Lecture15.pdfbbñvfkhdjfuhdjgdhffjfdygdhjfg

Chapter 20
Entropy & the 2nd Law
of Thermodynamics

Section 20.1
• Topic I :
➢Entropy
• Topic II:
➢Second law of thermodynamics

One – Way Processes
• What is a one-way process?
• It is a process that proceeds in one direction and
can not be reversed.
• Examples: Hands and coffee – Clay dropping.
• The law of conservation of energy is obeyed.
• Thus, changes in energy do not set the direction of
processes.
• What, then, sets the direction of processes?
• Let’s first make some definitions.

Reversible & Irreversible Processes
• Irreversible Process
➢Can not be reversed
➢Occurs in one direction
➢Occurs very quickly
➢Proceeds through non-equilibrium values
➢Can not be represented on a p – V diagram
• Reversible Process
➢Can be reversed
➢Occurs very slow
➢Proceeds through equilibrium values
➢Can be represented on a p – V diagram
• Closed system
A system that is isolated from its environment
(e.g. hand + coffee).

Direction of Processes
• The direction of processes is set by the change in
entropy { DS }.
• Entropy: degree of disorder (solid < liquid < gas).
• Entropy Postulate:
If an IRREVERSIBLE process occurs in a
CLOSED SYSTEM, the entropy { S } of the system
always increases.
Therefore, DS is called the arrow of time.

State Function
• A property of the system that depends on the state of
the system and not on how the system reached that
state.
• State function = intrinsic property.
• p , V , T , Eint are state functions.
• Q and W are not state functions.
• Entropy ( S ) is a state function.

MATHEMATICAL
DEFINITION OF
ENTROPY CHANGE

• To be able to evaluate the integral, you
must have a reversible process (path of
integration).
• Then, how do we deal with irreversible
processes?
• Let’s take an example.

Irreversible Process – Free Expansion
▪ This is an irreversible process.
▪Can not be represented on a p – V
diagram.
▪Thus, the integral cannot be
evaluated (path).
▪Does S increase? (Note that Q = 0)
▪How do we calculate DS?

Solution
• Replace it with a reversible process that
has the same initial and final states.
• Thus, DS is the same (because it is a state
function).
• Since T is constant, the easiest replace-
ment is an isothermal process.

1.
Set
p,
V
and
T
to
match
i.
2.
Remove
shot
slowly
until
you
reach
f.

)
ln(
)
ln(
1 int
i
f
i
f
V
V
nR
T
V
V
nRT
T
W
T
E
W
T
Q
dQ
T
T
dQ
S =
=
=
D
+
=
=
=
=
D  
▪Thus, DS for free expansion is DS = n R ln (Vf / Vi ).
▪It is positive, as expected.
To find the entropy change for an irreversible process
occurring in a closed system, replace the process with
any reversible process that connects the same initial and
final states and calculate DS for this reversible process.
0

Topic II
The second Law of
Thermodynamics

• Suppose you go from b to a (add shot slowly).
• Heat must be removed.
• Thus, Q is negative.
• So, DS is negative!
• Does this violate the
entropy postulate?
• NO! because:
i. Process is not irreversible.
ii.System is not closed.
• What happens if we consider the system to be:
system = gas + reservoir ?
• Then, DSgas = – Q/T and DSres = + Q/T.
• Thus, DSsystem = 0.
Entropy Postulate:
If an IRREVERSIBLE process
occurs in a CLOSED SYSTEM,
the entropy { S } of the system
always increases.

Second Law of Thermodynamics
• If a process occurs in a closed system, the entropy of
the system increases for irreversible processes and
remains constant for reversible processes.
It never decreases.
▪ DS  0
▪ DS > 0 – irreversible process
▪ DS = 0 – reversible process
▪ For a system of many objects: some of them may
have DS < 0 , others may have DS > 0 ; but the
overall DSsystem  0.

Example Eq. (20-4)
• Consider an ideal gas undergoing an arbitrary
reversible change from state i (pi , Vi , Ti ) to a
final state f (pf , Vf , Tf ). Find DS.

Checkpoint 1
Water is heated on a stove.
Rank the entropy changes of the water as
its temperature rises
(a) from 20 oC to 30 oC,
(b) from 30 oC to 35 oC,
(c) from 80 oC to 85 oC.

Checkpoint 2
An ideal gas has temperature T1 at the initial state i.
The gas has a higher temperature T2 at final states a
and b, which it can reach along the paths shown.
Which path has the larger change in entropy?

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