Chapter 6 Practice Problems

To accompany Introductory Chemical Engineering Thermodynamics © J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/6/2001) 1 of 2

(P6.2) 15,30 == rr TP (a) Use virial equation of state.

rr TPBBZ /)(1 10 ω++= …………………………………………..Eqn. 6.6

Where,

2.41

6.10

172.0139.0

422.0083.0

r

r

TB

TB

−=

−= ………………………………. Eqns. 6.8 & 6.9

)(041.0&

138998.0&

077459277.01

0

book

B

B

−==

=⇒

ω

14.115

30*)138998.0*041.0077459277.0(1 =−+=⇒ Z

(b) ??=ρ

MPaPPP

KTTT

Cr

Cr

59.7930*653.2*

6664.44*15*

======

3

3

/254.093.3

11

/93.3179.20*59.79

666*314.8*14.1

*

**,

cmgV

gcmMWP

TRZV

RT

PVZ

===⇒

===⇒=

ρ

(P6.4)

KCFT

MPaatmP

KT

OO 15.2982577

1.01

111

2

1

1

===

≈==

Use PREOS.XLS, ⇒ Use Solver, and set target cell on the volume and make it equal to

33.639114*2 = 67.278228, Then by changing the cell of pressure, making sure that T2 = 298.15K

Current State Roots

T (K) 298.15 Z V

P (MPa) 33.839895 cm3/gmol

Current State Roots

Stable Root has a lower fugacity

T (K) 111 Z V fugacity

P (MPa) 0.1 cm3/gmol MPa answers for three 0.9670679 8924.6249 0.096803

root region 0.0263855 243.49944 0.0036451 33.639114 0.093707

Chapter 6 Practice Problems

To accompany Introductory Chemical Engineering Thermodynamics © J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/6/2001) 2 of 2

answers for three #NUM! #NUM! root region #NUM! #NUM!

#NUM! #NUM!

& for 1 root region 0.9184568 67.278228

MPaP 84.332 =⇒