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Thermodynamics
THERMODYNAMICS
Thermodynamics is
the study of energy
relationships that
involve heat,
mechanical work,
and other aspects of
energy and heat
transfer.
Central Heating
A THERMODYNAMIC SYSTEM
• A system is a closed environment in
which heat transfer can take place. (For
example, the gas, walls, and cylinder of
an automobile engine.)
Work done on
gas or work
done by gas
INTERNAL ENERGY OF SYSTEM
• The internal energy U of a system is the
total of all kinds of energy possessed by
the particles that make up the system.
Usually the internal energy consists
of the sum of the potential and
kinetic energies of the working gas
molecules.
TWO WAYS TO INCREASE THE
INTERNAL ENERGY, U.
+U
WORK DONE
ON A GAS
(Positive)
HEAT PUT INTO
A SYSTEM
(Positive)
TWO WAYS TO DECREASE THE
INTERNAL ENERGY, U.
Wout
Qout
-U
Decrease
hot
WORK DONE BY
EXPANDING GAS:
W is positive
hot
HEAT LEAVES A
SYSTEM
Q is negative
THERMODYNAMIC PROCESS
Increase in Internal Energy, U.
Wout
Qin
Initial State:
P1 V1 T1 n1
Heat input
Final State:
Work by gas
P2 V2 T2 n2
The Reverse Process
Decrease in Internal Energy, U.
Win
Qout
Initial State:
P1 V1 T1 n1
Work on gas
Loss of heat
Final State:
P2 V2 T2 n2
THE FIRST LAW OF
THERMODYAMICS:
• The net heat put into a system is equal to
the change in internal energy of the
system plus the work done BY the system.
Q = U + W
final - initial)
• Conversely, the work done ON a system is
equal to the change in internal energy plus
the heat lost in the process.
SIGN CONVENTIONS
FOR FIRST LAW
• Heat Q input is positive
+Wout
+Qin
U
• Work BY a gas is positive
-Win
U
• Work ON a gas is negative
• Heat OUT is negative
Q = U + W
-Qout
final - initial)
APPLICATION OF FIRST
LAW OF THERMODYNAMICS
Example 1: In the figure, the
Wout =120 J
gas absorbs 400 J of heat and
at the same time does 120 J
of work on the piston. What
is the change in internal
energy of the system?
Qin
Apply First Law:
Q = U + W
400 J
Example 1 (Cont.): Apply First Law
Q is positive: +400 J (Heat IN)
Wout =120 J
W is positive: +120 J (Work OUT)
Q = U + W
U = Q - W
Qin
400 J
U = Q - W
= (+400 J) - (+120 J)
= +280 J
U = +280 J
Example 1 (Cont.): Apply First Law
Energy is conserved:
The 400 J of input thermal
energy is used to perform
120 J of external work,
increasing the internal
energy of the system by
280 J
The increase in
internal energy is:
Wout =120 J
Qin
400 J
U = +280 J
FOUR THERMODYNAMIC
PROCESSES:
• Isochoric Process:
V = 0, W = 0
• Isobaric Process:
P = 0
• Isothermal Process: T = 0, U = 0
• Adiabatic Process:
Q = 0
Q = U + W
ISOCHORIC PROCESS:
CONSTANT VOLUME, V = 0, W = 0
0
Q = U + W
so that
Q = U
QIN
+U
QOUT
No Work
Done
-U
HEAT IN = INCREASE IN INTERNAL ENERGY
HEAT OUT = DECREASE IN INTERNAL ENERGY
ISOCHORIC EXAMPLE:
No Change in
volume:
P2
B
P1
A
PA
TA
=
PB
TB
V1= V2
400 J
Heat input
increases P
with const. V
400 J heat input increases
internal energy by 400 J
and zero work is done.
ISOBARIC PROCESS:
CONSTANT PRESSURE, P = 0
Q = U + W
But
W = P V
QIN
QOUT
Work Out
+U
-U
Work
In
HEAT IN = Wout + INCREASE IN INTERNAL ENERGY
HEAT OUT = Win + DECREASE IN INTERNAL ENERGY
ISOBARIC EXAMPLE (Constant Pressure):
P
A
B
VA
TA
400 J
Heat input
increases V
with const. P
V1
=
VB
TB
V2
400 J heat does 120 J of
work, increasing the
internal energy by 280 J.
ISOBARIC WORK
P
A
B
VA
TA
400 J
V1
V2
=
TB
PA = PB
Work = Area under PV curve
W
orkPV
VB
ISOTHERMAL PROCESS:
CONST. TEMPERATURE, T = 0, U = 0
Q = U + W
AND
QIN
U = 0
Q = W
QOUT
Work Out
U = 0
Work
In
NET HEAT INPUT = WORK OUTPUT
WORK INPUT = NET HEAT OUT
ISOTHERMAL EXAMPLE (Constant T):
PA
A
B
PB
U = T = 0
PAVA = PBVB
V2
V1
Slow compression at
constant temperature:
----- No change in U.
ISOTHERMAL EXPANSION (Constant T):
PA
A
B
PB
U = T = 0
VA
VB
400 J of energy is absorbed
by gas as 400 J of work is
done by gas.
T = U = 0
PAVA = PBVB
TA = TB
Isothermal Work
VB
W  nRT ln
VA
ADIABATIC PROCESS:
NO HEAT EXCHANGE, Q = 0
Q = U + W ; W = -U or U = -W
U = -W
W = -U
U
Work Out
Q = 0
+U
Work
In
Work done at EXPENSE of internal energy
INPUT Work INCREASES internal energy
ADIABATIC EXAMPLE:
PA
A
B
PB
V1
Insulated
Walls: Q = 0
V2
Expanding gas does
work with zero heat
loss. Work = -U
ADIABATIC EXPANSION:
PA
A
B
PB
Q = 0
PAVA
TA
VA
400 J of WORK is done,
DECREASING the internal
energy by 400 J: Net heat
exchange is ZERO. Q = 0
=
PBVB
TB
VB


P
AV
A P
BV
B
ADIABATIC EXPANSION/COMPRESSION
According to the 1st Law of
Thermodynamics: ΔU = Q – Wby
but Q = 0, since walls are insulated
ΔU = – Wby
For a monatomic ideal gas:
Wby  U  32 nRTi  T f 
The red curve shows an adiabatic expansion of
an ideal gas.
The blue curves are isotherms at Ti and Tf.
Adiabatic curves can be approximated as linear.
To Be Continued…