Senior High School
General Physics 1
Module 16:
Gas Laws
LU_General Physics I_Module16
AIRs - LM
STEM – GENERAL PHYSICS 1
Module 16: Gas Laws
Second Edition, 2021
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LU_General Physics I_Module16
Senior High School
General Physics 1
Module 16:
Gas Laws
LU_General Physics I_Module16
Introductory Message
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LU_General Physics I_Module16
Target
An ideal gas is an idealized model for real gases that have
sufficiently low densities. The condition of low density means that the molecules of
the gas are so far apart that they do not interact (except during effectively elastic
collision).
The concepts of work, heat, and the internal energy of a system are related by
the first law of thermodynamics, which is an expression of the law of conservation of
energy.
A person’s body contains internal energy. When this person undergoes
rigorous exercises, part of this energy is used for the workout, and part is lost in the
form of heat carried off by evaporating sweat.
Ice cream melts when left out on a warm day. During hot days, a cold can of
soft drink warms up. These two never become colder when left in a hot environment,
because heat always flows spontaneously from hot to cold, and never from cold to
hot. The spontaneous flow of heat can be explained by the most profound law in all
of science, the second law of thermodynamics.
At the end of this lesson you are expected to:
1. Enumerate the properties of an ideal gas. STEM_GP12GLTIIh-57
2. Solve problems involving gas equations in contexts such as, but not limited
to,
the
design
of
metal
containers
for
compressed
gases.
STEM_GP12GLTIIh-58
3. Interpret PV diagrams of a thermodynamic process. STEM_GP12GLTIIh60
4. Compute the work done by a gas using dW=PdV. STEM_GP12GLTIIh-61
5. State the relationship between changes internal energy, work done and
thermal energy supplied through the First Law of Thermodynamics.
STEM_GP12GLTIIh-62
6. Differentiate the following thermodynamic processes and show them on a
PV diagram: isochoric, isobaric, isothermal, adiabatic, and cyclic.
STEM_GP12GLTIIh-63
7. Calculate the efficiency of a heat engine. STEM_GP12GLTIIi-67
8. Describe reversible and irreversible processes. STEM_GP12GLTIIi-68
9. Explain how entropy is a measure of a disorder. STEM_GP12GLTIIi-69
10. State the 2nd Law of Thermodynamics. STEM_GP12GLTIIi-70
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Jumpstart
Before we move on to the concepts of ideal gas law, which relates the pressure,
volume, and temperature of an ideal gas in one compact equation, we must review
the history of gas laws. Three famous names, in particular are associated with gas
laws, those being Robert Boyle, Jacques Charles, and J.L. Gay-Lussac.
Direction: Below are equations of three gas laws. Write the gas law that corresponds
to the equation in column 1(Equation of Gas Law) to column 2(Gas Law).
Equation of Gas law
Gas law
P1V1 = P2V2
V1 / T1 = V2 / T2
P1 / T1 = P2 / T2
Discover
Some concepts are related to chemistry, but which are equally relevant to
physics. The mole and Avogadro's number are included in the list. A mole is a
number like a dozen, but is much bigger. A dozen means 12; a mole means 6.022 x
1023. That number, 6.022 x 1023, is known as Avogadro's number.
A mole is better used to elements. A mole of tin and a mole of silicon both have
6.022 x 1023 atoms. The mole of tin has more mass, though, because a mole of an
element has a mass in grams equal to the atomic mass listed in the periodic table of
elements. A mole of silicon, then, has a mass of 28.09 g while a mole of tin has a
mass of 118.7 g.
A mole of an element has a mass conveniently measured in grams. A single
atom, on the other hand, has a mass that is just a small fraction of a gram. A single
atom of tin, for example, has a mass of:
118.7 g / 6.022 x 10
-23
g = 1.97 x 1024 g
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Note that this is the mass of an average atom of tin, accounting for different
isotopes.
The mass of individual atoms is more conveniently measured in atomic mass
units (u). 1 u = 1.66 x 10-24 g = 1.66 x 10-27 kg. For tin, then, which has an atomic
mass of 118.7, one mole has a mass of 118.7 g and an average atom has a mass of
118.7 u.
Gas Laws
The relationship between the volume of a gas(fixed amount), at a constant
temperature, is known as Boyle’s law. The pressure and volume are inversely
proportional to one another. In other words:
Boyle's law: P1V1 = P2V2.
Charles’ law shows a direct relationship between the volume of a gas and the
temperature of a gas when the temperature is given in the Kelvin scale. It is the
pressure that is kept constant. The volume is proportional to the temperature. This
can be expressed as:
Charles' law: V1 / T1 = V2 / T2
The pressure – temperature relationship shows a direct relationship between
the pressure of a gas and its temperature, when the temperature is given in the kelvin
scale, Gay- Lussac law. When the volume is kept constant, it is the pressure of the
gas that is proportional to temperature:
Gay-Lussac's law : P1 / T1 = P2 / T2
All of the above laws are combined in the ideal gas law. But what is an ideal gas?
An ideal gas has several properties; real gases often exhibit behavior very close
to ideal. The properties of an ideal gas are:
1. An ideal gas consists of a large number of identical molecules.
2. The volume occupied by the molecules themselves is negligible compared to
the volume occupied by the gas.
3. The molecules obey Newton's laws of motion, and they move in random
motion.
4. The molecules experience forces only during collisions; any collisions are
completely elastic and take a negligible amount of time.
The behavior of the gas molecules can be explained by the Kinetic Molecular
Theory. It states that:
a. Gases are composed of molecules. The distances from molecule to the molecule
are far greater than the molecules’ dimensions. These molecules can be considered
as spherical bodies which possess negligible mass and volume
b. Gas molecules are always in constant random motion and they frequently collide
with each other and with the walls of the container. Collisions among molecules are
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perfectly elastic, that is, energy may transfer from molecule to molecule as the result
of a collision, but the total energy of all the molecules in the system remains the
same/constant.
c. There is a negligible attractive or repulsive force between or among gas molecules.
d. Movement of gas molecules is affected by temperature. The average kinetic of the
molecules is directly related to the temperature of the gas
Ideal Gas Law
An ideal gas is an idealized model of real gases; real gases follow ideal gas
behavior if their density is low enough that the gas molecules don't interact much,
and when they do interact they undergo elastic collisions, with no loss of kinetic
energy.
The behavior of an ideal gas, that is, the relationship of pressure (P), volume
(V), and temperature (T), can be summarized in the ideal gas law:
Ideal gas law : PV = nRT
where n is the number of moles of gas, and R = 8.31 J / (mol K) is known as the
universal gas constant.
An alternate way to express the ideal gas law is in terms of N, the number of
molecules, rather than n, the number of moles. N is simply n multiplied by
Avogadro's number, so the ideal gas law can be written as:
Ideal gas law:
PV = NkT
Where k (Boltzmann’s constant) =R/NA=1.38 x10-23J/K
Sample problem: What is the volume of a container that can hold 0.50 mole of gas
at 25.0°C and 1.25atm?
The given are:
Pressure: 1.25 atm
Temperature: 25.0°C + 273 = 298 K
No. of moles: 0.50 mole
We are asked to calculate the volume so let’s substitute the given values to this
equation:
PV
V
= nRT
= nRT
P
= (0.50 mole) (0.0821 L atm/mol. K) (298K)
1.25 atm
= 9.8L
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First Law of Thermodynamics
Suppose that a system gains heat Q and that this is the only effect occurring.
The internal energy of the system increases from an initial value to a final value, ∆U
= Uf –Ui = Q. That is consistent with the law of conservation of energy. Work can also
change the internal energy of a system. If a system does work W on its surroundings
and there is no heat flow, energy conservation indicates that the internal energy of
the system decreases from initial to final, ∆U = U f –Ui = ─ W. A system can gain or
lose energy simultaneously in the form of heat Q and work W. The change in internal
energy due to both factors is given by the equation:
∆U = Uf - Ui = Q – W
Thus, the first law of thermodynamics is just the conservation of energy
principle applied to heat, work, and the change in internal energy.
Q is positive when the system gains heat and negative when it loses heat.
W is positive when work is done by the system and negative when work is done on
the system.
So positive Q adds energy to the system and positive W takes energy from the
system. Thus ΔU=Q−W. Note also that if more heat transfer into the system occurs
than work done, the difference is stored as internal energy. Heat engines are a good
example of this—heat transfer into them takes place so that they can do work.
PV Diagram
Consider a gas sealed in a container with a tightly fitting yet movable piston
as seen below. We can do work on the gas by pressing the piston downward, and we
can heat up the gas by placing the container over a flame or submerging it in a bath
of boiling water. When we subject the gas to these thermodynamics processes, the
pressure and volume of the gas can change.
Figure 2
https://www.khanacademy.org/science/physics/thermodynamics/laws-ofthermodynamics/a/what-are-pv-diagrams
A convenient way to visualize these changes in the pressure and volume is by
using a Pressure-Volume diagram or PV diagram for short. Each point on a PV
diagram corresponds to a different state of the gas. The pressure is given on the
vertical axis and the volume is given on the horizontal axis.
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Every point on a PV diagram represents a different state for the gas (one for
every possible volume and pressure). As gas goes through a thermodynamics
process, the state of the gas will shift around in the PV diagram, tracing out a path
as it moves (as shown in the figure below).
Figure 3
https://www.khanacademy.org/science/physics/thermodynamics/laws-ofthermodynamics/a/what-are-pv-diagrams
Being able to decode the information shown in a PV diagram allows us to make
statements about the change in internal energy ΔU, heat transferred Q, and work
done W on a gas. In the sections below, we'll explain how to decipher the hidden
information contained in a PV diagram.
THERMAL PROCESSES
A system can interact with its surroundings in many ways, and the
heat and work that come into play always obey the first law of
thermodynamics. There are four common thermal processes. In each case,
the process is assumed to be quasi-static, which means that it occurs
slowly enough that a uniform pressure and temperature exist throughout
all regions of the systems at all times.
An Isobaric process is a thermodynamic process in which the pressure stays
constant: ΔP = 0. The heat transferred to the system does work but also changes the
internal energy of the system.
An everyday example of an isobaric process is boiling water in an
open container. When heat energy is supplied to the water, the
temperature rises and turns into steam. The steam has a higher
temperature and occupies a greater volume, however, the pressure remains
constant. From the start, the pressure is equal to atmospheric pressure.
PV graph is a horizontal line.
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Sample problem
1. The PV diagram below shows an ideal gas that undergoes an
isobaric process. Calculate the work done by the gas in the process
AB.
Known:
Pressure (P) = 5 x 105 N/m 2
Initial volume (V i ) = 2 m 3
Final volume (V f ) = 6 m 3
Wanted: Work (W)
Solution:
W = P (V f – V i )
W = (5 x 10 5 )(6 – 2) = (5 x 10 5 ) (4)
W = 20 x 10 5 = 2 x 10 6 Joule
A thermodynamic process taking place at constant volume is known as the
isochoric process. It is also sometimes called an isometric process or constantvolume process. The term isochoric has been derived from the Greek words “iso”
meaning “constant” or “equal” and “choric” meaning “space” or “volume.”
An isochoric process is one for which,
Vf = Vi(∆V =0, dV = O)
In such a process, the work done is zero (since dW = P dV = 0 when V =
constant). Hence from the first law of thermodynamics
dQ = dU (isochoric process)
The total heat supplied or rejected is also equal to the increase or decrease in
the internal energy of the system.
PV diagram is a vertical line.
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Sample problem
1. PV diagram below shows an ideal gas undergoes an isochoric process. Calculate
the work done by the gas in the process AB.
Solution:
Process AB is an isochoric process (constant volume). The volume is constant so that
no work is done by the gas.
dW = PdV
dW = 3atm (0m3) = 0
The "isothermal process", is a thermodynamic process in which the
temperature of a system remains constant. The transfer of heat into or out of the
system happens so slowly that thermal equilibrium is maintained. "Thermal" is a
term that describes the heat of a system. "Iso" means "equal", so "isothermal" means
"equal heat", which is what defines thermal equilibrium.
During an isothermal process, there is a change in internal energy, heat
energy, and work, even though the temperature remains the same. Something in the
system works to maintain that equal temperature. One simple ideal example is the
Carnot Cycle, which describes how a heat engine works by supplying heat to gas. As
a result, the gas expands in a cylinder, and that pushes a piston to do some work.
The heat or gas has to then be pushed out of the cylinder (or dumped) so that the
next heat/expansion cycle can take place. This is what happens inside a car engine,
for example. If this cycle is completely efficient, the process is isothermal because
the temperature is kept constant while pressure changes.
To understand the basics of the isothermal process, consider the action of
gases in a system. The internal energy of an ideal gas depends solely on the
temperature, so the change in internal energy during an isothermal process for
an ideal gas is also 0. In such a system, all heat added to a system (of gas) performs
work to maintain the isothermal process, as long as the pressure remains constant.
Essentially, when considering an ideal gas, work done on the system to maintain the
temperature means that the volume of the gas must decrease as the pressure on the
system increases.
PV diagram is a rectangular hyperbola.
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Sample problem
1. PV diagram below shows an ideal gas that undergoes an isothermal process.
Calculate the work done by the gas in the process AB.
Solution
Work done by a gas is equal to the area under the PV curve
AB = triangle area + rectangle area
W = [½ (8 x 105–4 x 105)(3-1)] + [4 x 105 (3-1)]
W = [½ (4 x 105)(2)] + [4 x 105 (2)]
W = [4 x 105] + [8 x 105]
W = 12 x 105 Joule
The work done by the gas in the process AB = 1.2 x 106 Joule
An adiabatic process is defined as the thermodynamic process in which there
is no exchange of heat from the system to its sorroundings neither during expansion
nor during compression.
PV diagram is a steep hyperbola.
The adiabatic process can be either reversible or irreversible. Following are the
essential conditions for the adiabatic process to take place:


The system must be perfectly insulated from the surrounding.
The process must be carried out quickly so that there is a sufficient amount
of time for heat transfer to take place.
Figure 4
https://byjus.com/physics/adiabatic-process/
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Following is the adiabatic process equation:
PVγ = constant
Where,



P is the pressure of the system
V is the volume of the system
γ is the adiabatic index and is defined as the ratio of heat capacity at constant
pressure Cp to heat capacity at constant volume Cv
Sample problem
We often have the experience of pumping air into bicycle tires, using a hand
pump. Consider the air inside the pump as a thermodynamic system having volume
V at atmospheric pressure and room temperature, 27°C. Assume that the nozzle of
the tire is blocked and you push the pump to a volume 1/4 of V. Calculate the final
temperature of air in the pump? (For air , since the nozzle is blocked air will not flow
into tyre and it can be treated as an adiabatic compression).
Solution:
Here, the process is an adiabatic compression. The volume is given and temperature
is to be found.
TiViγ-1 = TfVfγ-1.
Ti = 300 K (273+27°C = 300 K)
T2 ≈ 522 K or 2490C
This temperature is higher than the boiling point of water. So it is very
dangerous to touch the nozzle of blocked pump when you pump air.
Cyclic process is a thermodynamic process in which the thermodynamic
system returns to its initial state after undergoing a series of changes. Since the
system comes back to the initial state, the change in the internal energy is zero. In
the cyclic process, heat can flow into ystem and heat flow out of the system. From
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the first law of thermodynamics, the net heat transferred to the system is equal to
the work one by the gas.
Q
net
= Qin – Qout = W
(for a cyclic process)
In the PV diagram, the cyclic process is represented by a closed curve.
Let the gas undergo a cyclic process in which it returns to the initial stage after an
expansion and compression as shown in Figure 5.
Figure 5: PV diagram for a cyclic process
Let Wi be the work done by the gas during expansion from volume V i to volume Vf. It
is equal to the area under the graph CBA as shown in Figure 6.
Figure 6: W for path ABC
Figure 7: W for path ADC
Let Wf be the work done on the gas during compression from volume V f to volume Vi.
It is equal to the area under the graph ADC as shown in Figure 7.
The total work done in this cyclic process = W i – Wf = Green shaded area inside the
loop, as shown in Figure 8.
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Figure 8: Net work done in a cyclic process
Thus the net work done during the cyclic process shown above is not zero. In
general, the net work done can be positive or negative. If the net work done is positive,
then work done by the system is greater than the work done on the system. If the
net work done is negative then the work done by the system is less than the work
done on the system.
Sample problem
1. The PV diagrams for a thermodynamical system are given in the figure below.
Calculate the total work done in each of the cyclic processes shown.
Solution:
In case (a) the closed curve is anticlockwise. So the net work done is negative,
implying that the work done on the system is greater than the work done by the
system. The area under the curve BC will give work done on the gas (isobaric
compression) and the area under the curve DA (work done by the system) will give
the total work done by the system.
Area under the curve BC = Area of rectangle BC12 = 1 × 4= − 4J
Area under the curve DA = 1 × 2= + 2J
Net work done in cyclic process = −4 + 2= −2 J
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In case (b) the closed curve is clockwise. So the net work done is positive,
implying that the work done on the system is less than the work done by the system.
The area under the curve BC will give work done on the gas (isobaric compression)
and area under the curve AB will give the total work done by the system.
Area under the curve AB = rectangle area+ triangle area = (1×2) + 1/ 2 × 1×2 = +3J
Area under the curve BC = rectangle area = 1 × 2 = − 2J
Network has done n the cyclic process = 1 J, which is positive.
In case (c) the closed curve is anticlockwise. So the net work done is negative,
implying that the work done on the system is greater than work done by the system.
The area under the curve AB will give the work done on the gas (isobaric compression)
and the area under the curve CA (work done by the system) will give the total work
done by the system.
The area under the curve AB =Rectangle of area = 4 × 1 = - 4J
The area under the curve CA = Rectangle area + triangle area = (1×2) + 1/2 × 1×2 =
+3J The total work in the cyclic process = -1 J. It is negative.
Second Law of Thermodynamics
The second law of thermodynamics states that heat flows spontaneously from a
substance at a higher temperature to a substance at a lower temperature and does
not flow spontaneously in the reverse direction.
Both the first and second laws of thermodynamics are needed by many of
important devices because they depend on heat and work in their operation. An
example is the automobile engine, which is a type of heat engine that uses heat to
produce work. Heat engines have three essential features:
1. Heat is supplied to the engine at a relatively high input temperature from
a place called the hot reservoir.
2. Part of the input heat is used to perform work by the working substance of
the engine, which is the material within the engine that does the work (e.g.
the gasoline – air mixture in an automobile engine).
3. The remainder of the input heat is rejected to a place called the cold
reservoir, which has a temperature lower than the input temperature.
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W
Figure 1: Schematic diagram of a standard heat engine.
https://aether.lbl.gov/www/classes/p10/heat-engine.html
The symbol Q1 refers to the magnitude of the input heat, and subscript 1
indicates the hot reservoir. The symbol Q2 refers to the magnitude of the rejected
heat, and subscript 2 indicates the cold reservoir. The symbol W stands for the
magnitude of the work done. These three symbols refer to magnitudes only, without
reference to algebraic signs. Therefore, when these symbols appear in an equation,
they do not have negative values assigned to them.
For a heat engine to be highly efficient, it must produce a relatively high
amount of work from a small input heat as possible. Therefore, the efficiency e of a
heat engine is defined as the ratio of the work W done by the engine to the input heat
Q1:
e = Work done = W
Input heat
Q1
Equation 1
If all of the heat input Q1 were converted into work, the engine would have an
efficiency of 1.00, Since W = Q1. Such an engine would be 100 % efficient. Efficiencies
are often given in percentages obtained by multiplying the ratio W/Q 1 x 100. So, an
efficiency of 80 % would mean that a value of 0.80 is used for the efficiency in
equation 1.
An engine like any device should obey the conservation of energy principle.
Some of the engine’s heat input Q1 is converted into work W, and the remaining Q2
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is rejected to the cold reservoir. If there are no losses in the engine the conservation
of energy principle requires that
Q1 = W + Q2
Equation 2
Using this equation for W and substituting the result into equation 1 gives us
the following alternative expression for the efficiency e of a heat engine:
e = Q1 – Q2 = 1 – Q2
Q1
Q1
Equation 3
Sample problem:
An automobile engine has an efficiency of 25% and produces 2810 J of work. How
much heat is rejected by the engine?
From equation 2: Q1 = W + Q2 but we need a value for the heat input Q1
Finding for the value of Q1, from the equation of efficiency (equation 1), we find that
Q1 = W/e, substituting this result into equation 2, the rejected heat is
Q2 = Q1 – W = W – W = (2810J) (1/ 0.25 – 1) = 8430 J (amount of heat that is not
e
converted to work)
What should be done to allow a heat engine to operate with maximum
efficiency? Sadi Carnot, a french engineer proposed that a heat engine could have a
maximum efficiency when a reversible process happens within the engine. When
both the system and its environment can be returned into their riginal states before
the process occurred, it is referred to as a reversible process. In a reversible process,
both the system and its environment can be returned to their initial states. So, a
process that involves energy – dissipating mechanism, such as friction, cannot be
reversible because the energy wasted due to friction would alter the system or the
environment or both. Aside from friction, there are other reasons why a process may
not be reversible. The spontaneous heat flow from a hot substance to a cold
substance is irreversible though friction is not present. For heat to flow in a reverse
direction, work must be done. The agent doing such work must be located in the
environment of both the hot and cold substances. In that case, the environment must
change while the heat is moved back from cold to hot and, therefore, it is irreversible
because the system and its environment cannot go back to their original state. All
spontaneous processes are in fact, irreversible.
The idea that the efficiency of a heat engine is a maximum when the engine
operates reversibly is called Carnot’s principle. No real engine operates reversibly,
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however, the idea of a reversible engine provides a useful standard for judging the
performance of real engines. Carnot engine is particularly useful as an idealized
model. An important feature of the Carnot engine is that all input heat Q1 originates
from a hot reservoir at a single temperature T1 and all rejected heat Q2 goes into a
cold reservoir at a single temperature T2. Carnot’s principle implies that the efficiency
of a reversible engine is independent of the working substances of the engine, and
therefore can depend only on the temperatures of the hot and cold reservoirs. Since
e = 1- Q2/Q1 according to equation 3, the ratio Q2/Q1 can only depend on the reservoir
temperatures. So, Lord Kelvin through his observation proposed a thermodynamic
temperature scale. As a result, the ratio of the rejected heat Q 2 to the input heat Q1
is
Q2 = T2
Q1 T1
Equation 4
Where T1 and T2 must be expressed in kelvins.
So, the efficiency of a Carnot engine can be written as:
Efficiency of a
Carnot engine
= 1- T2/ T1
Equation 5
A Carnot engine has the maximum possible efficiency for its operating
conditions because the process occurring within it are reversible. Irreversible
processes such as friction, cause real engines to operate at less than maximum
efficiency, for they reduce the ability to use heat to perform work. In general,
irreversible processes cause us to lose some but not necessarily all, of the ability to
perform work. This partial loss can be expressed in terms of a concept called
entropy.
To introduce the idea of entropy, remember the relation Q2/Q1 = T2/T1 that
applies to a Carnot engine. This can be rearranged as Q2/T2 = Q1/ T1, which focuses
on the heat Q divided by the Kelvin temperature T. The quantity Q/T is called the
change in the entropy ∆S.
∆S = (Q/T)R
T must be in Kelvins
R refers to the word reversible
J/K is the SI unit for entropy
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Entropy like internal energy is a function of the state or condition of the
system. Only the state of a system determines the entropy that a system has.
Therefore, the change in entropy is equal to the entropy of the final state of the system
minus the entropy of the initial state.
What happens to the entropy of a Carnot engine? As the engine operates, the
entropy of the hot reservoir decreases, sine heat Q 1 departs at a Kelvin temperature
T1. The change in the entropy of the hot reservoir is ∆S1 = -Q1/T1, where the minus
1 is needed to indicate a decrease in entropy since the symbol Q1 denotes only the
magnitude of the heat. In contrast, the entropy of the cold reservoir increases by an
amount ∆S2 = +Q2/T2, for the rejected heat enters the cold reservoir at a Kelvin
temperature T2. The total change in entropy is
∆S2 + ∆S1 = Q2/T2 – Q1/T1 = 0
because Q2/T2 = Q1/ T1 according to equation 4
Entropy is related not only to the unavailability of energy to do work—it is also
a measure of disorder. This notion was initially postulated by Ludwig Boltzmann in
the 1800s. For example, melting a block of ice means taking a highly structured and
orderly system of water molecules and converting it into a disorderly liquid in which
molecules have no fixed positions. (See Figure 5.) There is a large increase in entropy
in the process, as seen in the following example.
Figure 2
When ice melts, it becomes more
disordered and less structured. The
systematic arrangement of molecules in a
crystal structure is replaced by a more
random and less orderly movement of
molecules without fixed locations or
orientations.
Its
entropy
increases
because heat transfer occurs into it.
Entropy is a measure of disorder.
https://courses.lumenlearning.c
om/physics/chapter/15-6entropy-and-the-second-law-ofthermodynamics-disorder-andthe-unavailability-of-energy/
Sample problem
A 1200 J of heat flowing spontaneously through a copper rod from a hot reservoir
at 650K to a cold reservoir at 350 K. Determine the amount by which this irreversible
process changes the entropy of the universe, assuming that no other changes occur.
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LU_General Physics I_Module16
Solution: The total entropy change of the universe is the algebraic sum of the entropy
changes for each reservoir:
∆S = - 1200 J
650 K
Entropy lost by
Hot reservoir
+
1200 J
350 K
entropy gained
by cold reservoir
= + 1.6 J/K
Explore
Activity 1
DIRECTION: Solve the following problem using the ideal gas law/ equation.
Show your complete solution in the box below.
1. A sample of liquid acetone is placed in a 25.0 mL flask and vaporized by
heating to 75°C at 1.02 atm. The vapor weighs 5.87 g. Calculate the
number of moles of the acetone.
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LU_General Physics I_Module16
Activity 2
Direction: Compute for the difference of the work done by the gas in process
AB and process CD.? Interpret the PV diagram in process AB and process CD.
Activity 3. Problem-solving. Show your complete solution
1. An engine has an efficiency of 64% and produces 5500 J of work. Determine the
input heat and the rejected heat. Show your solution.
2. On a cold day, 24,500 J of heat leaks out of a house. The inside temperature is
21ºC, and the outside temperature is - 15ºC. What is the increase in the entropy
of the universe that this heat loss produces?
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LU_General Physics I_Module16
Deepen
Activity 1
DIRECTION: Solve the following problem using the ideal gas law/ equation.
Show your complete solution in the box below.
1. Calculate the pressure exerted by a 0.25 mole sulfur hexafluoride in a
steel vessel having a capacity of 1250 mL at 70.0°C.
Activity 2
DIRECTION: Interpret the PV diagram and compute for the work done by the
gas in process ABC.
Activity 3
DIRECTION: Problem-solving: Show your complete solution
1. In doing 16, 600J of work an engine rejects 9700 J of heat. What is the efficiency
of the engine?
2. A process occurs in which the entropy of a system increases by 125 J/K. During
the process, the energy that becomes unavailable for doing work is zero.
A) Is this process reversible or irreversible?
B) Determine the change in the entropy of the surroundings.
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LU_General Physics I_Module16
Gauge
A. DIRECTION: Choose the letter of the best answer. Write your answer on a separate
sheet.
1. A piston-cylinder device initially contains air at 150 kPa and 27°C. At this state,
the volume is 400 liters. The mass of the piston is such that a 350 kPa pressure
is required to move it. The air is now heated until its volume has doubled.
Determine the total heat transferred to the air.
A.747kJ
B. 757kJ
C. 767kJ
D. 777 kJ
2. A piston- cylinder contains 0.5 kg of air at 500 kPa and 500 K. The air expands
in a process so the pressure is linearly decreasing with volume to a final state of
100 kPa and 300 K. Find the work in the process.
A. 56.1 kJ
B. 66.1 kJ
C. 76.1 kJ
D. 86.1 kJ
3. Which of the following variables controls the physical properties of a perfect gas?
A. pressure
B. temperature
C. volume
D. all of the above
4. The unit of temperature in S.I. units is _____.
A. Celsius
B. Fahrenheit
C. Kelvin
D. Rankine
5. A closed system is one in which__________.
A. mass does not cross boundaries of the system, though energy may do so
B. mass crosses the boundary but not the energy
C. neither mass nor energy crosses the boundaries of the system
D. both energy and mass cross the boundaries of the system
6. Temperature of a gas is produced due to
A. its heating value
B. kinetic energy of molecules
C. repulsion of molecules
D. attraction of molecules
7. In an isothermal process, the internal energy of gas molecules
A. increases
B. decreases
C. remains constant
D. shows unpredictable behavior
8. A gas is compressed in a cylinder by a movable piston to a volume one-half its
original volume. During the process, 300kJ heat left the gas and internal energy
remained the same. The work done on gas in Nm will be __________.
A. 300 Nm
B. 300,000 Nm
C. 30 Nm
D. 3000 Nm
9. Isochoric process is one in which __________.
A. free expansion takes place
B. very little mechanical work is done by the system
C. no mechanical work is done by the system
D. all parameters remain constant
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LU_General Physics I_Module16
10. The first law of thermodynamics states that__________.
A. work done by a system is equal to heat transferred by the system
B. total energy of a system during a process remains constant
C. internal energy, work and heat remains constant
D. total energy of a system remains constant
11. First law of thermodynamics furnishes the relationship between __________.
A. heat and work
B. heat, work, and properties of the system
C. various properties of the system
D. various thermodynamic processes
12. Energy can neither be created nor destroyed but can be converted from one form
to another is inferred from__________.
A. zeroth law of thermodynamics
B. first law of thermodynamics
C. second law of thermodynamics
D. basic law of thermodynamics
13. In an isothermal process, the internal energy __________.
A. increases
B. decreases
C. remains constant
D. first increases then decreases
14. Change in internal energy in a closed system is equal to the heat transferred if
the reversible process takes place at constant __________.
A. pressure
B. temperature
C. volume D. internal energy
15. Addition of heat at constant pressure to a gas results in __________.
A. raising its temperature
B. raising its pressure
C. raising its volume
D. raising its temperature and doing external work
B. DIRECTION: Solve the following problems. Show your complete solution.
Required: a. Given data
b. unknown c. equation
d. solution
e. final answer
An ideal gas at 15.5 ºC and a pressure of 1.72 x 10 5 Pa occupies a volume of
2.81 m3.
a. How many moles of gas are present?
b. If the volume is raised to 4.16 m3, and the temperature is raised to 28.2
ºC, what will be the pressure of the gas?
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LU_General Physics I_Module16
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LU_General Physics I_Module16
Jumpstart
1.
Boyle’s law
2.
Charle’s law
3.
Gay-Lussac’s law
Explore
Activity 1
n = PV /RT
n = (1.02 atm) (0.0250 L) / (0.0821 L.atm./mole.K)( 75 + 273.15K)
= 8.92 x 10-4 mole of acetone
Activity 2
A. 1. Isobaric
2. Isochoric
3. Isothermal
4, Adiabatic
5.Cyclic
B.
Known :
Isobaric process AB :
Pressure (P) = 6 atm = 6 x 105 N/m2
Initial volume (V1) = 1 liter = 1 dm3 = 1 x 10-3 m3
Final volume (V2) = 3 liters = 3 dm3 = 3 x 10-3 m3
Isobaric process CD :
Pressure (P) = 4 atm = 4 x 105 N/m2
Initial volume (V1) = 2 liters = 2 dm3 = 2 x 10-3 m3
Final volume(V2) = 5 liters = 5 dm3 = 5 x 10-3 m3
Wanted : Difference of the work is done by the gas in process AB and CD.
Solution :
Work is done by the gas in process AB :
W = P (V2 – V1)
W = (6 x 105)(3 x 10-3 – 1 x 10-3)
W = (6 x 105)(2 x 10-3)
W = 12 x 102 = 1200 Joule
Work is done by the gas in process CD :
W = P (V2 – V1)
W = (4 x 105)(5 x 10-3 – 2 x 10-3)
Activity 3
1. A. 8600 J
2. 11.6 J/K
Answer Key
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LU_General Physics I_Module16
Deepen
Activity 1
In solving Ideal Gas Law related problem, it is important that the given unit for
volume is in liter to cancel the units of the constant (R). If not, it must be converted.
Since 1L = 1000 mL, therefore, 1250 mL = 1.25 L
Then substitute the values to this equation:
P = nRT / V
= (0.25 mole) ( 0.0821 L.atm./mole.K)(70.0 + 273.15 K) / 1.25L
= 0.56 atm
Activity 2
Known :
Pressure 1 (P1) = 6 x 105 Pa = 6 x 105 N/m2
Pressure 2 (P2) = 3 x 105 Pa = 3 x 105 N/m2
Volume 1 (V1) = 2 cm3 = 2 x 10-6 m3
Volume 2 (V2) = 6 cm3 = 6 x 10-6 m3
Wanted : Work done in process ABC.
Solution :
In process AB, the volume is kept constant so that no work is done by the gas.
Work was done by the gas in the process BC.
W = P2 (V2 – V1)
W = (3 x 105)(6 x 10-6 – 2 x 10-6)
W = (3 x 105)(4 x 10-6)
W = 12 x 10-1
W = 1.2 Joule
Work done in the process ABC = work done in the process AB = 1.2 Joule.
Activity 3
1. 0.631
2. a. reversible
b. – 125 J/K
GAUGE
A.
1. C
2. D
3. D
4. C
5. A
6. B
7. C
8. B
9. C
10.D
11.B
12.B
13.C
14.C
15.D
B.
1. 201 mol
2. 1.21 x 105 Pa
References
Science 10 LM
Physics Sixth Edition by Cutnell & Johnson 2004
Physics Sixth Edition by Cutnell & Johnson 2004
Physics Sixth Edition by Cutnell & Johnson 2004
Website:
https://www.google.com/search?ei=0UmZX4vTMaaUmAW44LvICw&q=PV+diagram
+of+an+isobaric+process&oq=PV+diagram+of+an+isobaric+process&gs_lcp=CgZwc3
ktYWIQAzIJCAAQyQMQFhAeMgYIABAWEB46BAgAEEdQ2SBY2SBgjRoAHACeACAAZoBiAGkApIBAzAuMpgBAKABAaoBB2d3cy13aXrIAQjAAQE&sclient
=psy-ab&ved=0ahUKEwjL_5b1i9fsAhUmCqYKHTjwDrkQ4dUDCA0&uact=5
https://physics.gurumuda.net/isochoric-thermodynamics-processes-problemsand-solutions.htm
https://physics.gurumuda.net/isothermal-thermodynamic-processes-problemsand-solutions.htm
http://www.brainkart.com/article/Cyclic-processes-and-PV-diagram-for-a-cyclicprocess_36256/
https://www.khanacademy.org/science/physics/thermodynamics/laws-ofthermodynamics/a/what-are-pv-diagrams
https://aether.lbl.gov/www/classes/p10/heat-engine.html
https://courses.lumenlearning.com/physics/chapter/15-6-entropy-and-thesecond-law-of-thermodynamics-disorder-and-the-unavailability-of-energy/
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LU_General Physics I_Module16
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LU_General Physics I_Module16