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ESCI 341 - Atmospheric Thermodynamics
Lesson 9 - Entropy
A Review of State Variables
• The thermodynamic state of a one-component, closed system in equilibrium can be
completely described by any two state variables (other than mass or moles).
• State variables have two important properties:
– The change in any of the state variables (say U) does not depend on the path of
the system on a thermodynamic diagram. It only depends on the endpoints.
Z b
dU = U (b) − U (a)
a
– The integral of a state variable around a closed path is zero.
I
dU = 0
• Mathematically, this means that differentials of state variables are exact differentials.
Heat is Not a State Variable
• Imagine that an air parcel starts out at Point 1 on the skew-T diagram shown in Fig. 1.
The initial pressure and temperature are p1 and T1 . The parcel moves in a reversible,
closed cycle around the path shown. The three legs of the path are:
Leg A: Adiabatic expansion from pressure p1 to p2 .
Leg B: Isothermal compression from pressure p2 back to p1 .
Leg C: Isobaric expansion back to the original state.
Figure 1: Three-legged closed cycle on a Skew-T diagram.
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The total heat transfer during this closed cycle is
Z 3
Z 1
I
Z 2
dqrev +
dqrev +
dqrev = qrev,A + qrev,B + qrev,C
dq =
1
2
3
Leg A: This leg is reversibly adiabatic, so the net heat for this leg is zero,
qrev,A = 0.
(1)
The temperature at Point 2 is found using the Poisson relation,
T2 = T1
p2
p1
Rd /cp
.
(2)
Leg B: This leg is isothermal, so the heat for this leg (see Lesson 6) is
p1
qrev,B = −Rd T2 ln
.
p2
(3)
Leg C: This leg is isobaric, so the heat for this leg (see Lesson 6) is
qrev,C = cp (T1 − T2 ).
(4)
• For the path shown in Fig. 1 we have p1 = 1000 mb, p2 = 700 mb, and T1 = 313.15K.
From (2) we calculate T2 = 282.81K. Using these values in (6) and (7) yields
qrev = qrev,A + qrev,B + qrev,C = 0 kJ/kg − 28.96 kJ/kg + 30.50 kJ/kg = 1.54 kJ/kg.
• Since the result is non-zero, we have demonstrated that
heat is not a state variable.
H
dqrev 6= 0, and therefore,
• We can also show that the work integrated around the closed path is also not zero.
This follows from integrating the First Law of Thermodynamics around a closed path,
I
I
I
du = dq + dw = 0,
which shows that
I
I
dw = −
dq.
• Since neither dw nor dq are exact differentials, some texts write their differentials with
a slash through the d, as dw
¯ or dq.
¯
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Entropy
H
H
• What if, instead of evaluation dqrev we evaluate dqTrev around the closed path?
I
Z 2
Z 3
Z 1
dqrev
dqrev
dqrev
dqrev
=
+
+
.
T
T
T
T
1
2
3
Leg A: This leg is reversibly adiabatic, so the integral along this path is still zero,
Z 2
dqrev,A
= 0.
T
1
(5)
Leg B: This leg is isothermal, so from the enthalpy form of the First Law we know that
dq = −αdp. The integral is then
Z 3
Z 3
Z 3
αdp
dp
p1
dqrev,B
=−
= −Rd
= −Rd ln
.
(6)
T
T
p
p2
2
2
2
Leg C: This leg is isobaric, so from the enthalpy form of the first law we know that
dq = cp dT , and the integral is
Z 1
Z 1
dqrev,C
cp dT
T2
=
= cp ln
.
(7)
T
T
T1
3
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• Using the same pressures and temperatures as before, the integral evaluates to
I
p1
T2
dqrev
= 0 − Rd ln
+ cp ln
= −102.4 J/kg K + 102.4 J/kg K = 0.
T
p2
T1
H
• We have demonstrated that dqTrev is an exact differential. The extensive form (multiH
plied by mass), dQTrev , is also an exact differential.
– This means that dQTrev is the differential of a state variable.
– We call this state variable entropy.
• Entropy is defined as
dQrev
T
• Note that entropy is defined in terms of reversible heat!
dS ≡
(8)
Mathematical Excursion
• A differential of the form dz = M dx + N dy is an exact differential if
∂M
∂N
=
.
∂y
∂x
(9)
• Sometimes an inexact differential can be made into an exact differential by multiplying
by a suitable function, called an integrating factor. In the case of the inexact differential
dQrev , the function T −1 is an integrating factor that turns it into the exact differential,
dS.
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Entropy and the Fundamental Equation
• From the definition of entropy, (8), we can write dQrev = T dS. We can therefore
replace dQ in the First Law of Thermodynamics with T dS, to get
dU = T dS − pdV,
(10)
dH = T dS + V dp.
(11)
or in enthalpy form,
– Note that (10) and (11) are restricted to reversible processes, since entropy is
defined in terms of reversible heat.
• Recall from Lesson 2 that the fundamental equation relates internal energy to entropy,
volume, and the amounts of the constituents,
U = f (S, V, ni ).
(12)
Taking the differential of (12) for a system in material equilibrium (ni constant) gives
∂U
∂U
dS +
dV.
(13)
dU =
∂S V
∂V S
• Comparing (10) with (13) we obtain the following identities,
T =
∂U
∂S
(14)
V
and
p=−
∂U
∂V
.
(15)
S
• Equation (14) is the thermodynamic definition of temperature, while (15) if the thermodynamic definition of pressure.
Entropy of an Ideal Gas
• The first-law in internal energy form for an ideal gas is
Cv dT = T dS − pdV,
which rearranges to
dS = Cv
4
dT
dV
+ nR
.
T
V
(16)
• Integrating (16) from a reference state (T0 , V0 ) to arbitrary state (T, V ) results in the
following expression for the entropy of an ideal gas,
S(T, V ) = S0 + Cv ln
V
T
+ nR ln ,
T0
V0
(17)
where S0 = S(T0 , V0 ).
• Performing a similar integration using the first law in enthalpy form results in
p
T
− nR ln ,
S(T, p) = S0 + Cp ln
T0
p0
(18)
where S0 = S(T0 , p0 ).
Entropy Change in Adiabatic Free Expansion
• Consider two scenarios:
– An ideal gas reversibly and adiabatically expands from volume Vi to Vf .
∗ The change in entropy is zero, dS = n0, because dQrev = 0.
– An ideal gas freely and adiabatically expands from volume Vi to Vf .
∗ No work is done in a free expansion.
∗ Since this is adiabatic, and no work is done, the temperature remains constant.
V
∗ The change in entropy is equal to ∆S = nR ln Vfi .
• In both scenarios the process is adiabatic, so why does the free expansion have a
non-zero change in entropy?
– Because entropy is defined in terms of reversible heat, dQrev , and the free expansion is not reversible!
• Figure 2 shows a p-V diagram with the initial state of the gas (i ), and the final states
for the reversible adiabatic expansion (f ) and the free expansion (f 0 ).
Figure 2: p-V diagram showing reversible adiabatic and free adiabatic expansion.
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• Since the temperature does not change during a free, adiabatic expansion, points i
and f 0 lie along an isotherm. In order to connect these two points using a reversible
process, heat would have to be added, and so the reversible heat, dQrev , is non-zero.
– This is why there is an entropy change for the free expansion, even though it is
adiabatic.
Gibbs’ Theorem
• The entropy of a mixture of ideal gases is given by Gibbs’ Theorem, which states that
the entropy of a mixture of ideal gases is the sum of the entropies that each gas would
have if it alone were to occupy the entire volume, V , at temperature T .
• Mathematically, Gibbs’ Theorem is given by the formula
S(T, V ) =
N
X
S0i +
i=1
N
X
i=1
N
X
V
T
+
ni R ln .
Cvi ln
T0 i=1
V0
(19)
where S0i is the entropy of the ith component in the reference state, (T0 , V0 ).
• Equation (19) can be written in terms of temperature and pressure by using the ideal
gas law and substituting for V and V0 ,
V =
ni RT0
ni RT
and V0 =
,
pi
pi
for each component. The result is
S(T, p) =
N
X
i=1
S0i +
N
X
i=1
N
Cpi ln
X
pi
T
−
ni R ln .
T0
p0
i=1
(20)
• In this form we state that:
– The entropy of a mixture of ideal gases is the sum of the entropies that each gas
would have if it were by itself at a pressure equal to its partial pressure in the
mixture, pi , at temperature T .
Entropy of Mixing
• Imagine N gases separated by rigid, impermeable partitions, as shown in Fig. 3. Each
gas is at the same temperature and pressure.
Figure 3: Gases separated by partitions.
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• The total entropy of the gases in this configuration is
S(T, p) =
N
X
S0i +
i=1
N
X
i=1
N
X
T
p
Cpi ln
−
ni R ln .
T0
p0
i=1
(21)
– NOTE! Equation (21) is not the same as (20). The pressure in (21) is the total
pressure, not partial pressure.
• If the partitions are removed, the gases will mix.
– The total pressure and temperature will not change.
– After mixing the entropy will be given by (20).
• The change in entropy due to mixing is found by subtracting (21) from (20), which
yields
N
N
X
p
pi X
ni R ln .
(22)
∆Smixing = −
ni R ln +
p
p
0
0
i=1
i=1
• Since partial pressure can be written as
pi =
ni
p,
n
(22) becomes
∆Smixing = −
N
X
ni R ln
i=1
ni p
n p0
+
N
X
i=1
ni R ln
p
,
p0
(23)
which expands and simplifies to
∆Smixing = −R
N
X
ni ln
i=1
ni
.
n
(24)
• Equation (24) can be written in terms of mole fraction, defined as χi = ni /n. In this
form, the entropy of mixing is
∆Smixing = −nR
N
X
χi ln χi .
(25)
i=1
• Note that even though there is a negative sign in (25), ∆Smixing will be a positive
value, because teh mole fractions are less than one, and ln of a number less than one
will be negative.
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Exercises
1. Start with the first law of thermodynamics in enthalpy form and show that the specific
entropy of an ideal gas is
s(T, p) = s0 + cp ln
p
T
− R0 ln .
T0
p0
2. What are the units for entropy? For specific entropy?
3. Two moles of helium (M = 4 g/mol) are at 15 degrees C and 1000 mb.
(a) If 6240 Joules of heat are added while keeping volume constant, what is the final
temperature and pressure?
(b) The helium is now allowed to expand isothermally to twice its initial volume.
What is the new pressure?
(c) What is the total change in entropy and specific entropy for the helium?
4. 100 grams of helium is mixed with 100 grams of molecular nitrogen (N2 , M = 28 g/mol)
are at 20 degrees C and 1000 mb.
(a) What is the entropy of mixing?
(b) The mixing process is adiabatic, yet the entropy change is not zero. Why?
5. The First Law of Thermodynamics for an ideal gas is
dqrev = cp dT − αdp.
(26)
(a) Show that (26) can also be written as
dqrev = cp dT −
R0 T
dp.
p
(27)
(b) Equation (27) has the form of (9) where x → T , y → p, M → cp , and N → −R0 T /p.
Show that (27) is not an exact differential by showing that the condition of (9) is
not met.
(c) Show that dqrev /T is an exact differential by multiplying (27) by 1/T and then
showing that condition (9) is met.
6. Derive the temperature-pressure form of Gibbs’ Theorem, (20), from the temperaturevolume form, (19).
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