LECTURE 4

TOPIC 3

KINETIC THEORY AND

IDEAL GAS LAW (CONT’)

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4. Heat and internal energy

OUTLINE

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4: Heat and internal energy

If the gas is molecular rather than atomic ,rotational and vibrational kinetic energy needsto be taken into account as well. 3

kTmvrms 232

21KE ==

nRTkTNmvNU rms 23

232

21 )( ===

The internal energy of a monoatomic ideal gas

The sum total of all the energy of all themolecules in a substance is its internal (orthermal) energy.

Temperature: measures molecules’ averagekinetic energy

Internal energy: total energy of all molecules 4

Problem 1A vessel with the volume 2.0 x 10 –3 m3 containsideal gas at pressure 101 kPa, at temperature 27ºC.

(a) Determine the average translational kineticenergy per molecule of the gas.

(b) The internal energy of gas if the molecule donot have rotational and vibrational energy.

(c) The number of molecules in the gas.

[Answer: (a) 6.21 x 10 –21 J; (b) 0.30 kJ; (c) 4.8 x 10 22]5

1. The Zeroth law of thermodynamics

2. The First law of thermodynamics

OUTLINE

3. Thermal processes and the first law

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Thermodynamics is the branch ofphysics that is built upon thefundamental laws that heat and workobey.

The collection of objects on which

Thermodynamic Systems and Their Surroundings

The collection of objects on whichattention is being focused is called thesystem , while everything else in theenvironment is called the surroundings .

To understand thermodynamics, it isnecessary to describe the state of asystem (P, V and T). 8

Thermodynamics is the name given to the study ofprocesses in which energy is transferred as heat andas work.

Heat is defined as a transfer of energy due to adifference in temperature, whereas work is a transferof energy that is not due to a temperature difference.

The Laws of Thermodynamics

of energy that is not due to a temperature difference.

The great laws of thermodynamics:0. The zeroth law of thermodynamics1. The first law of thermodynamics2. The second law of thermodynamics3. The third law of therodynamics

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The zeroth law of thermodynamics says that iftwo objects are each in equilibrium with a thirdobject , they are also in thermal equilibrium witheach other.

B CCA

1: The Zeroth Law of Thermodynamics

Two systems individually in thermal equilibriumwith a third system are in thermal equilibriumwith each other.

A B

If A is in thermal equilibrium with C, and B is in thermal equilibrium with C

Then A is thermal equilibrium with B

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Consider a cylinder with amoveable piston . A force isapplied to slowly compress

Work can be done on a deformable system,such as a gas.

2: The First Law of Thermodynamics

applied to slowly compressthe gas.

The compression is slowenough for all the systemto remain essentially inthermal equilibrium .

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∆⋅=⋅=

yF

ntdisplacemeForceW

Work done in Thermodynamics

Therefore, the work done on the gas is

∫=

∆=∆=

PdVW

VP

yPA)(

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WUUU if −=−=∆

If a system does work W on its surroundings and there isno heat flow, conservation of energy indicates that theinternal energy of the system will decrease:

Work is positive when it is done by the system andnegative when it is done on the system !

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Suppose that a system gains heat Q and that is the onlyeffect occurring. Consistent with the law of conservation ofenergy, the internal energy of the system changes:

QUUU if =−=∆

Heat is positive when the system gains heat and negativewhen the system loses heat!

THE FIRST LAW OF THERMODYNAMICS statethat the internal energy of a system changesdue to heat and work.

WQUUU if −=−=∆

(+) ve (–) veWhen heat is added When heat is

QWhen heat is addedto the system

When heat istransferred out ofthe system

WWhen work is doneby the system tosurrounding

When work is doneon the system bysurrounding

∆U Increase Decrease14

Example 1:

In part a of figure, the system gains1500 J of heat and 2200 J of work isdone by the system on itssurroundings.

In part b, the system also gains 1500 JIn part b, the system also gains 1500 Jof heat, but 2200 J of work is done onthe system .

In each case, determine the change ininternal energy of the system .

[Answer: –700 J; + 3700 J]15

Example 2:

The temperature of three moles of a

monatomic ideal gas is reduced from 267 ºC to

77 ºC as 5500 J of heat flows into the gas. Find

(a) the change in internal energy and

(b) the work done by the gas.

[Answer: –7100 J; 12600 J]

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An isothermal process is a process that occursat constant temperature.

An isobaric process is a process that occurs atconstant pressure.

3. Thermal processes and the First Law

An isochoric process is a process that occursat constant volume.

An adiabatic process is a process during whichno energy is transferred to or from the systemas heat.

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An isothermal process is one where thetemperature does not change (T constant) .

Isothermal Process

PV diagram for an ideal gas undergoingisothermal processes.

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If isothermal process is carried out on an idealgas, then PV = nRT becomes PV = constant.

Since the temperature is kept constant ( ∆T = 0),the internal energy also does not change:

03 =∆=∆ TnRU

Hence, by the 1 st law of thermodynamics, ∆U =Q – W = 0, so W = Q. Therefore, the work doneby the gas in an isothermal process equals theheat added to the gas.

02

3 =∆=∆ TnRU

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Isothermal Process

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Example 3:

Two moles of the monatomic gas argonexpand isothermally at 25 °C from an initialvolume of 0.025 m 3 to a final volume of 0.050m3. Assuming that argon is an ideal gas, find

(a) the work done by the gas,

(b) the change in internal energy of the gas,and

(c) the heat supplied to the gas.

Answer: (a) 3.4 kJ; (b) 0; (c) 3.4 kJ 23

An isobaric process occurs at constant pressure(P constant); so the process is represented by astraight horizontal line on the PV diagram.

Isobaric Process

In an isobaric process, the values of the heat Qand the work W are generally both non-zero.

VPW ∆=

The work done is W = P (Vf – Vi) where P is the constantpressure.

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VPW ∆=

Example 4:

One gram of water isplaced in the cylinder andthe pressure is maintainedat 2.0 x 10 6 Pa. Thetemperature of the water israised by 5.0 K. The wateris in the liquid phase andexpands by the smallamount of 1.0 cm 3.

Find the work done andthe change in internalenergy.

Answer: 2.0 J, 19 J 27

Isochoric Process

An isochoric (also called isovolumetric)

process is one in which the volume does not

change (V constant).

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Since the volume does not change ( ∆V = 0) in an

isochoric process, the work done W = P∆V = 0.

From the first law, ∆U = Q.

If energy is added by heat to a system kept at

constant volume, all of the transferred energy

remains in the system as an increase in its

internal energy.30

Adiabatic ProcessAn adiabatic process is one where there is no

heat flow into or out of the system (Q = 0).

PV diagram for adiabatic ( AC) and isothermal

(AB) processes on an ideal gas. 31

In practice, to carry out an

adiabatic process:

The gas must be carried out

fast so that there is not enough

time for heat to flow throughtime for heat to flow through

the walls of the container

The gas must be held in a

container with thick walls made

of a good insulator 32

For an adiabatic expansion (W positive) of idealgas; since Q = 0, thus ∆U = –W.

That is, internal energy decreases if the gasexpands. Hence the temperature decreases aswell, because

∆=∆ 3

In the reverse operation, for an adiabaticcompression, work is done on the gas. Hence theinternal energy increases and the temperaturerises.

TnRU ∆=∆2

3

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Examples of Adiabatic Processes

Some important examples of adiabatic processes

related to engineering are:

� The expansion of hot gases in an internal

combustion engine

� The liquefaction of gases in a cooling system

� The compression stroke in a diesel engine

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