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Laws of Thermodynamics and Their Application

1 Carnot’s Reversible Cycle

Carnot’s Reversible engine consists of

1) Source: at a ﬁxed temperature T1from which heat engine can absorb heat.

2) Sink: at a ﬁxed temperature T2where heat engine can reject heat.

3) A cylinder with non-conducting sides and conducting bottom with a non-

conducting frictionless piston containing perfect gas as a working substance.

It is provided with a non-conducting stand.

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The cylinder is placed over the stand. Its pressure is P1and volume V1at A

shown in PV diagram. When the cylinder is placed over the source at tem-

perature T1, the gas expands isothermally absorbing heat Q1from the source.

The pressure decreases to P2and volume increase to V2at B as shown in PV

diagram.

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Workdone by the gas from A to B is

W1=ZV2

P dV =RT1ln V2

V1(1)

Then the cylinder is placed on the stand again. The gas is expanded adia-

batically decreasing its temperature to T2. The pressure decreases to P3and

volume increase to V3st C as shown in PV diagram.

Workdone by the gas from B to C is

W2=ZV3

P dV =R

γ−1(T1−T2) (2)

Again the cylinder is placed on the sink at temperature T2, the gas is com-

pressed isothermally rejecting heat Q2to the sink at temperature T2. The

pressure increases to P4and volume decreases to V4at D shown in PV dia-

gram.

Workdone on the gas from C to D is

W3=ZV4

P dV =−RT2ln V3

V4(3)

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At last, the cylinder is placed on the stand again. The gas is compressed

adiabatically to its initial state at temperature T1at point A shown in ﬁgure.

Workdone on the gas from D to A is

W4=ZV1

P dV =−R

γ−1(T1−T2) (4)

Therefore, net workdone in one complete cycle ABCDA is

W=W1+W2+W3+W4

=RT1ln V2

γ−1(T1−T2)−RT2ln V3

V4−R

γ−1(T1−T2)

∴W=RT1ln V2

−RT2ln V3

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∴W=Q1−Q2= net heat absorbed by the working substance.

and eﬃciency,

η=net workdone

heat absorbed =Q1−Q2

Q−1

∴η= 1 −Q2

= 1 −

RT2ln V3

RT1ln V2

= 1 −T2

2 Q. State and prove Carnot’s Theorem. Show how Kelvin used this theorem to deﬁne a

new scale of temperature which is independent of the nature of the working substance.

It states that, ”No engine working between two given temperature can be more

eﬃcient than a reversible carnot’s engine working between the same limits of

temperature and that all the reversible engine working between same limits

of temperature have the same eﬃciency whatever be the working substance.”

Proof:

Consider two engines A and B (reversible and irreversible) working between

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the same temperature limits T1and T2. The working substance is so adjusted

that they perform equivalent amount of work.

Let Q1be the amount of heat absorbed by A from the source at temperature

T1. It does external work W and transfer it to B. The heat Q2is rejected tot

he sink at temperature T2. The eﬃciency of engine A is given by

ηA=W

=Q1−Q2

Let Q0

2be the amount of heat absorbed by B from the sink at temperature

T2and W amount of work is done on the working substance. The heat given

to source at temperature T1be Q0

Then eﬃciency of engien B is given by

ηB=W

=Q0

1−Q0

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Let engine A is more eﬃcient than B.

i.e., ηA> ηB

or, W

or, Q0

1> Q1

Also, W =Q1−Q2=Q0

1−Q0

So, Q0

2> Q2

Let us couple the two engines A and B. Then Q0

2−Q2is the amount of that

taken from sink at temperature T2and Q0

1−Q1is the quantity of heat given to

the source at temperature T1. It means that heat ﬂows from the sink at lower

temperature T2to source at higher temperature T1. But no external work has

been done on the system. This violates the 2nd law of thermodynamics.

Hence, ηA> ηBis wrong

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Therefore, no engine can be more eﬃcient than the reversible engine working

between the same limit of temperature.

3 Absolute Temperature Scale

The eﬃciency of a reversible Carnot’s engine depends upon the two temper-

ature between which it works and is independent of the properties of the

working substance. Kelvin used this concept and deﬁned the temperature

scale which doesnot depend upon the properties of any particular substance.

This is the Kelvin’s absolute thermodynamic scale of temperature.

Consider a reversible carnot engine working between the temperature θ1and

θ2measured on any arbitrary scale. Let Q1is the heat absorbed at θ1and Q2

is the heat rejected at temperature θ2. Then, eﬃciency of the engine can be

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written as a function of these two temperatures.

η= 1 −θ2

θ1

=f(θ1, θ2)

or, θ2

θ1

= 1 −f(θ1, θ2)

∴Q1

1−f(θ1, θ2)=F(θ1, θ2)...(1)

Similarly, if the engine works between the temperature θ2and θ3and Q2and

Q3be the heat absorbed and rejected at that temperatures then

=F(θ2, θ3)...(2)

Again for the temperature θ1and θ3, the heat absorbed is Q1and rejected is

Q3.

∴Q1

=F(θ1, θ3)...(3)

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From equation (1), (2) and (3), we get

=Q1

×Q2

or, F(θ1, θ3) = F(θ1, θ2)×F(θ2, θ3)...(4)

Here, LHS is the functional value of θ1and θ3where as RHS is the function of

θ1,θ2and θ3. So, to eliminate θ2from RHS, we choose some another function

of temperature as

F(θ1, θ2) = φ(θ1)

φ(θ2), F (θ2, θ3) = φ(θ2)

φ(θ3), F (θ1, θ3) = φ(θ1)

φ(θ3)

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putting this functional value in equation (4), we get

F(θ1, θ3) = φ(θ1)

φ(θ2)×φ(θ2)

φ(θ3)=φ(θ1)

φ(θ3)

or, F(θ1, θ3) = φ(θ1)

φ(θ3)

∴Q1

=F(θ1, θ2) = φ(θ1)

φ(θ2)...(5)

Since the function φ(θ1)> φ(θ2) and also, θ1> θ2.

Thus, function φ(θ) is linear function of θand can be used to measure tem-

perature. Let us denote φ(θ1) as T1and φ(θ2) as T2, then

=T1

...(6)

From (6), it is clear that the ratio of two temperature on this scale is equal to

the ratio of the heat absorbed to the heat rejected. This temperature scale is

called absolute scale or Kelvin’s thermodynamic scale of temperature.

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4 Q. Explain what do you mean by entropy of a substance. Show that (a) for any reversible

cycle change of a system the total change of entropy is zero. (b) entropy increases in

an irreverible process.

Entropy is a thermodynamic variable just like other thermodynamic variables

pressure, volume, temperature and internal energy. It measures the disorder-

ness of the system.

Consider an inﬁnitesimal isothermal expansion of an ideal gas. Let dQ be

the amount heat added at constant temperature T and the gas expands. The

internal energy being dependent of temperature remains constant. Thus, from

1st law, the workdone dW by the gas is equal to the heat added dQ.

i.e., dQ =dW =P dV =nRT

VdV

or, dV

V=1

nR dQ

T...(1)

when gas expands, the randomness and thus the disorder state of the gas

increases. Hence, the fractional volume change dV

Vis a measure of increase of

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disorder. From equn(1),

increase of disorder ∝dQ

The ratio dQ

Tis deﬁned as the entropy of the system

i.e., dS =dQ

For an adiabatic process dQ = 0, therefore change in entropy is zero or entropy

all along the adiabatic is constant. In any process, if the heat is absorbed,

there is increase in entropy and when heat is rejected during a process there

is decrease in entropy.

4.1 Change in entropy in reversible process

Consider a complete reversible process ABCDA. From A to B heat Q1is

absorbed by the working substance at temperature T1. The increase in entropy

of the working substance from A to B = Q1

BC is an adiabatic so there is no change in entropy from B to C. From C to D,

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heat Q2is rejected by the working substance at temperature at temperature

T2. The decrease in entropy of working substance from C to D = Q2

T2. Again,

from D to A, there is no change in entropy.

Thus, total increase in entropy by the working substance on the cycle ABCDA

=Q1

−Q2

But for a complete reversible process

=Q2

=constant

Hence, total increase in entropy is

dS =Q1

−Q2

= 0

i.e., entropy remains constant during reversible process.

4.2 Change in Entropy in a Irreversible Process

Consider an irreversible engine as an example of irreversible change. Let the

working substance absorbs Q1heat from source at temperature T1and rejects

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Q2heat to sink at temperature T2.

Now, Loss in entropy of source = Q

and, gain in entropy of sink = Q

The eﬃciency of irreversible engine is

ηi= 1 −Q1

...(1)

And for a reversible engine working between the same limits of temperature,

the eﬃciency is

ηr= 1 −T2

...(2)

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According to Carnot’s theorem,

ηr> ηi

or, 1 −T2

>1−Q1

or, T2

<Q1

or, Q2

>Q1

Therefore, total increase in entropy

=Q2

−Q1

Hence, entropy always increases during irreversible change.

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5 Principle of Increase of Entropy

The entropy of an isolated system either increases or remains contact according

as the processes it undergoes are irreversible or reversible. In practice there

is no realization of reversible process. Mostly the processes are irreversible.

Hence entropy increases in each processes and attains a maximum value when

the equilibrium conduction reaches. This is Principle of increase of entropy.

6 Q. Explain the term entropy with reference to the second law of thermodynamics. Show

that the entropy of a substance in a process can never decrease.

The second law of thermodynamics is rather stated in a descriptive statement

of impossibility. But it can also be stated as a quantitative relation with the

concept of entropy.

According to Kelvin Plank statement of second law of thermodynamics ”There

is no any engine which can convert total quantity of heat Q taken from the

source at a single temperature T into work”. If it is to be so then there would

be decrease in entropy of source by an amount Q

T, which is against principle

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of increase of entropy.

In the same way acording to Clausius Statement ”There will be no any re-

frigerator which can transfer heat from a colder body at temperature T2to a

hot body at temperature T1without any external agency”.If it is to be so, the

entropy of cold body will decrease and the entropy of hot body will decrease.

Deacrease in entropy = Q

Increase in entropy = Q

Here, T1> T2

So, Q

Therefore, total entropy Q

−Q

T1would decrease which is against principle

of increase of entropy.

In both the statements, entropy of the system is restricted to decrease. i.e.,

dS ≮0

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7 Q. Represent Carnot cycle on T-S diagram. Derive an expression for the eﬃciency of

Carnot engine from this diagram.

A graphical representation with entropy along x - axis and temperature along

y-axis to study a thermodynamic processes is called T-S diagram. A T-S

diagram of a carnot cycle ABCDA is shown below:

A carnot cycle consist of two reversible isothermal processes and two adiabatic

process. In above ﬁgure, AB represents the isothermal expansion at a constant

temperature T, of the source, the vertical line BC is the adiabatic expansion

during which there is no change in entropy but a fall of temperature from

T1to T2, the temperature of the sink. CD is second isothermal representing

compression involving a rise of temperature from T2to T1, entropy remaining

the same.

Now, the heat absorbed during isothermal expansion is

Q1= Area under the line AB

=T1(S2−S1)

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where S1and S2are the entropy value during adiabatic process BC and DA

respectively.

The heat rejected to sink during isothermal expansion is

Q1= Area under the line CD

=T2(S2−S1)

Heat energy converted to work

= Heat absorbed −Heat rejected

= Area ABEF −Area CDFE

= Area ABCD

= (T1−T2)(S2−S1)

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Therefore, the eﬃciency of the engine is

η=Heat energy converted into work

Total heat absorbed

=(T1−T2)(S2−S1)

T1(S2−S1)

=T1−T2

∴η= 1 −T2

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8 Q. Calculate the change in entropy when 1 kg of ice at 0oCis converted into steam at

100oC.

9 10gm of water at 40oCis mixed with 20gm of water at 70oC. Calculate the change in

entropy of the whole system.

10 For Silver, the molar speciﬁc heat at constant pressure in the range 50 to 100K is given

Cp= 0.076T−0.00026T2−0.15cal/K

where T is the temperature in Kelvin. If 2 mole of silver is heated from 50 to 100K,

Calculate the heat required and change in entropy