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

2. Isothermal processIsothermal process
• P,V may change but temperature isP,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 heatIt must happen very slowly so that heat
produced during compression isproduced during compression is
absorbed by surroundings and heat lostabsorbed by surroundings and heat lost
during compression is supplied byduring compression is supplied by
surroundings.surroundings.

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

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

5. • Work done when PV = nRT = constantWork done when PV = nRT = constant  P = nRT / VP = nRT / V
Isothermal processesIsothermal processes
∫ −=−=
final
initial
)curveunderarea(dVpW
∫∫ −=−=
f
i
f
i
V
V
V
V
/nRT/nRT VdVVdVW
)/VV(nRT ifnW −=
p
V
3 T1
T2
T3T4

6. AdiabaticAdiabatic 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
PVγ
= const.
All real processes are
not.
p
V
2
1
3
4
T1
T2
T3T4

7. 8
Adiabatic ProcessAdiabatic Process
•For an ideal gas, and most real gasses,For an ideal gas, and most real gasses,
•đQ = dU + PdVđQ = dU + PdV
•đQ = CđQ = CVVdT + PdV,dT + PdV,..
•Then, whenThen, when đQđQ = 0,= 0,
VC
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
+




 +
=
+





+=
+
+=
+
==
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
=
+=+





=
=+
+




 +
=
γ
γ
where,
0
0

11. 12
Adiabatic ProcessAdiabatic Process
( )
constant
constantlnlnln
constantlnln
,integratedbecanwhich,0
0
=
==+
=+
=+
=+
γ
γγ
γ
γ
γ
PV
PVPV
PV
P
dP
V
dV
VdPPdV

12. 13
Adiabatic ProcessAdiabatic Process
constant
constant
as,expressedbealsocanthis
ofhelpWith the
constant
1
1
=
=
=
=
−
−
γ
γ
γ
γ
P
T
TV
nRTPV
PV

13. 14
γγ 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
=+=
=+=
=+=
+=
γ
γ
γ
ν
γ

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

15. ISOTHERMAL PROCESS:ISOTHERMAL PROCESS:
CONST. TEMPERATURE,CONST. TEMPERATURE, ∆∆T = 0,T = 0, ∆∆U =U =
00
NET HEAT INPUT = WORK OUTPUTNET HEAT INPUT = WORK OUTPUT
∆∆Q =Q = ∆∆U +U + ∆∆W ANDW AND ∆∆Q =Q = ∆∆WW
∆U = 0 ∆U = 0
QQOUTOUT
WorkWork
InIn
Work OutWork Out
QQININ
WORK INPUT = NET HEAT OUTWORK INPUT = NET HEAT OUT

16. ISOTHERMAL EXAMPLEISOTHERMAL EXAMPLE (Constant T):(Constant T):
PAVA = PBVB
Slow compression at
constant temperature:
----- No change in UNo change in U.
∆∆U =U = ∆∆TT = 0= 0
B
A
PA
V2 V1
PB

17. ISOTHERMAL EXPANSION (ISOTHERMAL EXPANSION (ConstantConstant
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
BB
AA
PA
VA VB
PB
PAVA = PBVB
TA = TB
ln B
A
V
W nRT
V
=
Isothermal Work

18. ∆∆Q =Q = ∆∆U +U + ∆∆W ;W ; ∆∆W = -W = -∆∆U orU 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
In−∆U +∆U
∆Q = 0
∆W = -∆U ∆U = -∆W

19. ADIABATIC EXAMPLE:ADIABATIC EXAMPLE:
Insulated
Walls: ∆Q = 0
B
A
PPAA
VV11 VV22
PPBB
Expanding gas doesExpanding gas does
work with zero heatwork with zero heat
loss.loss. Work = -Work = -∆∆UU

20. ADIABATIC EXPANSION:ADIABATIC EXPANSION:
400 J of WORK is done,
DECREASING the internal
energy by 400 J: Net heat
exchange is ZERO. ∆∆Q =Q =
00
∆Q = 0
B
A
PPAA
VVAA VVBB
PPBB
PPAAVVAA PPBBVVBB
TTAA TT BB
=
A A B BP V P Vγ γ
=

21. ADIABATIC EXAMPLE:
∆Q = 0
AA
BB
PPBB
VVBB VVAA
PPAA PAVA PBVB
TTAA TT BB
=
PPAAVVAA = P= PBBVVBB
γ γ
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? (γ = 1.4)

22. ADIABATIC (Cont.): FIND PADIABATIC (Cont.): FIND PBB
∆Q = 0
PB = 32.4 atm
or 3284 kPa
1.4
12 B
B A
B
V
P P
V
 
=  
 
1.4
(1 atm)(12)BP =
PPAAVVAA = P= PBBVVBB
γ γ
AA
BB
PPBB
VVBB 12VVBB
1 atm1 atm
300 K Solve for PSolve for PBB::
A
B A
B
V
P P
V
γ
 
=  ÷
 

23. 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 BB
==
AA
BB
32.4 atm32.4 atm
VVBB 1212VVBB
1 atm1 atm
300 K
Solve for TSolve for TBB
TTBB=?=? A A B B
A B
P V P V
T T
=

24. 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 K21.1 j/mol K
AA
B
32.4 atm32.4 atm
1 atm1 atm
300 K
810 K
SinceSince ∆∆Q =Q =
0,0,
∆∆W = -W = - ∆∆UU
8 cm8 cm33
96 cm96 cm33
Find n fromFind n from
point Apoint A PV = nRTPV = nRT
PVPV
RTRT
n =n =

25. ADIABATIC (Cont.):ADIABATIC (Cont.): If VIf VAA= 96 cm= 96 cm33
and Vand VAA= 8 cm= 8 cm33
, FIND, FIND ∆∆WW
AA
BB
32.4 atm32.4 atm
1 atm
300 K
810 K
8 cm8 cm33
96 cm96 cm33
PVPV
RTRT
n =n = ==
(101,300 Pa)(8 x10(101,300 Pa)(8 x10-6-6
mm33
))
(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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FOR
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FOR
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