ADIABATIC PROCESSES
A process that occurs in an isolated system so that no heat is exchanged with the
surroundings is an adiabatic process (i.e. q = 0). Adiabatic processes have practical
importance in connection with air conditioning and refrigeration systems.
Reversible adiabatic expansion of an ideal gas
In an adiabatic expansion of a gas, work is done at the expense of the internal energy of the
gas and the temperature drops.
U = q + w
U = w
(since q = 0)
In this case, the change in internal energy of the gas equals the work done by the gas.
Figure 2: Reversible isothermal and adiabatic expansions
From the P-V curves in figure 2, we see that when a gas expands adiabatically to a larger
volume and lower pressure, the volume is smaller than it would be after an isothermal
expansion to the same pressure.
The energy for doing work in adiabatic expansion of a gas comes from self-cooling of the
gas.
Consider one mol of an ideal gas that expands reversibly and adiabatically.
For an adiabatic process:
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U = w or
dU= dw =-pdV
We saw earlier that dU  Cv dT
Therefore Cv dT   PdV
Now from P = nRT/V
We get:
Cv
(i.e., ideal gas condition)
dT
dV
 R
T
V
Which on integration:
T2
V
2
dT
dV
gives:
Cv 
 R 
T
V
T1
V1
T 
V 
Cv ln  2   R ln  1 
 T1 
 V2 
 T2

 T1



CV
V
  1
 V2
T2  V1 
 
T1  V2 
where
R
CV



R
V
T
or 2   1
T1  V2



 1
for an adiabatic expansion
R C P  CV

  1
CV
CV
Since for one mol of an ideal gas, PV = RT
PV
T
R
V 
T
PV
then 2  2 2   1 
T1
P1V1  V2 
 1
which gives P2V2  P1V1
Or
PV   constant
Compare
for adiabatic process
PV = constant for isothermal process
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Note: The student should also be able to show that for adiabatic process
T2  P2

T1  P1



 1

T 
or  2 
 T1 
CP
R

P2
P1
Work Done on Ideal Gas along a Reversible Path
V2
Work, w    PdV
V1

PV  constant
Since
P
(= b) for adiabatic expansion, therefore
V
b
b
b

V21  V11 
 and w     dV 
V
 1
V V
2
1
gives:
w
P2V2  P1V1
 1
w
nRT2  T1 
 1
w  nCV T  nCV T2  T1 
Summary Table: Work done, internal energy change and enthalpy change of an ideal
monatomic gas due to different P-V-T processes.
Constant
Pressure
Constant
Volume
Isothermal
Adiabatic
Variable
Work done
Molar Heat Capacity,
Cm (J mol-1K-1)
nCv(T2 – T1)
Cp = (5/2)R
Internal Energy
change,
∆U = q + w
Enthalpy,
Cv = (3/2)R
nCv(T2 – T1)
nCp(T2 – T1)
or
or
nCv(T2 – T1)
nCp(T2 – T1)
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Examples
1. When a perfect monatomic gas is allowed to expand adiabatically from 22.4 L at 1 atm
and 273 K to a volume of 44.8 L, the pressure drops to 0.32 atm. Confirm this pressure
and calculate the final temperature and the work done. Answers: T2 = 172 K and w =
-1261 J.
2. A 2-mol sample of N2 gas at STP is expanded reversibly and adiabatically to a pressure
of 0.2 atm.
(i) Which one(s) of the quantities q, w, ∆U and ∆H is(are) zero for that process?
(ii) Calculate the values of each of the non-zero quantities in (i) if any.
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