Work done in Isothermal and adiabatic Process

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1. Work Done in Isothermal And Work Done in Isothermal And
Adiabatic ProcessAdiabatic Process
From: DEEPANSHU CHOWDHARYFrom: DEEPANSHU CHOWDHARY
Roll no: 05Roll no: 05
Class: 11Class: 11
thth
A A

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

3. Adiabatic processAdiabatic process
•In an adiabatic process, the system is In an adiabatic process, the system is
insulated from the surroundings and heat insulated from the surroundings and heat
absorbed or released is zero. Since there is absorbed or released is zero. Since there is
no heat exchange with the surroundings,no heat exchange with the surroundings,
•When expansion happens temperature falls When expansion happens temperature falls
•When gas is compressed, temperature When gas is compressed, temperature
rises.rises.

4. Kinds of ProcessesKinds of Processes
Often, something is held constant.
Examples:
dV = 0isochoric or isovolumic process
dQ = 0adiabatic process
dP = 0isobaric process
dT = 0isothermal process

5. •Work done when PV = nRT = constant Work done when PV = nRT = constant  P = nRT / V P = nRT / V
Isothermal processesIsothermal processes
ò
-=-=
final
initial
)curveunder area( dVpW
òò
-=-=
f
i
f
i
V
V
V
V
/ nRT/ nRT VdVVdVW
)/VV( nRT
if
nW -=
p
V
3
T
1
T
2
T
3T
4

6. 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 case
PV
g
= const.
All real processes are
not.
p
V
2
1
3
4
T
1
T
2
T
3T
4

7. 8
Adiabatic ProcessAdiabatic Process
•For an ideal gas, and most real gasses,For an ideal gas, and most real gasses,
•đQ = dU + PdV đQ = dU + PdV
•đQ = CđQ = C
VVdT + PdV,dT + PdV,..
•Then, when Then, when đQđQ = 0, = 0,
V
C
PdV
dT-=

8. 9
Adiabatic ProcessAdiabatic Process
nR
VdPPdV
C
PdV
nR
VdPPdV
dT
nR
PV
T
V
+
=-
+
==
Then,
and ,
For an ideal gas, PV=nRT,
so

9. 10
Adiabatic ProcessAdiabatic Process
nR
VdP
nRC
CnR
PdV
nR
VdP
nRC
PdV
nR
VdPPdV
C
PdV
nR
VdPPdV
dT
nR
pV
T
V
V
VV
+
ú
û
ù
ê
ë
é+
=
+
ú
û
ù
ê
ë
é
+=
+
+=
+
==
0
11
0
Then,
and ,

10. 11
Adiabatic ProcessAdiabatic Process
V
p
V
P
PV
V
V
C
C
VdPPdVVdP
C
C
PdV
CCnR
VdP
C
CnR
PdV
=
+=+
ú
û
ù
ê
ë
é
=
=+
+
ú
û
ù
ê
ë
é+
=
g
g
where,
0
0

11. 12
Adiabatic ProcessAdiabatic Process
( )
constant
constantlnlnln
constantlnln
,integrated becan which ,0
0
=
==+
=+
=+
=+
g
gg
g
g
g
PV
PVPV
PV
P
dP
V
dV
VdPPdV

12. 13
Adiabatic ProcessAdiabatic Process
constant
constant
as, expressed be alsocan this
of help With the
constant
1
1
=
=
=
=
-
-
g
g
g
g
P
T
TV
nRTPV
PV

13. 14
gg for “Ideal Gasses”for “Ideal Gasses”
33.1
6
2
1 :polyatomic
40.1
5
2
1 :diatomic
67.1
3
2
1 :monatomic
2
1
=+=
=+=
=+=
+=
g
g
g
n
g

14. Combinations of Isothermal & Combinations of Isothermal & Adiabatic Adiabatic ProcessesProcesses
All engines employ a thermodynamic cycle
W = ± (area under each pV curve)
W
cycle
= area shaded in turquoise
Watch sign of the work!

15. ISOTHERMAL PROCESS: ISOTHERMAL PROCESS:
CONST. TEMPERATURE, CONST. TEMPERATURE, DDT = 0, T = 0, DDU = U =
00
NET HEAT INPUT = WORK OUTPUTNET HEAT INPUT = WORK OUTPUT
DDQ = Q = DDU + U + DDW ANDW AND D DQ = Q = DDWW
DU = 0 DU = 0
QQ
OUTOUT
Work Work
InIn
Work OutWork Out
QQ
ININ
WORK INPUT = NET HEAT OUTWORK INPUT = NET HEAT OUT

16. ISOTHERMAL EXAMPLE ISOTHERMAL EXAMPLE (Constant T):(Constant T):
P
A
V
A
=

P
B
V
B
Slow compression at
constant temperature:
----- No change in UNo change in U.
DDU = U = DDTT = 0 = 0
B
A
P
A
V
2
V
1
P
B

17. 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.
DT = DU = 0
DU = DT = 0
BB
AA
P
A
V
A
V
B
P
B
P
A
V
A
= P
B
V
B
T
A = T
B
ln
B
A
V
W nRT
V
=
Isothermal Work

18. DDQ = Q = DDU + U + DDW ; W ; DDW = -W = -DDU or U or DDU = -U = -DDWW
ADIABATIC PROCESS: ADIABATIC PROCESS:
NO HEAT EXCHANGE, NO HEAT EXCHANGE, DDQ = 0Q = 0
Work done at EXPENSE of internal energy
INPUT Work INCREASES internal energy
Work Out Work
In
-DU +DU
DQ = 0
DW = -DU DU = -DW

19. ADIABATIC EXAMPLE:ADIABATIC EXAMPLE:
Insulated
Walls: DQ = 0
B
A
PP
AA
VV
11 V V
22
PP
BB
Expanding gas does Expanding gas does
work with zero heat work with zero heat
loss. loss. Work = -Work = -DDUU

20. ADIABATIC EXPANSION:ADIABATIC EXPANSION:
400 J of WORK is done,
DECREASING the internal
energy by 400 J: Net heat
exchange is ZERO. DDQ = Q =
00
DQ = 0
B
A
PP
AA
VV
AA V V
BB
PP
BB
PP
AAVV
A A PP
BBVV
BB
TT
A A TT
B B
=
A A B B
PV PV
g g
=

21. ADIABATIC EXAMPLE:
DQ = 0
AA
BB
PP
BB
VV
BB V V
AA
PP
AA
P
A
V
A
P
B
V
B
TT
A A TT
B B
=
PP
AAVV
AA = P = P
BBVV
BB
g g
Example 2: A diatomic gas at 300 K and
1 atm is compressed adiabatically, decreasing
its volume by 1/12. (V
A
= 12V
B
). What is the
new pressure and temperature? (g = 1.4)

22. ADIABATIC (Cont.): FIND PADIABATIC (Cont.): FIND P
BB

DQ = 0
P
B
= 32.4 atm
or 3284 kPa
1.4
12
B
B A
B
V
P P
V
æ ö
=ç ÷
è ø
1.4
(1 atm)(12)
BP=
PP
AAVV
AA = P = P
BBVV
BB
g g
AA
BB
PP
BB
VV
BB
12VV
BB
1 atm1 atm
300 KSolve for PSolve for P
BB::
A
B A
B
V
P P
V
g
æ ö
=ç ¸
è ø

23. ADIABATIC (Cont.): FIND TADIABATIC (Cont.): FIND T
BB

DQ = 0
T
B
= 810 K
(1 atm)(12V(1 atm)(12V
BB))
(32.4 atm)(1 V(32.4 atm)(1 V
BB))
(300 K)(300 K)
TT
B B
==
AA
BB
32.4 atm32.4 atm
VV
BB 12 12VV
BB
1 atm1 atm
300 K
Solve for TSolve for T
BB
TT
BB=?=?
A A B B
A B
PV PV
T T
=

24. ADIABATIC (Cont.): ADIABATIC (Cont.): If VIf V
AA= 96 cm= 96 cm
33

and Vand V
AA= 8 cm= 8 cm
33
, FIND , FIND DDWW
DQ = 0
DDW = - W = - DDU = - nCU = - nC
VV DDTT & & CC
VV== 21.1 j/mol K 21.1 j/mol K
AA
B
32.4 atm32.4 atm
1 atm1 atm
300 K
810 K
Since Since DDQ = Q =
0,0,
DDW = - W = - DDUU
8 cm8 cm
3 3
96 cm96 cm
3 3
Find n from Find n from
point Apoint A
PV = nRTPV = nRT
PVPV
RTRT
n =n =

25. ADIABATIC (Cont.): ADIABATIC (Cont.): If VIf V
AA= 96 cm= 96 cm
33

and Vand V
AA= 8 cm= 8 cm
33
, FIND , FIND DDWW
AA
BB
32.4 atm32.4 atm
1 atm
300 K
810 K
8 cm8 cm
3 3
96 cm96 cm
33
PVPV
RTRT
n =n = = =
(101,300 Pa)(8 x10(101,300 Pa)(8 x10
-6-6
m m
33
))
(8.314 J/mol K)(300 K)(8.314 J/mol K)(300 K)
nn = 0.000325 mol = 0.000325 mol & & CC
VV= 21.1 j/mol K= 21.1 j/mol K
DDTT = 810 - 300 = 510 K = 810 - 300 = 510 K
DDW = - W = - DDU = - nCU = - nC
VV DDTT
DW = - 3.50 J

26. TH
AN
KS FO
R

TH
AN
KS FO
R

W
ATC
H
IN
G
!!!
W
ATC
H
IN
G
!!!