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THERMODYNAMICS
Thermodynamics is
Thermodynamics the study of energy
relationships that
involve heat,
mechanical work,
and other aspects of
energy and heat
transfer.
Central Heating

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Objectives: A THERMODYNAMIC SYSTEM

• State and apply the first and • A system is a closed environment in
second laws of thermodynamics. which heat transfer can take place. (For
example, the gas, walls, and cylinder of
• Demonstrate your understanding an automobile engine.)
of adiabatic, isochoric, isothermal,
and isobaric processes.
• Write and apply a relationship for determining Work done on
the ideal efficiency of a heat engine. gas or work
done by gas
• Write and apply a relationship for determining
coefficient of performance for a refrigeratior.

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TWO WAYS TO INCREASE THE

INTERNAL ENERGY OF SYSTEM INTERNAL ENERGY, DU.
• The internal energy U of a system is the
total of all kinds of energy possessed by
the particles that make up the system.
+DU

Usually the internal energy consists

of the sum of the potential and WORK DONE HEAT PUT INTO
kinetic energies of the working gas ON A GAS A SYSTEM
molecules.
(Positive) (Positive)

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TWO WAYS TO DECREASE THE

INTERNAL ENERGY, DU. THERMODYNAMIC STATE
Wout The STATE of a thermodynamic
Qout
system is determined by four
-DU factors:
Decrease
hot
• Absolute Pressure P in
hot
Pascals
• Temperature T in Kelvins
WORK DONE BY HEAT LEAVES A • Volume V in cubic meters
EXPANDING GAS: SYSTEM • Number of moles, n, of working gas
DW is positive DQ is negative

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THERMODYNAMIC PROCESS The Reverse Process

Decrease in Internal Energy, DU.
Increase in Internal Energy, DU.

Wout W in

Q out
Q in

Heat input Work on gas

Initial State: Final State: Initial State: Final State:
P1 V1 T1 n1 P2 V2 T2 n2 P1 V1 T1 n1 Loss of heat P2 V2 T2 n2
Work by gas

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THE FIRST LAW OF

THERMODYAMICS: SIGN CONVENTIONS
FOR FIRST LAW +Wout
• The net heat put into a system is equal to +Q in
the change in internal energy of the • Heat Q input is positive +DU
system plus the work done BY the system.
• Work BY a gas is positive -W in
-DU
DQ = DU + DW D = (final - initial) • Work ON a gas is negative
• Heat OUT is negative -Q out
• Conversely, the work done ON a system is
equal to the change in internal energy plus
the heat lost in the process. DQ = DU + DW D = (final - initial)

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APPLICATION OF FIRST Example 1 (Cont.): Apply First Law

LAW OF THERMODYNAMICS
Example 1: In the figure, the DQ is positive: +400 J (Heat IN) Wout =120 J
gas absorbs 400 J of heat and Wout =120 J
DW is positive: +120 J (Work OUT)
at the same time does 120 J Qin
of work on the piston. What DQ = DU + DW
is the change in internal 400 J
DU = DQ - DW
energy of the system? Qin
400 J
Apply First Law:
DU = DQ - DW
DQ = DU + DW = (+400 J) - (+120 J) DU = +280 J
= +280 J

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Example 1 (Cont.): Apply First Law

FOUR THERMODYNAMIC
Energy is conserved: PROCESSES:
Wout =120 J
The 400 J of input thermal
• Isochoric Process: DV = 0, DW = 0
energy is used to perform Qin
120 J of external work, • Isobaric Process: DP = 0
400 J
increasing the internal
energy of the system by • Isothermal Process: DT = 0, DU = 0
280 J
• Adiabatic Process: DQ = 0
The increase in
internal energy is: DU = +280 J
DQ = DU + DW

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ISOCHORIC PROCESS: ISOCHORIC EXAMPLE:

CONSTANT VOLUME, DV = 0, DW = 0 No Change in
0 volume: P2 B
DQ = DU + DW so that DQ = DU PA PB
=
QIN QOUT P1 A TA TB

No Work
+DU -DU V 1= V 2
Done
400 J
HEAT IN = INCREASE IN INTERNAL ENERGY Heat input 400 J heat input increases
increases P internal energy by 400 J
HEAT OUT = DECREASE IN INTERNAL ENERGY
with const. V and zero work is done.

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ISOBARIC PROCESS: ISOBARIC EXAMPLE (Constant Pressure):

CONSTANT PRESSURE, DP = 0
A B
DQ = DU + DW But DW = P DV P VA VB
=
QIN QOUT TA TB

Work Out Work

+DU -DU 400 J V1 V2
In

Heat input 400 J heat does 120 J of

HEAT IN = Wout + INCREASE IN INTERNAL ENERGY
increases V work, increasing the
HEAT OUT = Wout + DECREASE IN INTERNAL ENERGY internal energy by 280 J.
with const. P

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ISOBARIC WORK ISOTHERMAL PROCESS:

CONST. TEMPERATURE, DT = 0, DU = 0
A B
P VA VB DQ = DU + DW AND DQ = DW
=
TA TB Q IN Q OUT
PA = PB Work Out Work
400 J V1 V2 DU = 0 DU = 0
In

Work = Area under PV curve

NET HEAT INPUT = WORK OUTPUT
Work = P DV WORK INPUT = NET HEAT OUT

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ISOTHERMAL EXAMPLE (Constant T): ISOTHERMAL EXPANSION (Constant T):

A A
PA PA P AVA = P B V B
B
B PB
PB TA = T B
VA VB
DU = DT = 0
DU = DT = 0 V2 V1
400 J of energy is absorbed
Isothermal Work
Slow compression at by gas as 400 J of work is
constant temperature: done on gas. VB
P AVA = P B V B W = nRT ln
----- No change in U. DT = DU = 0 VA

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ADIABATIC PROCESS: ADIABATIC EXAMPLE:

NO HEAT EXCHANGE, DQ = 0
A
DQ = DU + DW ; DW = -DU or DU = -DW PA
B
D W = -D U D U = -D W
PB
Work Out Work
-DU +DU
In V1 V2
DQ = 0

Work done at EXPENSE of internal energy Insulated Expanding gas does

INPUT Work INCREASES internal energy Walls: DQ = 0 work with zero heat
loss. Work = -DU

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ADIABATIC EXPANSION:
HEAT ENGINES
A
PA A heat engine is any
B P AVA P BV B Hot Res. T H
= device which through
PB TA TB Qhot a cyclic process:
Wout
DQ = 0 VA VB • Absorbs heat Qhot
Engine

400 J of WORK is done, Qcold • Performs work Wout

DECREASING the internal g g
energy by 400 J: Net heat PAVA = PBVB Cold Res. TC • Rejects heat Qcold
exchange is ZERO. DQ = 0

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THE SECOND LAW OF

THE SECOND LAW OF THERMODYNAMICS
THERMODYNAMICS
Hot Res. TH Hot Res. TH
Hot Res. T H It is impossible to construct an
400 J 100 J 400 J
Qhot engine that, operating in a 400 J
Wout cycle, produces no effect other Engine Engine
Engine than the extraction of heat
from a reservoir and the 300 J
Qcold performance of an equivalent Cold Res. TC Cold Res. TC
amount of work.
Cold Res. TC
• A possible engine. • An IMPOSSIBLE
Not only can you not win (1st law); engine.
you can’t even break even (2nd law)!

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EFFICIENCY OF AN ENGINE EFFICIENCY EXAMPLE

The efficiency of a heat engine An engine absorbs 800 J and
Hot Res. TH is the ratio of the net work Hot Res. TH wastes 600 J every cycle. What
QH W done W to the heat input Q H. 800 J W is the efficiency?
QC
Engine W Q H- Q C Engine e=1-
e= = QH
QC QH QH 600 J
Cold Res. TC Cold Res. TC 600 J
QC e=1- e = 25%
e=1- 800 J
QH
Question: How many joules of work is done?

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EFFICIENCY OF AN IDEAL Example 2: A steam engine absorbs 600 J

of heat at 500 K and the exhaust
ENGINE (Carnot Engine) temperature is 300 K. If the actual
For a perfect engine, the efficiency is only half of the ideal efficiency,
Hot Res. TH
quantities Q of heat gained
how much work is done during each cycle?
QH W and lost are proportional to
TC Actual e = 0.5ei = 20%
Engine
the absolute temperatures T.
e=1-
T H - TC TH W
QC e=
e= 300 K QH
Cold Res. TC TH
e=1-
500 K W = eQ H = 0.20 (600 J)
TC
e=1-
TH e = 40% Work = 120 J

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THE SECOND LAW FOR

REFRIGERATORS
REFRIGERATORS
A refrigerator is an engine
operating in reverse: It is impossible to construct a
Hot Res. T H
Work is done on gas Hot Res. TH refrigerator that absorbs heat
Qhot W in extracting heat from cold
Qhot from a cold reservoir and
reservoir and depositing
Engine heat into hot reservoir.
deposits equal heat to a hot
Engine
reservoir with DW = 0.
Qcold W in + Qcold = Qhot Qcold
Cold Res. TC If this were possible, we could
Cold Res. TC
W IN = Qhot - Qcold establish perpetual motion!

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COP EXAMPLE
COEFFICIENT OF PERFORMANCE
500 K A Carnot refrigerator operates
The COP (K) of a heat between 500 K and 400 K. It
Hot Res. TH
engine is the ratio of the Hot Res. TH extracts 800 J from a cold
QH W HEAT Q c extracted to the reservoir during each cycle.
net WORK done W.
QH W
Engine What is C.O.P., W and Q H ?
Eng
QC QC QH ine TC 400 K
K= = K= =
W Q H- Q C 800 J T H - TC
Cold Res. TC 500 K - 400 K
Cold Res. TC
For an IDEAL TH 400 K
K= C.O.P. (K) = 4.0
refrigerator: T H - TC

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COP EXAMPLE (Cont.)

COP EXAMPLE (Cont.)
500 K Next we will find Q H by
assuming same K for actual 500 K
Hot Res. TH Now, can you say how much
refrigerator (Carnot). Hot Res. TH
QH W work is done in each cycle?
Eng QC 1000 J W
ine K=
Q H- Q C Engine Work = 1000 J - 800 J
800 J
800 J 800 J
Cold Res. TC 4.0 =
Q H - 800 J Cold Res. TC Work = 200 J
400 K
400 K
Q H = 1000 J

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Summary
The First Law of Thermodynamics: The net Summary (Cont.)
heat taken in by a system is equal to the
sum of the change in internal energy and The Molar Units are:Joules
Q
the work done by the system. Specific Heat per mole per c = n DT
capacity, C: Kelvin degree
DQ = DU + DW D = (final - initial)
The following are true for ANY process:
• Isochoric Process: DV = 0, DW = 0
DQ = DU + DW PAVA PBVB
• Isobaric Process: DP = 0 =
TA TB
• Isothermal Process: DT = 0, DU = 0
DU = nCv DT PV = nRT
• Adiabatic Process: DQ = 0

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Summary (Cont.)
Summary (Cont.)
Hot Res. TH The Second Law of Thermo: It is The efficiency of a heat engine:
impossible to construct an engine
Qhot
Wout that, operating in a cycle, QC TC
Engine produces no effect other than the e=1- Q e=1-
H TH
extraction of heat from a reservoir
Qcold and the performance of an
Cold Res. TC equivalent amount of work. The coefficient of performance of a refrigerator:

QC QC TC
Not only can you not win (1st law); K= = K=
you can’t even break even (2nd law)! Win QH - QC TH - TC

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