Thermodynamics 101

Chapter 7

Thermodynamics

_____________________________________________

7.0 Introduction

Thermodynamics is a branch of physics, which deals with the energy and work

of a system, was first studied in the 19th

century as scientists were first

discovering how to build and operate steam engines. Thermodynamics deals

with the large-scale response of a system in microscopic change, which can be

observed and measured in experiments. Small-scale gas interactions are

described by the kinetic theory of gasses, which is a compliment to

thermodynamics.

In this chapter, the laws of thermodynamic shall be discussed in great detail.

7.2 Review the Basis of Thermodynamic

Let’s review some basis of thermodynamic. They are covered in the following

paragraphs.

Heat is the mechanism by which energy is transferred between system and

its environment because there is temperature difference between them.

Internal energy U is the energy associated with the microscopic

component of the system such as atom and molecule, when view from reference

frame of rest with respect to the system. It includes kinetic energy and potential

energy associated with the random translational, rotational, and vibrational

motion of the atoms or molecules and intermolecular forces.

For an ideal monoatomic gas, the internal energy is just the translational

kinetic energy of the linear motion of the atom. For polyatomic gases, there is

rotational and vibrational kinetic energy as well. In liquids and solids there is

potential energy associated with the intermolecular attractive forces. A

simplified visualization of the contributions to internal energy is shown in Fig.

7.1.

Thermodynamics 102

Figure 7.1: Internal energy contribution of gas, liquid, and solid

If the water is placed in a pot and heated by flame, the internal energy of water

increases due to the heat transferred from the flame to the water. Comparison of

internal energy for copper and water is shown in Fig. 7.2.

Figure 7.2: Comparison of internal energy for copper and water

To set-up a temperature scale, one has to pick a reproducible thermal

phenomenon and arbitrary assigns a certain Kelvin temperature to its

environment. This standard fixed-point temperature is select at the triple point

of water. The triple point temperature T3 of water is T3 = 273.16 K. The triple

point temperature is the temperature where liquid water, solid ice, and water

vapor can coexist in thermal equilibrium at a fixed set value of pressure and

temperature. The triple point temperature can be measured using a constant –

volume gas thermometer shown in Fig. 7.3, filled with different type of gas such

as hydrogen, helium or nitrogen. At constant volume, the pressure P of the in

thermometer is dependent on the temperature of the liquid where it is immersed.

Thermodynamics 103

Figure 7.3: Constant-volume gas thermometer

Thus,

T = Cp (7.1)

where C is a constant and P the pressure is is equal to P = Po - h g. Po is the

atmospheric pressure which is 1.01x105 pascal Pa (Newton/m

2) or 760 torr or

14.7 lb/in2. At triple temperature the pressure shall be P3, and then an equation

relating it is

!!"

#$$%

&'

3P

PK15.273T (7.2)

However, it was found that at boiling point of water, different gas gives

different result. Thus, it is necessary to measure the triple point temperature

when the volume in the thermometer approaches zero, where is now

independent of the gas type. Equation (7.2) shall then be changed to

!!"

#$$%

&'

(3

0V P

PlimK15.273T . Figure 7.4 shows the result of the temperature-pressure plot

of constant-volume thermometer filled with helium, nitrogen, and hydrogen.

Thermodynamics 104

Figure 7.4: Result of temperature-pressure plot of constant-volume thermometer filled with

three-gas type

There are two ways that the unit of heat is being defined. In British system, it is

the British thermal unit BTU, in which one BTU is defined as the heat required

to raise the temperature of 1.0 lb of water from 630 F to 64

0 F. Alternatively,

one calorie is the energy required to raise the temperature of 1.0 g water from

14.50 C and 15.5

0 C.

In 1948, scientist linked the heat like work defined as the energy transfer

and defined one calorie c as equals to 4.186 J, which is equal to 3.969x10-3

BTU.

This shall mean 1.0 BTU = 251.9 calorie = 1,954.67 J.

The specific heat cm of a material is defined as the amount of energy

necessary to raise the temperature of one kilogram of that material by one

degree Celsius. In mathematical expression it is equal to )TT(m

Qcm

if )' , where

m is the mass, Q is the heat or energy.

Figure 7.5 shows the specific heats of some materials at room temperature.

Specific Heat Substance

cal/g-K J/kg-K

Lead 0.0305 128

Tungsten 0.0321 134

Silver 0.0564 236

Copper 0.0923 386

Aluminum 0.215 900

Brass 0.092 380

Thermodynamics 105

Granite 0.19 790

Glass 0.20 840

Ice (-100C) 0.530 2220

Mercury 0.033 140

Ethyl alcohol 0.58 2430

Sea water 0.93 3900

Water 1.00 4190

Figure 7.5: specific heats of some materials at room temperature

7.3 Thermal Expansion

When the temperature of material increases, its internal energy increases. This

result its dimensions increase due to increase in inter-atomic or intermolecular

distance that can be caused by vibration, rotational moment, and etc.

If the temperature of metal rod of length L is raised by temperature *T, its

length is increased by an amount following equation (7.3).

*L = L+*T (7.3)

where + is the coefficient of linear expansion, which has unit per Kelvin or per

degree Celsius.

The volume of the material will also be increased with the increase of

temperature, whereby the increase of temperature follows equation (7.4).

*V = V,*T (7.4)

where , is the coefficient of volume expansion. The relationship between

coefficient of linear expansion and coefficient of volume expansion is , = 3+.

Example 1 On a hot day in Las Vegas, an oil tanker loaded 37,000 liters of diesel fuel. The

driver unloaded the entire load of diesel in Payson, Utah, where the temperature

is 25.0 K lower than in Las Vegas. How many liters did he deliver? What is the

percentage increase of the cost price per liter of diesel? Given that the

coefficient of volume expansion of diesel is 9.54x10-4

/0C.

Thermodynamics 106

Solution

The change in volume of diesel in Payson is 37,000x9.54x10-4

x(-25) = -882.45

liters. In Payson the driver delivers 37,000 – 882.45 = 36,117.55 liters of diesel.

The increase of the cost of diesel is (37,000/36,117.55 –1)100 % = 2.44 % per

liter.

7.4 Heat Transformation (Latent Heat)

When a matter absorbs energy, it is not necessary used to raise the temperature

of the material. The energy may be used to change from one phase or state to

another. All matters can exist in three common phases. i.e. solid, liquid, and gas

states. Take for an example, the temperature of solid ice and ice water is the

same, and the temperature of boiling water and steam is the same. The energy

absorbed by the solid ice is used to break the intermolecular force so that the

molecules can become loosely bonded. This energy is termed as heat of fusion.

It has unit J/kg. Boiling water absorbs energy to break the weak intermolecular

bond into free molecular gas. The energy required is termed as heat of

vaporization. Figure 7.6 shows the heat of fusion and heat of vaporization of the

some substances.

Melting Boiling

Substance Melting Point

K

Heat of Fusion

kJ/kg

Boling

Point K

Heat of

Vaporization kJ/kg

Hydrogen 14.0 58.0 20.3 455

Oxygen 54.8 13.9 90.2 213

Water 273 333 373 2,256

Mercury 234 11.4 630 296

Lead 601 23.2 2,017 858

Copper 1,356 207 2,868 4,730

Silver 1,235 105 2,323 2,336

Figure 7.6: Heat of fusion and heat of vaporization of some materials

From the table, it shows that the heat of vaporization is higher than heat of

fusion, which is logically true because it requires less energy to break the solid

bond into loose bond than to free the atom or molecule from inter-atomic or

intermolecular bonds.

Thermodynamics 107

Example 2

How much heat must be absorbed by ice of mass m = 720 g at – 100C to take it

to liquid state at 150C?

SolutionThree steps are to be considered which are shown in the diagram below.

The energy required to raise the temperature of solid ice from – 10

0C to 0

0C is

10x0.720x2,200J/kg-K = 15.84x103 J.

The energy required to break the intermolecular bond to loose bond is the heat

of fusion, which is equal to 0.72x333x103 = 239.8x10

3 J.

The energy required to heat ice water from 00C to 15

0C is 0.72x15x4,190 =

45.25x103 J.

The total energy required is 300.80x103 J.

7.5 Heat Transfer Mechanism

Heat can transfer between a system and its environment via three mechanisms,

which are conduction, convection, and radiation.

Conduction

Conduction is transfer of kinetic energy and vibrational energy of atoms to the

atom of lower kinetic energy and vibrational energy. The rate of conduction

Pcond, which is also equal to dt

dQ follows equation (7.5).

Thermodynamics 108

L

TTAP CH

cond

)' k (7.5)

where k is the thermal conductivity, L is the thickness of the transfer material,

and A is the face contact area. Figure 7.7 illustrates the conduction.

Figure 7.7: Thermal conduction between hot reservoir and cold reservoir via a slab

Thermal resistance R is defined as

R = k

L (7.6)

Equation (7.5) shall become R

TTAP CH

cond

)' after substituting equation (7.6). If

there are two insulating layer of difference thermal conductivity k, it can be

shown that the rate of conduction is 21

CHcond

RR

TTAP

-

)' , where R1 and R2 are

respectively the thermal resistance of material 1 and material 2.

If two materials are connected in parallel with the ends contact the hot and

cold reservoirs then the rate of conductor Pcond(total) shall be equal to sum of rate

of conduction Pcond(1) of material 1 and rate of conduction Pcond(2) of material 2.

Thus, equation (7.5), the rate of conduction Pcond(total) is

Pcond(total) = 1

CH1

R

TTA

) +

2

CH2

R

TTA

) (7.7)

Thermodynamics 109

Example 3

The wall of house is made of white pine, unknown materials, and a brick wall.

The thickness Ld of brick wall is two times the thickness of white pine La. The

thermal conductivity kd of the brick

is five times the thermal

conductivity ka of white pine. The

temperature of indoor T1 is = 250C.

The temperature T2 = 200C, T5 = -

100C. If steady state transfer of heat

is attained, what is the temperature

T4 of the interface between brick

and the unknown material?

Solution

Since steady state of transfer of heat is attained, the conduction rate of white

pine is same as the brick wall, which isa

21cond

L

TTAP

)' ak and

d

54

condL

TTAP

)' dk .

Thus,

a

21cond

L

TTAP

)' ak

a

54

d

54

L2

TTA5

L

TTA

)'

)' ad kk . The equation is now contains one

unknown T4 and T4 is found to be – 80C.

Convection

Convection is occurred when liquid or gases come in contact with object whose

temperature is higher than that of liquid or gas. The gas or liquid comes in

contact hot object. Its temperature increases and becomes less dense. It buoyant

force causes it to rise. The surround liquid or gas is then flowed in to take place

of rising warm liquid or gas. The process continues is termed convection. An

illustration is shown in Fig. 7.8.

Figure 7.8: An illustration of convection of thermal heat transfer

Thermodynamics 110

Radiation

The transfer of heat between a system and its environment via electromagnetic

waves is called thermal radiation. A very good example of thermal radiation is

visible light. The rate of thermal radiation Prad at which an object emits energy

via electromagnetic radiation depends on surface area A, temperature of that

area and emissivity . that has a value between 0 and 1, which surface dependent.

A surface that has maximum emissivity of 1 is said to be a blackbody radiator.

The rate of radiation Prad follows equation (7.8).

Prad = (7.8) 4AT/.

where / is the Stefan-Boltzmann constant that has value 5.67x10-8

J/s-m2-K

4.

Indeed Stefan-Boltzmann constant is a universal constant that applies to all

bodies, regardless of the nature of surface.

The rate of absorption Pabs of an object from its environment, which has

uniform temperature Tenv, is

Pabs = (7.9) 4

envAT/.

Owing to the fact that the object radiates energy to the environment and absorbs

from the environment, the net rate of radiation Pnet shall be

Pnet = 0 14

env

4 TTA )/. (7.10)

Example 4 The supergiant star Betelgeuse in constellation Orion has a surface temperature

about 2,900 K and emits a radiant power of approximately 4.0x1030

W. The

temperature is about half and the power is about 10,000 greater than that of the

sun. Assuming that Betelgeuse and sun have perfect emissivity and spherical

shape, calculate the radii of the supergiant and sun.

Solution

From Stenfan-Boltzmann equation, Prad = , the surface area A is A=4AT/.4

rad

T

P

/..

The radius R of supergiant shall be R = 4

rad

T4

P

2/.=

48

30

2900x10x67.5x4

10x0.4)2

=

2.82x1011

m, which is larger than the orbit of mar, which is 2.28x1011

m.

Thermodynamics 111

The radius of sun is R = 4

rad

T4

P

2/.=

48

26

5800x10x67.5x4

10x0.4)2

= 7.04x108 m.

Example 5 A wood-burning stove stands unused in a room where the temperature is 18

0C.

A fire is started inside the stove. Eventually the temperature of the stove surface

reaches a constant temperature of 1980C and the room warms to a constant

temperature of 290C. The stove has an emissivity of 0.9 and a surface area of

3.50 m2. Determine the net radiant power generated by the stove when the stove

(a) is unheated and has temperature equal to the room temperature (b) has

temperature of 1980C.

Solution

The power generated by unheated stove is Prad = = 5.67x104AT/. -

8x0.9x3.5x291

4 = 1,280.75 W.

The power absorbed from the surrounding by the stove is Prad = =

5.67x10

4AT/.-8

x0.9x3.5x2914 = 1280.75 W. The net power generated by the stove is

zero.

The net power generated by the stove when it is heated is Pnet =

0 14

env

4 TTA )/. = = 7304.1 W. )302471(10x67.5x5.3x9.0 448 ))

7.6 Thermodynamic Processes

A thermodynamic process is the way that a system changes from one state of

thermal equilibrium to another state such as the gas may be in thermodynamic

equilibrium for specified temperature, pressure, and volume. Heat transfer and

work done will change this equilibrium state to another.

Thermodynamic process can be divided into quasi-static (reversible)

processes and irreversible processes. If a thermodynamic system undergoes a

change from one state in thermal equilibrium to another state slowly enough

such that in any instant of time the system is in thermal equilibrium is said to be

reversible process. In reality this process does not exist. Any thermal process

that is not a quasi-static process is an irreversible process. In nature, all

thermodynamic processes are irreversible.

There are four special types of quasi-static processes that are useful in

equilibrium thermodynamics. The processes are characterized by the

Thermodynamics 112

thermodynamic parameter that is kept constant during process while other

parameters change. The types are: isothermal process where the temperature of

the system is kept constant, isobaric process where the pressure is kept constant,

isochoric process where the volume is kept constant and adiabatic process or

isentropic process where no heat is transfer in or out of the system from or to

the environment.

7.7 Laws of Thermodynamics

There are three principal laws of thermodynamics, which are zeroth law – the

thermodynamic equilibrium law, first law - internal energy law, second law -

entropy law. Each law leads to the definition of thermodynamic properties,

which help to understand and predict the operation of a physical system. There

is another law, which is third law state that it is impossible to reach zero Kelvin

temperature in finite number steps. This law shall not be discussed.

7.8 Zeroth Law of Thermodynamics

Zeroth law of thermodynamic involves some simple definitions of

thermodynamic equilibrium and temperature. It is observed that some properties

of an object, like the pressure in a volume of gas, the length of a metal rod, or

the electrical conductivity of a wire, can change when the object is heated or

cooled. Some of these phenomenons have been demonstrated in the earlier

section.

Zeroth law states that if two systems are at the same time in thermal

equilibrium with a third system, they are in thermal equilibrium with each other.

If the two objects are initially at different temperatures into physical contact,

they will eventually achieve thermal equilibrium. During the process of

reaching thermal equilibrium, heat is transferred between the objects and there

is a change in the property of both objects such as their internal energy.

Thermodynamic equilibrium leads to the large-scale definition of temperature.

When two objects are in thermal equilibrium they have the same temperature.

The details of the process of reaching thermal equilibrium are described in the

first and second laws of thermodynamics.

As an illustration in Fig. 7.5, object A and object B are in physical contact

and in thermal equilibrium. Object B is also in thermal equilibrium with object

C. There is initially no physical contact between object A and object C. But, if

object A and object C are brought into contact, it is observed that they are in

Thermodynamics 113

thermal equilibrium. This simple observation allows making of thermometers,

in which it can be calibrated to measure the change in a thermal property such

as the length of a column of mercury by putting the thermometer in thermal

equilibrium with a known physical system. If the thermometer is brought into

thermal equilibrium with any other system such as placing under tongue, the

temperature of the other system is known by noting the change in the thermal

property. Objects in thermodynamic equilibrium have the same temperature.

Figure 7.5: Illustration of Zeroth law of thermodynamic

7.9 First law of thermodynamics

First law of thermodynamics relates the various forms of energy, kinetic and

potential, in a system to the work, which a system can perform and to the

transfer of heat. It is used extensively in the discussion of heat engines.

The first law of thermodynamics defines the internal energy E is equal to

the difference of the heat transfer Q into a system and the work W done by the

system.

*U = U2 - U1 = Q – W (7.11)

Heat Q is absorbed by the system uses to increase the internal energy U2 – U1.

The internal energy is used to perform work done W, resulting decrease of

internal energy. This is the rationale how equation is formed.

Heat removed from a system would be assigned a negative sign in the

equation. Similarly work done on the system is assigned a negative sign. Figure

7.7 shows the interpretation of the law.

Thermodynamics 114

In chemistry textbook, first law is written as *U= Q + W. It is the same law,

of course - the thermodynamic expression of the conservation of energy

principle. It is just that W is defined as the work done on the system instead of

work done by the system.

In the context of physics, the common scenario is one adding heat to a

volume of gas and using the expansion of the gas to do work, as in case pushing

down of a piston in an internal combustion engine. In the context of chemical

reactions and process, it may be more common to deal with situations where

work is done on the system rather than by the system.

Figure 7.7: Interpretation of first law of thermodynamic

State 2 shows that there is decrease in volume, therefore work done is negative.

Work done W is also defined as . It is clearly shown that if V3'2

1

v

V

pdVW 2 < V1,

the work done is negative.

Example 6 The temperature of three moles of a monoatomic ideal gas is reduced from

temperature Ti = 540 K to Tf = 350 K by two different methods. In the first

method 5,500 J of heat flows into the gas, while in the second method, 1,500 J

of heat flows into it. In each case find the change of internal energy and (b)

work done by the gas.

SolutionSince the gas is monoatomic type, the internal energy of the gas is only the

translational kinetic energy, which follows equation nRT2

3. The change in

Thermodynamics 115

internal energy from temperature 540 K to 350 K shall be )TT(nR2

3i)f =

)540350(31.8x32

3) = -7,105 J.

First method shall yield W = Q - *U = 5,500 J +7,105 J = 12,605 J

Second method shall yield 1,500 J + 7,105 J = 8,605 J.

Let’s understand the relationship of specific heat capacity and first law of

thermodynamic and also look at some special process cases that applying first

Law of Thermodynamics.

7.9.1 Specific Heat Capacities and First Law of Thermodynamics

As mentioned in the earlier section the heat capacity is defined as Q = Cm*T.

The heat capacity for a gas shall then be Q = Cn*T. Instead of mass, it is now

replaced by number of mole.

The internal energy U of an ideal monoatomic gas is U = nRT2

3. The

change of internal energy *U from temperature Ti to Tf shall be *U =

)TT(nR2

3if ) .

For isobaric process, the work done is W = P(Vf – Vi). However, for ideal

gas PV = nRT. This shall mean that the work done is also W = nR(Tf – Ti). For

isochoric process, the work done is equal to zero since there is no volume

change.

Using the first law equation Q = *U + W, for isochoric process Q =

)TT(nR2

3if ) . This implies that the specific heat C is equal to R

2

3. We shall use

CV to denote specific heat for constant volume.

Similarly the heat supplied for isobaric process is equal to Q =

)TT(nR2

3if ) + nR(Tf – Ti). This shall mean that the specific heat capacity is

R2

5. We shall use CP to denote specific heat for constant pressure.

Thermodynamics 116

7.9.2 Thermodynamic Process

Let’s look at several thermodynamic processes in detail here.

Adiabatic process

The Adiabatic process is a process that occurs so rapidly that there is no transfer

of heat between the system and its environment. From first law of

thermodynamics with Q = 0 shows that the change in internal energy is in the

form of work done. Therefore, from equation (7.6) become U2 - U1 = – W.

Free expansion is an adiabatic process whereby there is no transfer of heat

between the system and its environment, no work done on and by the system.

Thus, from equation (7.6), Q = W = 0.

For adiabatic condition PV4 = constant K, where 4 is the ratio of Cp/Cv heat

capacity at constant pressure and constant volume. This implies that P = K/V4

and the work done shall be 3 4'

2

1

v

V

dVV

KW =

4)

) 4)4)

1

)VV(K 1

i

1

f . The pressure-volume

diagram is shown in Fig.7.8.

Figure 7.8: The work done graph of an adiabatic process

Note that for an ideal monatomic gas, 4 value is equal to 5/3.

Isochoric Process

Isochoric process is a constant volume process that there is no change in volume.

Thus, the work done is zero. Therefore, equation (7.6) becomes U2 - U1 = Q,

Thermodynamics 117

which is equal to CVn*T. From the ideal heat equation V*P = nR*T. This shall

mean Q = )PP(VR

CVif ) . The pressure-volume diagram is shown in Fig.7.9.

Figure 7.9: The work done graph of an isochoric process

Cyclical process

Cyclical process is a process after certain exchange of heat and work done, the

system restores back to its initial state. Thus, there is no change in its internal

energy. This implies that equation (7.6) becomes Q = W.

Isothermal Process

The isothermal process is a process whereby the temperature is kept constant

that is no change of internal energy. This implies that equation (7.6) becomes Q

= W. The work done is . Substituting P = 3'2

1

v

V

pdVWV

nRT into the equation

3'2

1

v

V

dVV

nRTW = nRT !!

"

#$$%

&

i

f

V

Vln . The pressure- volume diagram is shown in Fig.

7.10.

Figure 7.10: The work done graph of an isothermal process

Thermodynamics 118

Example 7

Two moles of the monatomic gas argon expand isothermally at temperature 298

K from an initial volume of Vi = 0.025 m3 to a final volume of Vf = 0.050 m

3.

Assuming that argon is an ideal gas, find (a) the work done by the gas, (b) the

change in the internal energy of the gas and (c) the heat supplied to the gas.

Solution

The work done by isothermal process is = nRTW !!"

#$$%

&

i

f

V

Vln = 2x8.31x298 !

"

#$%

&025.0

050.0ln

= 3,432.9 J.

The internal energy of U = nRT2

3. Since the temperature is constant, the change

of kinetic energy is zero.

Using first law equation, the heat supplied Q is 3,432.9 J.

Isobaric process

Isobaric process is process where the pressure is kept constant. The work done

W is W = P V = P(Vf – Vi). T = PV/(nR), thus the change in temperature T =

)VV(nR

Pi)f . The heat absorbed shall be Q = Cp )VV(

R

Pi)f . The PV diagram is

shown in Fig. 7.11.

Figure 7.11: The work done graph of an isobaric process

7.9.3 Thermodynamics Potentials

There are four thermodynamic potentials that are useful in the chemical

thermodynamics of reactions and non-cyclic processes. They are internal energy,

enthalpy, Helmholtz free energy, and Gibbs free energy.

Thermodynamics 119

Helmholtz free energy F states that F = U – TS, where T is the absolute

temperature and S is the final entropy. The system in an environment of

temperature T, energy can be obtained by spontaneous heat transfer to and fro

between environment and the system. Helmholtz free energy is a measure of the

amount of energy that put in to create a system once the spontaneous energy

transfer to the system from the environment is accounted.

Enthalpy is H defined as H = U + PV. It is analogue to first law of

thermodynamic Q = U + P V. It is a useful quantity for tracking chemical

reactions. In an exothermic reaction, energy is released to a system. It has

shown up in some measurable forms in terms of the state variables. An increase

in the enthalpy H = U + PV may be associated with an increase in internal

energy which can be measured by calorimeter, or with work done by the system,

or a combination of the two. If the process changes the volume, as in a chemical

reaction which produces gas, then work must be done to produce the change in

volume. For a constant pressure process the work done to produce a volume

change V is P V.

Gibbs free energy G states that G = U – TS + PV, where P is the absolute

pressure and V is the final volume. As discussed in enthalpy, an additional

amount of work PV must be done if the system is created from a very small

volume into large volume system. As discussed in Helmholtz free energy, an

environment at constant temperature T will contribute an amount TS to the

system, reducing the overall energy necessary for creating the system. This net

energy contribution for a system created in environment with temperature T

from an initial very small volume is Gibbs free energy.

The change in Gibbs free energy G, in a reaction is a very useful

parameter. It can be treated as the maximum amount of work obtainable from a

reaction. For example, in the oxidation of glucose, the change in Gibbs free

energy is G = 686 kcal = 2,870 kJ. This reaction is the main energy reaction in

living cells.

Example 8 Electrolysis of water in hydrogen and oxygen is a very good example for the

application of the above mentioned thermodynamic potentials. This process is

presumed is done at temperature 298K and one atmosphere pressure, and the

relevant values are taken from a table of thermodynamic properties shown

below.

Thermodynamics 120

The process must provide the energy for the dissociation and the energy to

expand the produced gases. Both of those are included in the change in enthalpy

which is shown in the table.

Quantity H2O H2 0.5 O2 Change

Enthalpy -285.83 kJ 0 0 H = 285.83 kJ

Entropy 69.91 J/K 130.68 J/K 0.5 x 205.14 J/K T S = 48.7 kJ

At temperature 298K and one atmosphere pressure, the system work is W =

P V = (101.3 x 103 Pa)(1.5 moles)(22.4 x 10

-3 m

3/mol)(298K/273K) = 3,715.0 J

Since the enthalpy H= U + PV, the change in internal energy U is then equal to

U = H - P V = 285.83 kJ - 3.72 kJ = 282.1 kJ.

This change in internal energy must be accompanied by the expansion of the

gases produced, so the change in enthalpy represents the necessary energy to

accomplish the electrolysis. However, it is not necessary to put in this whole

amount in the form of electrical energy. Since the entropy increases in the

process of dissociation, the amount T S can be provided from the environment

at temperature T. The amount must be supplied by the battery, which is actually

the change in the Gibbs free energy. i.e. G = H - T S = 285.83 kJ - 48.7 kJ =

237.1 kJ.

Since the electrolysis process results in an increase in entropy, the environment

contributes energy amounted to T S. Gibbs free energy tells the amount of

energy in other forms must be supplied to get the process to proceed.

Thermodynamics 121

7.10 Second Law of Thermodynamics

Ice cream melts and a cold can of soda get warm-up when they are left in hot

day. They never get colder when left in hot day. This spontaneous flow of heat

is the focus of one of the most profound law of science, which is the second law

of thermodynamics.

They are several ways that second law of thermodynamic can be defined.

Based on the above examples, the first statement of the second law of

thermodynamics - heat flows spontaneously from a hot to a cold body.

An explanation for this form of the second law can be obtained from

Newton's laws and the microscopic description of how heat flow through

conduction occurs when fast atoms collide with slow atoms, transferring some

of their kinetic energy in the process. One might wonder why the fast atoms

don't collide with the cool atoms and subsequently speed up, thereby gaining

kinetic energy as the cool atoms lose kinetic energy - this would involve the

spontaneous transfer of heat from a cool object to a hot object, in which it is the

violation of the second law. From the laws of conservation of momentum and

energy, in a collision between two objects, the faster object slows down and the

slower object speeds up.

It is possible to make a cool object in a warm place cooler like the

refrigerator but this involves the input of some external energy. As such, the

flow of heat is not spontaneous in this case.

The second form of the second law of thermodynamic states heat cannot be

completely converted into other forms of energy that places some practical

restrictions on the efficiency like internal combustion and steam powered

engines.

The third statement of second law of thermodynamic states that there exists

useful state variable called entropy, which is defined as the measure of number

of state variable for a system at a given time. The change in entropy *S is equal

to the heat transfer *Q divided by the temperature T.

*S = Sf – Si = *Q/T = 3f

iT

dQ 5 AvgT

Q (7.12)

Thermodynamics 122

The entropy of the system and the environment will remain a constant if the

process is reversible. If the initial and final states are denoted by Si and Sf then

Sf = Si for reversible system. An example of a reversible process would be

ideally forcing a flow through a constricted pipe. Ideal means no losses. As the

flow moves through the constriction, the pressure, temperature, and velocity

would change, but these variables would return to their original values

downstream of the constriction. The state of the gas would return to its original

conditions and the change of entropy of the system would be zero.

The second law also states that if the physical process is irreversible, the

entropy of the system and the environment MUST increase. The final entropy

must be greater than the initial entropy. An example of an irreversible process is

a hot object put in contact with a cold object. Eventually, they both attain the

same equilibrium temperature. If objects are separated after attaining thermal

equilibrium, they do not naturally return to their original different temperature

states. The process of bringing them to the same temperature is irreversible.

7.10.1 Heat Transfer from Hot Object to Cold Object

Let’s look at how the second law describes why heat is transferred from the hot

object with temperature T1 to the cold object with temperature T2. If heat is

transferred from the hot object to the cold object, the amount of heat transferred

is Q and the final equilibrium temperature for both objects is Tf. The

temperature of the hot object changes as the heat is transferred away from the

object. The average temperature Th of the hot object during the process is the

average of T1 and Tf. Similarly, for the cold object, the final temperature of the

cold object is Tf. The average temperature Tc during the process is the average

of Tf and T2. The entropy change for the hot object will be (-Q/Th), with the

minus sign applied because the heat is transferred away from the object. For the

cold object, the entropy change is (Q/Tc), positive value because the heat is

transferred into the object. So according to equation (7.12), the total entropy

change for the whole system would be given by the equation.

Sf - Si = -Q/Th + Q/Tc (7.13)

where Si and Sf being the final and initial values of the entropy.

Temperature Th is greater than temperature Tc, because T1 is greater than

T2. The term Q/Tc will always be greater than -Q/Th and therefore, Sf will be

greater than Si, as the second law predicts.

Thermodynamics 123

If the heat was being transferred from the cold object to the hot object, then

the final equation would be Sf = Si + Q/Th -Q/Tc. The signs on the terms would

be changed because of the direction of the heat transfer. Th would still be greater

than Tc, and this would result in Sf being less than Si. The entropy of the system

would decrease, which would violate the second law of thermodynamics.

7.10.2 Heat Engines

A heat engine is a device that uses heat to perform work. Essentially it has three

features. It has a hot reservoir that supplies heat. Part of the heat is used to

perform work by the working substance of the engine like gasoline- air mixture

in the automobile engine. The remaining part of the heat is rejected at

temperature lower than the input temperature called cold reservoir. The

schematic of a heat engine is shown in Fig. 7.12.

Figure 7.12: A schematic representation of a heat engine

For an engine to be efficient, it must produce a relatively large amount of work

from the input heat. The efficiency e of a heat engine is defined as the ratio of

the work done by the engine to the input heat QH. i.e. e = HQ

W. However, QH is

also equal to QH = QC + W, thus, the efficiency is

Thermodynamics 124

e = H

C

Q

Q1) (7.14)

Example 9 An automotive engine has an efficiency of 22.0% and produces 2,510 J of work.

How much heat is rejected?

Solution

From equation (7.12), The heat rejected is QC = QH(1- e). But QH is equal to QC

+ W. Thus, QC = (QC + W)(1 - e). This implies that QC = W/e - W =

2,510(1/0.22 - 1). = 8,899.0 J.

7.10.3 Carnot Engine

According to a French engineer Sadi Carnot, in order to get maximum

efficiency for an engine, the process within the engine must be reversible. This

shall mean that the process will return to its initial states with no wasteful

transfer of energy. Figure 7.13 shows the pressure-volume plot of the Carnot

engine.

Figure 7.13: Pressure-volume diagram of a Carnot engine

During the process step a to b, which is an isothermal expansion, heat QH is

absorbed by the working substance from the hot reservoir. Since the process is

an isothermal expansion process, the work done by the system is also equal to

heat absorbed. During the process step c to d, the working substance is releasing

heat QC to the cold reservoir. Since this is an isothermal compression, work is

Thermodynamics 125

done by the environment on the system. From concept of heat engine, there is

no heat transfer between hot and cold reservoir directly. Thus, the process step

bc and de must be the adiabatic process. The work done by the Carnot process

shall be the orange color area of the graph shown in Fig. 7.13.

According to first law of thermodynamic, *U = Q –W. Since the process is

reversible, therefore the change of internal energy *U is zero. Also for

differential change dQ = dW. The work done W shall be W = QH – QC.

From process step a to b there is a positive change of entropy *SH = QH/TH

and during the process step c to d, there is a negative change of entropy *SL =

QL/TL. There is no change of entropy for process step bc and da because there is

not heat absorbed or released. The change of entropy shall be *S = QH/TH -

QL/TL. Since Carnot engine is a reversible engine, the change of entropy shall

be zero. This implies that QH/TH = QL/TL. From the equation TH > TC, this shall

mean QH > QC. This analysis confirms the second statement of second law of

thermodynamic.

The efficiency of the Carnot engine shall be e = W/QH = (QH – QC)/QH = 1-

TL/TH.

7.10.4 Refrigeration

Refrigeration is a device that uses work to transfer energy from low temperature

reservoir to high temperature reservoir. Figure 7.14 shows the schematic of a

refrigerator.

Figure 7.14: The schematic of a refrigerator

Thermodynamics 126

A measure of the efficiency of a refrigerator is the coefficient of performance K,

which is defined as the ratio of heat extracted from the cold reservoir to the

work done. For an ideal Carnot refrigerator the coefficient of performance K

=LH

L

QQ

Q

). The change of entropy for ideal refrigerator shall be *S =

H

L

L

L

T

Q

T

Q-) .

Notice that TH >TL, implying the *S is a negative value, which is not allow for

second law of thermodynamic. Therefore, there is no perfect refrigerator.

Thermodynamics 127

Tutorials

1. A certain diet doctor encourages people to diet by drinking ice water. His

theory is that the body must burn off enough fat to raise the temperature of

water from 0.00C to the body temperature of 37.0

0C. How many liters of ice

water would have to be consumed to burn-off 454 g of fat, assuming that this

much fat required 3,500 Cal be transferred to ice water? Why this is not

advisable to follow this diet.

2. A tank of water has been outdoor in cold weather and a slab of ice 4.0 cm

thick has formed on its surface. The air above the ice is –100C. Calculate the

rate (centimeter per sec) of ice formation. Take the thermal conductivity of

the ice to be 0.0040 cal/s-cm-K and density 0.92 g/cm3. Assume the energy

is transferred through the wall or bottom of the tank is negligible.

3. Find the change of entropy when a 2.3 kg block of ice melts slowly

(reversible) at 273 K.

4. 1200 J of heat flowing spontaneously through a copper rod from a hot

reservoir at 650 K to a cold reservoir at 350 K. Determine the amount the

change in entropy for this irreversible process.

The change in entropy shall be -1200/650 + 1200/35 = 1.58 J/K.

5. A Carnot engine operates between temperature TH = 850 K and TL = 300 K.

The engine performs 1200 J of work for each cycle that takes 0.2 s.

Thermodynamics 128

(a). Find the efficiency of the engine.

(b). How much heat QH is extracted from high temperature reservoir for

each cycle? And QL delivered to cold reservoir?

(c). What is the entropy change of the working substance for the energy

transfer to it from the high temperature reservoir? From it to the low

temperature reservoir?

6. If an inventor claims that he has invented an engine that has an efficiency of

75% when operated between boiling point and freezing point. Is it possible?