4D. Step 4,
represented by curve DA, is an adiabatic compression from P4V4Tc to P1V1Th. Work is
again done on the system. This is negative work equal in magnitude to the area under
the curve DA, which is bounded by DV4V1A. This work increases the internal energy.
Notice that if all the work done on and by the system is added for one cycle the net
work is represented by the area inside of the curves that join the points ABCD.
As a result of his theoretical work, Carnot postulated a theorem which we can state
in the following way:
No engine working between two heat reservoirs can be more efficient than a Carnot engine
operating between those two reservoirs.
In other words the Carnot cycle defines the best possible heat engine permitted by the
laws of nature. What is the efficiency of a Carnot engine. We can use Equation 12.15 and
our knowledge of isothermal processes to calculate this efficiency in terms of the
temperatures of the reservoirs Th and Tc,
efficiency = 1 - Qc/Qh
(12.15)
For isothermal processes we know that the change in heat energy must be equal to the
work done (Equation 12.11), and we have stated that the work done by the system
during an isothermal expansion is given by Equation 12.12,
Qh = W =P1V1 ln (V2/V1)
(12.12)
Physics Including Human Applications
267
for the expansion AB. For the compression CD, Qc must be the negative of the work
done by the system since the system does negative work,
Qc = -WCD = -P3V3 ln (V4/V3) =P3/V3 ln (V3/V4).
efficiency = 1 - P3V3 ln (V3/V4) /P1V1 ln (V2/V1)
(12.16)
Next we use the ideal gas laws and the relationship for adiabatic processes for a
confined gas.
P1V1/Th =P3V3/Tc so P3V3/P1V1 =Tc/Th
For the adiabatic processes
γ
γ
γ
P2V2 =P3V3 and P1V1 = P4V4
γ
Dividing the first equation by the second, we obtain
γ
γ
γ
P2V2 /P1V1 =P3V /P4V
γ
For the isothermal processes
P2/P1 =V1/V2
and
P3/P4 =V4/V3
Then
(V2/V1)
γ-1
= (V3/V4)
γ-1
or
V2/V1 =V3/V4
Then we can reduce Equation 12.16 to
efficiency = 1 - Tc/Th
(12.17)
where Tc and Th are the absolute temperatures in Kelvin degrees. The efficiency of the
Carnot engine depends only upon the temperatures of the hot and cold heat reservoirs,
and no engine can be more efficient than its equivalent Carnot engine.
EXAMPLES
1. A person burns food at a temperature of 37øC and exhausts body heat to an
environment of 20øC. What is the maximum efficiency possible for that person?
efficiency = 1 - Tc/Th = 1 -293/310 = 5.48 x 10-2 = 5.48 percent
2. An automobile engine during combustion produces a source of heat energy with a
temperature of 400øC. What is the maximum efficiency of this engine when the
outside temperature is 20øC?
efficiency = 1 - 293/673 = 565 x 10-1 = 56.5 percent
The usual automobile engine has an efficiency of about 20 percent.
Physics Including Human Applications
268
12.12 Refrigerators
A refrigerator is, in thermodynamic terms, a heat engine that works in reverse. A
refrigerator absorbs heat from a low-temperature heat reservoir and exhausts heat to a
high-temperature reservoir. To accomplish this task, work must be done on the
refrigerator system. Of course, we know from the Clausius statement of the second law
that it is not possible to build a refrigerator that requires no work input. From the first
law we can write an equation for the heat exhausted to the hot reservoir Qh as the sum
of the work done on the refrigerator W and the heat absorbed from the cold reservoir
Qc,
Qh = W + Qc
(12.18)
Refrigerators are characterized by their coefficient of performance. The coefficient of
performance h of a refrigerator is defined as the ratio of the amount of heat energy
absorbed from the cold reservoir to the amount of work done on the refrigerator,
η = Qc/Qh - Qc
(12.19)
Typical refrigerators have performance coefficients of about 5. The larger the
performance coefficient, the better is the refrigerator. Is there a maximum possible
coefficient of performance for a refrigerator? For our results for the efficiency of a
Carnot engine we can transform Equation 12.19 into a relationship between the
temperatures of the hot and cold reservoirs. We can combine the information from the
equations for the efficiencies of heat engines,
efficiency = 1 - Qc/Qh = 1 -Tc/Th
to deduce that
Qc/Qh =Tc/Th
(12.20)
We combine this equation with Equation 12.19 to express the coefficient of performance
as the ratio of the cold temperature to the temperature difference between the reservoir
temperatures,
η = Tc/(Th -Tc )
(12.21)
EXAMPLE
A room air conditioner removes heat from a room at 20øC and exhausts it to the outside
at 37øC. What is the maximum possible coefficient of performances for this air
conditioner?
η = 293ø/(310ø - 293ø) = 17.2
12.13 Entropy
One way to characterize the properties of a thermodynamic system is in terms of its
orderliness. A system where everything is lined up in rows is highly ordered. A system
where everything is happening at random is highly disordered. One form of the second
law of thermodynamics is based on the concept of the order of a system. In this form the
second law states that real processes always involve an increase in disorder. To facilitate
the formulation of this law, we define the entropy of a system. Entropy is a measure of
the disorder of a system, and it is defined in the following equation,
Physics Including Human Applications
269
ΔS = ΔQ/T
(12.22)
in which ΔS = change in entropy, ΔQ = change in heat energy, and T is the absolute
temperature at which process takes place. Therefore, entropy, like internal energy, is a
state function. Entropy depends only on the thermodynamic state of the system, and
changes in entropy are independent of the changes involved in going from one state to
another. Reversible changes are idealized processes. For reversible processes there is no
increase in the disorder of the system. Irreversible processes are ones in which there is
an increase of entropy and an increase in disorder. Every state of a system has a definite
value of entropy. In terms of entropy, the second law of thermodynamics can be written
for any isolated system as:
ΔS ≥ 0
(12.23)
that is, a process that starts in one equilibrium state and ends in another will go in a
direction so the entropy of the system plus its environment increases or remains the
same. For ΔS > 0, the process is irreversible; for ΔS = 0, the process is reversible
(idealized); and for ΔS < 0, the process is impossible.
The entropy of an isolated system never decreases in going from one state to
another. It should be noted that the directedness of time (arrow of time) is associated
with our experience with phenomena that obey the second law of thermodynamics. For
example, if you noted a footprint appearing before your eyes on the beach, you would
surely conclude that something strange was happening. In fact, you might have the
feeling that time was running backward! Our experience has been conditioned by the
increase in disorder (the vanishing of a footprint in the sand, for example) associated
with the passage of time. Another example in this connection is the phenomena
associated with a drop of ink added to a glass of water. The ink drop breaks up and
diffuses throughout the water (going from a spatially ordered drop to a randomized
distribution of ink particles of considerable disorder). And again, if an ink drop
appeared in a glass of water before your eyes, you would certainly have reason to
wonder about your sense of time.
12.14 Entropy and Living System
Do living systems obey the second law of thermodynamics? Living systems can be
characterized as systems that take in energy and create order. These statements
certainly are counter to the second law of thermodynamics except that the second law
applies only to isolated or closed systems. Upon reflection it becomes apparent that
when a living system is isolated, it ceases to live, and then it certainly obeys the second
law of thermodynamics. Could you accept the following statement as a definition of
life? Any system that takes in energy and creates order is a living system. If this is an
unacceptable definition, what would you add?
12.15 The Third Law of Thermodynamics
The third law of thermodynamics is a statement about the changes in physical systems
that occur as the temperature approaches absolute zero. In one form, the third law may
be written, as T → 0, S → 0, or as the temperature approaches absolute zero, the entropy
approaches zero. This gives us some insight into low-temperature phenomena. In
Physics Including Human Applications
270
particular, the phenomena of superfluidity of liquid helium and superconductivity of
aluminum are examples of systems that show increased ordering as absolute zero is
approached. In terms of atoms and electrons this means that the systems are in their
lowest energy states. This allows exact knowledge of the state of the particles of the
system and thus perfect order. This form of the third law of thermodynamics allows us
to consider the realm of low temperature research in terms of increased order rather
than in terms of energy. Indeed, an important question to ask about a system being
considered for low temperature study might well be, "Will increased order make a
difference in the behavior of the system?"
ENRICHMENT 12.16 Calculus Derivations of Thermodynamics Equations
Let us derive the equation for an adiabatic expansion of an ideal gas. To obtain this
equation we need to obtain a general relationship between the two molar specific heats, cp
and cv. We can do this using the calculus form of Equation 12.9
cpdT = n cvdT + PdV
for an isobaric process, but for an ideal gas PV = BT where B is a constant.
(12.24)
For a constant pressure process
P dV = B dT
(12.25)
Now we combine these two equations to obtain the following general result,
n cp dT - n cv dT = B dT
cp - cv =B/n
This result was introduced in Section 12.6, Equation 12.10.
(12.26)
For an adiabatic process we can write the calculus form of Equation 12.13
n cv dT + P dV = 0
(12.27)
For an ideal gas,
P dV + V dp = B dT
(12.28)
Now we solve both equations for dT and set them equal to each other
dT = P dV/n cv = P dV + V dP/B = dT
(12.29)
0 = (n cv + B) P dV +ncv dP
but B = n(cp - cv); so
0 = (n cv + n cp -n cv) P dV +n cv V dP
0 = cp P dV + cv V dP
(12.30)
Let γ = cp / cv;
0 = γdV/V +dP/P
Now we integrate this equation
γ
constant = γ In V + ln P = lnV + ln P = lnPV
γ
Physics Including Human Applications
γ
PV = constant
271
(12.31)
This equation was introduced in Section 12.8.
In Section 12.3 you learned that external work done by a thermodynamic system is ΔW =
PΔV. In calculus notation we can express this relationship as
dW = P dV
(12.32)
If we have a curve for a given process plotted on a PV diagram and we wish to determine
the work done in going along the curve from A to B, then we integrate this expression,
W = ∫VAVB P dV
(12.33)
If the pressure is a function of volume, P = f(V), then the total work done by the change in
the volume of the system from VA to VB is,
W = ∫VAVB f(V) dV
(12.34)
EXAMPLES
ISOTHERMAL PROCESS
1. PV = constant = C
P = f (V) =C/V
C =PAVA =PBVB
The work done in an isothermal expansion for an ideal gas is given by
W = ∫VAVB C/V dV = C ln (V)|VAVB =C ln (VB/VA)
(12.35)
This equation was introduced in Section 12.7, Equation 12.12.
ADIABATIC PROCESS
γ
2. PV = constant where γ = cp/cv.
γ
PV = C1
γ
P =C1/( V )
γ
C1 = PAVA =PBVB
γ
The work done by an adiabatic expansion for an ideal gas is given by
W=
∫
VA
VB
γ
γ
γ
γ
γ
γ
C1dV/ V = {(C1)/(1 - γ)}(V1- )|VAVB = PB VB VB1- /(1 - γ) - PAVA VA1- /(1 - γ)
W = (PBVB - PAVA)/ (1 - γ)
This equation was introduced in Section 12.8, Equation 12.13.
(12.36)
Physics Including Human Applications
272
ENTROPY
3. ΔS = ΔQ/T, or in calculus notation dS = dQ/T.
∫
S1
S2
dS = ∫dQ/ T
In many processes the change in the quantity of heat dQ is a function of the
temperature, dQ =f(T)dT.
Thus,
S2 - S1 = ∫S1S2 f(T)dT/T
For example for an isochoric process
Q = ncvT, dQ = ncv dT
S2 - S1 =ncv ln (T2/T1)
(12.37)
SUMMARY
Use these questions to evaluate how well you have achieved the goals of this chapter.
The answers to these questions are given at the end of this summary with the number of
the section where you can find related content material.
Definitions
Write the correct word or phrase in the blank.
1. A ______ is a system that ______ heat from a ______ temperature reservoir, does
work upon the universe, and ______ heat to a temperature reservoir, and whose
performance is characterized by its ______.
2. A ______ is a system that _______ heat from a temperature reservoir, has work
done upon it by an external agent, and _______ heat to a ______ temperature
reservoir, and whose performance is characterized by its ______.
3. The _______ is an idealized process that can be used to calculate the greatest
possible ______ of a heat engine.
4. A PV diagram is a ______ on which the horizontal axis represents the ______ of
the system, the vertical axis represents the _______ of the system and the area
between a curve and the ______ axis represents the _______.
5. Define each of the following processes, describe the representation of the process
on a PV diagram and write the first law of thermodynamics equation for the
process:
a. adiabatic
b. isothermal
c. isobaric
d. isochoric
Physics Including Human Applications
273
Laws of Thermodynamics
6. State the laws of thermodynamics, and describe the operation of a confined
volume of helium in a Carnot engine in terms of these laws.
Thermodynamics Problems
7. A cylinder contains oxygen at a pressure of 2 atm. The volume is 3 liters and
temperature is 300øK. The oxygen is carried through the following processes:
a. Heating at constant pressure 500øK
b. Cooling at constant volume to 250øK
c. Cooling at constant pressure to 150øK
d. Heating at constant volume to 300øK.
Show the processes above in a PV diagram giving P and V at the end of each
process. Calculate the net work done by the oxygen.
Answers
1. heat engine, absorbs, high, rejects, low, efficiency, the absolute temperatures of the
reservoirs, efficiency (Section 12.10)
2. refrigerator, absorbs, low, rejects, high coefficient of performance (Section 12.12)
3. Carnot cycle, efficiency (Section 12.11)
4. graph, volume, pressure, volume axis, work done (Section 12.3)
γ
5. a. constant heat energy, PV = constant curve, ΔQ = 0; ΔU = -PΔV;
b. constant temperature, PV = constant, hyperbolic curve, ΔU = 0; ΔQ = P ΔV
c. constant pressure, a horizontal line, ΔQ = ΔU +P ΔV or ncpΔT = ncvΔT + PΔV
d. constant volume, a vertical line, ΔW = 0; ΔU = ΔQ = ncvΔT
12.7, 12.8)
(Sections 12.5, 12.6,
6. zeroth law: helium reaches same temperature as high temperature reservoir; first
law: the helium expands and ΔU = ΔQ - ΔW; second law: the heat absorbed by
the helium gas at high temperature is partially converted to work and the rest is
rejected at a lower temperature (Section 12.9)
7. See Figure 12.10; net work = area within rectangle = 2 liter-atm
12.5, 12.6)
(Section 12.3,
Physics Including Human Applications
274
ALGORITHMIC PROBLEMS
Listed below are the important equations from this chapter. The problems following the
equations will help you learn to translate words into equations and to solve singleconcept problems.
Equations
ΔW = P ΔV
ΔU = ΔQ - ΔW
ΔU = ncv ΔT
ΔQ = ncpΔT
cp - cv = R
W =P1V1 ln (V2/V1)
efficiency = W/Qh = 1 - Qc/Qh = 1 - Tc/Th
η = Qc/W = Qc/Qh - Qc = Tc/(Th -Tc )
ΔS = ΔQ/T
ΔS ≥ 0
γ
PV = constant
cp/cv = γ
(12.2)
(12.4)
(12.6)
(12.8)
(12.10)
(12.12)
(12.15)(12.17)
(12.19, 12.21)
(12.22)
(12.23)
(12.31)
(definition)
Problems
1. A cylinder is filled with a gas and 2-kg piston of 18 cm diameter closes the
cylinder and falls 10 cm in compressing the gas. How much work is done?
2. If 200 cal of heat are added to a system which does 500J of work, how much is the
internal energy of the system changed in this process?
3. It is known that the heat input into an engine to produce 45,000 J of work is 15 x
104 J. How much heat is lost through the exhaust?
4. What is efficiency of the engine of problem 3?
5. The intake temperature of a Carnot engine is 500øK, and the exhaust temperature
is 360øK. What is the efficiency of the engine?
Answers
1.96 J
2. 338 J
3. 105,000 J
4. 30 percent
5. 28 percent
Physics Including Human Applications
275
EXERCISES
These exercises are designed to help you apply the ideas of a section to physical
situations. When appropriate the numerical answer is given in brackets at the end of
each exercise.
Section 12.3
1. A compressed gas is allowed to expand from a volume of 1.0 m3 to 2.5 m3 against
the pressure of the atmosphere (P = 1.02 x 105 N/m2). What work does the gas
do? From where does the energy come to do this work? [1.50 x 105 J, from
internal energy or applied heat energy]
Section 12.4
2. One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a pressure
of 1.013 x 105 N/m2. The heat of vaporization at this pressure is 539 cal/g.
Compute the external work, the increase in internal energy, and the amount of
heat energy added to the system. [1.69 x 102 J, 2.09 x 103 J, 2.26 x 103 J]
3. When a enclosed system of an ideal gas is taken from A to C via the path ABC,
2000 cal of heat input into the system, and 750 cal of work are done (Figure
12.11).
a. How much heat is put into the system along the path ADC if the work is 250 cal?
b. When the system returns from C to A along AC the work is 500 cal. Does the
system absorb or liberate heat, and how much?
c. Assume that the internal energy at A is 250 cal and at C is 1500 cal. How much
heat is absorbed in process AB and DC? TB = 3 TA, TC = 3TB
d. What is the change in internal energy for path AC. [a. 1500 cal; b. 1750 cal
liberated; c. 500 cal, 1000 cal; d. 1250 cal]
Section 12.5
4. One mole of water vapor in a sealed pressure cooker is heated from 100øC so that
the absolute pressure doubles. What is the final temperature? How much heat
energy is added to the system? What is the change in the internal energy of the
water vapor? [473øC, 2514 cal, 2514 cal]
Physics Including Human Applications
276
Section 12.6
5. A confined quantity (9.6 moles) of superheated steam at 427øC drive a piston 30
cm in diameter a distance of 0.80 m at a pressure 5.00 x 105 N/m2. The final
temperature of the steam is 77øC. What is the work done by the steam on the
piston? How much heat energy is used during the process? What is the change in
the internal energy of the steam? [2.83 x 104J, 1.23 x 105 J, -9.49 x 104 J]
6. A given quantity of gas at a constant pressure of 10 N/cm2 expands from a
volume of 10 liters to a volume of 20 liters. How much work is done by the gas?
If the original temperature was 27øC, what was the final temperature? What else
would you need to know to calculate the heat added to the system? [1000 J,
600øK]
7. A steel cylinder of cross-sectional area of 20 cm2 contains 200 cm3 of mercury. The
cylinder is equipped with a tightly fitting piston which supports a load of 30,000
N. The temperature is increased from 15øC to 65øC. Neglecting expansion of the
steel cylinder, find
a. the increase in volume of the mercury
b. the mechanical work done against the force
c. the amount of heat added
d. the change of internal energy [a. 1.8 cm3; b. 27 J; c. 19,000 J; d. 19,000 J]
Section 12.7
8. Assume that you push down on the plunger of a hand tire pump slowly so that
the temperature of the pump remains constant. At the end of your push, the
volume of confined air has decreased from 1.05 liters to 0.150 liters. Assume the
starting conditions were a pressure of 1.00 atm and a temperature of -13.0øC.
What is the final pressure? How much work was done on the system? What is
the change in heat energy of the system? What is the change in internal energy of
the system? Where did the heat energy go? [7.0 atm, 2.04 atms-liters, 2.04 atmsliters, 0, the surroundings of the pump]
Section 12.8
9. Assume that you push down the plunger of a hand tire pump so rapidly that no
heat escapes from the system during this action. At the end of your push the
volume of the confined air has decreased from 1.05 liters to 0.150 liters. Assume
the starting conditions were a pressure of 1 atm at a temperature of - 13.0øC.
What is the final pressure? How much work was done on the system? What is
the change in heat energy of the system? What is the change in the internal
energy of the system? What is the final temperature of the air? [15.2 atm, 3.09 atm
- liters, 0, +3.3 atm - liter, 293øC ]
10. A gas at 27øC at 1 atm is compressed until its volume is one-tenth of its original
volume. This compression is done so fast as to be adiabatic (γ = 1.5).
a. find the final pressure
b. find the final temperature assuming the gas is ideal. [a. 32 atm; b. 960øK]
Physics Including Human Applications
277
Section 12.11
11. A Carnot engine whose high temperature reservoir is 127øC takes in 200 cal of
heat at this temperature, and it gives up 160 cal to a low-temperature reservoir.
What is the efficiency of this engine and the temperature of the exhaust
reservoir? [20 percent, 320øK]
12. A Carnot engine is operating between the two temperatures of 450øK and 300øK.
The heat furnished at a high temperature reservoir is 1350 cal. How much work is
done by the engine, and how much heat is given out in exhaust? [450 cal, 900 cal]
Section 12.12
13. A refrigerator requires 450 J of work to exhaust 1350 J of heat to the outside air.
How much heat energy is absorbed from the low-temperature reservoir? What is
the coefficient of performance for this refrigerator? [900 J, 2]
14. A room air conditioner can remove 160 J of heat from the room for every 200 J of
heat it exhausts to the outside air. What is the work that must be done on the air
conditioner to accomplish this? What is the coefficient of performance of the air
conditioner? [40 J, 4]
Section 12.13
15. Given the graph in Figure 12.12 for the molar specific heat of a system at constant
pressure near T = 0, find the entropy per mole of this material at 4øK.[1.5 x 10-4
J/mol øK]
16. Compare the entropy change for 1 mole (16 g) of water going from ice to water at
0øC and from water to steam at 100øC. Does this answer support the idea that
greater entropy changes accompany changes of greater disorder? [ΔSws/ΔSiw =
4.93]
Physics Including Human Applications
278
PROBLEMS
The following problems may involve more than one physical concept. The numerical
answer is given in brackets at the end of the problem.
17. The cycle of a heat engine is described as follows:
a. start with n moles of gas at P0V0T0
b. change to 2P0V0 at constant volume
c. change to 2P02V0 at constant volume
d. change to P02V0 at constant volume
e. change to P0V0 at constant pressure
Show this cycle on the PV diagram, and find the temperature for the end of each
process. What is the maximum possible efficiency for this engine? [2T0, 4T0,
2T0, T0, 75 percent]
18. Assume the gas in problem 17 is an ideal monatomic gas, γ = 5/3. The specific
heat of the gas at constant volume is 3.00 cal/mole - K at T0 = 300øK. Assume you
have 0.250 moles of the gas. Find
a. the heat input
b. the heat exhaust
c. the efficiency of the engine [a. 975 cal; b. 825 cal; c. 15.4 percent]
19. A cylinder contains air at a pressure of 30 N/cm2. The original volume at 27øC is
4.0 liters. The air is carried through the following processes:
a. heating at constant pressure to 227øC
b. cooling at constant volume to -23øC
c. cooling at constant pressure to -123øC
d. heating at constant volume to 27øC
Show these on a PV diagram, and give the coordinates at the end of each process.
Calculate the net work done by the gas. [a. 30 N/cm2, 4 liters; b. 30 N/cm2, 6
2/3 liters; c. 15 N/cm2, 6 2/3 liters; d. 15 N/cm2, 4 liters; W = 400 J]
20. Use the data given in problem 19.
a. How many moles of air were in the cylinder?
b. Find the heat input during heating at constant pressure to 227øC and heating at
constant volume to 27øC given that cp = 7.00 cal/mole- øC and cv = 5.00 cal/moleø
C.
c. Find the heat liberated during cooling at constant volume to -23øC and cooling
at constant pressure to -123øC.
d. What is the efficiency of this device as a heat engine?
e. What would be the efficiency of a Carnot engine operating in the same
temperature range? [a. 0.48 mole; b. Qp = 672 cal, Qv = 360 cal; c. Qv = 600 cal, Qp
= -336 cal; d. 9 percent; e. 70 percent]
21. Assume that you had a Carnot engine working between the same temperature as
an automobile motor. Estimate the temperatures. What efficiency did you get?
Physics Including Human Applications
279
22. Given the following data from five actual air conditioners, rank them in the order
of efficiency (1 British Thermal Unit (BTU) = 1054.8 joules).
a. cooling capacity = 24,000 BTU, power = 3600 watts
b. cooling capacity = 21,000 BTU, power = 2800 watts
c. cooling capacity = 13,000 BTU, power = 1380 watts
d. cooling capacity = 10,000 BTU, power = 1375 watts
e. cooling capacity = 7,800 BTU, power = 850 watts [c, e, b, d, a]
23. A racing cyclist is capable of sustaining a power output of 300 watts for extended
periods. The associated rate of change of internal energy is measured to be 1400
watts. Find the rate of heat production and the mechanical efficiency (defined as
efficiency =Pw/Pu where Pw = power of work or output power and Pu = power of
internal energy or input power). [1100 watts, 21.4 percent]
24. Given the graph in Figure 12.13 for the blood pressure
during heart contraction (systole), find the rate of work
done by the heart beating 72 beats per minute. P2 = 140
mm (systolic) Hg; P1 = 90 mm (diastolic) Hg; V2 - V1 =
80 cm3. What percentage of a metabolic rate of 85
kcal/hr is used for this heart action? [14.7 watts, 15
percent]
25. The enthalpy of a system is defined as H = U + PV. Thus, ΔH = ΔU +PΔV + VΔP.
For constant pressure processes most common for chemical reactions show that
ΔH is the heat energy evolved or absorbed.
26. Given the three double containers in Figure 12.14 filled with water at the
temperatures shown, which system has the greatest entropy? What conclusion
can you make about the entropy and the enthalpy? (See the definition of
enthalpy in problem 25.)
27. The Gibb's free energy is defined as G = H - TS. Show that the change in the
Gibb's free energy is a measure of the energy available for work in an isothermal
and isobaric process. (Note that TΔS is the energy that goes into increasing
disorder and measures the unavailable energy in the process.)
Footnote
1) A mole of any substance is the amount of that substance that contains 6.02 x 1023
molecules and has a mass in grams equal to the molecular of the substance in
atomic mass units.