Thermal Physics Lecture Note 6

8/10/2019 Thermal Physics Lecture Note 6

1/24

6. Combined first and second laws

First law

dWdUdQ for all processes

Second law

TdSdQ= for reversible processes

and PdVdW= for systemP-V-T

Combined

2

1

2

1

PdVdUTdS

PdVdUTdS

This is the combined first and second laws forP-V-T system

Although we have obtained this equation by assuming reversible processes , it is actually

generally valid for all processes between 2 equilibrium states.

2

1


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Example: Isothermal process ofP-V-T system

PdVdUdST

For this system, the equation of state is given by

=P function (V,T)

Example of other system : dielectric

ET

ba

V

: Polarization, V: volume, T: temperature, E: electric field


3/24

6.1 Tds equation

(a) u(T,v) s(T,v)

1+2 law : PdvduT

ds 1

dvv

udT

T

udu

Tv

=

dvPv

u

TdT

T

u

Tds

Tv

=

11

Also, dvv

sdT

T

sds

Tv

=

Compare,

T

c

T

u

TT

s v

vv=

=

1

=

Pv

u

Tv

s

TT

1


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5/24

vTP

=

PT

v

u

T

Hence,vT T

PPP

T

Tv

s

=

1

dvP

TdTcTds v

From Lecture Note 4,

PT

vPT

vP

v

ucc

=

we get

2Tvv

Tcc vP =

=

Compare with ideal gas, Rcc vP =

T v


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(b) u(T,P), s(T,P)

Enthalpy Pvuh h(T,P)

vdPPdvdudh

vdPdhPdvdu

Hence, from 1+2 law : vdPdhT

PdvduT

ds 11

Write dP

P

hdT

T

hdh

TP

=

dPP

sdT

T

sds

TP

=

=

vdPdPPhdT

Th

TdP

PsdT

Ts

TPTP

1

PP T

h

TT

s

=

1


7/24

=

vPh

TPs

TT

1

FromTP

s

PT

s

=

22

=

=

v

P

h

TT

v

PT

h

T

P

s

TPT

s

TP

PT

2

2

2

11

=

=

TP

h

TT

s

PTP

s

TP

22 1

Equate

=

TPh

Tv

Ph

TTv

PTh

T TP

2

2

2 111

=

vP

h

TT

v

T TP2

11


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vTvTvTv

Ph

PT

andPT T

vv

P

s

P

P

cT

h=

T

c

T

s P

P

=

Hence, dP

P

sdT

T

sds

TP

=

dPT

vdT

T

cds

P

P

ordPvTdTcTds

T PP


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(c) u(P,v), s(P,v)

Similarly, using

T

c

P

s v

v

=

andTv

c

v

s P

P

=

dPP

Tcdv

v

TcTds

v

v

P

P

=

We get,


10/24

6.2 Examples of application of Tds equations

(1) Adiabatic compression 0TdsdQ

T= ? from the volume is reduced from v1v2

From dvT

P

TdTcTdsv

v

dvc

T

dvT

P

c

T

dTvvv

Hence

dvcT

dT

v

[ ] 21

2

1

v

v

v

T

T v

cnT

l

(2) Change of volume vwith pressurePduring adiabatic process

0

= dPP

Tcdv

v

TcTds

v

v

P

P

0dPc

dvv

c vP


11/24

=

=

P

v

s c

c

P

v

v

1

Define ssP

v

v

1 as coefficient of adiabatic compression

=s


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13/24

wherePT T

vTvPh

and PP

cTh

=

dPT

vTvdTcdh

P

P

dPT

vTvdPdTchh

P

P P

P

P

T

T

Po

ooo

The experssion

P

P PodPT

v

, and

P

PovdP dPT

v

T

P

P Po

can be obtained from the equation of state

To obtain

T

T

P

odTT

c

and , we need the value of c

T

TP

odTc Pat Po.

cPocan be obtained as follows :


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FromPT T

v

P

s

PT

v

PT

s

2

22

andP

PT

sTc

= T

P

P

c

TTP

s

=

12

PT

P

T

v

P

c

T

2

21

P

P P

c

c

P

o

P

po

dP

T

vTdc

2

2

P

P P

PPo

o

dPT

vTcc

2

2


15/24

Example 1 : For ideal gas,

P

RTv=

HenceP

R

T

v

P

=

, 02

2

=

PT

v

PoP cc=

=

P

P o

P

P P ooP

PnRdP

P

RdP

T

vl dP

P

RTdP

T

vT

P

P

P

P P oo

=

dPP

RTvdP

P

P

P

P oo

=

and

T

T

Po

o

dTchho

T

T

Po

P

PnRdT

T

css

o

l

If cP is a constant for closeby temperature,

oPo TTchh


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oo

Po

P

PnR

T

Tncss ll

From h, internal energy RTdTchPvhuT

T

Po

o

and Rcc vP =

o

T

T

voo

T

T

vo RTdTchRTTTRdTchu

oo

)(

since

T

T

vo

o

dTcuu ooo RThu

For 2 nearby states so that cv is a constant,

)( ovo TTcuu


17/24

Example 2 : van der Waals gas RTbv

v

aP =

)(2

Using the Tdsequation for the case of u(T,v)and s(T,v)

dv

T

PdT

T

cds

v

v

From the equation of state,bv

R

T

P

v

Hence,

bvbvnRdT

Tcss

o

T

T

vo

o

l

Take cvas constant,

bv

bvnR

T

Tncss

oo

vo ll

Similarly, for the internal energy,

dvv

adTcdvP

T

PTdTcdu v

v

v 2


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o

ovovv

aTTcuu 11)(

From

2Tv

T

vP

v

ucc

PT

vP =

=

PT

PT

v

u

vT

=

We get

=

=

2

2)(21

RTv

bva

R

T

v

T

PTcc

PvvP

[Compare with Rcc vP = for ideal gas]


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Example 3 : Liquid or solid under the effect of hydrostatic pressure

Equation of state dPP

vdT

T

vdv

TP

=

For coefficient of volume expansion and coefficient of isothermal compression,

PT

v

v

=

1 and

TP

v

v

1

vdPvdTdv

Integrate

P

P

T

T

o

oo

vdPvdTvv

For liquid and solid, the change in vwithPand Tis considered to be negligibly small

This means that and is almost not change with Pand T

Hence, ]ooo PPTTvv 1

This is the approximate equation of state


20/24

Tdsequation : dPTvTdTcTds

P

P

dPT

vdT

T

css

P

P P

T

T

Po

oo

1: (integrate w.r.tPo) 2: (integrate w.r.t. T)

From the equation of state, oP

vT

v

02

2

=

PT

v

ooo

Po

P

P

o

T

T

Po PPv

T

TncsdPvdT

T

css

oo

l


21/24

6.4 Joule and Joule-Thomson coefficients

From the first law,

1

vuT u

T

T

v

v

u

u

vT v

Tc

v

u

Similarly,

h

P

T P

Tc

P

h

Joule Coefficient:uv

T

=

Joule-Thomson Coefficient:hP

T

=

PT

v

u

T

=


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vvT

P

h

T

Jadi

TP

cv

u

c vTv

11

dan vvTcP

hc PTP

11


23/24

6.5 Multivariable systems

Combined first and second laws forP, V, T system

PdVdUTdS

can be generalised as follows :

Consider a system that can be described by using parameters Y, X, T

We can write YdXdUTdS

For system with 5 parameters, Y1X1Y2X2T,

2211 dXYdXYdUTdS

The state of the system can be fixed by knowing 3 parameters


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Example : paramagnetic gas in an external magnetic field

5 parameters : magnetic fieldH (intensive)

megnetic moment -M(extensive)

pressureP (intensive)

volume V(extensive)

temperature T

HdMPdVdUTdS

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Thermal Physics Lecture Note 6