Chapter 1 - Section B - Non-Numerical Solutions
1.1 This system of units is the English-system equivalent of SI. Thus, gc = 1(lbm )(ft)(poundal)1 (s)2 1.2 (a) Power is power, electrical included. Thus, Nm kgm2 energy [=] [=] time s s3 (b) Electric current

10.35 a) The equation from NIST is: Mi = ki yi P The equation for Henry's Law is:i Hi = yi P x Solving to eliminate P gives: By definition:
Mi Hi
Eq. (1) Eq. (2) Eq. (3)
=
Mi
=
ni ns Ms
ki xi
where M is the molar mass and the subscript s refers to the sol

Problems 10.25 to 10.34 have been solved using MS-EXCEL 2000 We give the resulting spreadsheets. Problem 10.25 a) BUBL P T=-60 F (-51.11 C) P=200 psia P=250 psia P=215 psia (14.824 bar) ANSWER Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 5.600 0.560 4.600 0.460 5.

Chapter 16 - Section A - Mathcad Solutions
16.10 (Planck's constant) h := 6.626 10
34
J s
(Boltzmann's constant) 23 J k := 1.381 10 K V := R T P
(Avagodro's number) NA := 6.023 10 mol m mol
3 23 1
P := 1bar
T := 298.15K 39.948 gm mol
V = 0.025
a) For Arg

Chapter 13 - Section A - Mathcad Solutions
Note: For the following problems the variable kelvin is used for the SI unit of absolute temperature so as not to conflict with the variable K used for the equilibrium constant 13.4 H2(g) + CO2(g) = H2O(g) + CO(g

Chapter 11 - Section A - Mathcad Solutions
11.1 For an ideal gas mole fraction = volume fraction CO2 (1): N2 (2): i := 1 . 2 P n := x1 := 0.7 x2 := 0.3 P := 1bar V1 := 0.7m V2 := 0.3m
3 3
T := ( 25 + 273.15)K
Vi
i
R T
n = 40.342 mol
S := n R
( xi ln( xi

Chapter 8 - Section A - Mathcad Solutions
8.1 With reference to Fig. 8.1, SI units, At point 2: Table F.2, At point 4: Table F.1, At point 1: At point 3: Table F.1, x3 := 0.96 H2 := 3531.5 H4 := 209.3 H1 := H4 Hliq := H4 H3 := Hliq + x3 Hlv Sliq := 0.7035

Chapter 7 - Section A - Mathcad Solutions
7.1 u2 := 325 m sec R := 8.314 J mol K molwt := 28.9 gm 7 R CP := mol 2 molwt
With the heat, work, and potential-energy terms set equal to zero and with the initial velocity equal to zero, Eq. (2.32a) reduces to H

Chapter 6 - Section A - Mathcad Solutions
6.7 At constant temperature Eqs. (6.25) and (6.26) can be written: dS = V dP and dH = ( 1 T) V dP
For an estimate, assume properties independent of pressure. T := 270 K V := 1.551 10 P1 := 381 kPa
3 3 m
P2 := 1200

Chapter 5 - Section A - Mathcad Solutions
5.2 Let the symbols Q and Work represent rates in kJ/s. Then by Eq. (5.8) = TC Work = 1 QH TH TH := 798.15 K TC TH or By Eq. (5.1), Work = 148.78 kJ s QH := 250 kJ s
TC := 323.15 K Work := QH 1
Work = 148.78 kW

Chapter 4 - Section A - Mathcad Solutions
4.1 (a) T0 := 473.15 K For SO2: A := 5.699 T := 1373.15 K B := 0.801 10
3
n := 10 mol C := 0.0 D := 1.015 10
5
H := R ICPH ( T0 , T , A , B , C , D) H = 47.007 kJ mol Q := n H Q = 470.073 kJ T := 1473.15 K B := 28

Chapter 3 - Section A - Mathcad Solutions
3.1 = 1 d dT = P 1 d dP
T
At constant T, the 2nd equation can be written: d = dP ln ( 1.01) ln
2 1
= P
:= 44.18 10
6
bar
1
2 = 1.01 1 Ans.
P :=
P = 225.2 bar
3
P2 = 226.2 bar
3.4 b := 2700 bar
c := 0.125
V

Chapter 2 - Section A - Mathcad Solutions
2.1 (a) Mwt := 35 kg Work := Mwt g z (b) Utotal := Work g := 9.8 m s
2
z := 5 m Work = 1.715 kJ Ans. Utotal = 1.715 kJ dU + d ( PV) = CP dT Ans.
(c) By Eqs. (2.14) and (2.21): Since P is constant, this can be writ

Chapter 1 - Section A - Mathcad Solutions
1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this equation by setting t(F) = t(C). Guess solution: Given t := 0 Find ( t) = 40 F = mass g 2 D 4 F mass := g A := m s 1.6 By definitio

Bubble P
Bubble T(choose one)
1. Know xi, T. Calc i, Pisat.
Dew T
1. Know xi, P Guess T (e.g. eq 9.62). 2. Calc i, Pisat . 3. yi= xii P isat = xiKi P 7. Guess T
1. Know yi, P Calc Pisat. Assume Raoults law for first T, xi calculation, then calc i at xi. 2

Section 11.6
Local Composition Theory
381
Nevertheless, there are some obvious limitations to the assumption of a constant packing fraction. A little calculation would make it clear that the for liquid propane at Tr = 0.99 is significantly larger that for

Chapter 10 Practice Problem Solutions (P10.5) (a) Perform bubble P calculations 1-CO2, 2- ethylene. For kij = 0 Output from PRMIX.exe bp COMPONENT IS CARBON DIOXIDE ID NO. IS 909 COMPONENT IS ETHYLENE ID NO. IS 201 T(K)= 222.00 P(MPa)= .8725 ZL= .2138E-01

0.1 Enthalpy of Mixtures
1
0.1 Enthalpy of Mixtures
The enthalpy of a mixture is determined using an equation of state by combining the departure function for the mixture with the enthalpy of an ideal gas mixture H = (H H ) + H
ig ig
= ( H H ) + xi Hi
ig

Chapter 9 Practice Problems
(p9.01) The stream from a gas well consists of 90 mol% methane, 5% ethane, 3% propane and 2% n-butane. This stream is flashed isothermally at 233 K and 70 bar. Use the shortcut K-ratio method to estimate the L/F fraction and li

Section 8.8 CALCULATION OF FUGACITY (LIQUIDS)
8.8
CALCULATION OF FUGACITY (LIQUIDS)
Example S8.1 Vapor and Liquid Fugacities using the Virial Equation
Determine the fugacity (MPa) for acetylene at: (a) 250K and 10 bar; (b) 250K and 20 bar. Use the virial

P. 8.1) CO2 at 15MPa and 25C is throttled to 0.1Mpa. Determine the temperature and fraction vapor. (NOTE: there is a typo. Part (c ) makes no sense at 1.5MPa.) Solution: Ebal for a valve, H = 0 (a) assume ideal gas, H = 0 = Cp (T ) T2 = 298K (b) by PREOS.

Chapter 7 Practice Problems (P7.1) G H TS G G ig H H ig S S ig = Eqn. 7.21 RT RT R 3 d H H ig a Z d + Z 1 = T = T + Z 1 5/ 2 RT T 2 RT (1 + b ) 0 0
note:
dx 1 = log e (ax + b) ax + b a
3 1 a = ln (1 + b ) 3/ 2 2T b R SS R
ig
+ Z 1 =
0
3a 1 ln (1 + b ) +

Determination of stability when searching for entropy values. Often when solving process problems, the entropy and pressure of a state are known, and the remaining state variables are to be found. If this is the case, it is helpful to keep in mind the beh

Chapter 4 Practice Problems (P4.1) COP = coef. of performance = QC WS ,net Using state numbers of Fig 4.9-4.10. P-H plot will look like Fig 4.10 of pg 151. Use P-H chart pg 653 and table pg 654: state 2 is satV at 40C ! H2 = 372 kJ/kg (chart) state 3, out