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1 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL GOVT MUSLIM SCIENCE DEGREE COLLEGE HYD. 1
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CONTAIN:
1. THERMODYNAMICS.
2. THERMODYNAMIC SYSTEM AND KINDS
3. FIRST LAW OF THERMODYNAMICS.
4. APPLICATIONS OF FIRST LAW OF THEMODYNAMICS.
5. CP - CV = R
6. SECOMD LAW OF THERMODYNAMICS
7. CARNOT ENGINE.
8. ENTROPY
9. EQUATIONS
10. DIMENSIONS
11. SHORT QUESTIONS AND ANSWERS
Thermodynamics
Introduction: Thermodynamics is a branch of physics which deals with the energy and work of a
system. Thermodynamics deals only with the large scale response of a system which we can observe
and measure in experiments. Gases have various properties that we can observe with our senses,
including the gas pressure p, temperature T, mass n, and volume V that contains the gas. Careful,
scientific observation has determined that these variables are related to one another, and the values of
these properties determine the state of the gas. A thermodynamic process, such as heating or
compressing the gas, changes the values of the state variables in a manner which is described by the
laws of thermodynamics. The work done by a gas and the heat transferred to a gas depend on the
beginning and ending states of the gas and on the process used to change the state.
A thermodynamic system is a collection of matter and energy which has a distinct boundary. Matter
or energy (heat) can be or can not be transferred across the system boundary. A thermodynamic
system can be defined as macroscopic region of the universe, often called a physical system.
Closed system:
A closed system is one where energy can cross the boundary, but matter
cannot. eg a sealed test tube.
Isolated system:
An isolated system is one where neither matter nor energy can cross between
the system and the surroundings. The universe itself is an isolated system.
Open system:
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"No cyclic process is possible whose sole result is a flow of heat from a single reservoir and the performance of
equivalent work." Lord Kelvin Quote
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An open system is one where both matter and energy can freely cross from
the system to the surroundings and back. eg an open test tube.
Introduction:
The first law of thermodynamics was formulated by Mayer; a much
clearer formulation was provided by H. von Helmholtz in 1847. The First Law of
Thermodynamics deals the energy is always conserved, it cannot be created nor destroyed. The
energy can be converted from one form into another.
Statement:

A change in the internal energy of a closed thermodynamic system is equal to the
difference between the heat supplied to the system and the amount of work done by the system on
its surroundings.
ΔU
=
ΔQ
ΔW
change in internal energy
heat added to the system
work done by the system
Description:
The first law of thermodynamics provides the existence of a state variable for
a system, the U internal energy, and tells how it changes in thermodynamic
processes. In this law given internal energy of a system to be reached by any
combination of Q heat absorbed or rejected and work. It is important that internal
energy is a variable of state of the system whereas heat and work change the state of
the system.
 The change in the internal energy of a closed thermodynamic system is equal
to the difference between the work done on the system and the amount of heat
energy rejected by the system.
ΔU
=
ΔW

ΔQ
change in internal energy
work done on the system
heat subtracted to the system
If heat flows into a system or the surroundings to do work on it, the internal energy increases and the
Q or W is positive. Conversely, heat flow out of the system or work done by the system will be at
the expense of the internal energy, and will therefore be negative.
There are a number of different thermodynamic processes are as
follows:
1. Isobaric Process - The process in which “Pressure” is kept
constant.
The quantity of gas placed in a closed system by means of a
moveable piston. Weights placed on top of the piston exert a
force F over the cross-section area A, producing a
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F
, which is exactly countered by the pressure of the gas, so that the piston remains
A
stationary. Now suppose that we heat the gas by supplying heat Q ; the gas is allowed to expand,
so the piston will be forced upward by a distance ΔY. Since this motion is opposed by the force, a
quantity of work F ΔY will be done by the gas on the piston. The work done by the system on the
surroundings is negative, so the work is given by: W = F Y
W = P A Y
W = P V
According First law of thermodynamics
 Q =  U + W
 Q =  U + [P V ]
Q =U + P V
2. Isochoric Process – The process in which “Volume” kept constant is also called an isometric
process.
The quantity of gas placed in a closed system by means of with fixed walls. And fixed piston.
A gas enclosed in it, by adding ΔQ heat that will increase the internal energy ΔU.
Due to increase of internal energy, the work done is zero in an
isochoric process, say, ΔW = 0 by the system. According first
law of thermodynamics,
Mathematical Derivation:
U= Q + W
ΔU = Q + 0
ΔU = ΔQ. This equation is called Isochoric equation.
3. Isothermal Process- The process in which “Temperature” is kept constant.
The quantity of gas placed in a closed system by means of a moveable piston. The system is
placed over a hot body reservoir and the pressure on the system is decreased. Due to expansion, the
temperature of the system is decreased but at the same time Q amount of heat is absorbed from the
hot body reservoir and the temperature of the system is again
maintained T = 0 The internal energy of an ideal gas is
proportional to
the temperature, so if the temperature is kept fixed the
internal energy does not change.
- U= - Q + W
0 = - Q + W
Q = W
4. Adiabatic - In an adiabatic process, no heat is added or
removed from the system.
The quantity of gas placed in a closed system by means
pressure P =
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of a moveable piston. The system is placed over an insulator. When a gas expands, it
does work on the surroundings; compression of a gas to
a smaller volume similarly requires that the surroundings
perform work on the gas. If the gas is thermally isolated
from the surroundings, then the process is said to occur
adiabatically. In an adiabatic change,
Q = 0, so the First Law becomes
U= Q + W
The heat capacity at constant pressure of a gas is the amount of heat required to change the
temperature of a unit-mass of gas one temperature degree at constant pressure.
ΔQ p = c p n ΔT
The heat capacity at constant pressure of a
gas is the amount of heat required to change the temperature of a unit-mass of gas one temperature
degree at constant volume.
ΔQv = c v n ΔT
No external work is being done when a gas is heated at constant volume i.e.
gas uses all the heat which is given to it for increasing its internal energy. Hence if temperature of one
mole of a gas is raised through 1oC, the molar heat capacity is given itself at constant volume by
increase in internal energy.
ΔQv = c v n ΔT
But when a gas is heated at constant pressure there will be expansion of gas i.e. increase in volume
take place and some external work will b done. For this some extra heat is required which
ΔQ p = c p n ΔT
should be given to the gas to perform the external work.
ΔQ p > ΔQ v
cp n ΔT > c v n ΔT
Hence the Molar Heat capacity of a gas at constant pressure must be greater than Molar Heat
capacity of a gas at constant volume.
CP > CV
Hence shown
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Mathematical and Descriptive Proof:
Suppose “n” moles gas is heat up while keeping its volume constant. The result of
adding heat to the system is an increase of its T temperature.
ΔQv = c v n ΔT
Here, CV is the molar heat capacity at constant volume, ∆QV is the heat added, and ∆T is the resulting
increase in the temperature of the system. The first law of thermodynamics shows that,
ΔU = ΔQ - ΔW = cv n ΔT - ΔW
Since the volume is kept constant (∆W = 0) we conclude that,
ΔU = c v n ΔT
Again heat is added to the system, the volume is changed such that the gas pressure does not
change. The change in the internal energy of the system is given by first law of thermodynamics,
ΔU = ΔQ - ΔW = c p n ΔT - ΔW
where cp is the molar heat capacity at constant pressure. This expression can be rewritten as,
ΔU + ΔW = cp n ΔT
c v n ΔT + ΔW = c p n ΔT
c v n ΔT + P(V2 -V1 ) = c p n ΔT
c v n ΔT + P V = cp n ΔT
For an ideal gas PV = n R T, we can relate ∆V to ∆T, if we assume a constant pressure
P V = n R T
c v n ΔT + n R T = cp n ΔT
n ΔT  cv + R  = cp n ΔT
 cv
+ R  = cp
Or R=  cp - c v 
we see that Cp ≠ CV. This shows that, difference between molar heat capacity at constant pressure and
c
at constant volume is equal to the universal gas constant. The values for γ are γ = p =1.4 for diatomic
cv
c 5
gasses like air and its major components, and γ = p = for monatomic gasses like the noble gasses.
cv 3
3
5
The formulas for specific heats would reduce in these special cases: Monatomic: c v = R and c p = R
2
2
5
7
and for Diatomic: c v = R and c p = R
2
2
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The 1st Law does not tell us
everything we need to know
about thermodynamic
processes. We need to develop an understanding of the 2nd Law of
Thermodynamics in order to address issues relating to spontaneity,
equilibrium and efficiency. There are two statements of this law:
Clausius statement:
The Clausius statement expresses as follows:
No process is possible whose sole result is the transfer of heat from a body
of lower temperature to a body of higher temperature. Description:
Spontaneously, heat cannot flow from cold regions to hot regions without external work being
performed on the system, for example. In a refrigerator, heat flows from cold to hot, but only when
forced by an external agent, a compressor.
Kelvin statement
The Kelvin statement expresses as follows:
No process is possible in which the sole result is the
absorption of
heat from a reservoir and its complete conversion into work.
Description:
This means it is impossible to extract energy by heat from a
High-temperature energy source and then convert all of the
energy into
work. At least some of the energy must be passed on to heat a
low-temperature
energy sink.
Equivalence of
Kelvin’s and Clausius Statements:
Description:
In order to show the equivalence between two statements, it is
supposed an ideal heat engine which can convert heat energy
absorbed from high temperature reservoir into work without rejecting
heat in to cold body reservoir. Thus Kelvin’s statement is false. Such
ideal heat engine can combine, with an eclectic refrigerator so that the
work produced by the engine, is used by the refrigerator. Now, the
combined system is a heat engine and refrigerator which uses no
external work, violating the Clausius statement of the second law. Thus
we see that, the Kelvin and -Clausius statements are equivalent, and one necessarily implies the
other. Heat engine
Definition:
The engine has minimum heat losses, and friction with maximum efficiency than all other heat
engines, designed by French scientist Sadi Carnot, is called “Carnot Engine”. The carnot heat engine
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(the ideal imaginary heat engine)
Construction:
A Carnot engine consists a cylinder with fixed distinct non- conducting boundaries,
provided frictionless non-conducting piston. An ideal gas is kept init as the working
substance. It operates under the control of two heat reservoirs one of them at high
temperature T1 called source, while the other at lower temperature T2 called sink.
The base of the system is conducting; it works in a cycle called “Carnot cycle”.
Carnot Cycle:
A cycle occurs when a system is taken through a series of different states and
finally to its initial state. In the process of going through this cycle, the system
may perform work on its surroundings. It consists of four basic reversible
processes meaning that the cycle as a whole is also reversible. Such cycle,
is known as “Carnot cycle.
Operational Process:
The Carnot cycle when acting as a heat engine consists of the following steps:
Reversible isothermal expansion of the gas at the "hot" temperature, TH
The heat source at high temperature T1 is contact with the cylinder which lowers the pressure in the
gas by expansion. The temperature remains constant, but the volume increases. The expanding gas
makes the piston work on the surroundings. The gas expansion is propelled by absorption of
quantity Q1 of heat from the high temperature reservoir.
Reversible adiabatic expansion of the gas (isentropic work output).
The piston and cylinder are to be thermally insulated, thus a system can not gain heat. The gas
continues to expand, working on the surroundings. The temperature decreases from TH to TC and the
volume increases as the gas expands to fill the volume. The gas expansion causes it to cool to the
"cold" temperature, TC.
Reversible isothermal compression of the gas at the "cold" temperature, TC. (Isothermal heat
rejection)
Now the surroundings do work on the gas, causing quantity Q2 of heat to flow out of the gas to the
low temperature reservoir. The sink at low temperature TC is brought into contact with the cylinder,
which raises the pressure in the gas. The temperature remains constant, but the volume decreases.
Reversible adiabatic compression of the gas (isentropic work input).
Once again the piston and cylinder are to be thermally insulated. During this step, the surroundings
do work on the gas. The temperature increases TC to TH and the volume decreases as the gas is
compressed and it causing the temperature to rise to TH.
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Efficiency:
For a heat engine, the efficiency is the ratio of useful work performed to the heat
energy consumed from the high-temperature reservoir
Explanation:
Many heat engines converting part of the heat energy into useful work. There are
losses due to friction. Consider the example of a heat engine, which extracts heat
energy Q1 from the combustion
at temperature TH, and which expels some heat, Q2, into the
exhaust at temperature TC. By the conservation of energy, the difference (Q1 – Q2) is the amount of
Out Put
work (mechanical energy) done.
Efficiency =
In put
Useful work
η=
Heat energy expel
Q - Q  
η=  1 2 
 Q1


Q 
η = 1 - 2 
Q1 

Thus the maximum efficiency of an ideal engine is controlled by the source temperature TH, which
should be as high as possible, and the exhaust (sink) temperature TC, which should be as low as
possible. A real engine will be even less efficient because of internal friction, and other factors.

T 
Efficiency always measured in percentage.
η = 1 - C 
TH 

"Entropy" is defined as a measure of unusable energy within a closed or isolated
system (the universe for example). As usable energy decreases and unusable
energy increases, "entropy" increases. Entropy is also a gauge of randomness or chaos within a closed
system. As usable energy is irretrievably lost, disorganization, randomness and chaos increase .Thus
entropy is the change of heat energy per maintained temperature.
Description:
The concept of entropy is defined by the second law of thermodynamics, which states
that the entropy of an isolated system always increases or remains constant. Thus, entropy is also a
measure of the tendency of a process, such as a chemical reaction, to be entropically favored, or to
proceed in a particular direction. It determines that thermal energy always flows spontaneously from
regions of higher temperature to regions of lower temperature, in the form of heat. These processes
reduce the state of order of the initial systems, and therefore entropy is an expression of disorder or
randomness.
Mathematical derivation:
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Suppose a very small quantity of heat Q is added slowly to a system at fixed
maintained temperature T, then the entropy “s” of the system is increased by
ΔQ
, Δs =
T
Its unit is J K-1. The entropy will be decreased by the same amount if the same
quantity of heat were removed at that that same temperature. Entropy, like energy,
is an additive
quantity, but unlike
energy,
entropy is not a
conserved
quantity.
1. Q = U + W
2. Q = U
3. ΔQ = ΔU + P ΔV
5. - ΔU = ΔW
6. - ΔU = -ΔW
7. PΔV = ΔW
9.  cp - c v  = k N A
10. Work done = Q1 - Q2
12.
DIMENSION
 M L2 T -2 
 M L2 T -2 
Molar heat capacity at constant Volume cv and at Pressure cp
Internal energy increases or decreases
UNIT
J
U
J
 M L2 T -2 K 1 
 M L2 T -2 
K
W
Temperature
Second YEAR
13.
 ΔQ 
14. Δs = 
 T 
PHYSICAL QUANTITY & SYMBOL
Heat energy supplied or rejected Q
T
Pressure
8.  cp - c v  = R
 work done 
11. Eff.= 
 100
Q

1

 Q 
Eff.= 1- 2  100
 Q1 
 T 
Eff.= 1 - 2  100
T1 

Work done
4. ΔQ = ΔW
J(mol-1 K-1)
J
K
 M T -2 
P
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 L3 
V
m3
Universal Gas Constant R
J(mol-1 K-1)
Area
 M L2 T -2 K -1 
 L2 
A
m2
Entropy
COME FOR SUCCESS
s
Efficiency

Avogadro’s number NA
 M L2 T -2 K 1 
Dimensionless
Dimensionless
J K-1
no unit
per mol
Q: No.1 What is Perpetual-Motion Machines.
Answer:
Any process will not occur unless it satisfies both the first and the second laws of
thermodynamics. A lot of efforts were spent to create machines which violate the first law or the
second law. This kind of machines is called perpetual-motion machines. Devices that violate the first
law are called perpetual-motion machines of the first kind. Devices that violate the second law are
called perpetual-motion machines of the second kind
Q: No.2 Define First law of thermodynamics in terms of entropy?
Answer
The first law of thermodynamics is also called the Law of Conservation of Energy. This law suggests
that energy can be transferred from one system to another in many forms. Also, it can not be created
or destroyed. Thus, the total amount of energy available in the Universe is constant.
Q: No.3 What are the first and second laws of thermodynamics?
Answer:
For polyatomic gas, First Law of Thermodynamics: Conservation of Energy
Second Law of Thermodynamics: Entropy increases in all natural processes
Q: No.4
Why heating produced adiabatic compression?
Answer:
An adiabatic is a process in which, heat energy neither enters nor leaves the system. When a system is
compressed, the work is to be done on the system. The enclosed gas molecules become more
energetic and heating produced. Such heat energy can’t leave the system. Hence, "heating produced
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adiabatic compression”.
Q: No.5
Why adiabatic process is steeper than isotherm?
Answer:
In the adiabatic process the thermodynamic variables are temperature, volume and
pressure. And in the isothermal process volume and pressure are the variables at
constant temperature. So that, the graph determined in adiabatic is steeper than
isotherm.
Q: No.6
Define second law of thermodynamics in terms of entropy?
Answer:
The second law can be thought of as a qualitative law: the second law says that all natural processes
occur in such a way as to result in an increase in entropy.
To understand this law, it is first necessary to explain the concept of entropy. Entropy means
disorder. Consider the dissolving of a sugar cube in water. The sugar cube itself represents a highly
ordered state in which every sugar particle is arranged in an exact position within the sugar crystal.
The entropy of a sugar cube is low because there is little disorder. But consider what happens when
the sugar cube is dissolved in water. The cube breaks apart, and sugar molecules are dispersed
completely throughout the water. There is no longer any order among the sugar molecules at all. The
entropy of the system has increased because the sugar molecules have become completely
disorganized
Q: No.7
Entropy has often been called as “time’s arrow”. Explain?
Answer
The second law of thermodynamics is that it tells us in which direction processes go. The processes in
which order increases or entropy decreases, this process not happen in our daily life. Hence, entropy
has been called “times arrow”, for it tells in which direction time is going.
Q: No.8
Under what condition, is it possible for the entropy of system to decreases?
Answer:
When heat is removed from the system, the heat energy is negative. Thus, entropy of the system
decreases. The example is ice. When water is cooled at constant temperature, the heat is to be
extracted. Hence, entropy decreases.
Q: No.9
What are some factors that affect the efficiency of automobile engines?
Answer:
The efficiency of automobile engine depends onto the hot body and cold body temperature
T 
reservoirs. If the ratio  C  is greater than efficiency will be smaller or reverse.
 TH 
Q: No.10
Is it possible, according to second law of thermodynamics to construct an
engine that will be free from “thermal pollution”?
Answer:
No, it is not possible to construct an engine, which will be free from, thermal pollution. We can
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construct an engine, which minimize the thermal pollution.
Q: No.11 Can a given amount of mechanical energy is converted completely in
to energy? If so, give an example.
Answer:
Yes, it is possible that a given amount of mechanical energy be converted completely
in to heat energy. The example of such statement is ideal heat engine. According to
the first law of thermodynamics,
Q = U+ W
Q = 0 + W
Or,
Q = W
Q: No.12
What is meant by heat death of universe?
Answer:
The 'heat-death' of the universe is when the universe has reached a state of maximum entropy. This
happens when all available energy (such as from a hot source) has moved to places of less energy
(such as a colder source). Once this has happened, no more work can be extracted from the universe.
Since heat ceases to flow, no more work can be acquired from heat transfer. This same kind of
equilibrium state will also happen with all other forms of energy. Since no more work can be
extracted from the universe at that point, it is effectively dead, especially for the purposes of
humankind.
Q: No.13
Name a process in which volume remains constant?
Answer:
The name of process, in which volume of a system remains constant, is “Isochoric process”.
Q: No.14
Name a process in which heat energy is transferred to or from the system but the
temperature of the system does not change.
Answer:
The name of a process in which heat is transferred to or from the system but the temperature of the
system remains constant is called “isothermal process”.
Q: No.15
Why, does the temperature drop in adiabatic expansion process? Answer:
When a gas in isolated system is to be heated, heat energy doesn’t leave the system. The work is to be
done at the cost of internal energy. So that temperature of the system drops.
Q: No.16 A gas expands adiabatically. Does the gas perform any work? What is the source of
energy needed to do this work?
Answer:
When a gas expands adiabatically. Yes, the gas performs work. The source of energy, to do work is
internal energy.
Q: No.17
A thermos flask contains milk. The flask is shaken rapidly. Consider the milk as
the system. Does the temperature of milk rise? Has heat been added to it? Has work been done on
it? Has its internal energy changed?
Answer:
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A thermos flask contains milk. The milk is shaken rapidly. Consider the milk as the
system. Yes, the temperature of milk rises. No, heat has not been added to it. Yes,
work has to be done on it. Yes, internal energy of the system increases.
Q: No.18
A gas is allowed to expand at constant temperature. It does
external work. Does the internal energy of the gas change during this process? If
not, what is the source of energy needed to do this work?
Answer:
A gas is allowed to expand at constant temperature, cording to the application say isothermal process
of first law of thermodynamics. It does external work during the process.
No, the temperature of the gas does not change during this process.
The source of energy, needed to do this work is heat reservoir.
We know that,
Q = U +W
Q = 0 + W
Or,
Q = W
Q: No. 19
Comment the statement the heat engine covert disorder motion in to order motion.
Answer:
We know that heat engine operates in a cycle, it absorbs heat from hot body, coverts some of it into
work and part of energy rejects to cold body. When the heat engine works, it receives the heat from
hot body, which appears in disordered motion. This heat energy is converted in to mechanical work
by heat engine. It means, disordered motion is converted in to ordered motion. As we have
knowledge that heat increases the, disorder of molecules. When heat coverts work, the disordered
motion also converts in to ordered motion.
Q: No. 20
Why more work is to be done, when a gas is heated at constant pressure than at
constant volume?
Answer:
When gas is to be heated at constant volume then all the heat energy supplied to the system is to be
used to increase internal energy. Hence no work is to be done. Q = U + W Or,
Q = U
(say isochoric process)
And gas is to be heated at constant pressure then part of heat energy is used to increase the internal
energy and reaming part is used to perform external work. Q = U + W
Q = U + PV
(say isobaric process)
This show that more energy is to be required at constant pressure than at constant volume.
Q: No.21
Write the formula of entropy?
Answer
Q
S =
T
Q: No.22
Give an example of a process in which no heat transferred to or from the
system but the temperature of the system changes.
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Answer:
A process in the thermodynamics is adiabatic process, in which no heat enters
or leaves the system. During this process if we decrease the pressure on the system,
the volume increases i.e. expansion occurs and work is done by the internal energy
of the system. Due to decrease in internal energy hence, temperature falls.
Q: No.23
Is it possible to construct a heat engine that will not expel heat into
atmosphere?
Answer:
According second law of thermodynamics, when heat is converted into work, partially heat is
rejected to cold body. For a heat engine, surrounding behaves as a cold body. Hence, it is not possible
to construct an engine that will not heat into atmosphere.
Q: No.24
Is it possible to convert internal energy into mechanical energy?
Answer:
According to Kelvin’s statement of second law of thermodynamics,” We can not construct an
engine which absorbs heat from source and convert it completely into useful work without a cold
body”. That is why a given amount of heat can be converted completely into work.
Q: No. 25
Does entropy of the system increases or decreases due to friction?
Answer:
When work is done against friction, heat is produced which disturbs the molecular motion say
disorder ness is created. Hence, entropy increases.
Q: No. 26
An ideal reversible heat engine has 100% efficiency?
Answer:
No an ideal heat engine has not 100% efficiency, but it has highest efficiency than all other heat
engine.
Q: No. 27
Which one of the following process is irreversible?
a) Slow compression of an elastic spring b) low compression of gas c) A chemical explosion
d) Slow evaporation of a substance in an isolated vessel.
Answer:
A chemical explosion is the irreversible process.
Q: No. 28 An adiabatic change in which a) Boyle’s law inapplicable b) Pressure and volume
changes c) No heat is added to or taken out of a system d) No change in temperature takes place.
Answer:
An adiabatic change is the one in which in which, no heat is added to or taken out of a system.
Q: No. 29 Give an example of a natural process that involves increase in entropy.
Answer:
Consider two glass of water full of water at 273 K and another glass that has water 373K.The
molecules of water, in the glass are in order with their constant average velocities. Such water are
mixed in a container thus disorder is created in the water molecules and will start to move from
water at temperature 373K to 273 K.A engine can be operated due to this heat flow from hot water to
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cold water and hence work will be available . This disorder increases due to mixing
and difference of temperature decreases. When mixture comes to the thermal
equilibrium state, disorder i.e. entropy will be maximum.
Q: No. 30 . What you mean by reservoir?
Answer:
The thermodynamic definition of a reservoir is something large enough that it can
transfer heat into or out of a system without changing temperature
Q: No. 31
Why gases have two molar heat capacities?
Answer:
For gases, it is necessary to specify the conditions under which the change of temperature takes
place, since a change of temperature will also produce large changes in pressure and volume.
Thus, gas is heated at constant volume by providing fixed piston or at constant pressure by
providing free piston. Hence, gases have two molar specific heat capacities.
Q: No. 32
Show that, cp> cv
Answer:
The gas is heated by absorbing ΔQ v heat energy at constant volume. The heat energy
absorbed is completely used to increase the temperature ΔT . In this system no work is to be done.
Thus we know that,
ΔQv = c v n ΔT
The gas is heated by absorbing ΔQ p heat energy at constant pressure in a closed system. The
energy supplied is partially used to increase the temperature ΔT and partially used to perform work
by the system ΔW .
ΔQ p = c p n ΔT
This shows that, ΔQ p > ΔQ v .We know that, cp n ΔT > c v n ΔT
cp n ΔT > cv n ΔT
cp > c v
Q: No. 33
What is Heat Engine?
Answer:
A device that converts heat into work is called a heat engine. The engine generally has a
working substance (eg. gas or steam) that under goes thermodynamic change. The engine generally
operates in a cycle which implies
 U = 0 (for one cycle)
All heat engines can be characterized by the following: (1) They receive heat from a hightemperature source. (2)They convert part of this heat to work. (3)They reject the remaining waste heat
to a low-temperature sink. (4)They operate on a cycle.
Q: No. 34
What is Heat Pump
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Answer:
Heat Pump is cyclically operating device which absorbs energy form a low
temperature reservoir and reject energy as heat to a high temperature reservoir
when work is performed on the device. Its objective is to reject energy as heat to a
high temperature body (space heating in winter). The atmosphere acts as the low
temperature reservoir.
T 
How the ratio of temperature of cold body and hot body reservoirs  C  affect the
 TH 
efficiency of heat engine?
Answer:
T 
T 
If the  C  ratio is greater than efficiency will be smaller or  C  ratio is smaller than
 TH 
 TH 
Q: No. 35
efficiency will be greater. Because,
Second YEAR

T 
Efficiency =  1 - C  is for the heat engine
TH 
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