110. (20.6) Review of some thermal processes.
a)
adiabatic process - no heat is transferred
dU = - dW
b)
isochoric process - constant volume process
dU = dQ
c)
cyclical process - the system returns to the initial state
∆Q = ∆W
d)
isothermal process - constant temperature
dU = dQ - dW
(for an ideal gas dU = 0)
e)
isobaric process - constant pressure
dU = dQ - dW
f)
free expansion - adiabatic with no work done
dU = 0
121
111. (22.7, 22.9) Entropy.
a) (Theorem) For any cyclical quasistatic process
dQ r
=0
∫
process T
We can introduce a function S,
called entropy, that is a function of
state of the system.
P
a
b
T
b) Definition
The change in entropy, dS, between two equilibrium
states is given by the heat transferred, dQ r , divided by the
absolute temperature, T, of the system in a quasi-static
process leading from the initial to the final state. That is
dQ r
dS ≡
T
(The change in entropy from the initial state to the final
state does not depend on the process.)
[c) Microscopic (statistical) definition of entropy.
If for a system of N identical particles, the number of
possible configurations for a considered state is W
(statistical sum), the entropy S of the system in this state is
S ≡ kB ln W
where kB is a physical constant (Boltzmann’s constant).
(The greater the number of configuration for a state, the
higher the entropy of the system in the state.)]
122
V
Example. Determine the change in the total
entropy of two systems during heat transfer
between two systems with temperature
independent specific heat.
m1 = m2 = 1kg, T1 = 100°C, T2 = 0°C,
c1 = c2 = 4186 J/kgK (water)
T1
m1 , c1
dQ
T2
m2 , c2
From the definition of entropy, each system changes
its entropy during the transfer of heat
dQ x Tf c x m x dTx
T
∆Sx = ∫ dS x = ∫
= ∫
= cx m x ln f
T
Tx
process
process T
Tx
We can find the equilibrium temperature Tf assuming
that the systems exchange energy only between themselves.
c1m1 (Tf − T1 ) + c2 m 2 (Tf − T2 ) = 0
Therefore the final temperature is
Tf = c1m1T1 + c2m2T2 = 50o C
c1m1 + c2m 2
The change in entropy of the hot water is
T
J
(50 + 273)K
J
∆S1 = c1m 1 ln f = 4186
⋅ 1kg ⋅ ln
= −602.5
T1
kg ⋅ K
(100 + 273)K
K
The change in entropy of the cold water is
T
J
(50 + 273)K
J
∆S2 = c2m 2 ln f = 4186
⋅1kg ⋅ ln
= 704
T2
kg ⋅ K
(0 + 273)K
K
The total change in the entropy of the system is
J
J
J
∆S = ∆S1 + ∆S2 = −602.5 + 704 = 1015
.
K
K
K
123
112. (22.8) Second law of thermodynamics.
In any thermodynamic process that proceeds from one
equilibrium state to another, the total entropy of the system
and its environment (the Universe) cannot decrease.
113. Equivalent versions (theorems) and consequences of
the second law of thermodynamics.
a) (22.1)
It is impossible to
construct a heat engine which when
operating in a cycle, produces no
other effect than the absorption of
thermal energy from a reservoir and
the performance of an equal amount
of work
Th
∆Qh
∆W
engine
∆Qc
Tc
proof.
Suppose there was a cyclical engine capable of
converting all absorbed heat into work. In a full cycle, the
change in the entropy of the engine is zero. Therefore, only
the hot reservoir changes its entropy. The change in the
entropy of the universe would be
∆ S = ∆ Sh =
− ∆ Qh
<0
Th
Comment. (22.3, 22.5) The presence of the cold reservoir
(heat sink) allows engines to operate cyclically.
124
The condition for an increase of entropy relates the heat
absorbed and given up by the engine, and the temperatures
of the reservoirs.
0 ≤ ∆S = ∆ Sh + ∆ Sc =
− ∆ Qh ∆ Qc
+
Th
Tc
In a cycle, the internal energy of the engine does not
change, therefore the work done by the engine is related to
the heat absorbed and given up by the engine
∆ W = ∆Q h − ∆ Q c
The efficiency of a heat engine is defined as the ratio
of the work done by the engine to the heat absorbed from
the energy source. The temperatures of the source and the
sink of energy limit the efficiency. No real cyclical heat
engine can be more efficient that a Carnot engine.
∆ Qh − ∆ Qc
∆ Qc
e = ∆W =
=1−
≤ 1 − Tc
∆Qh
∆ Qh
∆ Qh
Th
125
b) (22.intro) It is impossible to transfer heat from one
body to another body at a higher temperature with no other
consequences in the universe.
proof.
T1
T2
Suppose heat dQ is transferred from a system at
temperature T1 to a system at temperature T2 and no other
changes in the universe are taking place. The change in the
total entropy is due to the change in the entropy of both
systems. Therefore
− dQ dQ
0 ≤ dS = dS1 + dS2 =
+
T1
T2
Hence
T1 ≥ T2
(This theorem indicates the direction of heat transfer.)
126
114. (22.8, 22.intro, 22.2) Reversible and irreversible
processes.
a) If, during a thermodynamic process, the entropy of the
Universe is not changed, the process is reversible. (It is
possible to return to the initial state of the entire Universe.)
b) If during a thermodynamic process the entropy of the
Universe is changed (increased), the process is irreversible.
(It is not possible to return to the initial state of the entire
Universe.)
Example. Free expansion of an ideal monoatomic gas.
The temperature of the gas is directly proportional to
the internal energy of the ideal gas (21.3). During free
expansion the gas does not perform work and does not
absorb any heat. Therefore the initial and the final
temperatures of the gas are equal. In order to find the
change in the entropy of the gas, we can consider an
isothermal quasi-static process. (Free expansion is not a
quasi-static process.) Nothing else changes its entropy,
therefore the change in the entropy of the universe is
Vf
∆S Universe = ∆S gas =
dQ r ∆Q r ∆Wr
=
=
=
T
T
process T
∫
127
∫
Vi
nRT
dV
V
T
= nR ln
Vf
>0
Vi
115. (21.1, 21.2) Internal energy of an ideal gas.
The internal energy of an ideal gas results from the
translational kinetic energy, rotational kinetic energy, and
vibrational energy of the molecules constituting the system.
According to the kinetic theory of gases, the internal energy
of an ideal gas (and only of an ideal gas) is a function of
temperature only
U = nCVT
proof.
In an isochoric process the heat delivered to the gas is totally converted into
internal energy
dU = dQ = nC VdT
With the reference for the internal energy at 0K
T
U = 0 + ∫ nC V dT = nC V T
0
Molar heat capacity of an ideal gas depends on the
type of molecules forming the gas. Molar heat capacity of
all monatomic gases have the same value
3
CV = R = 12.5 molJ ⋅K .
2
Molar heat capacity of polyatomic gases depends on
the internal interaction between the atoms forming the
molecules.
128
116. (21.2) Molar heat capacity of an ideal gas at constant
pressure.
Molar heat capacity of an ideal gas in an isobaric
process is related to the molar heal capacity in an isochoric
process by
CP = CV + R
proof.
From the equation of state from an ideal gas
PV = nRT
in an isobaric process the work performed by the gas is
related to the temperature change by
PdV = nRdT
The change in the internal energy of an ideal gas (in an
isobaric process) can be related to the change in
temperature directly,
dU = nCVdT
or by using the first law of thermodynamics
dU = dQ - dW = nCPdT - nRdT
Therefore,
CP = CV + R
129
117. (21.3) Adiabatic (quasistatic) process for an ideal gas
In an adiabatic quasistatic process the gas does not
exchange heat with its surroundings. In this process
PVγ = const. (or TVγ −1 = const.)
where γ = CP/CV
proof.
In an adiabatic process the change in the internal energy (and
temperature) of the gas result only from the work performed by the
gas
-PdV = -dW = dU = nCVdT
From the equation of state for an ideal gas
PdV + VdP = nRdT
Eliminating temperature differential from the above two equations,
we obtain a differential equation relating volume and pressure.
R
PdV + VdP = −
PdV
CV
Separating the variables, and substituting R = CP - CV, we obtain
dV
dP
γ
=−
V
P
Integrating both sides in appropriate limits
V
P
γ ln f = − ln f
Vi
Pi
from which
PiViγ = Pf Vfγ
From the equation of state for an ideal gas we can get the relationship
between volume and pressure.
130
118. (22.3) The Carnot cycle.
A theoretical Carnot engine operating in an ideal
reversible cycle is the most efficient engine possible.
W2
W3
W1
Qh
W4
Qc
Th
isothermal
expantion
Tc
adiabatic
expantion
isothermal
compression
adiabatic
compression
P
The working substance is
an ideal gas contained in a
Qh
cylinder with a movable
B
piston at one end. The
Th
W
Carnot cycle consists of
two adiabatic and two
D
isothermal processes (all
Tc
Qc
C
reversible).
T
Since in one cycle
the internal energy of the gas is not changed, the work W
done by the gas equals the net heat delivered to the gas in
the cycle
W = Qh - Qc
A
131
119. The four-stroke combustion engine (Otto cycle)
The Otto cycle represents
operation of a common P
gasoline engine. The cycle
includes two isochoric and
two adiabatic processes. If the
air-fuel mixture is assumed to
be an ideal gas, the efficiency
of the combustion engine is
determined by the change in
temperature in any of the two
adiabatic processes.
C
Qh
B
O
V1
W
D
Qc
A
V2
T
From the first law of thermodynamics,
W = Qh + Qc (Qc < 0)
For an ideal gas,
Qh = nCV(TC - TB)
Qc = nCV(TA - TD)
Therefore,
W
Q
T − TA
T
T
T
e=
= 1+ c = 1− D
= 1 − A = 1 − D > 1 − A = eC
Qh
Qh
TC − TB
TB
TC
TC
The efficiency of the Otto cycle is related to the compression
ratio V2/V1
1
e = 1−
( V1 / V2 ) γ −1
132
120. (22.6) Refrigerators
A refrigerator is a device that
Th
moves heat from a system at a lower
temperature to the system with higher
temperature. According to the second ∆Qh
∆W
law of thermodynamics it is necessary
pump
that work is done on the refrigerator
∆Qc
substance.
Tc
The effectiveness of a refrigerator
is described in terms of the coefficient
of performance (COP), which is defined by the ratio of the
heat removed from the cold system to the work performed
on the refrigerator substance.
∆Q c
COP =
∆W
P
A
The highest possible
coefficient of performance
is that of a refrigerator
whose working substance
is carried through a reverse
Carnot cycle
Qh
B
Th
W
D
Qc
C
Tc
COPC =
T
133
Tc
Th − Tc