Moran M.J., Shapiro H.N. Fundamentals of Engineering Thermodynamics

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The entropy of an ideal gas depends on two properties, not on temperature alone as

for internal energy and enthalpy. Accordingly, for the mixture

S 5 S

1 S

. . .

1 S

i51

(12.25)

where S

is the entropy of component i evaluated at the mixture temperature T and

partial pressure p

(or at temperature T and total volume V).

Equation 12.25 can be written on a molar basis as

5 n

1 n

. . .

1 n

i51

(12.26)

where s is the entropy of the mixture per mole of mixture and s

is the entropy of

component i per mole of i. Dividing by the total number of moles of mixture, n, gives

an expression for the entropy of the mixture per mole of mixture

s 5

i51

(12.27)

In subsequent applications, the specific entropies s

of Eqs. 12.26 and 12.27 are evalu-

ated at the mixture temperature T and the partial pressure p

12.3.4 Working on a Mass Basis

In cases where it is convenient to work on a mass basis, the foregoing expressions are

written with the mass of the mixture, m, and the mass of component i in the mixture,

, replacing, respectively, the number of moles of mixture, n, and the number of moles

of component i, n

. Similarly, the mass fraction of component i, mf

, replaces the mole

fraction, y

. All specific internal energies, enthalpies, and entropies are evaluated on a

unit mass basis rather than on a per mole basis as above. To illustrate, Table 12.2

Property Relations on a Mass Basis for Binary Ideal Gas Mixtures

Notation: m

5 mass of gas 1, M

5 molecular weight of gas 1

5 mass of gas 2, M

5 molecular weight of gas 2

m 5 mixture mass 5 m

1 m

, mf

5 (m

/m), mf

5 (m

/m)

T 5 mixture temperature, p 5 mixture pressure, V 5 mixture volume

Equation of state:

p 5 m(

R/M)T/V (a)

where M 5 (y

1 y

) and the mole fractions y

and y

are given by

5 n

/(n

1 n

), y

5 n

/(n

1 n

) (b)

where n

5 m

and n

5 m

Partial pressures: p

5 y

p, p

5 y

p (c)

Properties on a mass basis:

Mixture enthalpy: H 5 m

(T ) 1 m

(T ) (d)

Mixture internal energy: U 5 m

(T ) 1 m

(T ) (e)

Mixture specific heats: c

5 (m

/m)c

(T ) 1 (m

/m)c

(T )

5 (mf

(T ) 1 (mf

(T ) (f)

5 (m

/m)c

(T ) 1 (m

/m)c

(T )

5 (mf

(T ) 1 (mf

(T ) (g)

Mixture entropy: S 5 m

(T, p

) 1 m

(T, p

) (h)

TABLE 12.2

12.3 Evaluating U, H, S, and Specific Heats 713

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714 Chapter 12

Ideal Gas Mixture and Psychrometric Applications

provides property relations on a mass basis for binary mixtures. These relations are

applicable, in particular, to moist air, introduced in Sec. 12.5.

By using the molecular weight of the mixture or of component i, as appropriate,

data can be converted from a mass basis to a molar basis, or conversely, with relations

of the form

u 5 Mu,





h 5 Mh,



5 Mc



5 Mc



s 5 Ms (12.28)

for the mixture, and

5 M





h

5 M





c

p,i

5 M

p,i





c

y,i

5 M

y,i





s

5 M

(12.29)

for component i.

12.4 Analyzing Systems Involving Mixtures

To perform thermodynamic analyses of systems involving nonreacting ideal gas mixtures

requires no new fundamental principles. The conservation of mass and energy principles

and the second law of thermodynamics are applicable in the forms previously intro-

duced. The only new aspect is the proper evaluation of the required property data for

the mixtures involved. This is illustrated in the present section, which deals with two

classes of problems involving mixtures: In Sec. 12.4.1 the mixture is already formed, and

we study processes in which there is no change in composition. Section 12.4.2 considers

the formation of mixtures from individual components that are initially separate.

12.4.1 Mixture Processes at Constant Composition

In the present section we are concerned with the case of ideal gas

mixtures undergoing processes during which the composition remains

constant. The number of moles of each component present, and thus

the total number of moles of mixture, remain the same throughout the

process. This case is shown schematically in Fig. 12.2, which is labeled

with expressions for U, H, and S of a mixture at the initial and final

states of a process undergone by the mixture. In accordance with the

discussion of Sec. 12.3, the specific internal energies and enthalpies of

the components are evaluated at the temperature of the mixture. The

specific entropy of each component is evaluated at the mixture tem-

perature and the partial pressure of the component in the mixture.

The changes in the internal energy and enthalpy of the mixture

during the process are given, respectively, by

2 U

i51

22 u

(12.30)

2 H

i51

22 h

(12.31)

where T

and T

denote the temperature at the initial and final states. Dividing by

the number of moles of mixture, n, expressions for the change in internal energy and

enthalpy of the mixture per mole of mixture result

¢u 5

i51

22 u

(12.32)

¢h 5

i51

22 h

(12.33)

, n

, …, n

)

, p

State 1 State 2

, n

, …, n

)

, p

Fig. 12.2 Process of an ideal gas mixture.

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Similarly, the change in entropy for the mixture is

2 S

i51

, p

22 s

, p

(12.34)

where p

and p

denote, respectively, the initial and final partial pressures of com-

ponent i. Dividing by the total moles of mixture, Eq. 12.34 becomes

¢s 5

i51

, p

22 s

, p

(12.35)

Companion expressions for Eqs. 12.30 through 12.35 on a mass basis also can be

written. This is left as an exercise.

The foregoing expressions giving the changes in internal energy, enthalpy, and

entropy of the mixture are written in terms of the respective property changes of the

components. Accordingly, different datums might be used to assign specific enthalpy

values to the various components because the datums would cancel when the com-

ponent enthalpy changes are calculated. Similar remarks apply to the cases of inter-

nal energy and entropy.

Using Ideal Gas Tables

For several common gases modeled as ideal gases, the quantities u

and h

appearing

in the foregoing expressions can be evaluated as functions of temperature only from

Tables A-22 and A-23. Table A-22 for air gives these quantities on a mass basis. Table

A-23 gives them on a molar basis.

The ideal gas tables also can be used to evaluate the entropy change. The change

in specific entropy of component i required by Eqs. 12.34 and 12.35 can be deter-

mined with Eq. 6.20b as

¢s

5 s8

22 s8

22 R ln

Since the mixture composition remains constant, the ratio of the partial pressures in

this expression is the same as the ratio of the mixture pressures, as can be shown by

using Eq. 12.12 to write

Accordingly, when the composition is constant, the change in the specific entropy of

component i is simply

5 s8

22 s8

22 R ln

(12.36)

where p

and p

denote, respectively, the initial and final mixture pressures. The terms

of Eq. 12.36 can be obtained as functions of temperature for several common gases

from Table A-23. Table A-22 for air gives s8 versus temperature.

Assuming Constant Specific Heats

When the component specific heats c

y,i

and c

p,i

are taken as constants, the specific

internal energy, enthalpy, and entropy changes of the mixture and the components of

the mixture are given by

¢u 5 c

2 T

2, ¢u

5 c

y,i

2 T

2 (12.37)

¢h 5 c

2 T

2, ¢h

5 c

p,i

2 T

2 (12.38)

s 5 c

2 R ln

¢s

5 c

p,i

2 R ln

(12.39)

TAKE NOTE...

When mixture composition

remains constant, a ratio

of partial pressures, p

equals the ratio of mixture

pressures, p

12.4 Analyzing Systems Involving Mixtures 715

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716 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

where the mixture specific heats c

and c

are evaluated from Eqs. 12.23 and 12.24,

respectively, using data from Tables A-20 or the literature, as required.

The expression for ¢u can be obtained formally by substituting the above

expression for ¢u

into Eq. 12.32 and using Eq. 12.23 to simplify the result. Simi-

larly, the expressions for ¢

and ¢s can be obtained by inserting ¢h

and ¢s

into

Eqs. 12.33 and 12.35, respectively, and using Eq. 12.24 to simplify. In the equations

for entropy change, the ratio of mixture pressures replaces the ratio of partial

pressures as discussed above. Similar expressions can be written for the mixture

specific internal energy, enthalpy, and entropy changes on a mass basis. This is left

as an exercise.

Using Computer Software

The changes in internal energy, enthalpy, and entropy required in Eqs. 12.32, 12.33,

and 12.35, respectively, also can be evaluated using computer software. Interactive

Thermodynamics: IT provides data for a large number of gases modeled as ideal

gases, and its use is illustrated in Example 12.4 below.

The next example illustrates the use of ideal gas mixture relations for analyzing a

compression process.

Analyzing an Ideal Gas Mixture Undergoing Compression

c c c c EXAMPLE 12.3 c

A mixture of 0.3 lb of carbon dioxide and 0.2 lb of nitrogen is compressed from p

5 1 atm, T

5 540°R to p

3 atm in a polytropic process for which n 5 1.25. Determine (a) the final temperature, in 8R, (b) the work, in Btu,

SOLUTION

Known:

A mixture of 0.3 lb of CO

and 0.2 lb of N

is compressed in a polytropic process for which n 5 1.25.

At the initial state, p

5 1 atm, T

5 540°R. At the final state, p

5 3 atm.

Find: Determine the final temperature, in °R, the work, in Btu, the heat transfer, in Btu, and the entropy change

of the mixture in, Btu/°R.

Schematic and Given Data:

Fig. E12.3

States of the

mixture

n = 1.25

0.3 lb CO

0.2 lb N

Boundary

= 1 atm, T

= 540°R,

= 3 atm

Engineering Model:

As shown in the accompanying figure,

the system is the mixture of CO

and N

The mixture composition remains

constant during the compression.

2. The Dalton model applies: Each mix-

ture component behaves as if it were

an ideal gas occupying the entire sys-

tem volume at the mixture temperature.

The overall mixture acts as an ideal gas.

3. The compression process is a poly-

tropic process for which n 5 1.25.

4. The changes in kinetic and potential

energy between the initial and final

states can be ignored.

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Analysis:

(a)

For an ideal gas, the temperatures and pressures at the end states of a polytropic process are related by

Eq. 3.56

5 T

1n212

Inserting values

5 540a

0.2

5 6738R

(b) The work for the compression process is given by

W 5

p dV

Introducing pV

5 constant and performing the integration

W 5

2 p

1 2 n

With the ideal gas equation of state, this reduces to

W 5

m1R

M21T

2 T

1 2 n

The mass of the mixture is m 5 0.3 1 0.2 5 0.5 lb. The apparent molecular weight of the mixture can be

calculated using M 5 m/n, where n is the total number of moles of mixture. With Eq. 12.1, the numbers of moles

of CO

and N

are, respectively

0.3

5 0.0068 lbmol,



0.2

5 0.0071 lbmol

The total number of moles of mixture is then n 5 0.0139 lbmol. The apparent molecular weight of the mixture

is M 5 0.5/0.0139 5 35.97.

Calculating the work

W 5

10.5 lb2a

1545 ft ? lbf

35.97 lb ? 8R

b16738R 2 5408R2

1 2 1.25

1 Btu

778 ft ? lbf

5214

.69

where the minus sign indicates that work is done on the mixture, as expected.

Q 5 ¢U 1 W

where DU is the change in internal energy of the mixture.

The change in internal energy of the mixture equals the sum of the internal energy changes of the components.

With Eq. 12.30

➊ ¢U 5 n

22 u

241 n

22 u

This form is convenient because Table A-23E gives internal energy values for N

and CO

, respectively, on a

molar basis. With values from this table

¢U 5 10.0068213954 2 298421 10.0071213340 2 26782

5 11.3 Btu

Inserting values for DU and W into the expression for Q

Q 5111.3 2 14.69 523.39 Btu

where the minus sign signifies a heat transfer from the system.

12.4 Analyzing Systems Involving Mixtures 717

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718 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

The next example illustrates the application of ideal gas mixture principles for the

analysis of a mixture expanding isentropically through a nozzle. The solution features

the use of table data and IT as an alternative.

(d) The change in entropy of the mixture equals the sum of the entropy changes of the components. With Eq. 12.34

¢S 5 n

¢s

1 n

¢s

where ¢s

and ¢s

are evaluated using Eq. 12.36 and values of s8 for N

and CO

from Table A-23E.

That is

¢S 5 0.0068a53.123 2 51.082 2 1.986 ln

1 0.0071a47.313 2 45.781 2 1.986 ln

520.0056 Btu

Entropy decreases in the process because entropy is transferred from the system

accompanying heat transfer.

➊ In view of the relatively small temperature change, the changes in the inter-

nal energy and entropy of the mixture can be evaluated alternatively using

the constant specific heat relations, Eqs. 12.37 and 12.39, respectively. In

these equations, c

and c

are specific heats for the mixture determined using

Eqs. 12.23 and 12.24 together with appropriate specific heat values for the

components chosen from Table A-20E.

➋ Since the composition remains constant, the ratio of partial pressures equals

the ratio of mixture pressures, so Eq. 12.36 can be used to evaluate the

component specific entropy changes required here.

Recalling that polytropic processes are internally reversible,

determine for the system the amount of entropy transfer accompanying

heat transfer, in Btu/°R. Ans. 20.0056 Btu/°R.

➋

Considering an Ideal Gas Mixture Expanding Isentropically Through a Nozzle

c c c c EXAMPLE 12.4 c

A gas mixture consisting of CO

and O

with mole fractions 0.8 and 0.2, respectively, expands isentropically and

at steady state through a nozzle from 700 K, 5 atm, 3 m/s to an exit pressure of 1 atm. Determine (a) the tem-

perature at the nozzle exit, in K, (b) the entropy changes of the CO

and O

from inlet to exit, in kJ/kmol ? K,

SOLUTION

Known:

A gas mixture consisting of CO

and O

in specified proportions expands isentropically through a nozzle

from specified inlet conditions to a given exit pressure.

Find: Determine the temperature at the nozzle exit, in K, the entropy changes of the CO

and O

from inlet to

exit, in kJ/kmol ? K, and the exit velocity, in m/s.

Ability to…

❑

analyze a polytropic pro-

cess of a closed system for

a mixture of ideal gases.

❑

apply ideal gas mixture

principles.

❑

determine changes in inter-

nal energy and entropy for

ideal gas mixtures using

tabular data.

✓

Skills Developed

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Fig. E12.4

= 700 K

= 3 m/s

= 5 atm

= 700 K

1 atm

= 5 atm

= ?

= 1 atm

States of the mixture

Schematic and Given Data:

Engineering Model:

The control volume shown by the dashed line on the accompanying figure operates at steady state.

2. The mixture composition remains constant as the mixture expands isentropically through the nozzle.

3. The Dalton model applies: The overall mixture and each mixture component act as ideal gases. The state

of each component is defined by the temperature and the partial pressure of the component.

4. The change in potential energy between inlet and exit can be ignored.

Analysis:

(a)

The temperature at the exit can be determined using the fact that the expansion occurs isentropically: s

2 s

5 0.

As there is no change in the specific entropy of the mixture between inlet and exit, Eq. 12.35 can be used to write

2 s

5 y

¢s

1 y

¢s

5 0 (a)

Since composition remains constant, the ratio of partial pressures equals the ratio of mixture pressures. Accord-

ingly, the change in specific entropy of each component can be determined using Eq. 12.36. Equation (a) then

becomes

cs 8

22 s8

22 R ln

d1 y

cs8

22 s8

22 R ln

d5 0

On rearrangement

21 y

25 y

21 y

21 1y

1 y

2R ln

The sum of mole fractions equals unity, so the coefficient of the last term on the right side is

1 y

5 1.

Introducing given data, and values of s8 for O

and CO

at T

5 700 K from Table A-23

0.2s8

21 0.8s8

25 0.21231.35821 0.81250.66321 8.314 ln

0.2s8

21 0.8s8

25 233.42 kJ

kmol ? K

To determine the temperature T

requires an iterative approach with the above equation: A final temperature

is assumed, and the s8 values for O

and CO

are found from Table A-23. If these two values do not satisfy

the equation, another temperature is assumed. The procedure continues until the desired agreement is attained.

In the present case

at T 5 510 K:



0.21221.20621 0.81235.70025 232.80

at T 5 520 K:



0.21221.81221 0.81236.57525 233.62

Linear interpolation between these values gives T

5 517.6 K.

12.4 Analyzing Systems Involving Mixtures 719

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720 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Alternative Solution:

Alternatively, the following IT program can be used to evaluate T

without resorting to iteration with table data.

In the program, yO2 denotes the mole fraction of O

, p1_O2 denotes the partial pressure of O

at state 1, s1_O2

denotes the entropy per mole of O

at state 1, and so on.

T1 5 700 // K

p1 5 5 // bar

p2 5 1 // bar

yO2 5 0.2

yCO2 5 0.8

p1_O2 5 yO2 * p1

p1_CO2 5 yCO2 * p1

p2_O2 5 yO2 * p2

p2_CO2 5 yCO2 * p2

s1_O2 5 s_TP (“O2”,T1,p1_O2)

s1_CO2 5 s_TP (“CO2”,T1,p1_CO2)

s2_O2 5 s_TP (“O2”,T2,p2_O2)

s2_CO2 5 s_TP (“CO2”,T2,p2_CO2)

// When expressed in terms of these quantities, Eq. (a) takes the form

yO2 * (s2_O2 2 s1_O2) 1 yCO2 * (s2_CO2 2 s1_CO2) 5 0

Using the Solve button, the result is T

5 517.6 K, which agrees with the value obtained using table data. Note that

IT provides the value of specific entropy for each component directly and does not return s8 of the ideal gas tables.

➊ (b) The change in the specific entropy for each of the components can be determined using Eq. 12.36. For O

¢s

5 s8

22 s8

22 R ln

Inserting s8 values for O

from Table A-23 at T

5 700 K and T

5 517.6 K

¢s

5 221.667 2 231.358 2 8.314 ln10.225 3.69 kJ

kmol ? K

Similarly, with CO

data from Table A-23

¢s

5 s8

22 s8

22 R ln

5 236.365 2 250.663 2 8.314 ln10.22

➋ 520.92 kJ

kmol ? K

0 5 h

2 h

2 V

where h

and h

are the enthalpy of the mixture, per unit mass of mixture, at the inlet and exit, respectively.

Solving for V

5 2V

1 21h

2 h

The term (h

2 h

) in the expression for V

can be evaluated as

2 h

1 y

2 h

where M is the apparent molecular weight of mixture, and the molar specific enthalpies of O

and CO

are from

Table A-23. With Eq. 12.9, the apparent molecular weight of the mixture is

M 5 0.814421 0.213225 41.6 kg

kmol

Then, with enthalpy values at T

5 700 K and T

5 517.6 K from Table A-23

2 h

41.6

30.2121,184 2 15,32021 0.8127,125 2 18,46824

5 194.7 kJ

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Finally,

➌ V

1 2a194.7

1 kg ? m

1 N

N ? m

1 kJ

`5 624 m

➊ Parts (b) and (c) can be solved alternatively using IT. These parts also can

be solved using a constant c

together with Eqs. 12.38 and 12.39. Inspection

of Table A-20 shows that the specific heats of CO

and O

increase only

slightly with temperature over the interval from 518 to 700 K, and so suit-

able constant values of c

for the components and the overall mixture can

be readily determined. These alternative solutions are left as exercises.

➋ Each component experiences an entropy change as it passes from inlet to

exit. The increase in entropy of the oxygen and the decrease in entropy of

the carbon dioxide are due to entropy transfer accompanying heat transfer

from the CO

to the O

as they expand through the nozzle. However, as

indicated by Eq. (a), there is no change in the entropy of the mixture as it

expands through the nozzle.

➌ Note the use of unit conversion factors in the calculation of V

Ability to…

❑

analyze the isentropic expan-

sion of an ideal gas mixture

flowing through a nozzle.

❑

apply ideal gas mixture

principles together with

mass and energy balances

to calculate the exit veloc-

ity of a nozzle.

❑

determine the exit tem-

perature for a given inlet

state and a given exit pres-

sure using tabular data and

alternatively using IT.

✓Skills Developed

12.4.2

Mixing of Ideal Gases

Thus far, we have considered only mixtures that have already been formed. Now let

us take up cases where ideal gas mixtures are formed by mixing gases that are initially

separate. Such mixing is irreversible because the mixture forms spontaneously, and a

work input from the surroundings would be required to separate the gases and return

them to their respective initial states. In this section, the irreversibility of mixing is

demonstrated through calculations of the entropy production.

Three factors contribute to the production of entropy in mixing processes:

The gases are initially at different temperatures.

The gases are initially at different pressures.

The gases are distinguishable from one another.

Entropy is produced when any of these factors is present during a mixing process.

This is illustrated in the next example, where different gases, initially at different

temperatures and pressures, are mixed.

Investigating Adiabatic Mixing of Gases at Constant Total Volume

c c c c EXAMPLE 12.5 c

Two rigid, insulated tanks are interconnected by a valve. Initially 0.79 lbmol of nitrogen at 2 atm and 4608R fills

one tank. The other tank contains 0.21 lbmol of oxygen at 1 atm and 5408R. The valve is opened and the gases are

allowed to mix until a final equilibrium state is attained. During this process, there are no heat or work interactions

between the tank contents and the surroundings. Determine (a) the final temperature of the mixture, in 8R, (b) the

final pressure of the mixture, in atm, (c) the amount of entropy produced in the mixing process, in Btu/8R.

SOLUTION

Known:

Nitrogen and oxygen, initially separate at different temperatures and pressures, are allowed to mix with-

out heat or work interactions with the surroundings until a final equilibrium state is attained.

What would be the exit velocity, in m/s, if the isentropic nozzle

efficiency were 90%? Ans. 592 m/s.

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722 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Find: Determine the final temperature of the mixture, in 8R, the final pressure of the mixture, in atm, and the

amount of entropy produced in the mixing process, in Btu/8R.

Schematic and Given Data:

Valve

Initially 0.21 lbmol

of O

at 1 atm

and 540°R

Initially 0.79 lbmol

of N

at 2 atm

and 460°R

Insulation

Engineering Model:

The system is taken to be the nitrogen and the oxygen together.

2. When separate, each of the gases behaves as an ideal gas.

3. The final mixture acts as an ideal gas and the Dalton model applies:

Each mixture component occupies the total volume and exhibits the

mixture temperature.

4. No heat or work interactions occur with the surroundings, and there

are no changes in kinetic and potential energy.

Fig. E12.5

Analysis:

(a)

The final temperature of the mixture can be determined from an energy balance. With assumption 4, the

closed system energy balance reduces to

¢U 5 Q

2 W



2 U

5 0

The initial internal energy of the system, U

, equals the sum of the internal energies of the two gases when

separate

5 n

21 n

where T

5 4608R is the initial temperature of the nitrogen and T

5 5408R is the initial temperature of the

oxygen. The final internal energy of the system, U

, equals the sum of the internal energies of the two gases

evaluated at the final mixture temperature T

5 n

21 n

Collecting the last three equations

22 u

241 n

22 u

245 0

The temperature T

can be determined using specific internal energy data from Table A-23E and an iterative

procedure like that employed in part (a) of Example 12.4. However, since the specific heats of N

and O

vary

little over the temperature interval from 460 to 5408R, the solution can be conducted accurately on the basis of

constant specific heats. Hence, the foregoing equation becomes

y, N

2 T

21 n

y,O

2 T

25 0

Solving for T

y, N

1 n

y, O

y, N

1 n

y, O

Selecting c

values for N

and O

from Table A-20E at the average of the initial temperatures of the gases,

5008R, and using the respective molecular weights to convert to a molar basis

y, N

5 a28.01

lbmol

ba0.177

Btu

lb ? 8R

b5 4.96

Btu

lbmol ? 8R

y, O

5 a32.0

lbmol

ba0

.156

Btu

lb ? 8R

b5 4.99

lbmol ? 8R

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