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“ Physics will never be the same again ”
Theory Notes
On
“ ENTROPY ”
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YT : Indian school of physics
NITIN SACHAN
ENTROPY (En → Inside; Trope → Transformation):
It is denoted by S in the name of Sadi Carnot.
→ The second law leads to the concept of entropy.
→ It is the property of a state of a system.
→ For example, an engine will lose some energy to the surrounding. This puzzle of loss leads to entropy.
From the Carnot engine (reversible engine)
Heat
Engine
(Reversible)
Q1 T1
=
Q 2 T2
Q1 Q 2
=
T1 T2
Q1  Q2 
+−  = 0
T1  T2 
w.r.t heat engine, the algebraic sum of
Q
a cyclic process for a reversible heat engine is zero.
T
Mathematically,

The ratio
Q R
=0
T
Q
= entropy (S)
T
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YT : Indian school of physics
NITIN SACHAN
It can also be proved that a reversible cyclic engine operates in a cycle that does not take the heat at a
constant temperature. The source is also not at a constant temperature. Because if the heat engine
(system) runs not at a constant temperature, the surrounding temperature also varies.
In this case, we can derive the same equation as above by dividing the given cycle into an infinite number of
Carnot cycles, as shown in the figure.
P
V
Let us consider a thermodynamical plane P – V. In this case; we can divide this thermodynamically reversible
cycle into an infinite number of small elemental Carnot's cycles consisting of reversible isothermal and
reversible adiabatic processes, as shown in the figure.
For each elemental cycle
Q1 Q 2
=
T1
T2
Similarly
Q1 Q2
=
T1
T2
Q1 Q2
=
T1
T2
 By
adding all the equations, we get
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YT : Indian school of physics
NITIN SACHAN

Q1
Q 2
=
T1
T2
 Q1 Q2 
−
= 0
T2 
cycle  T1

Hence
 Q1
 Q 2
+−
cycle  T1
 T2
 

Q R
=
T


  = 0

Q R
= 0 (for an infinite number of cycles
T
→ 
)
Now for a reversible heat engine

Q R
= 0 …. (*)
T
Some portions QR are positive, and some QR are negative and
Q R
= exact differential. So, it is defined as a new quantity S where
T

 dS = 0
b
dQ R
=  dS = Sb − Sa
T
a
S=
QR
= entropy of the system
T
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YT : Indian school of physics
NITIN SACHAN
Clausius Inequality
Entropy for a reversible process:
P
C
2
b
a
1
V
Along the path
1 − a − 2 − b − 1 (Re versible) , we can write

Q
=0 
T
 Q 
 Q  
 +
 =0
T 1a 2  T 2b1 
 
1 − a − 2 − c −1
(Reversible)
 Q 
 Q 
 + 
 =0
T 1a 2
 T 2c1
 

… (1)
Q
Q
+
=
T 1a 2
T 2c1

… (2)
Q
Q
+
T 1a 2
T 2c1
 Q 
 Q 
 + 
 =0
T 1b2
 T 2c1
 
 Q 
 = 0 (for a reversible cycle)
T 
 
Q
Property  
 T  Re v
= Entropy
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YT : Indian school of physics
NITIN SACHAN
Entropy for Irreversible Process
P
c
2
b
a
1
V
For the cycle 1 − a − 2 − b − 1 (Re versible process)
 Q 
 Q  
 +
 =0
T 1a 2  T 2b1 
 
… (1)
For the cycle 1 − a − 2 − c − 1 (Irr e versible process)
 Q 
 Q 
− 
 + 
 0
 T 2b1
 T 2c1
….(2)
From (1) and (2)
 Q 
 Q 
 T    T 
2c1
2b1
 Q 
dS   
 T  2C1
…(3)
From (1) and (3),
For any general process
dS 
Q
T
….(4)
The above equation is said to be Clausius Inequality.
For internal reversible cycle
 Q 
dS =  
 T  int ernally reversible
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YT : Indian school of physics
NITIN SACHAN
 Q 
dS    [General path]
T 
Hence,
 Q 
dS =   + ( S)generation
T 
dS is a state function.
( S)generation Is a path function
SOME IMPORTANT POINTS
1.
A process is reversible if it is both internally as well as externally reversible
2.
Internally reversible means no entropy is generated within the system, whereas externally reversible
means no entropy is generated outside the system.
3.
Entropy is the measure of unavailable energy
4.
Though entropy is a property (state function), entropy generation is not a property & is a path
function.
The entropy of a system can change in the following three ways.
1. Heat interaction 
 → External Interaction
2. Mass interaction 
3. Entropy generation → Internal Irreversibility
*Work, kinetic energy & potential energy are available energies & hence no entropy is
associated with energy interactions in these forms.
We know that,
dS =
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Q
+ ( S)gen
T
YT : Indian school of physics
NITIN SACHAN
Case (1): Internally reversible  ( S)gen = 0


(i)
Heat is added to the system Q > 0 (positive)
Hence, dS >0 (positive)
(ii) Heat rejected by the system, Q → negative
dS =
Q
→ negative
T
(iii) Adiabatic system Q →zero
dS =
Q
= zero
T
Case (2): Internally Irreversible ( S)gen  0
dS =
(i)
Q
+ ( S)gen
T
If some heat is added to the system, then Q →positive
We know that
dS =
Q
+ ( S)gen
T
Q
 0 and (S)gen  0 (always)  dS  0
T
(ii)
If heat is rejected by the system, then Q →negative , (S)gen  0 (always)
dS =
Q
+ ( S)gen
T
dS can be positive, negative or zero
(iii)
For adiabatic system, Q = 0
dS =
Q
+ ( S)gen
T
Q
= 0  dS = (S) gen
T
For a reversible adiabatic process (S)gen = 0  dS = 0
For an irreversible adiabatic process (S)gen is positive (always)  dS > 0
Therefore dS  0
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YT : Indian school of physics
NITIN SACHAN
ENTROPY CHANGE OF THE UNIVERSE
We know that
dS =
Q
+ ( S)generation
T
The universe is an isolated system, So Q it is zero.
o
Q
dS =
+ ( S)generation  0
T
i.e., according to the second law, all the processes are possible for which dS  0 . This is an increase in the
entropy principle.
Now we consider two systems A and B interact with each other (in everything)
Moreover, the two systems do not interact with anything else. Hence the combined system itself is
considered an isolated system. Considering the number of systems like the above, we can call it the
universe (in general).
Universe
A
B
C
Z
Within the universe, one interacts with the other. However, one universe does not interact with the other.
(S)universe  0
(S)universe = (S)A + (S)B + (S)C + ...
In the case of A and B systems, if A is a system and B is surrounding
(S) universe = (S)system + (S)surrounding
In irreversible heat transfer
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YT : Indian school of physics
NITIN SACHAN
Let us consider the irreversible heat transfer process
The thermal capacity of reservoir = c = 
T1 > T 2
Thermal heat capacity of the reservoir =
Thermal heat capacity of the reservoir =
(S) A = −
Q
T1
( S) B = +
Q
T2
 1 1
(S)universe = Q  − +   0 (
 T1 T2 
T1 >T2 )
Ex1:
Consider two bodies, A and B, of finite heat capacities having temperatures T1 and T2, respectively; when
heat transfer (Q) occurs between A and B, the temperature of A decreases, and that of B increases. Finally,
the process is naturally stopped and will be at a mean temperature. If we consider the heat capacities of A
and B are identical.
Then mean temperature
=
T1 + T2
(the process is natural or irreversible)
2
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NITIN SACHAN
(S)universe = (S)A + (S)B
SA = 
T
T
Q f cp dT
=
= cp ln f  0 (
T T1 T
T1
Tf  T1 )
(reversible heat transfer)
SA  0 (obviously, it loses heat)
Similarly
T
T
Q f cp dT
SB = 
=
= cp ln f  0
T T2 T
T2
SB  0
 (S)universe = Cp ln
Tf2
T1T2
(S)universe = Cp ln
(T1 + T2 ) 2
0
T1T2
(
T1T2 
T1 + T2
) ( GM  AM)
2
If a reversible heat engine connects two bodies, then the final temperature of the combined system
Tf =?
Suniverse = ?
Engine
Taken internal irreversibility = 0 for A, B.
The entire system (A, B, H.E.R.) is reversible if the heat engine is reversible.
i.e., (S)universe = (S) A + (S) HE + (S) B
R
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NITIN SACHAN
A heat engine is reversible or irreversible; the entropy change = 0 (always)
(
Entropy is a property, and the heat engine operates in a thermodynamic cycle)
SA , SB they operate on a reversible process so that SA + SB = 0 (summation = 0)
(S)universe = 0
(S) universe = C p ln
Tf
T
− C p ln f
T1
T2
= Cp ln
Tf 2
=0
T1T2
Tf 2
= 1  Tf = T1T2
T1T2
Conclusion: All natural processes in this universe, considered an isolated system, occur in such a way
(S)universe  0 . This is the direction of constraint, as mentioned earlier. A directional constraint for any natural
process occurs in such a way that it makes a permanent change in the surroundings, i.e., the algebraic sum of
the entropy change of the process and the entropy change of the interacting system and surroundings must
be greater than zero. If the processes are natural, i.e., the entropy of the universe monotonically increases
(i.e., always increases). This is the directional; constraint (in general).
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YT : Indian school of physics
NITIN SACHAN
Ex2:
The coefficient of performance of The heat pump is 50% of the theoretical maximum. It maintains a house
at 200C, which leaks energy of 0.6kW per degree of temperature difference to the ambient. For a maximum
of 1.0kW, power is ignored. Find the minimum outside temperature for which the heat pump is a good heat
source.
Sol:
Surrounding temperature Ta =?
Leaking energy by house = Q.L.
To maintain the house at 2930K
QH (taken from pump) = QL (leaked energy)
(to maintain a steady state)
i.e., QH = QL = ( 0.6)( TH − Ta ) kW
Q.H. = (coefficient of performance)Heat pump W =
( c.o.p )HP
(
W = 1)
But (c.o.p)HP = (0.5) (c.o.p)ideal
(c.o.p) =
QH
QH
=> (c.o.p) =
W
QH − Qa
( c.o.p )ideal =
( TH − Ta )
2
 T 
TH

H
=>QH = 0.5 

TH − Ta
 TH − Ta 
= 0.6 (T.H. – Ta)
 0.5 
T
= 
 H
 0.6 
T.H. = 293 K
Ta = 277.4 K
 40 C
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NITIN SACHAN
Ex3:
We wish to produce refrigeration at -300C. A reservoir is available at 2000C, and the ambient temperature is
300CThus the work is done by a cyclic heat engine [C.H.E.] operating between the 2000C reservoir and the
surroundings. This work will be used to drive the refrigerator. What will be the ratio of the heat transferred
from the 2000C reservoir to the heat transferred from the –300C reservoir? Assuming all processes are
reversible.
Sol:
Drives
Refrigerator
Heat Engine
W
Q1
=?
QL
From heat engine
 T 
W = Q1 = Q1 1 − a 
 TH 
From refrigerator
c.o.p =
QL
W
W=
Q2
QL
TL
(reversible heat pump)
=
=
c.o.p Q H − Q L Ta − TL
 T −T 
QL
L
= Q L  a

1/ c.o.p
 TL 

 T −T 
T 
L
 Q1  1 − a  = Q L  a

T
T
H
L



Q1
= 0.687
QL
W=
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NITIN SACHAN
T- S Diagram
dS =
Q
+ ( S)generation
T
If a process is internally reversible, then ( S)generation it is zero.
dS =
Q
T
TdS = Q or Q =  TdS
T
2
P
3
=
1
V
4
S
Area under the T-S diagram gives heat transfer for an internally reversible process when projected
on the entropy axis.
rev = 1 −
QL
T
; rev = 1 − L
QH
TH
If arbitrary = reversible [Re versible]
If arbitrary  reversible [Irreversible]
If arbitrary  reversible [Im possible]
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NITIN SACHAN
CLAUSIUS INEQUALITY
The value of the cyclic integral Q / T is always less than or equal to zero.

Q
0
T
This inequality is valid for all cycles, reversible or irreversible.
For Reversible cycle
Heat
Energy

rev
Q
Q
Q
=+ H − L
T
TH TL
Q H TH Q H Q L
=

=
Q L TL
TH
TL

Q
= zero
T
rev
Note
If a device is shown exchanging heat with more than two reservoirs, then Clausius inequality is advisable.

Q
0
T
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NITIN SACHAN
PROBLEMS
1.
An ideal gas goes from an initial state ( V1 ,T1 ) to a final state ( V2 ,T2 ) by an unspecified process. The
entropy change is
(A) nCp ln
Ans:
Sol:
T2
T1
(B) nC V ln
T2
T1
(C) nCp ln
B
From the first law of thermodynamics,
dQ = dU + dW
=>
dQ
dT PdV
= nC v
+
T
T
T
T1
T2
(D) nC V ln
T1
T2
dQ = nCv dT + PdV
… (1)
The entropy change is
f
dQ
i T
f
dT
= nC v 
+
i T
S = 
… (2)
f
i
P
dV
T
… (3)
From the ideal gas equation
P nR
=
T V
Substituting this into the second term of equation (3), we have
f
dT
+ nR
i T
T
V
= nC v ln 2 + nR ln 2
T1
V1
S = nC v 
f
i
dV
V
We have calculated S without consideration of the path followed. Since entropy is a state variable,
the entropy change will be the same regardless of the process between the two states.
For an isothermal process, T2 = T1
ln
T2
= ln (1) = 0
T1
=>
S = nR ln
V2
V1
For an isochoric process,
V2
= ln (1) = 0
V1
T
S = nC v ln 2
T1
ln
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NITIN SACHAN
2.
Figure shows a container with a membrane partition. The left-hand side of the container, of volume
V1 , is filled with an ideal gas; the right-hand side is initially vacuumed. The total volume of the
container is V2 . The membrane is punctured and free expansion of gas, which is an irreversible
process. The entire system is thermally insulated from its surroundings.
Vacuum
Gas
Membrane
The entropy change associated with this free expansion of the ideal gas is
(A) nR ln
Ans:
Sol:
V1
V2
(B) 2nR ln
V1
V2
(C) nR ln
V2
V1
(D) 2nR ln
V2
V1
C
From the first law of thermodynamics,
Q = U + W
=>
In free expansion, W = 0
The temperature of an ideal gas is unchanged during the free expansion; internal energy is a function
of temperature only; hence. U = 0
for isolated system; there is no heat transfer to or from the surroundings; Q = 0.
dQ = 0 It seems that S = 
2
1
dQ
is zero, but this is not the case.
T
We know that
S = nCp ln
T2
V
+ nR ln 2
T1
V1
This expression applies to any reversible or irreversible process between the same initial and final
states. Since the temperature of the gas is constant (free expansion),
S = nR ln
V2
V1
Since the final volume Vf is greater than the initial volume, the entropy change is positive; entropy
has increased. This is one of the examples in which dS  0 but dQ = 0
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3.
Figure shows a well-insulated container with a thin membrane partition. Each of the two chambers
has two different liquids. The membrane is punctured, and liquids are allowed to mix. Consider
m1 = m2 and C1 = C2 .
The entropy change associated with this free mixing of liquids is
Ans:
Sol:
 T1 + T2 2

(A) mC ln 
 / T1T2 
 2 

 T1 − T2 2

(B) mC ln 
 / T1T2 
 2 

 T1 + T2 2

(C) mC ln 
 / T1T2 
 4 

 T1 − T2 2

(D) mC ln 
 / T1T2 
 4 

A
We first compute the final temperature of the mixture
Q net = 0
m1C1 ( Tf − T1 ) + m2C2 ( Tf − T2 ) = 0
m C T + m 2 C 2 T2
Tf = 1 1 1
m1C1 + m 2 C 2
Change in entropy of liquid 1,
f
dQ
i T
Tf
dT
= m1C1 
T1 T
Since Tf  T1 this entropy change is negative,
S1 = 
The entropy change of liquid 2,
S2 = m 2 C 2 ln
Tf
T2
Since Tf  T2 this entropy change is positive, the total entropy change is the sum of the individual
entropy changes.
Stotal = S1 + S2
T
T
= m1C1 ln f + m 2C 2 ln f
Tl
T2
The first term is negative, and the second is positive, but total entropy change is positive. Let us see
it in a particular case.
m1 = m2 , C1 = C2
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 T + T2 

Tf =  1

2


Stotal
Tf2
= mC ln
T1T2
 T1 + T2  2

= mC ln 
 / T1T2 
 2 

T +T
Since 1 2 the arithmetic -mean of two temperatures T1 T2
2
GM
=> Stotal  0
T1T2 is the geometric mean, and AM >
This proves that total entropy change will be positive in this irreversible mixing process.
4.
Heat Q is transferred from a hot reservoir at temperature T2 to a cold reservoir at temperature T1 . The
entropy change is
T2 + T1
T1T2
(A) Q
Ams:
Sol:
(B) Q
T2 − T1
2T1T2
(C) Q
T2 + T1
2T1T2
(D) Q
T2 − T1
T1T2
D
The process of heat transfer is irreversible.
Entropy change of hot reservoir
S = 
f
i
Q
dQ
=−
T
T2
Entropy change of cold reservoir,
S = 
f
i
Q
dQ
=+
T
T1
The total entropy change of both reservoirs is
Stotal =
Q
Q Q
−
T1 T2
T2 − T1
T1T2
Since T2  T1 the total entropy change is positive.
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YT : Indian school of physics
NITIN SACHAN
5.
Ans:
Sol:
An electric current of 10 amp is maintained for 1 sec. In a resistor of 25 ohms, the resistor's
temperature is kept constant at 27°C. What are the resistor's entropy change and the entropy change
of the universe?
(A) 0 and 8.33 J/K respectively
(B) 8.33 J/K and 0 respectively
(C) 4.33 J/K and 0 respectively
(D) 0 and 4.33 J/K respectively
A
We know that heat dissipated in a resistor is given by
2
Q = i 2 Rt = (10 )  25  1 = 2500 joules
As the resistor is maintained at 27 o C (300 K), the heat Q is not absorbed by the resistor but taken
away by the surroundings. Hence the entropy – change of the resistor
( S)resistor =
Q
=0
T
The entropy change of the surroundings is
( S)surroundings =
6.
Ans:
Sol:
Q 2500
=
T 300
= 8.33 joules / K
This is also the change of entropy of the universe.
Find the change in entropy when 10kg of ice 2 o C is changed to water at 2 o C . (Latent heat of ice is 80
k. cal. Per kg.).
(A) 1.001 kcal/ K (B) 2.001 kcal/ K (C) 3.001 kcal/ K (D) 4.001 kcal/ K
C
Change of entropy
T 
L
+ 2.302 ms log10  2 
T
T
 1
 275 
80

=
+ 2.303  1  1  log10 

273
 273 
= 0.2930 + 0.007137 = 0.3001 cal./K
= ( S2 − S1 ) =
 change in entropy for 10kg (=10,000 gm)
= 10,000  0.3001 = 3001cal / o K = 3.001 k cal / o K
= 3.001 k cal / o K
7.
Ans:
Sol:
Find the change in entropy when 0.1 kg. of ice −10o C is completely converted into steam at 100o C ,
given specific heat of ice = 2100 J kg −1K −1 . The ice has Latent heat of fusion = 80 cal. gm −1 and latent
of steam = 540 cal. gm −1 .
(A) 147 cal / C
(B) 167 cal / C
(C) 187 cal / C
(D) 207 cal / C
D
Change of entropy
T  L
2
1
 +
 T1  T1
T  L
+2.3026 ms log10  2  + 2
 T1  T2
= 2.3026 ms log10 
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NITIN SACHAN
 273   80 
 +

= 2.3026  1  0.5  log10 
  273 
263

 

 373   540 
 +

+2.3026  1  1log10 
  373 
273

 

 273 
 + 0.2930
= 2.3026  0.5  log10 

 263 
+2.3026  1 1 0.1335 +1.447
= 2.3026  0.5  0.0162 + 0.2930
+2.3026  0.1335 + 1.447
 273 
 + 0.2930
= 2.3026  0.5  log10 

 263 
+2.3026  1 1 0.1335 +1.447
= 0.01865 + 0.2930 + 0.3119 + 1.447
= 2.07065cal. / o K
 Change in entropy for 0.1 kg. ( = 100gm) = 100  2.07065
= 207 cal. / o K
8.
Ans:
Sol:
Calculate the increase in helium's standard entropy when heated from 298K to 1000K at constant
pressure.
(A) 25.16 Jk −1 mol−1 (B) 18.8 Jk −1 mol−1
(C) 12.9 Jk −1 mol−1
(D) 8.17 Jk −1 mol−1
A
The change of entropy is given by
dQ
where dQ = 1.Cp .dT
T
C dT

dS = p
T
T2
T 
dT
= Cp log e  2 
or S2 − S1 = Cp 
T1 T
 T1 
T 
= 2.303Cp log10  2 
 T1 
 1000  

5
  C = 5 R
= 2.303  R  log10 
p


2
2 
 298  
 1000 
5

= 2.303   8.31  log10 

2
 298 
Solving we get, S2 − S1 = 25.16J K −1mol−1
dS =
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YT : Indian school of physics
NITIN SACHAN
9.
Ans:
Sol:
Calculate the change in entropy when 1kg of ice −10o C is wholly converted into steam 100o C .
( ( Cp )ice = 2100 J / kg.K . Latent heat of fusion of ice = 336000 joules/kg, ( Cp )water = 4200J / kg / K
Latent heat of steam = 22.68  105 joule /kg.)
(A) 7946.38 J / K
(B) 8166.38 J / K
(C) 8699.38 J / K
(D) 8978.38 J / K
C
Consider the change of entropy as follows:
(i) Change is the entropy of 1kg of ice at - 10o C in converting into ice at 0 o C is given by
 273 

= mCp loge ( T2 / T1 ) = 1  2100  2.303 log10 

263


= 2100  2.303  0.0162 = 78.38 Joule /K
(ii) Change of entropy when 1kg of ice 0 o C is converted into the water at 0 o C is given by
mL 1  336000
=
= 1.23  103 Joule / K
T
273
(iii) Change of entropy when 1kg of water 0 o C is converted into the water it 100o C is given
T 
 373 

= mCp log e  2  = 1  4200  2.303log10 

273
T


 1
= 4200  2.303  0.1355 = 1311 Joule / K
(iv) Change of entropy when 1kg of water 100o C is converted into steam 100o C is given by
mL
=
= (1  22.68  105 / 273)
T
= 6.08  103 joule / K
The total change of entropy is given by
78.38 + 1.23 103 + 1311 + 6.08 103  joule/K
= 8699.38 Joule / K .
10.
Ans:
Silver's molar specific heat capacity at constant pressure in the range 40 K to 120K varies as
Cp = 0.076T - 0.0026 T2 − 0.15cal / ( mole − K ) . 5 moles of silver are heated from 40K to 120 K,
calculate the change in entropy.
(A) 11.25 cal/K
(B) 16.25 cal/K
(C) 21.25 cal/K
(D) 26.25 cal/K
C
Sol:
S = 
120
120
40
T
40
S = 5 
= 5
Cp dT
120
40
 0.076T − 0.00026 T 2 − 0.15 

 dT


T


 0.076 − 0.00026T − ( 0.15 / T ) dT


= 5 0.076 (120 − 40 ) − 0.00013 (1202 − 402 ) − 0.15  2.3026 log10 (120 / 40 ) 
= 50.076  80 − 0.0013  12800 − 0.15  2.3026  0.4771
= 21.25 cal/K
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NITIN SACHAN
11.
A mass m of a liquid at temperature Tx K is mixed with an equal mass m of the same liquid at


temperature Ty K . The entropy change of the universe will  mClog e ( Tx + Ty ) /  ( Tx Ty ) be. Find
 + ?
Ans:
Sol:
(A) 2
(B) 4
(C) 6
(D) 8
B
When equal masses at temperature Tx and Ty are mixed, let the temperature of the mixture be T.
Then
{ Heat lost = heat gained}
m C ( Tx − T ) = mC ( T − Ty )
 Tx + Ty 

 2 
 Temperature of mixture T = 
Entropy changes when the temperature of m units of the liquid changes from Tx T.
S1 = mCloge ( T / Tx )
Similarly, entropy changes when the temperature of m units of the liquid change from Ty
S2 = m C log e ( T / Ty )
A total change of entropy of the universe
S = S1 + S2 = mC log e ( T / Tx ) + mC log e ( T / Ty )
= mC log e T − log e Tx + log e T − log e Ty 
= mC  2 log e T − log e Tx − log e Ty 

 T 
= mC log e  y  
 T T 

 x y 
2

Tx + Ty ) 
(
  T = ( Tx + Ty ) / 2
= mC log e

4Tx Ty 



 T + Ty  

= mC log e  x
 2 Tx Ty  




= 2mClog e ( Tx + Ty ) / 2 Tx Ty 
12.
Ans:
1 gm mole of an ideal gas expands isothermally from V0 to 4V0 . Find the change in its entropy in
terms of gas constant R
(A) 1.128 R J/K
(B) 1.218 R J/K
(C) 1.386 R J/K
(D) 1.428 R J/K
C
Sol:
W =
V2
V1
PdV =
V2
V1
RT
dV
dT
(
PV = RT )
= RTloge ( V2 / V1 ) = 2.3026 RTlog10 ( V2 / V1 )
W
= 2.3026 R log10 (4) ]
 Change in entropy =
T
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NITIN SACHAN
13.
The formula gives the variation of specific heat of a substance with temperature Cp = a + bT + cT2 .
The change in entropy as the system goes from T1 to T2 is
 T2 
1 2
2
 − b ( T2 − T1 ) + ( T2 − T1 )
C
 T1 
(A) 2.3026a log10 
 T2 
1 2
2
 + b ( T2 − T1 ) − ( T2 − T1 )
C
 T1 
(B) 2.3026a log10 
 T2 
1 2
2
 − b ( T2 − T1 ) − ( T2 − T1 )
C
 T1 
(C) 2.3026a log10 
 T2 
1 2
2
 + b ( T2 − T1 ) + ( T2 − T1 )
T
C
 1
(D) 2.3026a log10 
Ans:
D
[Hint: dQ = CP dT and dS = CP
T2
S =  CP
T1
dT
T
dT
T
T2
a

=   + b + cT  dT
T

T1 
=  a loge T + bT +12 cT2 
14.
T2
T1
Calculate the change in entropy when heat flows between two bodies at temperatures T1 T2 and until
both reach the standard temperature T0. (The heat capacities of the bodies are C1 and C2 .)

 T0 
 T0  
 + C 2 log10   
 T1 
 T2  
(A) 2.306 C1 log10 


T 
 T 
(B) 2.306 C1 log10  0  − C2 log10  0  
 T1 
 T2  


 T0 
 T0  
 + C 2 log10   
 T1 
 T2  
(C) 2.306  2C1 log10 


T 
 T 
(D) 2.306 C1 log10  0  + 2C2 log10  0  
 T1 
 T2  

Ans: A
Hint : C1 ( T1 − T0 ) = C2 ( T0 − T2 )
 C T +C T 
1 1
2 2
 C + C  S1 = C1 loge ( T0 / T4 ) and S2 = C2 loge ( T0 / T2 )
1
2



T 
 T 
S = 2.306 C1 log10  0  + C2 log10  0  
 T1 
 T2  

 T = 
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0.2 kg ice 0C is in 0.5kg water 20C .
Calculate the change in entropy of the ice as it melts 0C .
Lf= 3.33 105 J / kg
Ans: +244.0 J/K
Sol: Convert all temperatures into Kelvin.
f dQ
m L
Smelt = Sf − Si = 
= ice f
i T
T
15.
=
16.
Sol:
17.
Sol:
18.
( 0.2 kg ) ( 3.33 105 J / kg )
273K
= +244.0 J / K
0.1 kg ice 0C is in 0.5 kg water 20C .
Calculate the entropy change of the ice-water mixture as it moved to the final temperature.
f dQ
f mice water c w dT
f dT
Sice water = Sf − Si = 
=
= mice water c w 
i T
i
i T
T
T 
 276.4K 
= mice water c w ln  f  = ( 0.1kg )( 4186J / kg  K ) ln 
 = +5.2J / K
 273K 
 Ti 
Suppose 0.1 kg ice 0C is in 0.5 kg water at 20C .
Find the entropy change of the water as it melted the ice.
f m
f dT
dQ
water c w dT
=
= m water c w 
i T
i
i T
T
T 
 277.1K 
= mwater c w ln  f  = ( 0.5kg )( 4186J / kg  K ) ln 
 = −116.8J / K
 293K 
 Ti 
Swater melt = Sf − Si = 
f
Suppose 0.1 kg ice at 0C is in 0.5 kg water at 20C .
Find the entropy change of the water as it cooled down to the final temperature.
Ans:
Sol:
19.
f m
f dT
c dT
dQ
=  water w
= m water c w 
i T
i
i T
T
T 
 276.4K 
= mwater c w ln  f  = ( 0.5kg )( 4186J / kg  K ) ln 
 = −5.3J / K
T
277.1K


 i
Suppose 0.1 kg ice at 0C is in 0.5 kg water at 20C .
Determine the change in entropy of the ice/water system in thermal equilibrium.
Swater cooled = Sf − Si = 
Ans:
Sol: Stotal = Smelt
+
Sice water
= (+122.0 + 5.2
= +5.1 J / K
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f
+ Swater melt + Swater cooled
+ – 116.8
+ – 5.3) J/K
YT : Indian school of physics
NITIN SACHAN
20.
One mole of an ideal, monatomic gas is initially at point A with pressure ( pA ) , volume ( VA ) , and
temperature ( TA = TH ) . After that, it is allowed to expand isothermally to point B
Calculate the change in entropy of the gas SAB going from A to B.
A
Q AB
p
B T
H
V
Sol:
21.
Eint = Q − W = 0  QAB = WAB = nRTH ln ( VB / VA )
B dQ
1 B
Q AB
SAB = 
=
dQ
=
A T
TH A
TH
SAB = nR ln ( VB / VA )  0
1 mole of an ideal, monatomic gas is initially at point A with pressure ( pA ) , volume ( VA ) , and
temperature ( TA = TH ) . Then, the gas expands isothermally to point B and cools isochronally to
point C. What is the change in entropy of the gas SBC going from B to C?
A
B T
H
p
C
TL
QBC
V
Ans:
Sol: Eint = QBC  dEint = dQBC = 3 / 2nR dT
C dQ
TL 3 / 2 nR dT
SBC = 
=
= 3 / 2 nR ln ( TL / TH )
B T
TH
T
3
3
SBC = nR ln ( TL / TH ) = − nR ln ( TH / TL )  0
2
2
22. 1 mole of an ideal, monatomic gas is initially at point A with pressure ( pA ) , volume ( VA ) , and
temperature ( TA = TH ) . After that, it is allowed to expand isothermally to point B, then cool
isochronally to point C, and then contract isothermally to point D
Find the entropy change of the gas SCD going from C to D.
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A
p
B T
H
D
QCD
C
TL
V
Sol:
23.
Eint = Q − W = 0  QCD = WCD = nRTL ln ( VA / VB )
D dQ
Q
1 D
SCD = 
=
dQ = CD

C T
TL C
TL
SCD = nR ln ( VA / VB ) = −nR ln ( VB / VA )  0
1 mole of an ideal, monatomic gas is initially at point A with pressure ( pA ) , volume ( VA ) , and
temperature ( TA = TH ) . Then, it expands isothermally to point B, cools isochronally to point C,
contracts isothermally to point D, and warms up isochronally back to the initial point A.Calculate
the change in entropy of the gas SDA going from D to A.
A
p
Q DA
B
D
C
TH
TL
V
Sol:
SDA
24.
Eint = QDA  dQDA = 3 / 2nR dT
A dQ
TH 3 / 2 nR dT
SDA = 
=
= 3 / 2 nR ln ( TH / TL )
D T
TL
T
3
= nR ln ( TH / TL )  0
2
1 mole of an ideal, monatomic gas is initially at point A with pressure ( pA ) , volume ( VA ) , and
temperature ( TA = TH ) . It then expands isothermally to point B, cools isochronally to point C,
contracts isothermally to point D, and warms up isochronally back to the initial point A.
Calculate the net entropy change in one cycle.
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A
p
B
D
C
TH
TL
V
Sol:
25.
Easy way : S cycle = 0 !
1 mole of an ideal, monatomic gas is initially at point A with pressure ( pA ) , volume ( VA ) , and
temperature ( TA = TH ) . It then expands isothermally to point B, cools isochronally to point C,
contracts isothermally to point D, and warms up isochronally back to the initial point A.
Find the net entropy change in one cycle.
A
p
B
D
C
TH
TL
V
Ans:
Sol: Harder way : Snet = SAB + SBC + SCD + SDA
V  3
T 
V  3
T 
= nR ln  B  − nR ln  H  − nR ln  B  + nR ln  H  => Snet = Scycle = 0 !
 VA  2
 TL 
 VA  2
 TL 
26.
A Carnot engine that operates between the temperatures TH = 1700K and TL = 600K , the engine
performs 1200 J of work each cycle
TH
QH
QL
TL
What is the efficiency of this engine?
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TL
300
= 1−
= 0.647
TH
850
A Carnot engine that operates between the temperatures TH = 1700K and TL = 600K performs 1200
J of work each cycle.
 = 1−
27.
TH
QH
QL
TL
Calculate the heat extracted from the high-temperature reservoir each cycle.
W
W 1200 J
=
= 0.647  Q H =
=
= 1854 J
QH
 0.647
28.
A Carnot engine that operates between the temperatures TH = 1700K and TL = 600K , the engine
performs 1200 J of work each cycle
TH
QH
QL
TL
Calculate the heat expelled into the low-temperature reservoir each cycle.
QL = QH − W = 1854J − 1200J = 654J
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NITIN SACHAN