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Page 1: Main ISM Ch07

7-1

Chapter 7: The Properties of Real Gases Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions on Concepts Q7.1) Explain why the oscillations in the two-phase coexistence region using the Redlich-Kwong and van der Waals equations of state (see Figure 7.4) do not correspond to reality. The oscillations predict that as V increases, P will increase. No real gas exhibits this behavior. Q7.2) Explain the significance of the Boyle temperature. The Boyle temperature provides a way to classify the way in which z varies with P at low values of P for different gases. If T > TB, z increases with increasing P; if T < TB, z decreases with increasing P. Q7.3) The value of the Boyle temperature increases with the strength of the attractive interactions between molecules. Arrange the Boyle temperatures of the gases Ar, CH4, and C6H6 in increasing order. Ar < CH4 < C6H6 Q7.4) Will the fugacity coefficient of a gas above the Boyle temperature be less than one at low pressures?

No. The integral 0

1P z dPP− ′′∫ is always greater than zero for this case. Therefore,

γ > 1 for all P. Q7.5) Using the concept of the intermolecular potential, explain why two gases in corresponding states can be expected to have the same value for z. Two different gases will have different values for the depth of the intermolecular potential and for the distance at which the potential becomes positive. By normalizing P, T, and V to their critical values, the differences in the intermolecular potential are to a significant extent also normalized. Q7.6) By looking at the a and b values for the van der Waals equation of state, decide whether 1 mole of O2 or H2O has the higher pressure at the same value of T and V.

Page 2: Main ISM Ch07

Chapter 7/The Properties of Real Gases

7-2

A is significantly larger for H2O. Therefore, the attractive forces between H2O molecules are greater than between O2 molecules. Consequently, O2 will have a higher pressure. Q7.7) Consider the comparison made between accurate results and those based on calculations using the van der Waals and Redlich-Kwong equations of state in Figures 7.1 and 7.5. Is it clear that one of these equations of state is better than the other under all conditions? The Redlich-Kwong gives more accurate results for almost all of the values of pressure shown in these figures. However, it is not better under all conditions. Q7.8) Why is the standard state of fugacity, f º, equal to the standard state of pressure, Pº? If this were not the case the fugacity would not become equal to the pressure in the limit of low pressures. Q7.9) For a given set of conditions, the fugacity of a gas is greater than the pressure. What does this tell you about the interaction between the molecules of the gas? If the fugacity is greater than the pressure, the repulsive pat of the potential dominates the interaction between the molecules. Q7.10) A system containing argon gas is at pressure P1 and temperature T1. How would you go about estimating the fugacity coefficient of the gas? Use the critical constants of the gas to determine the reduced pressure and temperature. With these results, estimate the fugacity coefficient using Figure 7.11. Problems P7.1) A sample containing 35.0 g of Ar is enclosed in a container of volume 0.165 L at 390 K. Calculate P using the ideal gas, van der Waals, and Redlich-Kwong equations of state. Based on your results, does the attractive or repulsive contribution to the interaction potential dominate under these conditions?

Page 3: Main ISM Ch07

Chapter 7/The Properties of Real Gases

7-3

?

? ? ?

?

? ? ? 2 ?

2 ? ?

0.165 L 39.95 g 0.1883 L mol35.0 g mol

8.314 10 L bar K mol 390 K = 172 bar0.1883 L mol

8.314 10 L bar K mol 390 K 1.355L bar mol0.1883 L mol 0.0320 L mol 0.18

m

ideal gasm

vdWm m

V

RTPV

RT aPV b V

= × =

× × ×= =

× ×= − = −

− − ( )

( )

( )

2?

? ? ?

? ?

12 ? 2

? ? ?

83 L mol

169 bar1

8.314 10 L bar K mol 390 K0.1883 L mol 0.02219 L mol

16.86 L bar mol K 1 0.1883 L mol 0.1883L mol + 0.02219 L mol390 K

174 bar

RKm m m

RK

RT aPV b V V bT

P

=

= −− +

× ×=

= Because PvdW < Pideal gas, the attractive part of the interaction appears to dominate using the van der Waals equation of state. However, because PRK > Pideal gas, the repulsive part of the interaction appears to dominate using the Redlich-Kwong equation of state. This is a case where a more accurate equation of state is needed to answer the question. P7.2) Calculate the density of O2(g) at 375 K and 385 bar using the ideal gas and the van der Waals equations of state. Use a numerical equation solver to solve the van der Waals equation for Vm or use an iterative approach starting with Vm equal to the ideal gas result. Based on your result, does the attractive or repulsive contribution to the interaction potential dominate under these conditions?

( )

( )

? ? ??

? ? ? 2 ?

22 ?

?

8.314 10 L bar K mol 375 K = 8.10 10 L385 bar

8.314 10 L bar K mol 370 K 1.382 L bar mol0.0319 L mol

The three solutions to this equation are0.0131 0.0339 L mol

m

vdWm m m m

m

RTVP

RT aPV b V V V

V i

× ×= = ×

× ×= − = −

− −

= ± ?

??

?

??

?

and 0.0867 L molOnly the real solution is of significance.

32.0 g mol = 395 g L0.0810 L mol

32.0 g mol = 369 g L0.0867 L mol

m

idealgasm

vdWm

V

MV

MV

ρ

ρ

=

= =

= =

Page 4: Main ISM Ch07

Chapter 7/The Properties of Real Gases

7-4

Because the van der Waals density is less than the ideal gas density, the repulsive part of the potential dominates. P7.3) At 500 K and 400 bar, the experimentally determined density of N2 is 7.90 mol L–1. Compare this with values calculated from the ideal and Redlich-Kwong equations of state. Use a numerical equation solver to solve the Redlich-Kwong equation for Vm or use an iterative approach starting with Vm equal to the ideal gas result. Discuss your results. For the ideal gas,

( )

( )

?? ? ?

? ? ?

?

12 ? 2

?

1 400 bar = 9.62 mol L8.314 10 L bar K mol 500 K

1

8.314 10 L bar K mol 500 K400 bar = 0.02208 L mol

17.40 L bar mol K 1 + 0.02208 L mol500 K

The

m

RKm m m

m

m m

PV RT

RT aPV b V V bT

V

V V

= =× ×

= −− +

× ×−

( ) ? ?

??

three solutions to this equation are0.00529 0.0186 L mol and 0.1145 L mol

Only the real solution is of significance.1 1 = 8.73 mol L

0.1145 L mol

m m

m

V i V

V

= − ± =

=

The ideal gas density is greater than that calculated with the Redlich-Kwong equation of state and the experimental result showing that the repulsive part of the potential dominates. The Redlich-Kwong result is in error by +10%. P7.4) The observed Boyle temperatures of H2, N2, and CH4 are 110, 327, and 510 K, respectively. Compare these values with those calculated for a van der Waals gas with the appropriate parameters.

Page 5: Main ISM Ch07

Chapter 7/The Properties of Real Gases

7-5

( )

3 3 3

6 ?

2 ? ? ? 3

for a van der Waals gas.

In solving this problem, keep in mind that 1 dm 1 L = 10 m .

0.2452 dm bar molH 111 K8.314 10 L bar mol K 0.0265 dm mol

This value is slightly higher

B

B

aTRb

T

=

=

= =× ×

( )

( )

6 ?

2 ? ? ? 3 ?

6 ?

4

than the experimental value.

1.370 dm bar molN 426 K8.314 10 L bar mol K 0.0387 dm mol

This value is substantially larger than the experimental value.

2.303 dm bar molCH8.314

B

B

T

T

= =× ×

= ? ? ? 3 ? 643 K 10 L bar mol K 0.0431 dm mol

This value is substantially larger than the experimental value.

=× ×

P7.5) Calculate the van der Waals parameters of methane from the values of the critical constants.

( ) ( )

? 3 ? ?3 ?

2 2? 3 ? ?2 26 ?

8.314 10 dm bar K mol 190.56 K = 0.0431 dm mol8 8 45.99 bar

27 8.314 10 dm bar K mol 190.56 K27 = 2.303 dm bar mol64 64 45.99 bar

c

c

c

c

RTbP

R TaP

× ×= =

×

× × ×= =

×

P7.6) Calculate the Redlich-Kwong parameters of methane from the values of the critical constants.

( )( ) ( )

( )( ) ( )

525 ? 3 ? ? 2 12 26 22

1 13 3

1 1 ? 3 ? ?3 3

3 ?

8.314 10 dm bar K mol 190.56= = 32.20 dm bar K mol

9 2 1 9 45.99 bar 2 1

2 1 2 1 8.3145 10 dm bar K mol 190.56 K= 0.02985 dm mol

3 3 45.99 bar

c

c

c

c

R TaP

RTb

P

−× ×

=− × −

− − × × ×= =

× P7.7) Use the law of corresponding states and Figure 7.8 to estimate the molar volume of methane at T = 285 K and P = 180 bar.

285 K 180 bar1.5 3.91.190.56 K 45.99 barr rT P= = = = Therefore, z ≈ 0.8.

Page 6: Main ISM Ch07

Chapter 7/The Properties of Real Gases

7-6

? 3 ? ?

?

8.314 10 dm bar K mol 285 K0.8; 0.8 0.8180 bar

0.105 L mol

mm

m

PV RTVRT P

V

× ×= = = ×

=

P7.8) Calculate the P and T values for which H2(g) is in a corresponding state to Xe(g) at 450 K and 85.0 bar.

2 2

2 2

450 K 1.55; 1.55 1.55 32.98 K = 51.2 K289.74 K

85.0 bar 1.46; 1.46 1.46 12.93 bar = 18.8 bar58.40 bar

XeH HXe

R CXeC

XeH HXe

R CXeC

TT T TT

PP P PP

= = = = = ×

= = = = = ×

P7.9) Assume that the equation of state for a gas can be written in the form

( )( ) .mP V b T RT− = Derive an expression for 1 1 and P T

V VV T V P

β κ∂ ∂⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ for such

a gas in terms of ( ) ( ), , , and .m

db Tb T P V

dT

( ) ( )

( )

( ) ( )

2 2

;

1 1 1

1 1P m

T m

V V RTP b T RT b Tn n P

nRTV nb TP

ndb T db TV nR RV T V dT P V dT P

V nRT RTV P V P V P

β

κ

⎛ ⎞− = = +⎜ ⎟⎝ ⎠

= +

⎛ ⎞ ⎛ ⎞∂⎛ ⎞= = + = +⎜ ⎟ ⎜ ⎟⎜ ⎟∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠∂⎛ ⎞ ⎛ ⎞= − = − − =⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠

P7.10) 1 mol of Ar initially at 298 K undergoes an adiabatic expansion against a pressure Pexternal = 0 from a volume of 20.0 L to a volume of 65.0 L. Calculate the final temperature using the ideal gas and van der Waals equations of state. w = q = 0. ΔU = 0 for an ideal gas and ΔT = 0 because U is a function of T only. Using the results of Example Problem 3.5 for a van der Waals gas,

Page 7: Main ISM Ch07

Chapter 7/The Properties of Real Gases

7-7

,, ,

5 ? 36 ?

6 ? 3 ? ? 3 ?

?,

? ?,

1 1

10 Pa 10 m 1 11.355 dm bar molbar dm 65.0 10 m mol 20.0 10 m mol

4.69 J4.69 J mol 0.376 K

12.5 J K mol297.6 K

T mm i m f m

T m

V m

f

U aV V

UT

CT

⎛ ⎞Δ = −⎜ ⎟⎜ ⎟

⎝ ⎠

⎛ ⎞= × × × −⎜ ⎟× ×⎝ ⎠= −

ΔΔ = = − = −

=

P7.11) 1 mol of Ar undergoes an isothermal reversible expansion from an initial volume of 1.00 L to a final volume of 65.0 L at 298 K. Calculate the work done in this process using the ideal gas and van der Waals equations of state. What percentage of the work done by the van der Waals gas arises from the attractive potential? For the ideal gas,

? ? 365.0 Lln 1 mol 8.314 J mol K 298 K ln 10.34 10 J1.00 L

f

i

Vw nRT

V= − = − × × × = − ×

For the van der Waals gas,

( )

( )3

3

2 2

2

0.065 m6? ? ? 3

0.001 m

3 3

3

ln

0.1355 Pa m1mol 8.314 J mol K 298 K ln 3.20 10 m +

10.42 10 J 133 J = 10.41 10 J133 J100 1.3%

10.41 10 J

ff

i i

VV

V V

nRT n a anw dV nRT V nbV nb V V

VV

⎛ ⎞ ⎡ ⎤= − − = − − +⎢ ⎥⎜ ⎟−⎝ ⎠ ⎣ ⎦

⎡ ⎤= − × × × − ×⎢ ⎥

⎣ ⎦= − × + − ×

× =×

P7.12) For a van der Waals gas, m

m m

V azV b RTV

= −−

. Expand the first term of this

expression in a Taylor series in the limit Vm >> b to obtain 11m

az bRT V

⎛ ⎞≈ + −⎜ ⎟⎝ ⎠

.

( ) ( ) ( ) ( )0

10 ... In this case, and 1 mx

m

df x bf x f x f x xbdx VV

=

⎛ ⎞= + + = =⎜ ⎟

⎝ ⎠ −

Page 8: Main ISM Ch07

Chapter 7/The Properties of Real Gases

7-8

1

1

1

1

1 11 1 1 1

1

11 1

m

m

m

m m m

m

m

bm m mV

m bmV

m m m

V a az bV b RTV RTVV

bdVb b b

b V V VbdV V

b a az bV RTV V RT

=

=

= − = −− −

⎡ ⎤⎛ ⎞⎡ ⎤−⎢ ⎥⎜ ⎟⎢ ⎥⎛ ⎞⎡ ⎤ ⎣ ⎦⎢ ⎥⎝ ⎠≈ − + = +⎜ ⎟⎢ ⎥ ⎢ ⎥⎛ ⎞⎣ ⎦⎝ ⎠− ⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

⎛ ⎞≈ + − = + −⎜ ⎟⎝ ⎠

P7.13) Show that ln1P

zT TT

β ∂⎛ ⎞= + ⎜ ⎟∂⎝ ⎠and that ln1 .

T

zP PP

κ ∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠

2 2

2

1 1;

;

ln 1 1

ln 1Therefore 1 1

P T

P P

P P

P

V VV T V P

PV z VP P V VP PVzRT T RT RT T RT RT

z z RT VP PVT z T VP RT RT T

zT T TT T

β κ

β

β β

β β

∂ ∂⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂ ∂⎛ ⎞ ⎛ ⎞= = − + = − +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = − + = − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∂⎛ ⎞ ⎛ ⎞+ = + − + =⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠

ln 1 1

ln 1Therefore 1 1

T T

T T

T

z V P V V PVP RT RT P RT RT

z z RT V PVP z P PV RT RT P

zP P PP P

κ

κ κ

κ κ

∂ ∂⎛ ⎞ ⎛ ⎞= + = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∂⎛ ⎞ ⎛ ⎞− = − − =⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠

P7.14) A van der Waals gas has a value of z = 1.00084 at 298 K and 1 bar and the Boyle temperature of the gas is 125 K. Because the density is low, you can calculate Vm from the ideal gas law. Use this information and the result of Problem P7.12 to estimate a and b.

Page 9: Main ISM Ch07

Chapter 7/The Properties of Real Gases

7-9

? 3 ? ?

3 ? 5 3 ?

? ? 5 3 ?

11 ;

1 1

1 0.00084 8.314 10 dm bar mol K 298 K 125 K 1 bar1 1298 K

= 0.0359 dm mol 3.59 10 m mol8.314J mol K 3.59 10 m mol 125 K = 3.7

Bm

B

m

B

B

a az b TV RT Rbb Tz

V T

z RTb T PT

a RbT

⎛ ⎞− = − =⎜ ⎟⎝ ⎠

⎛ ⎞− = −⎜ ⎟⎝ ⎠

− × ×= = ×

− −

= ×

= = × × × 2 6 ?3 10 m Pa mol−×

P7.15) The experimental critical constants of H2O are Tc = 647.14 K, Pc = 220.64 bar, and Vc = 55.95 × 10–3 L. Use the values of Pc and Tc to calculate Vc. Assume that H2O behaves as (a) an ideal gas, (b) a van der Waals gas, and (c) a Redlich-Kwong gas at the critical point. For parts (b) and (c), use the formulas for the critical compression factor. Compare your answers with the experimental value. Assuming an ideal gas,

2 1 18.314 10 L bar mol K 647.14 K = 0.2438L220.64 bar

cc

c

RTVP

− − −× ×= =

For a van der Waals gas, 33 3 3; 0.2438 L 91.4 10 L

8 8 8c c c

c cc c

PV RTz VRT P

−= = = = × = ×

For a Redlich-Kwong gas, 30.333; 0.333 0.333 0.2438 L 81.2 10 Lc c c

c cc c

PV RTz VRT P

−= = = = × = ×

Although the agreement with experiment is better for the van der Waals and Redlich-Kwong models than for the ideal gas model, all results differ significantly from the true value.

P7.16) Another equation of state is the Bertholet equation, 2mRT aV bP RT

= + − . Derive

expressions for 1 1 and P T

V VV T V P

β κ∂ ∂⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠from the Bertholet equation in terms

of V, T, and P.

Page 10: Main ISM Ch07

Chapter 7/The Properties of Real Gases

7-10

2

3 3

2 2 2

1 1 2 1 2

1 1P m

T m

nRT naV nbP RT

V nR na R aV T V P RT V P RT

V nRT nRT RTV P V P P V P V

β

κ

= + −

∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠∂⎛ ⎞ ⎛ ⎞= − = − − = =⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠

P7.17) For the Bertholet equation, 2mRT aV bP RT

= + − , find an expression for the

Boyle temperature in terms of a, b, and R.

2 3

2 3

2 3

= 1

and at , 0

Therefore 0

m

BT T

B

B

PV Pb PazRT RT R T

z a Zb T TP R T P

abR T

aTRb

= + −

∂ ∂⎛ ⎞ ⎛ ⎞= − = =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

− =

=

P7.18) The experimentally determined density of H2O at 1200 bar and 800 K is 537 g L–1. Calculate z and Vm from this information. Compare this result with what you would have estimated from Figure 7.8. What is the relative error in using Figure 7.8 for this case?

?? ?

?

? ?

? ? ?

18.02 g mol = 3.34 10 L mol537 g L1200 bar 3.34 10 L mol 0.602

8.314 10 L bar mol K 800 K

m

m

MV

PVzRT

ρ= = ×

× ×= = =

× ×

Because 1200 bar 5.44220.64 barrP = = and 800 K 1.24,

647.14 KrT = = Figure 7.7 predicts z = 0.75.

The relative error in z is 25%. P7.19) The volume of a spherical molecule can be estimated as V = b/4NA where b is the van der Waals parameter and NA is Avogadro’s number. Justify this relationship by

considering a spherical molecule of radius r, with volume 343

V rπ= . What is the volume

centered at the molecule that is excluded for the center of mass of a second molecule in terms of V? Multiply this volume by NA and set it equal to b. Apportion this volume

Page 11: Main ISM Ch07

Chapter 7/The Properties of Real Gases

7-11

equally among the molecules to arrive at V = b/4NA. Calculate the radius of a methane molecule from the value of its van der Waals parameter b.

The excluded volume is ( )34 2 8 ,3 moleculer Vπ = or 4 moleculeV per molecule. Therefore,

b = 4NA moleculeV and moleculeV = b/4NA.

3

1 15 33 3

? 023

163

3 3 4.31 10 m 1.63 10 m16 16 6.022 10

A

A

N r b

brN

π

π π

=

⎛ ⎞ ⎛ ⎞× ×= = = ×⎜ ⎟ ⎜ ⎟× ×⎝ ⎠⎝ ⎠

P7.20) At what temperature does the slope of the z versus P curve as P → 0 have its maximum value for a van der Waals gas? What is the value of the maximum slope?

, 0

2 3 20

maxmax

1 for a van der Waals gas

1 1 1 2

Setting this derivative equal to zero gives 2 20

The maximum slope

T P

T P

Z abP RT RT

Z a ab bT P RT RT RT RT RT

a ab TRT Rb

∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠

⎛ ⎞∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − + = − −⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

− = =

2

max max

1is 2 2 4

a b b bb b aRT RT a a a

⎛ ⎞ ⎡ ⎤⎛ ⎞− = − =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎝ ⎠

P7.21) Show that the van der Waals and Redlich-Kwong equations of state reduce to the ideal gas equation of state in the limit of low density.

2m m

RT aV b V

−−

: In the limit of low density, Vm is large and Vm >> b. The second term in

the van der Waals equation can be neglected because it goes as 2

1

mV, and in the first term,

m m

RT RTV b V

→−

.

Page 12: Main ISM Ch07

Chapter 7/The Properties of Real Gases

7-12

( )1

m m m

RT aPV b V V bT

= −− +

: In the limit of low density, Vm is large and Vm >> b. As

for the van der Waals equation, the second term in the Redlich-Kwong equation can be

neglected because it goes as 2

1

mV. In the limit, the first term becomes

m m

RT RTV b V

→−

.

P7.22) Show that the second virial coefficient for a van der Waals gas is given by

( ) 11

m T

z aB T bRT RT

V

⎛ ⎞⎜ ⎟∂⎜ ⎟= = −⎜ ⎟∂⎜ ⎟⎝ ⎠

( ) ( ) ( )

( ) ( )

22

200 0

2

20

22

1Let . The virial expansion takes the form

1 ...2

1Therefore, the second virial coefficient is 2

1

m

uu u

u

m m

uV

dP u d P uP RT P u u u

du du

d P uB T

RT du

RT a RTP auV b V u b

== =

=

=

⎛ ⎞⎡ ⎤⎡ ⎤= + + +⎡ ⎤⎜ ⎟⎢ ⎥⎢ ⎥⎣ ⎦⎜ ⎟⎣ ⎦ ⎣ ⎦⎝ ⎠

⎡ ⎤= ⎢ ⎥

⎣ ⎦

= − = −− −

( ) ( ) ( )

( )

( )

( ) ( )

2 2 22

2

32

30

2

20

1 2 2 21 1 1

2 21

2lim 2 2 2 and 1

12

u

u

dP RT d u RT RTau au audu duu b u u b ub

d P bRT adu ub

bRT a bRT aub

d P u aB T bRT du RT

=

= − − = − = −− − −

= −−

− = −−

⎡ ⎤= = −⎢ ⎥

⎣ ⎦

P7.23) For a gas at a given temperature, the compressibility is described by the

empirical equation 2

3 51 9.00 10 4.00 10P PzP P

− − ⎛ ⎞= − × + × ⎜ ⎟⎝ ⎠o o

, where Pº = 1 bar.

Calculate the activity coefficient for P = 100, 200, 300, 400, and 500 bar. For which of these values is the activity coefficient greater than one?

Page 13: Main ISM Ch07

Chapter 7/The Properties of Real Gases

7-13

( )( )23 5

0 03 5 2

1 9.00 10 4.00 10 11ln

ln 9.00 10 2.00 100.497, 0.368, 0.406, 0.670, and 1.65 at 100, 200, 300, 400, and 500 bar, respectively.

P P P Pz dP dPP P

P P

γ

γγ

− −

− −

′ ′− × + × −− ′ ′= =′ ′

= − × + ×=

∫ ∫

P7.24) For values of z near one, it is a good approximation to write ( ) 1 .T

zz P PP∂⎛ ⎞= + ⎜ ⎟∂⎝ ⎠

If z =1.00054 at 0ºC and 1 bar, and the Boyle temperature of the gas is 220 K, estimate the values of Vm, a and b for the van der Waals gas.

42 1 1 2 1 1

From Example Problem 7.2, 1

We can write three equations in three unknowns:

1

1 bar5.4 108.314 10 L bar mol K 273.15 K 8.314 10 L bar mol K

T

z abP RT RT

a Pz bRT RT

ab−− − − − − −

∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠

⎛ ⎞− = −⎜ ⎟⎝ ⎠

⎛ ⎞× = − ×⎜ ⎟× × ×⎝ ⎠

2 1 1

2 1 1

2 2 1

273.15 K

220 K8.314 10 L bar mol K

8.314 10 L bar mol K 273.15 K 1 bar

Using an equation solver, the results are 1.15 L bar mol , 0.0630 L mol , 22.72

B

m m m m

m

a aTRb bRT a aP

V b V V b V

a b V

− − −

− − −

− −

×

= = =× ×

× ×= − = − =

− −

= = = L. P7.25) Calculate the critical volume for ethane using the data for Tc and Pc in Table 7.2 (see Appendix B, Data Tables) assuming a) the ideal gas equation of state and b) the van der Waals equation of state. Use an iterative approach to obtain Vc from the van der Waals equation, starting with the ideal gas result. How well do the calculations agree with the tabulated values for Vc? Using the ideal gas law,

2 ? ?8.314 10 L bar mol K 305.32 K 0.5211 L48.72 bar

cc

c

RTVP

−× ×= = =

Using the van der Waals equation of state,

? ? ? 2 ?

2 ? 2

8.314 10 L bar K mol 305.32 K 5.580 L bar mol0.0651 L mol

cc

c c c c

RT aPV b V V V

× ×= − = −

− −

Page 14: Main ISM Ch07

Chapter 7/The Properties of Real Gases

7-14

Setting Vc = 0.5211 L gives Pc = 35.13 bar, which is below the experimental value Pc = 48.72 bar. Setting Vc = 0.300 L gives Pc = 46.06 bar, and setting Vc = 0.1784 L gives Pc = 48.72 bar. The relative error of this value for Vc is

0.1874 L 0.1455 L100 29%.0.1455 L

−× = +