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What is Thermodynamics  1. Understanding why things happens 2. Concerning heat, work, related temperature, pressure, volume and equilibrium

3. Equations relate macroscopic properties

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The laws of thermodynamics 

 Number 

Basis Property

Zeroth Law 

Thermal Equilibrium

Temperature

First Law Relation between work, energy and heat

Internal Energy U

Second Law 

Spontaneous process

Entropy

Third Law Absolute Zero Degree of Temperature

Entropy S0 as T0 Kelvin

 

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       Study of heat engines

       Being studied by all students in physical science and engineering

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Concept of State       ),( TPVV

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From Avogadro’s hypothesis the volume per mole of all ideal gases at 0oC and 1atm pressure is 22.414 litres. 

molereejoulesmolereeatmlitresR

RTPV

molereeatmlitremolereelitresatm

oToVoP

.deg/3144.8.deg/.082057.0

.deg/.082057.0.deg16.273

414.221

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For n mole gas

PV=nRT

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For water vapour, the number of moles for Kg water is obtained by 

molmolg

g

M

massn 56.55

.18

10001

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Thermodynamics

Process

Work and Energy

Heat

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1. Open system- material and energy exchange

2. Closed system- energy exchange only

3. Isolated system- no material and energy exchange

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What we learn from this module?   

1.Internal energy U and entropy S 2.Combining U and S with P, T and V gives enthalpy H=U + PV and Gibbs energy G=H-TS  3.H is related to heat adsorption or release at constant pressure

4. G controls the position of equilibrium in closed systems at constant temperature and pressure. 

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Why is Thermodynamics useful? 1.Qualitative explanation of materials behaviour 2.Quantitatively understanding of materials status.

3. Physical significance of thermodynamic functions.

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Applications of Thermodynamics  1.Extraction, refining 2.Corrosion 3.Phase transformation-phase diagram calculation  4.Materials processing 5.Design of new materials.

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The First Law of Thermodynamics      Conservation of Energy Principle     Same principle in mechanics, physics and chemistry  

wqiUf

UU

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Work done in an Expansion (or contraction) against an External Pressure 

 

dVfinalV

initialVextPw

VVextPVextPw

xAextPw

AextPFAF

extP

xFw

)(

)(

)12

.(

.

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Expansion against a constant external pressure   

final

v

initialv

VextPinitial

vfinal

vextPdVextPw )(

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W12

Q12

Reversible process

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Reversible process and Maximum Work

Reversible process for a closed system  

 W system-environment =W environment-system

Q environment-system =Q system-environment

 

 

final

v

initialv

dVextPesw

max

int

wsewesw

dVfinal

v

initalv

Psew

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For an ideal gas     

VnRTP

int

dVfinal

v

initialv V

nRTw

)()(lnmax initialv

finalvnRTrevqw

For isothermal process,

ie. T=constant

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Questions*:

1. How to calculate W for a constant pressure process?

2. How to calculate W for an isothermal reversible process?

3. Is a constant pressure process an reversible process? Explain why?

* All of these questions are concerned with ideal gas systems.

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Example:Gas compress during quasi-equilibrium processing, with PV=constant. The system is the gas P1=101325 N/m2, V1=0.01m3, V2=0.005m3

 Find W

W=-702J

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Enthalpy U=q-w     Q Heat is transferred due to the presence of a temperature difference.

    Work here is considered as the work of expansion.

    U results from the oscillation of atoms or ions in solid and movement of the particles in gas and liquid.

    Q and w depend on how the change is carried out where the difference between them does not.

    At constant volume, w=0 and U=q

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The Enthalpy Function

PVUH

VPU

initialPV

initialU

finalPV

finalU

initialH

finalHH

)()(

VPw

wqU

.

qVpVPqVPwqVPUH

....

When P=constant

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     At constant pressure, the change in enthalpy is equal to the heat

    The change of enthalpy is independent of path.Q: Does q or W depend on path?

    For the change involving solids and liquids, HU, but for gases, HUQ:explain why?

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Example 1: Given p=constant=101.3 kPaV1=1m3, V2=2m3

Q12=200kJ

 Find a) Ub) an expression for Q12 in terms of

thermodynamics properties for a quasi-equilibrium process.

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Example 2. Given T=T1=T2=constant for the

processP1=200 kPa, T1=300K

V1=2m3, V2=4m3

Find a) W and b) Q

W=277KJ

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Heat Capacities (Cp and Cv)

  a)                      Under constant volume conditions Cv- all heat supplied increases energy of sampleb)                     Under constant pressure conditions Cp- Heat supplied increases energy of sample and provides energy for work performed. 

TqC

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Relation between Cv and U  The 1st Law  When V=constant  Therefore  

T

qvC

0

w

wqU

qU

vT

U

vT

UvC

TvCU

.

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For n moles of materials over small ranges in temperature Cvconstant

TvnCiTf

TvnCU )(

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Relation between Cp and H At constant pressure      

Over small range of T for n moles of materials 

T

qpC

pT

H

T

HpC

T

HpC

TpCH

.

TpCH

TpnCiTf

TpnCH )(

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Summary At constant volume  At constant pressure  Molar heat capacity at constant pressure  Molar heat capacity at constant volume 

Uq

Hq

PT

HpC

vT

UvC

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Questions:

1. For a constant temperature process of an ideal gas, prove H=U.

2. For a gas system, explain why Cp is larger than Cv?

3. For a solid/liquid system, explain why Cp is close to Cv?

4. What are the equations for calculating change of enthalpy and internal energy due to temperature change?