32. a) Define Adiabatic process
b) Obtain the expression for work done in an adiabatic process
Answers
Answer:
In an adiabatic process, change in pressure, volume and temperature takes place under thermal isolation. ... For an adiabatic process, PV. =K where γ=CPCV is known as the ratio of specific beats of a gas.
Explanation:
^_________^
Answer:
Adiabatic process is the process in which there is no heat transfer of system or surroundings.
For an adiabatic process of ideal gas equation we have
P
V
γ
=
K
Where
γ
is the ratio of specific heat (ordinary or molar) at constant pressure and at constant volume
γ
=
C
p
C
v
Suppose in an adiabatic process pressure and volume of a sample of gas changes from (P1, V1) to (P2, V2) then we have
Thus,
P
=
K
V
γ
Work done by gas in this process is
W
=
∫
V
2
V
1
P
d
V
where limits of integration goes from V1 to V2
Putting for
P
=
K
V
γ
, and integrating we get,
W
=
∫
V
2
V
1
K
V
γ
d
V
W
=
C
1
−
γ
(
1
V
γ
−
1
2
−
1
V
γ
−
1
1
)
Now we have
P
1
V
γ
1
=
P
2
V
γ
2
=
K
W
=
1
1
−
γ
(
P
2
V
2
−
P
1
V
1
)
Now Since
P
1
V
1
=
n
R
T
1
and
P
2
V
2
=
n
R
T
2
, Work done can be expressed as
W
=
n
R
1
−
γ
(
T
1
−
T
2
)
In and adiabatic process if W>0 i.e., work is done by the gas then T2< T1
If work is done on the gas (W<0) then T2 > T1 i.e., temperature of gas rises.
Solved Examples
Question 1
A gas (
γ
=
1.4
) of 2m3 Volume and at a pressure of
4
×
10
5
N
/
m
2
is compressed adiabatically to a volume .5 m 3. Find its new pressure. Calculate the work done in the process?
Given
4
1.4
=
6.96
Solution
For an adiabatic Process
P
V
γ
=
K
Here
P
1
=
4
×
10
5
N
/
m
2
,
V
1
=
2
m
3
,
V
2
=
.5
m
3
,
P
2
=
?
Now
P
1
V
γ
1
=
P
2
V
γ
2
or
P
2
=
P
1
(
V
1
V
2
)
γ
=
(
4
×
10
5
)
(
2
.5
)
γ
=
2.78
×
10
6
N
/
m
2
Now Work done in an adiabatic process
W
=
1
1
−
γ
(
P
2
V
2
−
P
1
V
1
)
Substituting all the values , we have
W
=
−
14.75
×
10
5
J
Question 2
Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in below figure.
(a) Find the work done when the gas is taken from state 1 to state 2.
(b) What is the ratio of temperature
T
1
T
2
, if
V
2
=
2
V
1
?
(c) Given the internal energy for one mole of gas at temperature T is
3
2
R
T
, find the heat supplied to the gas when it is taken from state 1 to 2, with
V
2
=
2
V
1
Work done in an adiabatic Process
Solution
This is clearly an adiabatic Process with
γ
=
1
2
P
V
1
/
2
=
C
o
n
s
t
a
n
t
a. Now work done is given by
W
=
1
1
−
γ
(
P
2
V
2
−
P
1
V
1
)
or
W
=
2
(
P
2
V
2
−
P
1
V
1
)
b. Now since
P
V
=
R
T
or
P
=
R
T
V
So,
P
V
1
/
2
=
C
o
n
s
t
a
n
t
becomes
R
T
V
V
1
/
2
=
C
o
n
s
t
a
n
t
T
V
−
1
/
2
=
c
o
n
s
t
a
n
t
Therefore
T
1
V
−
1
/
2
1
=
T
2
V
−
1
/
2
2
T
1
T
2
=
(
V
2
V
1
)
−
1
/
2
or
T
1
T
2
=
1
√
2
c. Change in Internal Energy will be given by
Δ
U
=
U
2
−
U
1
=
3
2
R
T
2
−
3
2
R
T
1
=
3
2
R
(
T
2
−
T
1
)
=
3
2
R
T
1
(
√
2
−
1
)
Δ
W
=
n
R
1
−
γ
(
T
1
−
T
2
)
=
2
R
T
1
(
√
2
−
1
)
Now
Δ
Q
=
Δ
U
+
Δ
W
=
3
2
R
T
1
(
√
2
−
1
)
+
2
R
T
1
(
√
2
−
1
)
=
7
2
R
T
1
(
√
2
−
1