Thermodynamics week 2 (2).pdf

Physics (PHY 102)Thermodynamics
Mohammad Rashik Zaman
Lecturer in Physics, Basic Science Division
World University of Bangladesh

What is System?
 A thermodynamic system is defined as a quantity of matter or a region in space that is of interest.
 A system can exchange exclusively mass, exclusively energy, or both mass and energy with its surroundings

Internal Energy
The internal energy is the mean value of the
system's total energy in the macroscopic scale i.e.,
Kinetic Energy + Potential Energy
The Change in Internal Energy Formula is:
ΔU = Q - W.
Here,
U = the total change in internal energy within the
system.
Q = the heat exchanged between a system and its
surroundings (outside the system)
W=Workdone

First law of Thermodynamics
o The first law states that the change in the internal energy of a system is
equal to the sum of the heat gained or lost by the system and the work
done by or on the system.
o It can also be stated this: The change in the internal energy of a closed
thermodynamic system is equal to the sum of the amount of heat energy
supplied to or removed from the system and the work done on or by the
system.

First law of Thermodynamics-Mathematical Equation
o Mathematical expression
Then according to the first law of thermodynamics:
Heat supplied = increase in internal energy +work done
DQ = DU+ DW
DQ = (E2 - E1) + DW
DQ = (E2 - E1) + P DV
DU =DE = E2 - E1
DW = P DV

First law of Thermodynamics-Example
o What is an example of the first law of thermodynamics?
 A bicycle pump provides a good example. when we pump on the handle rapidly, it
becomes hot due to mechanical work done on the gas, raising their by its internal
energy. one such simple arrangement is shown in the figure.
Try to find out another example!!!!
The first law can be extended to include changes in mechanical energy =
kinetic energy K + potential energy U
As, ΔU = Q - W.

First law of Thermodynamics-Applications
o Isobaric Process
 It is a process which occurs at
constant pressure.
The heat transferred to the
system does work but also
changes the internal energy (U)
of the system.
o Isochoric process
 A thermodynamic process taking place at
constant volume is known as the isochoric
process. It is also called an isometric
process or constant-volume process. In
such a process, the work done is zero
Since there is no work done (since dW = P
dV = 0 when V = constant,W = ∫pdV=0)
on or by the system, based on the first law
of thermodynamics
dQ = dU (isochoric process)
o Isothermal process
 An isothermal process is
a thermodynamic process in which
the temperature of a system remains
constant. The transfer of heat into or
out of the system happens so slowly
that thermal equilibrium is
maintained.
Isothermal process (dU = 0):
=> dU = 0 = Q – W
=> W = Q

First law of Thermodynamics-Applications
o Adiabatic Process
 An adiabatic process is one in which no heat is gained or lost
by the system. An adiabatic process may be accomplished by
thermal insulation or by making rapid changes in volume so
that there is no time for heat to be exchanged as shown in the
animation.
o Limitations of first law of thermodynamics
 The limitation of the first law of thermodynamics is that
it does not say anything about the direction of flow of
heat.
 It does not say anything whether the process is a
spontaneous process or not.
 The reverse process is not possible. In actual practice, the
heat doesn’t convert completely into work. If it would
have been possible to convert the whole heat into work,
then we could drive ships across the ocean by extracting
heat from the water of the ocean.

First law of Thermodynamics-Mathematical Problems
1. 3000 J of heat is added to a system and 2500 J of work is
done by the system. What is the change in internal
energy of the system?
2. 3.0-g bullet traveling at a speed of 400 m/s enters a tree and
exits the other side with a speed of 200 m/s. Where did the
bullet’s lost kinetic energy go, and how much energy was
transferred?
Known :
Heat (Q) = +3000 Joule
Work (W) = +2500 Joule
Wanted: the change in internal energy of the system
Solution :
The sign conventions :
Q is positive if the heat added to the system
W is positive if work is done by the system
Q is negative if heat leaves the system
W is negative if work is done on the system
The change in internal energy of the system :
The equation of the first law of thermodynamics is
Δ U = Q-W
=>ΔU = 3000-2500
=>ΔU = 500 Joule
Internal energy increases by 500 Joule.
Known :
vi = 400 m/s, vf = 200 m/s
K.Ei – K.Ef = ½ m (vi
2-vf
2)
Solution :
K.Ei – K.Ef = ½ m (vi
2-vf
2)
= ½ (3x10-3 )(16-4)x104 J
= 180 J
We know that,
 A 5.00-gm lead bullet traveling at 300 m/s is stopped by a large tree. If
half the kinetic energy of the bullet is transformed into internal energy and
remains with the bullet while the other half is transmitted to the tree, what
is the increase in temperature of the bullet?
o Try yourself!!!

Specific Heat
Specific Heat Capacity Formula
Q = C m ∆t
Or, C =
𝐐
𝐦∆𝐭
Where,
• Q = quantity of heat absorbed by a body
• m = mass of the body
• ∆t = Rise in temperature
• C = Specific heat capacity of a substance depends on the
nature of the material of the substance.
• S.I unit of specific heat is J kg-1 K-1.
Specific heat is the amount of heat required to change the heat content of exactly 1
gram of a material by exactly 1°C.

Inroduction of Cp & Cv
It is the amount of thermal energy that is
exchanged between a system and its surroundings
without any change in the volume of the system
being transferred.
Specifically, if this change in internal energy occurs
while the volume remains constant, then it is known
as the specific heat at constant volume (Cv).
Cv=dU/dt
Where, dU= change in internal energy.
dt.= change in temperature
It is the energy transmitted between a system and
its environment when the system is subjected to
continuous pressure.
As the temperature of the system changes, the
enthalpy of the system will change as well. As a result,
the following can be presented.
Cp=dH/dt
where, Cp= specific heat at constant pressure,
dH= change in enthalpy ,
dt= change in temperature.
Cv Cp

Relation between Cp & Cv
Consider an ideal gas. Let dq be the amount of heat given to the
system to raise the temperature of an ideal gas by dT, and change
in internal energy be du. Then, According to the first law of
thermodynamics
The above relation between Cp & Cv is true only for an ideal gas

Mathematical Problem
1.Find the molar specific heat of carbon dioxide gas at constant volume and constant
pressure. Given that 𝜸=1.33 and R =8.31 J
Solution : We know,
Cp-Cv=R
and, 𝛾 =
𝐶𝑝
𝐶𝑣
So, Cp= 𝛾 Cv
∴ 𝛾Cv –Cv= R
=> Cv=
𝑅
𝛾−1
=
8.31Jmol
−
1
K
−
1
1.33−1
=25.18 Jmol-1K-1
Again, Cp-Cv=R
Or, Cp= Cv+R
=25.18 Jmol-1K-1+ 8.31Jmol
− 1K
− 1
=33.49 𝐉𝐦𝐨𝐥
− 𝟏𝐊 − 𝟏

Thermodynamics week 2 (2).pdf

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