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1. 1
Recap
In previous class we have studied about
•List of thermodynamic processes
•Expression for work done in constant volume process
•Problems on constant volume process

2. 2
OBJECTIVES
On completion of this period, you would
be able to understand
• Constant pressure process
• Derive an expression for work done in
Constant pressure process
• Solving problems on constant pressure
process

3. 3
•In a constant pressure process,
pressure remains constant
throughout the process.
•The gas contained in cylinder
and piston arrangement is the
example of constant pressure
process.

Constant Pressure or
Isobaric Process

4. 4
Fig 1
Constant Pressure Process (Contd.,)

5. 5
•Consider heat is added to the
system from an out side source.
• This increases the volume and
temperature of gas with no
change in pressure.
Constant Pressure
Process (Contd.,)

6. 6
Fig 2
Constant Pressure process (Contd.,)
system
Piston
Cylinder

7. 7
•Similarly, if heat is removed
from the gas to the
surroundings of the system
there will be decrease in
volume and temperature of
the gas and there is no
change in pressure.
Constant Pressure
Process (Contd.,)

8. 8
Expression for Work done

9. 9
•As heat is supplied, the gas will expand from state 1 to
state 2 at constant pressure
•Work done by gas w
1-2
=
=p ∫ dv
•(Since pressure remains constant)
•Therefore, Work done, W
1-2
= p[ V ]
= p(V
2-V
1)
P
v1
v2
2
1
V
V
pdv
1
2
W
V
V
1
V
2
Work done Expression
Fig.3 P – V diagram

10. 10
W
1-2
= p(V
2
-V
1
)
= pV
2-pV
1
= mRT
2
-mRT
1

= mR(T
2-T
1)

Work done Expression(contd.,)

11. 11
We know that NFEE,
Q
1-2
= (U
2
– U
1
) + W
1-2

= mC
v (T
2-T
1) + mR(T
2-T
1)
= m(T
2
-T
1
)(C
v
+R)
= m(T
2
-T
1
)C
p

= mC
p
(T
2
-T
1
)
Heat Supplied in Constant Pressure Process

12. 12
Change in Enthalpy in
constant pressure process
•From the definition, H = U + pV
•At initial state 1 , H
1
= U
1
+ p
1
V
1
•At final state 2 , H
2
= U
2
+ p
2
V
2

•Change in enthalpy ( H
2 – H
1 ) = (U
2+
p
2
V
2
) - (U
1
+ p
1
V
1
)

13. 13
• Change in enthalpy ( H
2 – H
1 )= (U
2 – U
1) + p
2V
2-p
1V
1

=mC
v (T
2-T
1) + mR(T
2-T
1)
= m(T
2-T
1)(C
v+R)
= m(T
2-T
1)C
p
= mC
p (T
2-T
1)
Thus in a constant pressure process,
Heat transferred = Change in enthalpy
Change in enthalpy in
Constant Pressure
process (Contd.,)

14. 14
Pressure-Volume-Temperature (P-V-T)
Relationship:
• We know that the general gas equation is
• P
1
V
1
P
2
V
2
T
1
T
2
=
Constant Pressure (Contd.,)

15. 15
• Since the gas is heated at constant pressure,
( i.e, P
1
= P
2
)
•
V
1 V
2
T
1 T
2
• or
• V
T
• Thus, the constant pressure process is governed by
Charles law.
=
=Constant
Constant Pressure (Contd.,)

16. 16
•If the gas is compressed
work is done on the gas
andheat is rejected
• Work done on the gas( W
1-2
) = P (V
1
-V
2
)
• = mR (T
1
-T
2
)
• Heat rejected ( Q
1-2
) = mC
p
(T
1
-T
2
)
Constant Pressure (Contd.,)

17. 17
Worked Examples

18. 18
1.One kg of nitrogen contained in a
cylinder at a pressure of 7bar and
temperature of 300k expands 4 times of
its original volume. The process of
expansion is assumed to take place at
constant pressure. Calculate
•Initial volume.
•Final temperature.
•Work done
•Heat added.
•Change in internal energy.
for nitrogen assume, C
p
=1.045KJ/kgK,R=296KJ/kgk

19. 19
•Given data :
• m = 1 Kg
• P
1
= 7 bar = 7x100 = 700KN/m
2
• T
1
= 300 K
• V
2
= 4V
1
 V
2
= 4
V
1
• Constant pressure process,
P
2
= P
1
= P=700KN/m
2

• C
p= 1.045KJ/kg K
• R = 296J/kg K = 0.296KJ/kg K
Solution

20. 20


a) Initial volume (V
1
) = ?
• We know that the gas equation, P
1
V
1
= mRT
1
•
• = 0.1272m
3
V1 =
mRT
1
P
1
1X0.296X300
700
=
Solution (Contd.,)

21. 21
b)Final temperature (T
2
)=?
• From constant pressure process,

• = 4X300 = 1200 K
V1
T1
V2
T2
=
V2T
1
V1
T
2=
Solution (Contd.,)

22. 22
• c) Work done ( W
1-2
) = ?
• W
1-2
= P(V
2
- V
1
)
• = 700 (4V
1 - V
1)
• = 700 (3V
1
)
• = 700 x 3x 0.1272
• =267.12 KJ
Solution (Contd.,)

23. 23
• d ) Heat added ( Q
1-2 ) = ?
• Q
1-2
= mC
p
(T
2
-T
1
)
• = 1x 1.045 (1200 – 300 )

• = 940.5 KJ
Solution (Contd.,)

24. 24
• e) Change in internal energy (U
2
-U
1
) = ?

• From NFEE , U
2
-U
1
= Q
1-2
- W
1-2

• = 940.5 – 267.12

• = 673.38KJ
Solution (Contd.,)

25. 25
Example.2
During a constant pressure process, the internal
energy of one kg system increases 28.5 KJ and
enthalpy increases 44.3 KJ. The pressure is 620
KN/m
2
.
•a) What is the work which accompanies this process ?
•b) If the initial volume is 0.793m
3.
What is the final
volume?.

26. 26
• Given data :
•Constant pressure process, P = P
1
= P
2
= 620 KN/m
2

•

m = 1 Kg
• U
2 – U
1 = 28.5 KJ
• H
2 – H
1 = 44.3 KJ
Solution

27. 27
–For Constant pressure process,
• Heat added ( Q
1-2
) = Change in enthalpy ( H
2
-H
1
)
• = 44.3 KJ
Solution (Contd.,)

28. 28

• Work done ( W
1-2
) = ?
• From NFEE, W
1-2 = Q
1-2 - ( U
2 - U
1
)
• = 44.3 – 28.5
• = 15.8KJ
Solution (Contd.,)

29. 29
• V
1
= 0.793m3 ,
• Final volume( V
2 ) = ?

• We know that, W
1-2
= P ( V
2
- V
1
)
• 15.8= 620 ( V
2 – 0.793 )
• => V
2 = 0.82m
3

Solution (Contd.,)

30. 30
QUIZ

31. 31
1. For constant pressure process, heat transfer is equal to
a)Change in internal energy.
b)Work done.
c)Change in enthalpy
d)Change in entropy

32. 32
2. Constant pressure process is governed by which law
a)Boyle’s law
b)Gay-lussac law
c)Charle’s law
d)Avogadro's law
Ans: b

33. 33
3. When a gas is heated at constant pressure
a)Its temperature will increase
b)Its volume will increase
c)Both temperature and volume will increase
d)Neither temperature nor pressure will increase
Ans:c

34. 34
4. In constant pressure process, the change in enthalpy
is 20KJ and work done is 15 KJ, what is the heat
transfer ?
•15KJ.
•5KJ
•20KJ
•0

35. 35
5. During constant pressure process at 10KN/m
2
, the
volume changes from 1m
3
to 5m
3
. What is the work
done ?
•10KJ
•30KJ
•40KJ
•20KJ

36. 36
Summary
In this period we have learnt
•Constant pressure process .
•Expression for work done on Constant Pressure
Process.
•Solving problems on constant pressure process