thermodynamics 1 law-closed-system

ThermodynamicsChapter: First Law of Thermodynamics (Control Mass System)

Prof. Vinod M. MakwanaAPME, GEC, Bhuj

Objectivediscuss energy conservation principle and mode of energy

transferdescribe the boundary work for various thermodynamic

processesdefine the first law of thermodynamics (cyclic process and

process, enthalpy mechanical equivalent of heat (Joulesconstant), PMM-I, specific heat, internal energy andenthalpy.

explain about the Joules experiment and internal energy.prove that energy is property of system.create the difference between two specific heatdetermine the internal energy and enthalpy for solid, liquid

and gas.

CONSERVATION OF MASS PRINCIPLE

• The conservation of mass principle states the following:

• Net mass transfer to or from a system during a process isequal to the net change in the total mass of the systemduring that process

Conservation Of Mass Principle

massinchangeNet

system

transfermasstotalNet

outin mmm D

in the rate form, as

massinchangeofrate

system


outin dt/dmmm

min = total mass enter in to systemmout = total mass leave from systemDmsystem = net change of mass within system

MECHANISMS OF ENERGY TRANSFEREIN AND EOUT

1. Heat Transfer, Q Heat transfer to a system (heat gain) increasesthe energy of the molecules (system) and thus the internalenergy of the system, and heat transfer from a system (heat loss)decreases energy of the molecules (system).

2. Work Transfer, W An energy interaction that is not caused by atemperature difference between a system and its surroundings iswork. Work transfer to a system (i.e., work done on a system)increases the energy of the system, and work transfer from asystem (i.e., work done by the system) decreases energy of thesystem.

3. Mass Flow, m

Mass flow in and out of the system serves as an additional mechanismof energy transfer. When mass enters a system, the energy of thesystem increases because mass carries energy with it (in fact, mass isenergy). Likewise, when some mass leaves the system, the energycontained within the system decreases energy of system because theleaving mass takes out some energy with it.

TYES OF HEAT TRANSFER AND WORK TRANSFER

ENERGY CONSERVATION PRINCIPLE

Consider a room whose door and windows aretightly closed, and whose walls are well-insulatedso that heat loss or gain through the walls isnegligible.Now let’s place a refrigerator in the middle of theroom with its door open, and plug it into a walloutlet (Fig.). You may even use a small fan tocirculate the air in order to maintain temperatureuniformity in the room.Now, what do you think will happen to theaverage temperature of air in the room? Will itbe increasing or decreasing? Or will it remainconstant?

Energy Conservation Principle

Entire room—including the air and the refrigerator—as the system,which is an adiabatic closed system since the room is well-sealed andwell-insulated, the only energy interaction involved is the electricalenergy crossing the system boundary and entering the room.

EE Flow path is from Electric Motor to Air of Room viaRefrigerator

The refrigerator or its motor does not store this energy. Therefore, thisenergy must now be in the room air, and it will manifest itself as arise in the air temperature. The temperature rise of air can becalculated on the basis of the conservation of energy principle usingthe properties of air and the amount of electrical energy consumed.

1) “Energy can neither be created nor destroyed during aprocess, but can be converted from one form to anotherform.”

2) “Energy of the universe is constant”.

Example


Noting that energy can be transferred in the forms of heat,work, and mass, and that the net transfer of a quantity is equalto the difference between the amounts transferred in and out.

systemout,massin,massoutinoutinoutin EEEWWQQEE D


Energy balance can be written more explicitly as

Energy balance for any system undergoing any kind of processwas expressed as

energiesetcpotentialkineticernalinChange

system

massandworkheatbytransferenergyNet

outin EEE

.,int

,

D



system

massandworkheatbytransferenergyNet

outin dtEEE

.,int

,

/D


Energy balance can also be expressed in the differential formas

systemoutin dEEE

systemout,massin,massoutinoutinoutin EEEWWQQEE D

TYPES OF WORK

Displacement (boundary) work 2

1

PdVWb

Electrical work 2

1

VIdWe

Shaft work 2

1

TdWsh 2

1

TwWsh

VIWe

Work in elastic solid rod 2

1

AdxW nelasic

We have already noted that there are a variety of ways in which work can be done on or by a system.

Work in with the Stretchingof a Liquid Film

2

1

dAW ssurface

BOUNDARY WORK

The work associated with a movingboundary is called boundary work.Therefore, the expansion and compressionwork is often called moving boundarywork, or simply boundary work (Fig.).Some call it the P dV work.

The moving boundary work associated with real engines or compressorscannot be determined exactly from a thermodynamic analysis alonebecause the piston usually moves at very high speeds, making it difficultfor the gas inside to maintain equilibrium. Then the states through whichthe system passes during the process cannot be specified, and no processpath can be drawn. Work, being a path function, cannot be determinedanalytically without a knowledge of the path. Therefore, it is determinedby direct measurements in real engines or compressors

Boundary Work

In this section, we analyze the moving boundary work for aquasi-equilibrium process, a process during which the systemremains nearly in equilibrium at all times. A quasi-equilibrium process, also called a quasistatic process, isclosely approximated by real engines, especially when thepiston moves at low velocities.

Under identical conditions, the work output of the engines isfound to be a maximum, and the work input to thecompressors to be a minimum when quasi-equilibriumprocesses are used in place of nonquasi-equilibrium processes.

Boundary Work

Consider the gas enclosed in the piston–cylinderdevice shown in Fig.. The initial pressure of thegas is P, the total volume is V, and the cross-sectional area of the piston is A. If the piston isallowed to move a distance ds in a quasi-equilibrium manner, the differential work doneduring this process is

PdVPAdsFdsWb

Note in Eq. that P is the absolute pressure, which is always positive.However, the volume change dV is positive during an expansion process(volume increasing) and negative during a compression process (volumedecreasing). Thus, the boundary work is positive during an expansionprocess and negative during a compression process.

Boundary Work

PdVPAdsFdsWb The total boundary work done during theentire process as the piston moves isobtained by adding all the differentialworks from the initial state (1) to the finalstate (2):

2

1

PdVWb

This integral can be evaluated only if we know the functionalrelationship between P and V during the process. That is, P = f (V)should be available. Note that P = f (V) is simply the equation of theprocess path on a P-V diagram.

Boundary Work

A gas can follow several different paths asit expands from state 1 to state 2. Ingeneral, each path will have a differentarea underneath it, and since this arearepresents the magnitude of the work, thework done will be different for eachprocess (Fig.). This is expected, sincework is a path function (i.e., it dependson the path followed as well as the endstates).

If work were not a path function, no cyclic devices (car engines, powerplants) could operate as work-producing devices. The work produced bythese devices during one part of the cycle would have to be consumedduring another part, and there would be no net work output.

Boundary Work

The quasi-equilibrium expansion processdescribed is shown on a P-V diagram inFig.. On this diagram, the differential areadA is equal to P dV, which is thedifferential work. The total area A underthe process curve 1–2 is obtained byadding these differential areas:

2

1

2

1

PdVdAAWb

Eq. reveals that the area under theprocess curve on a P-V diagram is equal,in magnitude, to the work done during aquasi-equilibrium expansion orcompression process of a closed system.(On the P-v diagram, it represents theboundary work done per unit mass.)

GENERALIZED BOUNDARY WORK

Quasi-static Process: System is quasi-equilibrium states. Itbecomes equal to the pressure of the gas in the cylinder andthus the entire gas in the cylinder is at the same pressure at anygiven time. Pi = Ps = Pg = P

Strictly speaking, the pressure P in Eq. is the pressure atthe inner surface of the piston.

Nonquasi-static Process: The pressure at the inner face of thepiston is used for P. (Pressure of gas and system pressure cannot measured during nonquasi-equilibrium process becausesystem are not in equilibrium states)Pi ≠ Ps ≠ Pg

Generalized Boundary Work

we can generalize the boundary work relation by expressing itas

2

1

dVPW ib

Quasi-static ProcessPi = Ps = Pg = P

Nonquasi-static ProcessPi ≠ Ps ≠ Pg

Use Pi

BOUNDARY WORK FOR VARIOUS THERMODYNAMIC PROCESS

1. Reversible Constant Volume (or Isochoric) Process (v = constant) :In a constant volume process the working substance is contained in arigid vessel, hence the boundaries of the system are immovable and nowork can be done on or by the system, other than paddle-wheel workinput. It will be assumed that ‘constant volume’ implies zero work unlessstated otherwise.

02

1

PdVWb

2. Reversible Constant Pressure (or Isobaric) Process (p =constant).

Boundary Work For Various Thermodynamic Process

12

2

1

VVPPdVWb

12

12

vvmPW

VVPW

b

b

2. Reversible Constant Pressure (or Isobaric) Process (p =constant).

3. Reversible Temperature (or Isothermal) Process (pv = constant, T =constant) :A process at a constant temperature is called an isothermal process. When aworking substance in a cylinder behind a piston expands from a high pressure toa low pressure there is a tendency for the temperature to fall. In an isothermalexpansion heat must be added continuously in order to keep the temperature atthe initial value.


2

1

2

1

dVV

CPdVWb

CmRTPV

Equation of state for ideal gas, For isothermal process,T=Const. T1 = T2

V

CP

1

2

2

1V

VlnC

V

dVCWb

CVPVPPV 2211

CPV

1

211

V

VlnVPWb

Such a process can be reversible or irreversible. Thereversible adiabatic non-flow process will be considered inthis section.

4. Reversible Adiabatic Process ( pvγ = constant) :An adiabatic process means no heat is transferred to or fromthe fluid during the process.


In an adiabatic expansion, the work done W by the fluid is atthe expense of a reduction in the internal energy of the fluid.Similarly in an adiabatic compression process all the workdone on the fluid goes to increase the internal energy of thefluid.

For an adiabatic process, provide the perfectthermal insulation for the system

Reversible Adiabatic Process

2

1

2

1

dVV

CPdVWb

V

CPCPV

2

1

dVVCWb

11

1

1

1

2 VVCWb

11

1

1

1

2 VVCWb

11

1

1

1

2 CVCVWb

11

1

111

1

222 VVPVVPWb

111122 VPVP

Wb

12211

VPVPWb

1

21

TTmRWb

5. Polytropic Reversible Process (pvn = constant) :It is found that many processes in practice approximate to areversible law of form pvn = constant, where n is a constant.Both vapours and perfect gases obey this type of law closely inmany non-flow processes. Such processes are internallyreversible.

2

1

2

1

dVV

CPdVW

nb

nV

CP CPV n

2

1

dVVCW nb

11

1

1

1

2

n

V

n

VCW

nn

b

n

CVCVW

nn

b1

1

1

1

2

5. Polytropic Reversible Process (pvn = constant) :

n

CVCVW

nn

b1

1

1

1

2

n

VVPVVPW

nnnn

b1

1

111

1

222

n

VPVPWb

11122

12211

n

VPVPWb

1

21

n

TTmRWb

Property relation in reversible polytropic and adiabaticprocesses

n

V

V

P

P

2

1

1

2

n

n

P

P

T

T1

1

2

1

2

1

2

1

1

2

n

V

V

T

T

2

1

1

2

V

V

P

P

1

1

2

1

2

P

P

T

T

1

2

1

1

2

V

V

T

T

Closed System

CONSERVATION OF MASS PRINCIPLE


massinchangeofrate

system


outin dt/dmmm

0dt

dmsystem Constmsystem

energies.etcpotentialkinetic,ernalintinChange

system

massandwork,heatbytransferenergyNet

outin EEE D

Consider energy conservation principle


Noting that a closed system does not involve any mass flowacross its boundaries hence the possible energy transfer atboundary is heat and work


system

workandheatbytransferenergyNet

outin EEE

.,int

D

For a closed system undergoing a cycle, the initial and final states are identical, and thus

012 D EEE system

0

.,int

D


system

workandheatbytransferenergyNet

outin EEE

0 outin EE outin EE

out,netin,net WQ OR in,netout,net WQ

The net work output during a cycle is equal to net heat input

outinin,net QQQQ inoutoutnet WWWW ,


1. For a closed system undergoing a cycle

OR

in,netout,net QW

QW

For a system (control mass) undergoing a cycle , the cyclicintegral of the work is proportional to the cyclic integral ofthe heat.

QJW

0 QJW

1 LAW FROM ENERGY PRINCIPLE

1-Law for Cyclic ProcessThe first law states that when heat and work interactions take place

between a closed system and the environment, the algebraic sum ofthe heat and work interactions for a cycle is zero.

OR“When a system undergoes a thermodynamic cycle then the net heatsupplied to the system from the surroundings is equal to net work doneby the system on its surroundings.

(ΣδQ)cycle ∝ (ΣδW)cycle or (ΣδW)cycle= J (ΣδQ)cycle

J is called the mechanical equivalent of heat or Joule’s equivalent; itexpresses the number of work units, which are equivalent to one unit.For SI unit the value of J is unity. If kJ and kcal are chosen as the units ofwork and heat respectively, then the value of J is 4.1868 kJ/kcal.

PROOF: JOULE’S EXPERIMENT

He used a paddle to stir an insulated vessel filled with fluid.The amount of work done on the paddle was noted (the workwas done by lowering a weight, so that work done =mgx).Mechanical work converted into heat energy via kineticenergy. Hence temperature of fluid increased.

TCpwaterofmassQmgx DQ value is small hence we repeated number of times (n) araising and falling of this weight.

TCpwaterofmassmgxn D

Proof: Joule’s Experiment

Cp is specific of heat.

If we take 1 kg of water in vessel. This experiment repeatedtill the temperature of water is increased by one unit ( 10

degree Celsius)

TCpwaterofmassmgxn D

Twaterofmass

mgxnCp

D


This process carried out number of times till temperature offluid increased by 10 degree for unit mass of water.

Joule found that when falling weight lost 4.186 mechanicalenergy lost, it raised 10 degree temperature for unit mass ofwater.

ttanConsJoulekJ.

Cp

11

1864

kJ.kcal 18641


Joule’s Mechanical Equivalent of Heat

F

m

Dx

This proved 1 kcal = 4,186 J

1 kg H2O

DT = 1oC

E = FDx = 4.186 kJ


Later, this vessel was placed in a cool bath and to be cooled.Heat Q was transferred to the fluid bath as seen throughthermometer until the original state of the water system wasreached.

The amount of heat transferred was measured from theincrease in energy of fluid. The energy involved in increasingthe temperature of the fluid was shown to be equal to thatsupplied by the lowered weight.


Energy lost by lowered weight = Energy gain by waterEnergy lost by water = Energy gained by fluid.

System reached back to its original condition and watersystem is undergone to a complete cycle.

He then repeated the experiments by taking different systemsand surroundings. Joule found that ratio of work done on thesystem to heat generated was a constant. Mathematically.

ttanConsJoules,JQ

W


ttanConsJoules,JQ

W

0 QJW

If work and heat is expressed in the same unit (Joule) then 1J = 1Nm.

0 QW OR QW


E = U + K.E. + P.E.+M.E.+E.E.+…… and is called the total energy of the system.

systemEWQ D

2. For a closed system undergoing a process (non cyclic process)

outinin,net QQQQ

inoutoutnet WWWW ,

where

The net change (increase ordecrease) in the total energy ofthe system during a process isequal to the difference betweenthe total energy entering and thetotal energy leaving the systemduring that process. That is, Ein –Eout =ΔEsystem

1 LAW FROM ENERGY PRINCIPLE

1-Law for Process (Closed system)

It is possible to state the first law for a non-cyclic process.

So there are 2 energy interactions to the gas,it will increase by a quantity Q, because it isabsorbing energy. And it will decrease by aquantity W since it is losing energy by doingsome work. So you can write change inenergy (E) of gas as follows.

Q - W = DE

In differential form as follows. It is in form ofrate of change of quantities per unit time.

WQdt

dE dt

dE

dt

W

dt

Q

(Q - W) is the property of system.

00 WQWQ

(Q - W) = dEdE =Q - W

E is the total energy of system and it is the property of system.Integrating between state 1 to 2

2

1

2

1

2

1

WQdE122121 EEWQ Q - W = DE

PROOF: JOULE’S EXPERIMENT

PROOF: ENERGY IS PROPERTY OF SYSTEM

Consider a system which changes its state from state 1 to state 2 byfollowing the path L, and returns from state 2 to state 1 by following thepath M (Fig). So the system undergoes a cycle.

Writing the first law for path L ,QL = ΔEL + WL ...(1)and for path M , QM = Δ EM + WM ...(2)The processes L and M together constitute acycle, for which (dW)C = (dQ)C

WL + WM = QL + QM

hence QL – WL = WM – QM ...(3)From equations (1), (2) and (3), it yields

Δ EL = – Δ EM ...(4)

Similarly, had the system returned fromstate 2 to state 1 by following the path Ninstead of path M

Δ EL = – Δ EN ...(5)

From equations (4) and (5),

Δ EM = Δ EN ...(6)

Thus, it is seen that the change in energy between two states ofa system is the same, whatever path the system may follow inundergoing that change of state. Therefore, energy has adefinite value for every state of the system. Hence, it is apoint function and a property of the system.

Proof: Energy Is Property Of System

EXPANSION OF FIRST LAW

According to the first law, for a process Q - W = dE

2

1

2

2121221212

1CCmzzmgUUWQ

If the electric, magnetic and chemical energies are absent andchanges in potential and kinetic energy for a closed systemare neglected, the above equation can be written asso theimportant term is the internal energy function U.

12 UUUWQ D Non-Flow Energy Equation

INTERNAL ENERGY

Internal energy is defined earlier as the sum of all themicroscopic forms of energy of a system. It is related to themolecular structure and the degree of molecular activityand can be viewed as the sum of the kinetic and potentialenergies of the molecules.

The molecules of a gas move throughspace with some velocity, and thuspossess some kinetic energy. This isknown as the translational energy.The atoms of polyatomic moleculesrotate about an axis, and the energyassociated with this rotation is therotational kinetic energy.

Internal Energy

The electrons in an atom rotate about thenucleus, and thus possess rotational kineticenergy. Electrons also spin about their axes, andthe energy associated with this motion is thespin energy. Other particles in the nucleus of anatom also possess spin energy.

The portion of the internal energy of a systemassociated with the kinetic energies of themolecules is called the sensible energy (Fig.).

The atoms of a polyatomic molecule may also vibrate about theircommon center of mass, and the energy associated with this back-and-forth motion is the vibrational kinetic energy. For gases, the kineticenergy is mostly due to translational and rotational motions, withvibrational motion becoming significant at higher temperatures.

Internal EnergyThe internal energy associated withthe atomic bonds in a molecule iscalled chemical energy.

The tremendous amount of energyassociated with the strong bondswithin the nucleus of the atom itselfis called nuclear energy.

A chemical reaction involves changes in thestructure of the electrons of the atoms, but anuclear reaction involves changes in the core ornucleus. Therefore, an atom preserves its identityduring a chemical reaction but loses it during anuclear reaction.

Internal Energy is Molecular Energy

INTERNAL ENERGY VS. HEAT

• The term heat refers is the energy that is transferred from onebody or location due to a difference in temperature. This issimilar to the idea of work, which is the energy that istransferred from one body to another due to forces that actbetween them. Heat is internal energy when it is transferredbetween bodies.

• Technically, a hot potato does not possess heat; rather itpossesses a good deal of internal energy on account of themotion of its molecules. If that potato is dropped in a bowl ofcold water, we can talk about heat: There is a heat flow (energytransfer) from the hot potato to the cold water; the potato’sinternal energy is decreased, while the water’s is increased bythe same amount.

TEMPERATURE VS. INTERNAL ENERGY

Which has more internal energy, a bucket of hot water or abucket of cold water? answer:

The bucket of hot water has more internal energy, at least ifthe buckets contain the same amount of water.

Internal energy depends on the amount (mass) of substanceand the kinetic energy of the molecules of the substance.

Temperature only depends on the molecules’ kinetic energy;it is independent of mass.

Corollary 1.

• The internal energy of a closed system remainsunchanged if the system is isolated from itssurroundings.

• If the system is isolated from the surroundings, Qand W are both zero and hence, ΔU must be zero.The system represented in Fig. below is anexample of an isolated system. All that happens inthis case is a spontaneous redistribution of energybetween parts of the system, which continues untila state of equilibrium is reached; there is nochange in the total quantity of energy within thesystem during the process.

• Corollary 1 is often called the Law ofConservation of Energy.

Corollary 2.

But as per First Law no net amount of workcan be delivered by the system if a netamount of heat is not supplied by thesurroundings to system during a cycle. Thusthe First law implies that a perpetualmachine of first kind is impossible.

A perpetual motion machine of the first kind (PMM I) isimpossible.

Definition: PMM1 is a device which delivers workcontinuously without any energy input.

ENTHALPYEnthalpy (h).h = u + pv— The enthalpy of a fluid is the property of the fluid, since itconsists of the sum of a property and the product of the twoproperties. Since enthalpy is a property like internal energy,pressure, specific volume and temperature, it can beintroduced into any problem whether the process is a flow or anon-flow process.

The total enthalpy of mass, m, of a fluid can beH = U + pV, where H = mh.

(Note that, since it has been assumed that u = 0 at T = 0, then h = 0 at T= 0).

THE PERFECT GAS

The Characteristic Equation of State

At temperatures that are considerably in excess of critical temperatureof a fluid, and also at very low pressure, the vapour of fluid tends toobey the equationPv/T = constant = R

In practice, no gas obeys this law rigidly, but many gases tend towards

it.

An imaginary ideal gas which obeys this equation of a state is called aperfect gas, and the equation Pv/T = R, is called the characteristicequation of a state of a perfect gas.

MOLAR CHARACTERISTIC EQUATION OF A STATE FOR A PERFECT GAS

The constant R is called the gas constant. Each perfect gas has adifferent gas constant.

Units of R are Nm/kg K or kJ/kg K.

Usually, the characteristic equation is written aspv = RT ...(1)or for m kg, occupying V m3

pV = mRT ...(2)

As per definition of the kilogram-mole, for m kg of a gas, we havem = nM ...(3) where n = number of moles.

Note. Since the standard of mass is the kg, kilogram-mole willbe written simply as mole.

Substituting for m from eqn. (3) in eqn. (2) gives pV = nMRTor MR = pV/nTAccording to Avogadro’s hypothesis the volume of 1 mole of any gasis the same as the volume of 1 mole of any other gas, when thegases are at the same temperature and pressure. Therefore, V/n is thesame for all gases at the same value of p and T. That is the quantity(pV)/(nT) is a constant for all gases. This constant is calleduniversal gas constant, and is given the symbol, R0.i.e., MR = Ru = pV/nTorpV = nRuT ...(4)

Molar Characteristic Equation of a State for Perfect Gas

pV = nRuT ...(4)

It has been found experimentally that the volume of 1 mole of anyperfect gas at 1 bar and 0°C is approximately 22.71 m3.

Therefore from eqn. (4),Ru = pV/nT = (1*105 *22.71)/(1* 273.15) = 8314.3 Nm/mole K

The gas constant for any gas can be found when the molecularweight is known.

Example. For oxygen which has a molecular weight of 32, thegas constantR = Ru/M = 8314/32 = 259.8 Nm/kg K.

Molar Characteristic Equation of a State for Perfect Gas

SPECIFIC HEAT

The specific heat is defined as the energy required to raise thetemperature of a unit mass of a substance by one degree.

In general, this energy depends onhow the process is executed.

In thermodynamics, we areinterested in two kinds of specificheats: specific heat at constantvolume Cv and specific heat atconstant pressure Cp.

Specific Heat

The specific heat at constant volumeCv can be viewed as the energyrequired to raise the temperature ofthe unit mass of a substance by onedegree as the volume is maintainedconstant.The specific heat at constantpressure Cp can be viewed as theenergy required to raise thetemperature of the unit mass of asubstance by one degree as thepressure is maintained constant.

Specific Heat at Constant Volume

First, consider a fixed mass in a stationary closed systemundergoing a constant-volume process (and thus no expansionor compression work is involved).

duwq

Apply the first law of process for closed system

1...duqBut amount of heat transfer during constant volume process is

2....dTCq vComparing both equation

dTCdu v cv

vT

uC

Specific Heat at Constant Pressure

First, consider a fixed mass in a stationary closed systemundergoing a constant-pressure process (and thus expansion orcompression work is involved).

duwq Apply the first law of process for closed system

dupdvq

1.....dhpvudpdvduq

But amount of heat transfer during constant pressure process is

2....dTCq p

Comparing both equation

dTCdh p cp

pT

hC

Specific Heat

Note that Cv and Cp areexpressed in terms of otherproperties; thus, they must beproperties themselves

JOULE’S LAW

Joule’s law states as follows :“The internal energy of a ideal gas is a function of the absolutetemperature only.”

u =u(T)

Using the definition of enthalpy and the equation of state of anideal gas, we have

RTuhRTPv

pvuh

Since R is constant and u = u(T)

h =h(T) Ideal gas

Ideal gas

JOULE’S LAW

Joule’s law states as follows :“The internal energy of a ideal gas is a function of the absolutetemperature only.”

u =u(T)

According to Joule’s law u = u(T), which means that internalenergy varies linearly with absolute temperature. Internalenergy can be made zero at any arbitrary referencetemperature.

For a perfect gas it can be assumed that u = 0 when T = 0.

Ideal gas

ENTHALPY FOR IDEAL GAS

Joule’s law states as follows :

u =u(T)

Using the definition of enthalpy and the equation of state of anideal gas, we have

RTuhRTPv

pvuh

Since R is constant and u = u(T)

h =h(T) Ideal gas

Ideal gas

DETRMINATION OF IE AND ENTHALPY

Since u and h depend only on temperature for an ideal gas, thespecific heats Cv and Cp also depend, at most, on temperatureonly. Therefore, at a given temperature, u, h, Cv, and Cp of anideal gas have fixed values regardless of the specific volumeor pressure.

u=u(T)h =h(T)Cv=Cv(T)Cp=Cp(T)

Ideal gas

Thus, u, h, Cv and Cp varies with temperature for ideal gases

Determination Of IE And Enthalpy

Thus, for ideal gases, the partial derivatives in above can be replaced byordinary derivatives.

cp

pT

hC

cv

vT

uC

Then the differential changes in the internal energy andenthalpy of an ideal gas can be expressed as

cv

vdT

duC

cp

pdT

dhC

dTTCdu v dTTCdh p


The change in internal energy or enthalpy for an ideal gasduring a process from state 1 to state 2 is determined byintegrating these equations:

kg/kJdTTCuuu vD2

1

12

kg/kJdTTChhh pD2

1

12

Valid for a ideal gas between two states for any process,reversible or irreversible.


In the ideal-gas tables given in the appendix, zero Kelvin ischosen as the reference state, and both the enthalpy and theinternal energy are assigned zero values at that state.

At low pressures, all real gases approach ideal-gas behaviour, andtherefore their specific heats depend on temperature only. The specificheats of real gases at low pressures are called ideal-gas specific heats, orzero-pressure specific heats, and are often denoted Cp0 and Cv0.

The integrations in Du and Dh are straightforward but rathertime-consuming and thus impractical.

kg/kJdTTCuuu vD2

1

12 kg/kJdTTChhh pD

2

1

12


Therefore the specific heat functions in Eqs. can be replacedby the constant average specific heat values.

First, the specific heats of gaseswith complex molecules (moleculeswith two or more atoms) are higherand increase with temperature.Also, the variation of specific heatswith temperature is smooth and maybe approximated as linear oversmall temperature intervals (a fewhundred degrees or less).


The ideal-gas specific heats of monatomic gases such as argon, neon,and helium remain constant over the entire temperature range. Thus, uand h of monatomic gases can easily be evaluated from above equations.

kg/kJTTCuuu avg,v 1212 D

kg/kJTTChhh avg,p 1212 D

The specific heat values for some commongases are listed as a function of temperaturein Table.The average specific heats Cp,avg andCv,avg are evaluated from this table at theaverage temperature (T1 + T2)/2, as shownin fig.


On the contrary, the relation is valid for any ideal gas undergoing anyprocess .

Above equation are valid for all processes.

kg/kJdTTCuuu vD2

1

12

kg/kJdTTChhh pD2

1

12

kg/kJTTCuuu avg,v 1212 D

kg/kJTTChhh avg,p 1212 D

Perfect gases has constant specific heat whereas ideal gas hasspecific heat which is function of temperature.For mass m, for a perfect gas, dU = m Cv dTFor a perfect gas, in any process between states 1 and 2,

Valid for a perfect gas between two states for any process,reversible or irreversible.

Determination Of IE And Enthalpy (Perfect gas)

kg/kJTTCdTTCuuu vv 12

2

1

12 D

kg/kJTTCdTTChhh pp 12

2

1

12 D

RELATIONSHIP BETWEEN TWO SPECIFIC HEATS

Consider a ideal gas

RTupvuh

differentiating the relation

RdTdudh

But dTCdu&dTCdh vp

RdTdTCdTC vp Kkg/kJRCC vp

Molar basisKkmol/kJRuCC vp

RATIO OF SPECIFIC HEAT

The ratio of specific heat at constant pressure to the specificheat at constant volume is given the symbol γ (gamma).

Since Cp = Cv + R, it is clear that Cp must begreater than Cv for any perfect gas.

v

p

C

C

The specific ratio also varies with temperature, but thisvariation is very mild.

Hence, Cp/Cv = γ is always greater than unity.

Ratio of Specific Heat

In general, the approximate values of γ are as follows :

For monoatomic gases such as argon, helium = 1.6.For diatomic gases such as carbon monoxide, hydrogen,nitrogen, oxygen and including air = 1.4.For triatomic gases such as carbon dioxide and sulphurdioxide = 1.3.

For some hydro-carbons the value of γ is quite low.[e.g., for ethane γ = 1.22, and for isobutane γ = 1.11]

SPECIFIC HEAT FOR SOLID AND LIQUID

Therefore, liquids and solids can be approximated asincompressible substances without sacrificing much inaccuracy.

A substance whose specificvolume (or density) is constant iscalled an incompressiblesubstance. The specific volumes ofsolids and liquids essentiallyremain constant during a process(Fig.).

Specific Heat For Solid And Liquid

CCC vp

The constant-volume and constant-pressure specific heats are identicalfor incompressible substances (Fig. ).

INTERNAL ENERGYCHANGE FOR SOLID AND LIQUID

Like those of ideal gases, the specific heats of incompressiblesubstances depend on temperature only. Thus, the partialdifferentials in the defining equation of Cv can be replaced byordinary differentials, which yield

dTTCdu

The change in internal energy for incompressible substancesduring a process from state 1 to state 2 is determined byintegrating these equations:

kg/kJdTTCuuu D2

1

12

Internal Energy change For Solid And Liquid

For small temperature intervals, a C value at the averagetemperature can be used and treated as a constant, yielding

kg/kJTTCuuu avg 1212 D

ENTHALPY CHANGE FOR SOLID AND LIQUID

Using the definition of enthalpy h = u + Pv and noting that v constant,the differential form of the enthalpy change of incompressible substancescan be determined by differentiation to be

vdppdvdudh

for incompressible substances dv = 0

vdpdudh

Integrating

pvuh DDD pvTTCh avg DD 12

For solid, vDp is insignificant 12 TTCh avg D

Enthalpy Change For Solid And Liquid

for liquid, two special cases are found

1. Constant pressure process (heater)

12 TTCuh avg DD Solid

12 TTCuh avg DD

pvTTCh avg DD 12

Liquid, P=C

2. Constant temperature process (pump)

Pvh DDLiquid,

T=C

thermodynamics 1 law-closed-system

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