Test Bank With Solutions For Introduction To Chemical Engineering Thermodynamic 7Th Editon By Smith Hendrick Latest Update 2025/2026 A+

, 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: t  0

Given t = 1.8t  32 Find(t)  40 Ans.

F
1.5 By definition: P= F = massg Note: Pressures are in
A gauge pressure.
 2
P  3000bar D  4mm A  D A  12.566 mm
2
4
m F
F  PA g  9.807 mass  mass  384.4 kg Ans.
2 g
s

F
1.6 By definition: P= F = massg
A
 2
P  3000atm D  0.17in A  D 2
4
ft F
F  PA g  32.174 mass  mass  1000.7 lbm Ans.
2 g
sec


1.7 Pabs = gh  Patm

gm m
  13.535 g  9.832 h  56.38cm
3
cm s2

Patm  101.78kPa Pabs  gh  Patm Pabs  176.808 kPa Ans.


gm ft
1.8   13.535 g  32.243 h  25.62in
3 2
cm s

Patm  29.86in_Hg Pabs  gh  Patm Pabs  27.22 psia Ans.

1

, gm m
1.10 Assume the following:   13.5 g  9.8
3 2
cm s
P
P  400bar h  h  302.3 m Ans.
g

1.11 The force on a spring is described by: F = Ks x where Ks is the spring
constant. First calculate K based on the earth measurement then gMars
based on spring measurement on Mars.
On Earth:
m
F = massg = Kx mass  0.40kg g  9.81 x  1.08cm
2
s
F N
F  massg F  3.924 N Ks  Ks  363.333
x m

On Mars:

 3
x  0.40cm FMars  Kx FMars  4  10 mK
FMars 
mK
gMars  gMars  0.01 Ans.
mass kg


d MP d MP
1.12 Given: P = g and: = Substituting: P= g
dz RT dz RT

P z
 Denver 1  Denver  Mg
Separating variables and integrating:  dP =   dz
 P 
P 0  RT 
sea



 PDenver Mg zDenver
After integrating: ln =
 Psea  RT


Taking the exponential of both sides   Mg z 
Denver
and rearranging: PDenver = Psea e RT 

gm m
Psea  1atm M  29 g  9.8
mol 2
s

2

, 3
cm atm
R  82.06 T  (10  273.15)K zDenver  1mi
molK
Mg
zDenver  0.194
RT

  Mg  z 
Denver
PDenver  Psea e
RT  PDenver  0.823 atm Ans.

PDenver  0.834 bar Ans.


1.13 The same proportionality applies as in Pb. 1.11.
ft ft
gearth  32.186 gmoon  5.32 lmoon  18.76
2 2
s s
gearth
learth  lmoon  learth  113.498
gmoon


M  learthlbm M  113.498 lbm Ans.


wmoon  Mgmoon wmoon  18.767 lbf Ans.


5.00dollars hr 0.1dollars hr
1.14 costbulb  10 costelec  10 70W
1000hr day kWhr day

dollars dollars
costbulb  18.262 costelec  25.567
yr yr
dollars
costtotal  costbulb  costelec costtotal  43.829
yr Ans.

ft
1.15 D  1.25ft mass  250lbm g  32.169
2
s



3