Thermodynamic Processes

Types of Thermodynamic Processes
• Isobaric process
The processes during which the pressure of the system remains
constant are called isobaric thermodynamic processes.

• Isochoric process:
The process, during which the volume of the system remains constant,
is an isochoric process. Heating of a gas in a closed cylinder is an
example of the isochoric process.
Work done in an isochoric process
W=∫ P dV
Here, dV =0
Therefore, W=0
• Isothermal Process
In an isothermal process, there is no change in temperature, that
means the temperature remains constant.
• Adiabatic Process:
The process, during which the heat content of the system remains
constant, is an adiabatic process. During this process heat neither
enters the system nor leaves the system.
For an adiabatic process,
ΔQ=0
Then according to the first law of thermodynamics,
ΔU+ΔW = ΔQ = 0
where, Q is the heat supplied to the system and W is the work done by
the system and U is the internal energy of the system.
• Quasi-Static Process
When a process in which system remains close to an equilibrium state
at each time, such process will be termed as the quasi-static
process or quasi-equilibrium process.
Question For You
Q. The process in which the internal energy of the system remains
constant is :
• a. Adiabatic b. Isobaric
• c. Isochoric d. Isothermal
THE REVERSIBLE PROCESS
• A process is reversible when its direction can be reversed at any point
by an infinitesimal change in external conditions.
OR
A thermodynamic process is reversible if the process can return back in
such a that both the system and the surroundings return to their
original states, with no other change anywhere else in the universe. It
means both system and surroundings are returned to their initial states
at the end of the reverse process.
Summary Remarks on Reversible
Processes
A reversible process:
∙ Can be reversed at any point by an minute change in external conditions
∙ Is never more than minutely removed from equilibrium
∙ Crosses a succession of equilibrium states
∙ Is frictionless
∙ Is driven by forces whose imbalance is minute in magnitude
∙ Proceeds infinitely slowly
∙ When reversed, retraces its path, restoring the initial state of system and
surroundings
Computing Work for Reversible Processes
•The
work done on the system is given by
dW = − P d V
Provided for a mechanically reversible process:
 Temperature and pressure are constant &

 Internal pressure balance the external force i.e. F = PA
For such process,
W=-
Isochoric & Isobaric Processes
For 1 mole of a homogeneous fluid contained in a closed system, the energy
balance is given by
dU = dQ + dW
The work for a mechanically reversible, closed-system process is given by
dW = −P dV
Substitution into the preceding equation yields:
dU = dQ − PdV
This is the general energy balance for one mole or a unit mass of
homogeneous fluid in a closed system undergoing a mechanically reversible
process.
Isochoric & Isobaric Processes (contd.)
For a constant-volume change of state

dU = dQ (constant V )
Integration yields:
ΔU = Q (constant V )
The internal energy change for a mechanically reversible, constant-
volume, closed-system process equals the amount of heat transferred
into the system.
Isochoric & Isobaric Processes (contd.)
For a constant-pressure change of state:
dU + PdV = d ( U + PV ) = dQ
The mathematical (and only) definition of enthalpy is:

H ≡ U + PV
where H, U, and V are molar or unit-mass values. The preceding energy balance
becomes:
dH = dQ (constant P)
Integration yields:
ΔH = Q (constant P)
• The enthalpy change in a mechanically reversible, constant-pressure, closed-system
process equals the amount of heat transferred into the system.
Q1. Calculate ΔU and ΔH for 1 kg of water when it is vaporized at the
constant temperature of 100°C and the constant pressure of 101.33 kPa. The
specific volumes of liquid and vapor water at these conditions are 0.00104
and 1.673 m3·kg−1, respectively. For this change, heat in the amount of
2256.9 kJ is added to the water.
Q2. Liquid water at 180 C and 1,002.7 kPa has an internal energy of 762.0
kJkg-1 and specific volume of 1.128 cm3g-1.
a) what is its Enthalpy
b) The water is brought to the vapour state at 300 C and 1,500 kPa where
its internal energy is 2,784.4 kJ kg-1 and its specific volume is 169.7 cm3g-1.
Calculate ΔU and ΔH
Heat Capacity at Constant Volume
•Different
substances respond to heat in different ways. If a metal chair
sits in the bright sun on a hot day, it may become quite hot to the
touch. An equal mass of water in the same sun will not become nearly
as hot. We would say that water has a high heat capacity
This means that smaller the temperature change in a body caused by
the transfer of a given quantity of heat, the greater its capacity. So Heat
Capacity may be define as
C=
At constant volume dQ=dU, therefore
at constant
volume
Heat Capacity at Constant Volume
(contd.)
Integration of the above equation yields
At constant volume ΔU = Q , therefore

Heat Capacity at Constant Pressure
• C=
At constant pressure dH = dQ (constant P)
Therefore
At constant pressure ΔH = Q (constant P)

THE REVESIBLE ABIABATIC PROCESS
•In adiabatic process, no heat is transferred from and to the system, so
the first law of thermodynamics equation will be
dU = dW= - PdV
For any process
dU = Cv dT
Therefore Cv dT = - PdV
But for one mole of a gas PV=RT-------- P=
Cv dT = - dV
……………… (1)
Let = and we have Cp= Cv + R
•
=
Therefore
After integration
ln = ln
Or
ln = ln ( )
Or ln = ln ()
Or ……………… (2)
•
Substituting in (1)
= ……… (3)
……………… (2)
•Equating
(2) & (3)
=

=
P1 V1 = P2 V2
dW = Cv dT
Integration will give
W =Cv (T2 – T1)
As Cv =
W = (T2 – T1) T1 = & T2 =
W= (- )
W = (- ) …………… (4)
•V2 usually not known. We can find it from the following equation
P1 V1 = P2 V2
V2 = or V2 = ( ) 1/γ
Substituting V2 in equation (4)

W= [-]
W = [(- 1]
W = [(- 1]
For monoatomic gases = 1.67
For diatomic gases = 1.4
For polyatomic gases = 1.3
One mole of an ideal gas with CP = (7/2)R and CV = (5/2)R expands from P1 =
8 bar and T1 = 600 K to P2 = 1 bar by each of the following paths:
(a) Constant volume;
(b) Constant temperature;
(c) Adiabatically.
Assuming mechanical reversibility, calculate W, Q, ΔU, and ΔH for each
process.
One mole of an ideal gas with CP = (5/2)R and CV = (3/2)R expands from
P1 = 6 bar and T1 = 800 K to P2 = 1 bar by each of the following paths:
(a) Constant volume
(b) Constant temperature
(c) Adiabatically
Assuming mechanical reversibility, calculate W, Q, ΔU, and ΔH for each
process.
One mole of air, initially at 150°C and 8 bar, undergoes the following
mechanically reversible changes. It expands isothermally to a pressure such
that when it is cooled at constant volume to 50°C its final pressure is 3 bar.
Assuming air is an ideal gas for which CP = (7/2)R and CV = (5/2)R, calculate
W, Q, ΔU, and ΔH.
One mole of an ideal gas, initially at 30°C and 1 bar, is changed to 130°C
and 10 bar by three different mechanically reversible processes:
1. The gas is first heated at constant volume until its temperature is

130°C; then it is compressed isothermally until its pressure is 10 bar.
2. The gas is first heated at constant pressure until its temperature is
130°C; then it is compressed isothermally to 10 bar.
3. The gas is first compressed isothermally to 10 bar; then it is heated
at constant pressure to 130°C.
Calculate Q, W, ΔU, and ΔH in each case. Take CP = (7/2)R and CV =
(5/2)R. Alternatively, take CP = (5/2)R and CV = (3/2)R.
• One cubic meter of an ideal gas at 600 K and 1000 kPa expands to five
times its initial volume as follows:
(a) By a mechanically reversible, isothermal process.
(b) By a mechanically reversible, adiabatic process.
(c) By an adiabatic, irreversible process in which expansion is against a
restraining pressure of 100 kPa.
For each case calculate the final temperature, pressure, and the work
done by the gas.
Take CP = 21 J ・ mol−1 ・ K−1.
• The state of an ideal gas with CP = (5/2)R is changed from P = 1 bar
and V1t  = 12 m3 to P2 = 12 bar and V2t  = 1 m3 by the following
mechanically reversible processes:
(a) Isothermal compression.
(b) Adiabatic compression followed by cooling at constant pressure.
(c) Adiabatic compression followed by cooling at constant volume.
(d) Heating at constant volume followed by cooling at constant pressure.
(e) Cooling at constant pressure followed by heating at constant volume.
Calculate Q, W, ΔUt, and ΔHt for each of these processes, and sketch the
paths of all processes on a single PV diagram.
One mole of air, initially at 30°C and 1 bar, undergoes the following
mechanically reversible changes. It is compressed isothermally to a
point such that when it is heated at constant volume to 120°C its final
pressure is 12 bar., calculate W, Q, ΔU, and ΔH. Take CP = (7/2)R and
CV = (5/2)R
A process consists of two steps:
(1) One mole of air at T = 800 K and P = 4 bar is cooled at constant volume
to T = 350 K.
(2) (2) The air is then heated at constant pressure until its temperature
reaches 800 K.
If this two-step process is replaced by a single isothermal expansion of the
air from 800 K and 4 bar to some final pressure P, what is the value of P
that makes the work of the two processes the same? Assume mechanical
reversibility and treat air as an ideal gas with CP = (7/2)R and CV = (5/2)R.
500 liters of oxygen is given with pressure of 106 Pa and temperature of
30 C. The oxygen will be compressed isothermally to the 1/5th of its
original volume. What is the work done by the outer neighboring
medium during the gas is being compressed? What is the heat
transmitted to the gas during the process? What is the change of the
internal energy of the gas? What is the new pressure of the gas at the
end of the process? (ΔU=0, W=804718.956J, Q=-W, p=5x106 Pa)
Air is compressed from 10 liters to 2 liters using two different ways.
First way is use of adiabatic and second way is use of isothermal way.
Work done by the neighbor medium during the adiabatic process is Wa.
Work done by the neighbor medium during the isothermal process is
Wi. What is the higher value, Wa or Wi? What is the ratio of Wa and
Wi? (Wa>Wi)

Thermodynamic Processes

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Thermodynamic Processes

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Types of Thermodynamic Processes

 Temperature and pressure are constant &

For a constant-volume change of state

For a constant-pressure change of state:

The mathematical (and only) definition of enthalpy is:

Integration of the above equation yields

At constant volume ΔU = Q , therefore

At constant pressure ΔH = Q (constant P)

Substituting V2 in equation (4)

1. The gas is first heated at constant volume until its temperature is

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