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