Second Law of Thermodynamics & Entropy | Kinetic Theory & Thermodynamics - Physics PDF Download

Second Law of Thermodynamics

Statement: It is impossible to transfer total heat into work in a cyclic process in the absence of other effect (by Lord Kelvin).

In another way, it is also stated that it is impossible for heat to be transferred by a cyclic process from a body to one which is warmer than itself, without producing other changes

at the same time.

Heat Engines

A machine that can convert heat into work is said to be heat engine.

It is a system that performs the conversion of heat or thermal energy to mechanical work.

Heat Reservoir

It is an effectively infinite pool of thermal energy at a given constant temperature. Ideally its heat capacity is large enough, that when it is in thermal contact with another system, its temperature remains constant.

All heat engines have mainly three essential components.

• A source: This is a hot region which is a part of the surrounding from which energy flows by heat transfer. Popularly it is known as hot reservoir eg. Nuclear reactor, Furnace, etc.
• The working agent: This under goes change at state as part of a continuous cycle e.g. steam-water.
• A sink: This is respectively a cold region, which is part of the surrounding into which heat is rejected by heat transfer e.g., cooler.

Efficiency of Heat Engine (η):

η =

Since, the process is cyclic dU = 0 ⇒dQ = dW

Heat given to system is Q₁ and heat rejected by system is Q₂ .

η =

Carnot Cycle

It is theoretical thermodynamic cycle proposed by Nicolas, Leonard Sadi Carnot.

It can be shown that, it is most efficient cycle for converting a given amount of thermal energy into work.

A system undergoing a Carnot cycle is called a Carnot heat engine.

Stages of the Carnot cycle:

1. Reversible isothermal expansion of the gas at hot temperature, T₁ (isothermal heat addition). During this step (1 to 2), the gas is allowed to expand and it does work on the surrounding. The gas expansion is propelled by absorption of Q₁ quantity of heat from the high temperature T₁ .

2. Reversible adiabatic expansion of the gas. For this step (2 to 3) on figure, the gas

continues to expand i.e., working on surrounding. The gas expansion causes it to cool to

the cold temperature T₂ . 

3. Reversible isothermal compression of the gas at the low temperature T₂ (isothermal heat rejection). The process is shown by 3→4 . Now the surroundings will do work on the gas, causing Q₂ quantity of heat to flow out of the gas to the low temperature reservoir. 

4. Adiabatic compression of the gas (4 to 1). During this step, the surrounding do work on the gas, compressing it and cause the temperature to rise to T₁ . At this point the gas is in the same state as that of the starting at step 1.

Efficiency of Carnot engine

η = W/Q₁

W is the work done during cycle and Q₁ is heat given to system.

W = W₁ + W₂ + W₃ + W₄

W₁  = work done during process 1 to 2 for isothermal process =

Since, the change in internal energy during the process i.e., dU = 0

Therefore, Q₁ = W₁ =  (From first law of thermodynamics)

W₂ is work done during process 2 to 3 in adiabatic process.

Here, dQ = 0 and W₂ =

W₃ is work done during process 3 to 4 in isothermal compression.

Here, W₃ =

Since, dU = 0 ⇒ Q₃ = W₃

W₄ is the work done during the adiabatic process from 4 to1, is given by

W₄ =

W₂ = - W₄

W =

Q₁ =

η =                                               ….(i)

From the figure,

                                               ….(ii)

                                                   ….(iii)

Dividing equation (ii) by (iii) gives

                            ….(iv)

From equation (i) and (iv)

η =                                                          ….(v)

For Carnot cycle,

This gives

η =                                                       ….(vi)

Example 1: An ideal gas engine operated in a cycle which when represented on a P-V diagram, is a rectangle. If we call P₁,P₂ as the lower and higher pressures respectively and V₁, V₂ as lower and higher volume respectively.

(a) Calculate the work done in complete cycle

(b) Indicate in which parts of the cycle heat is absorbed and in which part heat is librated.

(c) Calculate the quantity of heat following into the gas in one cycle

(d) Show that efficiency of the engine is

Work done during process, AB = P₂(V₂ - V₁)

Heat absorbed =  _nC_pΔT

=

=

Since, V₂ > V₁, so heat is absorbed in the process.

In isochoric process, B-C

W_B→C = 0 ⇒ dQ - dU = _nC_vΔT

=

Since, P₁ < P₂ , therefore heat is rejected during this process

In isobaric process, C-D

W_C→D = P₁(V₁ - V₂)

Heat exchange during the process

Since, V₁ < V₂, so heat is rejected during this process

In isochoric process, D→A

W_D→A = 0

Now, heat exchange during this process is,

Since, P₂ > P₁ therefore, heat is absorbed.

(a) Work done =

⇒ W =

(b) Heat absorbed during the process A→ B and D→ A is given by

Heat is liberated during the process B→C and C → D is given by

(c) Heat flowing into the cycle is,

Q =

=

=

⇒dQ = dW

(d) Efficiency = W/Q_absorbed =

Therefore, η =

Example 2: An ideal gas is expanded adiabatically from (P₁,V₁) to (P₂,V₂). Then it is

compressed isobarically to (P₂,V₁) . Finally the pressure is increased to P₁ at constant

volume V₁ .

(a) Show the process in P -V Diagram

(b) Show that the efficiency of the cycle is

(a) The cycle is shown in the figure

The work the system does in the cycle is

W =

Because AB is adiabatic and an ideal gas has the equations

During the CA part of the cycle the gas absorbs heat

Q =

Hence, the efficiency of the engine is

η = W/Q =

Entropy

Entropy is an extensive thermodynamic property that is measure of a system’s thermal energy per unit temperature, which is unavailable for doing useful work. Thermodynamic entropy is a non-conserved state function. For Isolated system, entropy never decreases.

In statistical mechanics, entropy is a measure of the number of ways in which a system may be arranged, often taken to be a measure of ‘disorder’ (the higher the entropy the higher the disorderness).

The infinitesimal change in the entropy (dS) of a system is the infinitesimal transfer of heat energy(∂Q) to a closed system, which is driving a reversible process, divided by temperature (T) of the system.

ΔS =

It has unit Joule/Kelvin

Laws of thermodynamics and entropy

According to first law of thermodynamics

∂Q = dU + PdV

From definition of Entropy

ΔS = ⇒ ∂Q = TΔS ⇒ TdS

which gives, TdS = dU + PdV

Comparative study between P -V diagram and T - S diagram for different process 

From first and second law of thermodynamics

As TdS = dU + pdV = C_vdT + PdV

dV = 0 for V = constant the slope ⇒

And TdS = dH - VdP

TdS = C_pdT -VdP,  dP = 0 for P = const) The slope ⇒

as C_p > C_{v ⇒}

_{Slope of a constant volume process is more than a constant pressure process,}

_{For Ideal Gas process is given by PV}ⁿ _{= C, For different value of n the} the slope is given by

n = 0 P = C Identified as Isobaric process

n = ∞ V = C Identified as Isochoric Process

n =1 T = C Identified as Isothermal Process

n = γ S = C Identified as Adiabatic Process

Inequality of Clausius

Consider an irreversible cyclic engine, working between T₁ and T₂ . If reversible engine is operating between same temperature, then from Carnot’s theorem.

Efficiency of irreversible engine (η_ir) will always be smaller than efficiency of reversible engine(η_r).

η_irr < η_rev

For irreversible cyclic Engine,

or,

This relation is known as inequality of Clausius.

Example 2: Two Carnot Engines A and B are operated in series. The first one, A receives heat at 900K and rejects to a reservoir at temperature T K . The second engine, B receives the heat rejected by the first engine and then rejects to a heat reservoir at 400K .

Calculate the temperature T for the situation.

(a) The work outputs of the two engines are equal

(b) The efficiency of the two engines are equal.

For Engine A take in heat Q₁ at temperature T₁ and rejected heat Q at temperature T; and the engine B taken in heat Q at temperature T and reject heat Q₂ at temperature T₂ .

(a) W_A = Q₁ - Q and W_B = Q - Q₂

W_A = W_B ⇒ Q₁ - Q = Q - Q₂

⇒  = 2

⇒ T =  = 650 K

(b) ηA = ηB ⇒  =  ⇒ T =  = 600 K

Example 3: Calculate the entropy change in isothermal expansion from an initial volume V_i to final volume V_f

For reversible process

TdS = dU + PdV

For isothermal process, dU = 0

⇒T =  = 600 K

As, P =  ⇒ dS =

Example 4: A mole of an ideal gas undergoes a reversible isothermal expansion from volume V₁ to 2V₁ .

(a) What is the change in entropy of the gas?

(b) What is the change in entropy of the universe?

Suppose the same expansion takes place as a free expansion:

(c) What is the change in entropy of the gas?

(d ) What is the change in the entropy of the universe?

(a) In the process of isothermal expansion, the external work done by the system is

W =

Because the internal energy does not change in this process the work is supplied by the heat absorbed from the external world. Thus the increase of entropy of the gas is

ΔS₁ =

(b) The change in entropy of the heat source ΔS₂ = -ΔS₁ thus the total change i entropy of the universe is ΔS = ΔS₁ + ΔS₂ = 0.

(c) If it is a free expansion, the internal energy of the system is constant. As its final state is the same as for the isothermal process, the change in entropy of the system is also the same as ΔS = R In 2

(d) In this case, the state of the heat source does not change, neither does its entropy. 

Therefore the change in entropy of the universe is ΔS = R In 2

Example 5: A mass of liquid at a temperature T₁ is mixed with an equal mass of the same liquid at a lower temperature T₂. If the system is thermally insulated, then

(a) Compute the entropy-change.

(b) Show that it is necessarily positive.

Let c be the specific heat of the liquid. On mixing equal mass m of the same liquid at temperature T₁ and T₂ , where T₁ > T₂.

Let T be the equilibrium temperature of mixture, then

T =

ΔS₁ = , is the entropy change of hotter liquid when cooled fromT₁ to T

ΔS₁ =

ΔS₂ ⇒ Entropy change of cold liquid when heated from T₂ to T.

ΔS₂ =

ΔS = Entropy change of the system.

ΔS = ΔS_{1 +} ΔS₂

= =  =

(b) We know that arithmetic mean is greater than geometric mean i.e., AM > G.M. So,

 ∴ ΔS > 0

Example 6: Compute the change in entropy when ice melts into steam. It is given that L₁ is latent heat of fusion, c is specific heat of water and L₂ is latent heat of vaporization.

Assume T₁ be the Kelvin temperature at which ice melts into water and T₂ the Kelvin temperature at which water is boiled to steam.

ΔS₁ is entropy change when ice is converted into water

ΔS₁ =

ΔS₂ is entropy change when water is heated from T₁ to T₂ i.e.,

ΔS₂ =

ΔS₃ is entropy change when water changes into vapour i.e.,

ΔS₃ =

Therefore, total change in entropy,  ΔS = ΔS_{1 +} ΔS₂ + ΔS₃ =

Example 7: A body of constant heat capacity C_p and a temperature T_i is put into contact with a reservoir at temperature T_f. Equilibrium between the body and the reservoir is established at established at constant pressure.

(a) Determined the change in Entropy for Body.

(b) Determined the change in Entropy for Reservoir.

(c) Determine the total entropy change

(a) We assume T_i ≠ T_f (because the change of entropy must be zero when T_i = T_f). The change of entropy of the body is

ΔS₁ =

(b) The change of entropy of the heat source is

ΔS₂ =

(c) Therefore the total entropy change is

ΔS = ΔS_{1 +} ΔS₂ =

when x > 0 and x ≠ 1 the function f(x) = x - 1 - In x > 0.Therefore

ΔS =

Example 8: Water powered machine, A self-contained machine only inputs are two equal steady streams of hot and cold water at temperatures T₁ and T₂. Its only output is a single high jet of water. The heat capacity per unit mass of water, C , may be assumed to be independent of temperature. The machine is in a steady state and the kinetic energy in the incoming streams is negligible.

(a) What is the speed of jet in  terms of T₁,T₂ and T, where T is the temperature of water in the jet?

(b) Find the minimum possible temperature T

(c) What is the maximum possible speed of the jet?

(a) The heat intake per unit mass of water is

dQ =

As the machine is in a steady state, v²/2 = ΔQ, giving

v =

(b) Since the entropy increase is always positive, i.e.

ΔS =

(b) We have T ≥  So T_min =

(c) For maximum speed T must be minimum

Thus,

v_max =

Example 9: Two identical bodies have internal energy U + NCT , with a constant C . The values of and C are the same for each body. The initial temperatures of the bodies are T₁ and T₂, and they are used as a source of work by connecting them to a Carnot heat engine and bringing them to a common final temperature T_l.

(a) What is the final temperature T_l ?

(b) What is the maximum work delivered?

(a) The bodies attain an equilibrium temperature of T .

T₁ > T > T₂

as, C_v = dU/dT = NC

Q₁ =

Q₂ =

W = Q₁ - Q₂ =

=

ΔS = ΔS_{1 +} ΔS₂

=

=  = 0

(b) Conservation of energies gives W =

For maximum work T =

W =