1. For more information about
the basics of heat
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‘Heat, Calorimetry and
Thermodynamics by Aditya
Abeysinghe’

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3. 1. Zeroth law of thermodynamics
If two systems are at the same time in equilibrium
with a third system, they are in equilibrium with each
other.
Practically, this means that all three systems are at the
same temperature.
A
B
C
Since A and B are at
equilibrium and B and
C are at
equilibrium, A and C
are also at equilibrium
according to the
zeroth law

4. 2. First law of thermodynamics
The change in internal energy of a system is equal
to the heat added to the system minus the work done
by the system.
dU = δQ – δW (U- internal energy of the system, Qheat added to the system, W- work done by the
system)
The first law of thermodynamics is the application of
the conservation of energy principle to heat and
thermodynamic processes.

5. 3. Second law of thermodynamics
When two initially isolated systems which are at
thermal equilibriums are brought into contact they
reach a common thermal equilibrium
However, the second law can also be expressed in
terms of the application in which it is used.
For example,
I.
Second law in terms of heat flow:
Heat flows spontaneously from hotter to colder
objects but not vice versa.
II.
Second law in terms of heat engines:
It is impossible to construct an engine which has
100% efficiency or a system in which the heat added to
the system is solely used to perform work.

6. 4. Third law of thermodynamics
The entropy of a system approaches a constant value
as the temperature approaches zero.

7. Kelvin-Planck statement:
It is impossible to extract an amount of heat QH from a hot
reservoir and use it all to do work W. Some amount of heat QC
must be exhausted to a cold reservoir.

8. Refrigerator and Air ConditionerClausius StatementIt is impossible for heat to flow from a cold body to a warm body without any
work ( or without the aid of an external agency) having been done to accomplish
the flow.
Hot Reservoir
QH
W
QC
Cold Reservoir

9. Internal energy of a gasThe internal energy of a gas is the kinetic energy of
thermal motion of its molecules.
Important:
When a cylinder of a gas is moving in a locomotive or
any other moving agent, the molecules inside the
cylinder move relative to that of the cylinder (molecules
travel in the same speed as the cylinder). Thus the
kinetic energy of each molecule is equivalent to the
kinetic energy experienced by the cylinder.
Therefore, the temperature and thus the internal energy
of the molecules relative to the cylinder is unchanged.

10. The internal energy of a gas is the sum of the kinetic
energy of the molecules and the potential
energy(due to intermolecular attractions) between
molecules.
But there are no intermolecular forces between
molecules in an ideal gas. Therefore, potential
energy in such an instance is zero.
Therefore, Internal Energy = Kinetic Energy of
molecules

11. But, PV = ⅓ mNc2 and E = ½mNc2
So, PV = ⅓ × 2 × ½ mNc2
E
Therefore, PV = ⅔E or, E=
PV
3
2
But, PV = nRT . Therefore, E=
nRT
Now that E = ΔU,
For a difference of temperature,
Internal Energy =
nRΔT
(ΔU)

12. Work done by gasWork done by gas is the work done to increase its volume
during expansion.
A
Δ W = Fx
Δ W = PAx
Therefore, Δ W = P(ΔV)
P
F
(PA
)
X

13. When heat is given to a substance it expands and
does external work. In the case of solids and liquids
the change in volume and hence the external work
done is negligible. Therefore, there is only one
specific heat for a substance.
Gases experience the effect of the change of volume
to a great extent. Since the volume can be
controlled, gases can be used to do variable amounts
of external work. Therefore, there are no. of specific
heats that can be defined for a gas.

14. The two most commonly used are1. Specific heat at constant volume (CV)
It is the amount of heat required to raise the temperature
of unit mass of a gas through 1°C, when volume is kept
constant.
2.
Specific heat at constant pressure (CP)
It is the amount of heat required to raise the temperature
of unit mass of a gas through 1°C, when the pressure of the
gas is kept constant
However, CP > CV and CP / CV = γ. γ is called the ratio of
specific heats.
At constant volume, ΔQ = n CV θ
At constant pressure, ΔQ = n CP θ

15. Increase /
Decrease
Ter
m
ΔQ
Positive
OR
Increasing
If heat is
given
from
outside
Negative
OR
Decreasin
g
If heat is
released
to
outside
ΔU
ΔU
ΔW
If internal
energy
increases
If a gas
does work
to outside
If internal
energy
decreases
If work is
done on
the gas

16. Isothermal Process (ΔU = 0)
1.
This usually occurs when a system is in thermal contact
with a reservoir. The change occurs very slowly and the
system will continually adjust to the temperature of the
reservoir through heat exchange.
ΔT = 0, therefore, ΔU = 0.
Thus, ΔQ = ΔW
The graph of such a process is as follows:
P
Temperature constant.
Therefore the system obeys
Boyle’s Law
V

17. 2. Adiabatic Process (ΔQ = 0)
An expansion in which no heat energy enters or leaves
the system.
Adiabatic processes can take place if the container in
which the process takes place has thermally-insulated
walls or the process happens in an extremely short
time.
Adiabatic CoolingOccurs when the pressure of a substance is decreased as
it does work on the surroundings
Adiabatic HeatingOccurs when the pressure of a gas is increased from
work done on it by the surroundings.

18. The graph of such a process is as follows:
P
A (T1)
B (T2)
V
Adiabatic
process
An adiabat is similar to an
isotherm, except that during
expansion an adiabat loses more
pressure than an isotherm, so it has
a steeper inclination
Density of isotherms
stays constant but the
density of adiabats
grow
Isothermal
Processes
Each adiabat
intersects each
isotherm exactly
once

19. 3. Isochoric Process (ΔW = 0)
Also called constant-volume process, iso volumetric
process, and isometric process.
Is a thermodynamic process during which the volume of
the closed system stays constant.
Since the system undergoes isochoric process, the volume
is constant.
Therefore,
Q= mCv ΔT
and Q = U (W=0 and thus PΔV = 0)
The graph for such a process is as follows:
and since
V is constant
T

20. 1.
Isobaric Process (Pressure Constant)
Since pressure stays
constant,
ΔQ = nCP ΔT
.
2. Isoentropic process (Entropy of system stays
constant)
3. Isoenthalpic process(Enthalpy of system
constant)
4. Quasistatic Process (Is a process that
happens infinitely slowly)

21. 1.
For an isothermal process,
P
A(T)
ΔU = 0
ΔQ = ΔW
Since T is constant,
ΔQ = +
By applying Boyle’s law, ΔW= +
P1V1 = P2 V2
B(T)
2.
V
ΔW = 0
ΔQ = - ΔU
ΔQ = +
ΔU= -
P1 / T1 = P2 / T2

22. 3. The work done in a process is the area under a P-V
graph
P
PΔV- the work done
by the process ( Area
under the graph)
P
V
ΔV
4. For an adiabatic process,
P
A (T1)
ΔQ = 0
ΔW = - ΔU
B (T2)
V
P1 V1 / T1 = P2 V2 /
T2

23. 5. If some system is closed the sum of internal
energies should equal to zero.
Since the system is closed sum of ΔU for
the whole process is zero.
However, the area shaded inside the
process represents the net work done.
6. In a P- V diagram the internal energy acquired or
released is independent of the path
ΔU
Internal energy between two points
for a closed system is independent of
the path
(Same value regardless of path)

24. 7. If some system is clockwise ΔQ > 0 for that
system. Similarly, if a system is counterclockwise,
ΔQ < 0 for that system
System clockwise.
Therefore, ΔQ > 0
System
counterclockwise.
Therefore, ΔQ < 0

25. Sample QuestionUsing 3 moles of air an engine goes through the
following changes in a single cycle.(ABCD)
i.
Isothermal expansion from a volume of 0.03m3 to
0.07m3 at 373K (A to B)
ii.
At constant volume cooling to 320K(B to C)
iii.
Isothermal compression at 320K to 0.03m3 (C to D)
iv.
At constant volume, compression to original
volume, pressure, and temperature.(D to A)
Show these changes in a P-V diagram and calculate the
following:
a.
The heat taken during process B to C
b.
The heat taken during the process D to A
c.
The total internal energy of the engine after process
completion

26. P
(Pa)
For point A,
PV= nRT
P × 0.03 = 3 × 8.314 × 373
P = 3.1 × 106 Pa
A
3.1 × 106
ΔU = 0
ΔQ =
ΔW
1.33 × 106
1.14 ×106
B
For point B,
PV=nRT
P × 0.07 = 3 × 8.314 × 373
P = 1.33 × 106 Pa
ΔW=0
ΔQ = ΔU
D
C
0.03
For point D,
At constant value,
P1 /T1 = P2 / T2
(1.33 × 106)/ 373 = P2 / 320
P2= 1.14 × 106 Pa
0.07
V
(m3
)

27. By considering process B to C,
ΔU = (3/2) nRΔT
= (3/2) × 3 × 8.314× 373
ΔU = 13955 J
However, during B to C,
ΔQ = ΔU as the process is isochoric.
Therefore, heat added to the system during B to C is
13995 J
Similarly, same heat is added to the system during D to A
as during B to C as the temperature difference(ΔT) is
same.(ΔU same)
Therefore, heat added to the system during D to A is also
13995 J.
Since this is a closed process, the sum of internal energies
should be zero.