Thermal Physics Lecture Note 6
Transcript of 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
-
8/10/2019 Thermal Physics Lecture Note 6
2/24
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
-
8/10/2019 Thermal Physics Lecture Note 6
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
-
8/10/2019 Thermal Physics Lecture Note 6
4/24
-
8/10/2019 Thermal Physics Lecture Note 6
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
-
8/10/2019 Thermal Physics Lecture Note 6
6/24
(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
-
8/10/2019 Thermal Physics Lecture Note 6
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
-
8/10/2019 Thermal Physics Lecture Note 6
8/24
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
-
8/10/2019 Thermal Physics Lecture Note 6
9/24
(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,
-
8/10/2019 Thermal Physics Lecture Note 6
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
-
8/10/2019 Thermal Physics Lecture Note 6
11/24
=
=
P
v
s c
c
P
v
v
1
Define ssP
v
v
1 as coefficient of adiabatic compression
=s
-
8/10/2019 Thermal Physics Lecture Note 6
12/24
-
8/10/2019 Thermal Physics Lecture Note 6
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 :
-
8/10/2019 Thermal Physics Lecture Note 6
14/24
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
-
8/10/2019 Thermal Physics Lecture Note 6
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
-
8/10/2019 Thermal Physics Lecture Note 6
16/24
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
-
8/10/2019 Thermal Physics Lecture Note 6
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
-
8/10/2019 Thermal Physics Lecture Note 6
18/24
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]
-
8/10/2019 Thermal Physics Lecture Note 6
19/24
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
-
8/10/2019 Thermal Physics Lecture Note 6
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
-
8/10/2019 Thermal Physics Lecture Note 6
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
=
-
8/10/2019 Thermal Physics Lecture Note 6
22/24
vvT
P
h
T
Jadi
TP
cv
u
c vTv
11
dan vvTcP
hc PTP
11
-
8/10/2019 Thermal Physics Lecture Note 6
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
-
8/10/2019 Thermal Physics Lecture Note 6
24/24
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