work done in isothermal and adiabatic process

Work Done in Isothermal Work Done in Isothermal And Adiabatic ProcessAnd Adiabatic Process

From: DEEPANSHU CHOWDHARYFrom: DEEPANSHU CHOWDHARYRoll no: 05Roll no: 05

Class: 11Class: 11thth A A

Isothermal processIsothermal process

• P,V may change but temperature is P,V may change but temperature is constant.constant.

• The cylinder must have conducting wallsThe cylinder must have conducting walls• It must happen very slowly so that heat It must happen very slowly so that heat

produced during compression is absorbed produced during compression is absorbed by surroundings and heat lost during by surroundings and heat lost during compression is supplied by surroundings. compression is supplied by surroundings.

Adiabatic processAdiabatic process

• In an adiabatic process, the system is insulated from the In an adiabatic process, the system is insulated from the surroundings and heat absorbed or released is zero. Since surroundings and heat absorbed or released is zero. Since there is no heat exchange with the surroundings,there is no heat exchange with the surroundings,

• When expansion happens temperature falls When expansion happens temperature falls • When gas is compressed, temperature rises.When gas is compressed, temperature rises.

Kinds of ProcessesKinds of Processes

Often, something is held constant. Examples:

dV = 0 isochoric or isovolumic processdQ = 0 adiabatic processdP = 0 isobaric processdT = 0 isothermal process

• Work done when PV = nRT = constant Work done when PV = nRT = constant P = nRT / V P = nRT / VIsothermal processesIsothermal processes

final

initial

)curveunder area( dVpW

f

i

f

i

V

V

V

V

/ nRT/ nRT VdVVdVW

)/VV( nRT ifnW

p

V

3 T1

T2T3T4

Adiabatic Adiabatic ProcessesProcesses An adiabatic process is process in which there is no

thermal energy transfer to or from a system (Q = 0)

A reversible adiabatic process involves a “worked” expansion in which we can return all of the energy transferred.

In this casePV = const.

All real processes are not.

p

V

2

13

4T1

T2

T3T4

8

Adiabatic ProcessAdiabatic Process

•For an ideal gas, and most real For an ideal gas, and most real gasses,gasses,•đQ = dU + PdV đQ = dU + PdV •đQ = CđQ = CVVdT + PdV,dT + PdV,..

•Then, when Then, when đQđQ = 0, = 0, VCPdVdT

9

Adiabatic ProcessAdiabatic Process

nRVdPPdV

CPdV

nRVdPPdVdT

nRPVT

V

Then,

and ,

For an ideal gas, PV=nRT, so

10

Adiabatic ProcessAdiabatic Process

nRVdP

nRCCnRPdV

nRVdP

nRCPdV

nRVdPPdV

CPdV

nRVdPPdVdT

nRpVT

V

V

VV

0

110

Then,

and ,

11

Adiabatic ProcessAdiabatic Process

V

p

V

P

PV

V

V

CC

VdPPdVVdPCCPdV

CCnR

VdPC

CnRPdV

where,

0

0

12

Adiabatic ProcessAdiabatic Process

constant

constantlnlnln

constantlnln

,integrated becan which ,0

0

PV

PVPV

PVP

dPVdV

VdPPdV

13

Adiabatic ProcessAdiabatic Process

constant

constant

as, expressed be alsocan this of help With the

constant

1

1

PT

TV

nRTPVPV

14

for “Ideal Gasses”for “Ideal Gasses”

33.1621 :polyatomic

40.1521 :diatomic

67.1321 :monatomic

21

Combinations of Isothermal & Combinations of Isothermal & Adiabatic Adiabatic ProcessesProcessesAll engines employ a thermodynamic cycleW = ± (area under each pV curve)Wcycle = area shaded in turquoise

Watch sign of the work!

ISOTHERMAL PROCESS: ISOTHERMAL PROCESS: CONST. TEMPERATURE, CONST. TEMPERATURE, T = 0, T = 0, U U

= 0= 0

NET HEAT INPUT = WORK OUTPUTNET HEAT INPUT = WORK OUTPUT

Q = Q = U + U + W ANDW ANDQ Q = = WW

U = 0

U = 0

QQOUTOUT

Work Work InIn

Work Work OutOut

QQININ

WORK INPUT = NET HEAT OUTWORK INPUT = NET HEAT OUT

ISOTHERMAL EXAMPLE ISOTHERMAL EXAMPLE (Constant (Constant T):T):

PAVA =

PBVB

Slow compression at constant temperature: ----- No change in UNo change in U.

U = U = TT = = 00

B

APA

V2 V1

PB

ISOTHERMAL EXPANSION (ISOTHERMAL EXPANSION (Constant Constant T)T)::

400 J of energy is absorbed by gas as 400 J of work is done on gas.

T = U = 0

U = T = 0

BBAA

PA

VA VB

PB

PAVA = PBVB

TA = TB

ln B

A

VW nRTV

Isothermal Work

Q = Q = U + U + W ; W ; W = -W = -U or U or U = -U = -WW

ADIABATIC PROCESS: ADIABATIC PROCESS: NO HEAT EXCHANGE, NO HEAT EXCHANGE, Q = 0Q = 0

Work done at EXPENSE of internal energy INPUT Work INCREASES internal energy

Work Out

Work InU +U

Q = 0

W = -U U = -W

ADIABATIC EXAMPLE:ADIABATIC EXAMPLE:

Insulated Walls: Q =

0

BA

PPAA

VV11 V V22

PPBB

Expanding gas Expanding gas does work with does work with zero heat loss. zero heat loss. Work = -Work = -UU

ADIABATIC EXPANSION:ADIABATIC EXPANSION:

400 J of WORK is done, DECREASING the internal energy by 400 J: Net heat exchange is ZERO. Q = 0Q = 0

Q = 0

B

APPAA

VVAA VVBB

PPBB

PPAAVVA A PPBBVVBB

TTA A TT B B

=

A A B BP V P V

ADIABATIC EXAMPLE:

Q = 0

AA

BBPPBB

VVBB V VAA

PPAA PAVA PBVB

TTA A TT B B

=

PPAAVVAA = P = PBBVVBB

Example 2: A diatomic gas at 300 K and 1 atm is compressed adiabatically, decreasing its volume by 1/12. (VA = 12VB). What is the new pressure and temperature? ( = 1.4)

ADIABATIC (Cont.): FIND PADIABATIC (Cont.): FIND PBB

Q = 0

PB = 32.4 atm or 3284 kPa

1.412 B

B AB

VP PV

1.4(1 atm)(12)BP

PPAAVVAA = P = PBBVVBB

AA

BBPPBB

VVBB 12VVBB

1 1 atmatm

300 K Solve for Solve for PPBB::

AB A

B

VP PV

ADIABATIC (Cont.): FIND TADIABATIC (Cont.): FIND TBB

Q = 0

TB = 810 K

(1 atm)(12V(1 atm)(12VBB)) (32.4 atm)(1 V(32.4 atm)(1 VBB)) (300 K)(300 K) TT B B

==

AA

BB32.4 32.4 atmatm

VVBB 12 12VVBB

1 1 atmatm

300 K

Solve for Solve for TTBB

TTBB=?=?A A B B

A B

P V P VT T

ADIABATIC (Cont.): ADIABATIC (Cont.): If VIf VAA= 96 cm= 96 cm33 and Vand VAA= 8 cm= 8 cm33, FIND , FIND WW

Q = 0

W = - W = - U = - nCU = - nCVV TT & & CCVV== 21.1 j/mol 21.1 j/mol KK

AA

B32.4 32.4 atmatm

1 1 atmatm

300 K

810 KSince Since Q = Q =

0,0,

W = - W = - UU 8 cm8 cm3 3 96 cm96 cm3 3

Find n Find n from point from point

AAPV = nRTPV = nRT

PVPV

RTRT n =n =

ADIABATIC (Cont.): ADIABATIC (Cont.): If VIf VAA= 96 cm= 96 cm33 and Vand VAA= 8 cm= 8 cm33, FIND , FIND WW

AA

BB32.4 32.4 atmatm

1 atm

300 K

810 K

8 cm8 cm3 3 96 cm96 cm33

PVPV

RTRT n =n = = = (101,300 Pa)(8 x10(101,300 Pa)(8 x10-6-6 m m33))

(8.314 J/mol K)(300 K)(8.314 J/mol K)(300 K)nn = 0.000325 mol = 0.000325 mol & & CCVV= 21.1 j/mol K= 21.1 j/mol K

TT = 810 - 300 = 510 K = 810 - 300 = 510 K

W = - W = - U = - nCU = - nCVV TT

W = - 3.50 J

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