The work of Reversible Adiabatic change for an ideal gas.
constant.not is T
because used bet cann' nRT/V P
V and P
between iprelationsh a need We
f
i
V
VPdVw
dTCdU
Hence
V
U
V
UWhere
dVdTCdU
dVV
UdT
T
UdU
TVfU
V
T
T
TT
TV
TV
,
0
gas, idealan of molecules the
between n interactio no is thereSince
sample. in the
forces cohesive of measure theisIt
pressure. internal called is
),(
)(
0
process, adiabatican For
mics,Thermodyna of LawFirst From
12 TTCw
dTCdw
dwdU
dq
dqdwdU
V
V
What if the final temperature is not available.
constTV
constTV
constVC
RT
constVnRTC
gIntegratin
dVV
nR
T
dTC
dVV
nRTdTC
PdVdTC
dwdU
dq
mV
mV
C
R
C
R
mV
V
V
V
V
,
,ln
lnln
lnln
0
process, adiabatican For
,
constPV
constPV
constPV
constPV
constR
PVV
mV
mp
mv
mVmp
mV
mV
C
C
C
CC
C
R
C
R
,
,
,
,,
,
,
1
1
(a) Free expansion
pex=0, w=0(b) Expansion against constant pressure
(Expansion of a gas formed in a chemical reaction)
Work done in an Isothermal Expansion/compression of an ideal gas.
w =P2(V2-V1)
© Reversible isothermal Expansion/contr
1
2
2
1 lnln
d
d
2
1
2
1
p
pnRT
V
VnRTw
VV
nRTw
Vpw
V
V
V
V ex
Internal Energy Change in an Isothermal Expansion/compression of an ideal Gas
wq
dHdU
dT
dTCdHdTCdU
Hence
V
U
V
UWhere
dVdTCdU
dVV
UdT
T
UdU
TVfU
pV
T
TT
TV
TV
and
0 ,0
0
process isothermalan For
,
,
0
gas, idealan For
),(
1
)(
0
process, adiabatican In
1
2
11,
1
2
1
1
2
1
12
,
V
VTCw
V
V
T
T
constTVTV
TTCw
dTCdw
dwdU
dq
mV
C
R
V
V
mV
Thermodynamics changes in an adiabatic process of an ideal gas.
1
law, gas ideal theFrom
1
1
21,
1
1
2
1
2
12
21
2
1
1
2
1
1
2
P
PTCw
P
P
T
T
PT
PT
V
V
V
V
T
T
mV
Thermodynamic changes for a real gas
(a) Free expansion
pex=0, w=0(b) Expansion against constant pressure
w =P2(V2-V1)
© Reversible isothermal Expansion/contr
12
2
1
2
2
2
11ln
d
d
2
1
2
1
VVan
nbV
nbVnRTw
VV
an
nbV
nRT
VPw
V
V
V
V
T
TT
TV
TV
V
U
V
UWhere
dVdTCdU
dVV
UdT
T
UdU
TVfU
get toHow
0
),(
Internal energy change in an isothermal process for a real gas.
Unfamiliar Quantities
TTTT V
H
P
H
P
U
V
U
,, ,
VTP T
P
P
V
T
V
,,
Quantities that can be recognized, interpreted, or measured.
Change of volume with temperature.
EXPANSION COEFFICIENT ()
PT
V
V
1
Thermal expansion is also used in mechanical applications to fit parts over one another.
There exist some alloys with a very small CTE, used in applications that demand very small changes in physical dimension over a range of temperatures. One of these is invar 36, with a coefficient in the 0.0000016 range. These alloys are useful in aerospace applications where wide temperature swings may occur.
Change of volume with Pressure
ISOTHERMAL COMPRESSIBILITY (T)
TT P
V
V
1
Oxygen compresses more than helium. Therefore, if two cylinders with the same internal volume are filled to the same pressure, one with oxygen and the other with helium, the oxygen cylinder will hold more cubic feet of gas than the helium cylinder.
Change of pressure with Temperature
TVT
P
Change of internal energy with Volume
TV
U
T ,
VTTV V
p
TT
p
V
Test of a state Function
(e.g. p=f(V,T) path cyclicevery for 0dp
VTTV
TV
TV
TV
pV
U
TTT
U
TV
T
dq
dVpV
U
TdT
T
U
TT
dq
dVpV
UdT
T
Udq
pdVdVV
UdT
T
Udq
dwdUdq
T
dq
11
function. state a is Since
11
amics, thermodynof lawfirst From
path) cyclic a(For 0
pV
U
T
pT
pV
U
TT
p
T
pV
U
T
T
p
TVT
U
T
pV
U
T
T
p
VT
U
TTV
U
T
pV
U
TTT
U
TV
TV
TV
T
V
T
V
VTTV
2
2
2
2
22
11
1
1
1
1
1
1
11
Internal energy change in an isothermal process for a real gas.
12
2
2
2
2
2
11
0
gas, der Waal van aFor
),(
VVanU
dVV
an
dVdTCdUV
an
nbV
nR
T
p
pT
pT
pV
U
T
pTSince
dVdTCdU
dVV
UdT
T
UdU
TVfU
TV
V
V
TV
TV
TV
Internal energy change in an adiabatic process for a real gas.
VCR
V
V
V
TV
nbV
nbVTT
dVnbV
nRTdTC
dVV
an
nbV
nRTdV
V
andTC
pdVdU
VVanTTCU
dVdTCdU
2
112
2
2
2
2
2
12
212
get, we thisSolving
T ofn Calculatio
11)(
Process, adiabatican For
Enthalpy changes in a thermodynamic Processes.
gas) idealan (For 0
process, isothermalan for thusand
0
gas, idealan For
),(
dH
dTCdH
p
H
dTCdpp
H
dTT
Hdp
p
HdH
TpfH
p
T
p
T
pT
Changes in enthalpy with pressure
dpVp
H
TdT
T
H
TT
dq
dpVp
HdT
T
Hdq
Vdpdpp
HdT
T
Hdq
VdpdH
pdVVdppdVdH
dwpVHd
dwdUdq
TP
TP
TP
11
)(
amics, thermodynof lawfirst From
t.Coefficien
Thomson-Joule isothermal called is
it, Solving
11
function. state a is Since
11
T
VT
PT
PTTP
TP
p
H
pT
pT
V
U
VT
VT
p
H
Vp
H
TTT
H
Tp
T
dq
dpVp
H
TdT
T
H
TT
dq
Changes in Internal energy with temperature at constant pressure
VP
V
PTVP
TV
CT
U
C
T
V
V
U
T
U
T
U
dVV
UdT
T
UdU
0 gas, idealan For
V
tcoefficienThomson -joule called is
or
1
rule, cyclic sEuler' from Also,
get y toAnother wa
p
PHT
pHT
T
C
T
H
p
T
p
H
H
T
T
p
p
H
p
H
H
P T
How to impose the constraint of constant enthalpy.
pi,,Vi,Ti
pf,Vf,Tf
if
iiifff
ffiiif
ffii
HH
VpUVpU
VpVpUUU
q
VpVpw
0
process, adiabatican is thisSince
p
P
H
p
T
p
T
C
VTV
T
P
T
C
pH
Cp
H
For an ideal gas, 0
Real gases have non-zero Joule-Thompson coefficient. Since p is always –ve in a Joule Thompson experiment, a positive corresponds to cooling on expansion, a negative , to warming.
bRT
a
C
bTR
abp
RT
a
C
pJT
pJT
21
raturehigh tempe and
pressure low of conditions Under the
32122
The sign of depends on conditions.The temperature corresponding to the boundary at a given pressure is the ‘inversion temperature’ of the gas at that pressure.
Changes in enthalpy with temperature at constant volume.
TV
p
T
pVTV
p
T
pT
T
p
Cp
H
CT
p
p
H
T
H
dTCdpp
H
dTT
Hdp
p
HdH
Also,
1
so
Tp
pT
pV
TV
p
T
pVTV
C
CCT
H
T
p
Cp
H
CT
p
p
H
T
H
Relation between Cp and CV
VT
VT
pT
VppT
pT
Vp
CVpVC
T
U
T
Vp
T
U
T
U
T
PVU
T
U
T
HCC
V
V
VV
Vpp
VP
VPVp
2
)(
2
3
32
2
21
2
gas, der Waals van aFor
so,
1
1
gas, idealan For
bVRTVaR
CC
Vab
Va
p
RV
bV
R
nRCC
p
T
VTCC
Vp
Vp
Vp