a bit of algebra leads to
3.5 κabP⋅+=a 3.9 10 6−
⋅atm 1−
⋅:= b 0.1−10 9−
⋅atm 2−
⋅:=
P11 atm⋅:= P23000 atm⋅:= V1ft
3
⋅:= (assume const.)
Combine Eqs. (1.3) and (3.3) for const. T:
Chapter 3 – Section A – Mathcad Solutions
3.1 β1−
ρT
ρ
d
d
⎛
⎜
⎝
⎞
⎠
⋅=κ1
ρP
ρ
d
d
⎛
⎜
⎝
⎞
⎠
⋅=
P T
At constant T, the 2nd equation can be written:
dρ
ρκdP⋅=ln
ρ2
ρ1
⎛
⎜
⎝
⎞
⎠κ∆P⋅=κ44.18 10 6−
⋅bar 1−
⋅:= ρ21.01 ρ1
⋅=
3.4 b 2700 bar⋅:= c 0.125 cm3
gm
⋅:= P11 bar⋅:= P2500 bar⋅:=
Since Work
V1
V2
VP
⌠
⎮
⌡d−=
21
P21 bar⋅:= T1600 K⋅:= CP
7
2R⋅:= CV
5
2R⋅:=
(a) Constant V: W0=and ∆UQ=CV∆T⋅=
T2T1
P2
⋅:= ∆TT
2T1
−:= ∆T 525−K=
3.6 β1.2 10 3−
⋅degC 1−
⋅:= CP0.84 kJ
kg degC⋅
⋅:= M5kg⋅:=
V1
1
1590
m3
kg
⋅:= P 1 bar⋅:= t10 degC⋅:= t220 degC⋅:=
With beta independent of T and with P=constant,
dV
VβdT⋅=V2V1exp βt2t1
−
()
⋅
⎡
⎣
⎤
⎦
⋅:= ∆VV
2V1
−:=
3.8 P18 bar⋅:=
22
Step 41: Adiabatic T4T1
P4
P1
⎛
⎜
⎝
⎞
⎠
R
CP
⋅:= T4378.831 K=
∆U41 CVT1T4
−
()
⋅:= ∆U41 4.597 103
×J
mol
=
∆H41 CPT1T4
−
()
⋅:= ∆H41 6.436 103
×J
mol
=
Q41 0J
mol
:= Q41 0J
mol
=
W41 ∆U41
:= W41 4.597 103
×J
mol
=
P23bar:= T2600K:= V2
RT
2
⋅
P2
:= V20.017 m3
mol
=
Step 12: Isothermal ∆U12 0J
mol
:= ∆U12 0J
mol
=
∆H12 0J
mol
⋅:= ∆H12 0J
mol
=
γCP
CV
:= T2T1
P2
P1
⎛
⎜
⎝
⎞
⎠
γ1−
γ
⋅:= T2331.227 K= ∆TT
2T1
−:=
∆UC
V∆T⋅:= ∆HC
P∆T⋅:=
3.9 P42bar:= CP
7
2R:= CV
5
2R:=
P110bar:= T1600K:= V1
RT
1
⋅
P1
:= V14.988 10 3−
×m3
mol
=
23
Step 34: Isobaric ∆U34 CVT4T3
−
()
⋅:= ∆U34 439.997−J
mol
=
∆H34 CPT4T3
−
()
⋅:= ∆H34 615.996−J
mol
=
Q34 CPT4T3
−
()
⋅:= Q34 615.996−J
mol
=
W34 R−T4T3
−
()
⋅:= W34 175.999 J
mol
=
3.10 For all parts of this problem: T2T1
=and
∆U∆H=0=Also Q Work−=and all that remains is
to calculate Work. Symbol V is used for total volume in this problem.
Q12 R−T1
⋅ln P2
P1
⎛
⎜
⎝
⎞
⎠
⋅:= Q12 6.006 103
×J
mol
=
W12 Q12
−:= W12 6.006−103
×J
mol
=
Step 23: Isochoric ∆U23 CVT3T2
−
()
⋅:= ∆U23 4.157−103
×J
mol
=
∆H23 CPT3T2
−
()
⋅:= ∆H23 5.82−103
×J
mol
=
Q23 CVT3T2
−
()
⋅:= Q23 4.157−103
×J
mol
=
W23 0J
mol
:= W23 0J
mol
=
P42 bar=T4378.831 K=V4
RT
4
⋅
P4
:= V40.016 m3
mol
=
24
PiP1
V1
V2
⎛
⎜
⎝
⎞
⎠
γ
⋅:= (intermediate P) Pi62.898 bar=
W1
PiV2
⋅P1V1
⋅−
γ1−
:= W17635 kJ=
(d) Step 1: heat at const V1 to P2W10=
Step 2: cool at const P2 to V2
(e) Step 1: cool at const P1 to V2
W1P1
−V2V1
−
()
⋅:= W11100 kJ=
(a) Work n R⋅T⋅ln P2
P1
⎛
⎜
⎝
⎞
⎠
⋅=Work P1V1
⋅ln P2
P1
⎛
⎜
⎝
⎞
⎠
⋅:=
(b) Step 1: adiabatic compression to P2
γ5
3
:= ViV1
P1
P2
⎛
⎜
⎝
⎞
⎠
1
γ
⋅:= (intermediate V) Vi2.702 m3
=
W1
P2Vi
⋅P1V1
⋅−
γ1−
:= W13063 kJ=
Step 2: cool at const P2 to V2
W2P2
−V2Vi
−
()
⋅:= W22042 kJ=
(c) Step 1: adiabatic compression to V2
25
P1100 kPa⋅:= P2500 kPa⋅:= T1303.15 K⋅:=
CP
7
2R⋅:= CV
5
2R⋅:= γ CP
CV
:=
Adiabatic compression from point 1 to point 2:
Q12 0kJ
mol
⋅:= ∆U12 W12
=CV∆T12
⋅=T2T1
P2
P1
⎛
⎜
⎝
⎞
⎠
γ1−
γ
⋅:=
∆U12 CVT2T1
−
()
⋅:= ∆H12 CPT2T1
−
()
⋅:= W12 ∆U12
:=
Cool at P2 from point 2 to point 3:
T3T1
:= ∆H23 CPT3T2
−
()
⋅:= Q23 ∆H23
:=
∆U23 CVT3T2
−
()
⋅:= W23 ∆U23 Q23
−:=
Step 2: heat at const V2 to P2W20=
3.17 (a) No work is done; no heat is transferred.
∆Ut∆T=0=T2T1
=100 degC⋅=Not reversible
(b) The gas is returned to its initial state by isothermal compression.
Work n R⋅T⋅ln V1
V2
⎛
⎜
⎝
⎞
⎠
⋅=but nR⋅T⋅P2V2
⋅=
V14m
3
⋅:= V2
4
3m3
⋅:= P26 bar⋅:=
⎜
3.18 (a)
26
(b) If each step that is 80% efficient accomplishes the same change of state,
all property values are unchanged, and the delta H and delta U values
are the same as in part (a). However, the Q and W values change.
Isothermal expansion from point 3 to point 1:
∆U31 ∆H31
=0=P3P2
:= W31 RT
3
⋅ln P1
P3
⎛
⎜
⎝
⎞
⎠
⋅:=
Q31 W31
−:=
FOR THE CYCLE: ∆U∆H=0=
QQ
12 Q23
+Q31
+:= Work W12 W23
+W31
+:=
27
(b) Adiabatic: P2P1
V1
V2
⎛
⎜
⎝
⎞
⎠
γ
⋅:= T2T1
P2
P1
⋅V2
V1
⋅:=
(c) Restrained adiabatic: Work ∆U=Pext
−∆V⋅=
nP1V1
⋅
RT
1
⋅
:= ∆UnC
V
⋅∆T⋅=
FOR THE CYCLE:
QQ
12 Q23
+Q31
+:= Work W12 W23
+W31
+:=
3.19 Here, V represents total volume.
P11000 kPa⋅:= V11m
3
⋅:= V25V
1
⋅:= T1600 K⋅:=
CP21 joule
mol K⋅
⋅:= CVCPR−:= γ CP
CV
:=
(a) Isothermal: Work n R⋅T1
⋅ln V1
V2
⎛
⎜
⎝
⎞
⎠
⋅=P2P1
V1
V2
⋅:=
⎜
28
W23 0kJ
mol
⋅:= ∆U23 CVT3T2
−
()
⋅:=
Q23 ∆U23
:= ∆H23 CPT3T2
−
()
⋅:=
Q23 2.079−kJ
mol
= ∆U23 2.079−kJ
mol
= ∆H23 2.91−kJ
mol
=
3.20 T1423.15 K⋅:= P18 bar⋅:= P33 bar⋅:=
CP
7
2R⋅:= CV
5
2R⋅:= T2T1
:= T3323.15 K⋅:=
Step 12: ∆H12 0kJ
mol
⋅:= ∆U12 0kJ
mol
⋅:=
If rV1
V2
=
V1
V3
=Then rT1
T3
P3
P1
⋅:= W12 RT
1
⋅ln r()⋅:=
W12 2.502−kJ
mol
=Q12 W12
−:= Q12 2.502 kJ
mol
=
Step 23:
29
P11 bar⋅:= P310 bar⋅:=
∆UC
VT3T1
−
()
⋅:= ∆HC
PT3T1
−
()
⋅:=
Each part consists of two steps, 12 & 23.
(a) T2T3
:= P2P1
T2
T1
⋅:=
W23 RT
2
⋅ln P3
⎛
⎜
⎞
⋅:= Work W23
:=
3.21 By Eq. (2.32a), unit-mass basis: molwt 28 gm
mol
:= ∆H1
2∆u2
⋅+ 0=
But ∆HC
P∆T⋅=Whence ∆Tu22u12
−
()
−
2C
P
⋅
=
CP
7
2
R
molwt
⋅:= u12.5 m
s
⋅:= u250 m
s
⋅:= t1150 degC⋅:=
3.22 CP
7
2R⋅:= CV
5
2R⋅:= T1303.15 K⋅:= T3403.15 K⋅:=
30
Q23 ∆H23
:=
∆U23 CVT3T2
−
()
⋅:= W23 ∆U23 Q23
−:=
For the second set of heat-capacity values, answers are (kJ/mol):
∆U 1.247=∆U 2.079=
(a) Work 6.762=Q 5.515−=
(b) Work 6.886=Q 5.639−=
(c) Work 4.972=Q 3.725−=
(b) P2P1
:= T2T3
:= ∆U12 CVT2T1
−
()
⋅:=
∆H12 CPT2T1
−
()
⋅:= Q12 ∆H12
:=
W12 ∆U12 Q12
−:= W12 0.831−kJ
mol
=
W23 RT
2
⋅ln P3
P2
⎛
⎜
⎝
⎞
⎠
⋅:= W23 7.718 kJ
mol
=
(c) T2T1
:= P2P3
:= W12 RT
1
⋅ln P2
P1
⎛
⎜
⎝
⎞
⎠
⋅:=
∆H23 CPT3T2
−
()
⋅:=
31
For the process: Work W12 W23
+:=
3.24 W12 0=Work W23
=P2V3V2
−
()
−=R−T3T2
−
()
⋅=
But T3T1
=So… Work R T2T1
−
()
⋅=
Also WRT
1
⋅ln P
P1
⎛
⎜
⎝
⎞
⎠
⋅=Therefore
ln P
P1
⎛
⎜
⎝
⎞
⎠
T2T1
−
T1
=T2350 K⋅:= T1800 K⋅:= P14 bar⋅:=
⎜
3.23 T1303.15 K⋅:= T2T1
:= T3393.15 K⋅:=
P11 bar⋅:= P312 bar⋅:= CP
7
2R⋅:= CV
5
2R⋅:=
For the process: ∆UC
VT3T1
−
()
⋅:= ∆HC
PT3T1
−
()
⋅:=
Step 12: P2P3
T1
T3
⋅:= W12 RT
1
⋅ln P2
P1
⎛
⎜
⎝
⎞
⎠
⋅:=
W12 5.608 kJ
mol
=Q12 W12
−:= Q12 5.608−kJ
mol
=
Step 23: W23 0kJ
mol
⋅:= Q23 ∆U:=
32
TBfinal()T
B
=
nAnB
=Since the total volume is constant,
2n
A
⋅R⋅T1
⋅
P1
nAR⋅TATB
+
()
⋅
P2
=or 2T
1
⋅
P1
TATB
+
P2
=(1)
(a) P21.25 atm⋅:= TBT1
P2
P1
⎛
⎜
⎝
⎞
⎠
γ1−
γ
⋅:= (2)
TA2T
1
⋅P2
P1
⋅TB
−:= Qn
A∆UA∆UB
+
()
⋅=
Define qQ
nA
=qC
VTATB
+2T
1
⋅−
()
⋅:= (3)
3.25 VA256 cm3
⋅:= Define: ∆P
P1
r=r 0.0639−:=
Assume ideal gas; let V represent total volume:
P1VB
⋅P2VAVB
+
()
⋅=From this one finds:
3.26 T1300 K⋅:= P11 atm⋅:= CP
7
2R⋅:= CVCPR−:= γ CP
CV
:=
The process occurring in section B is a reversible, adiabatic compression. Let
P final()P
2
=TAfinal()T
A
=
33
(d) Eliminate TATB
+from Eqs. (1) & (3):
⎜
(b) Combine Eqs. (1) & (2) to eliminate the ratio of pressures:
TA425 K⋅:= (guess) TB300 K⋅:=
Given TBT1
TATB
+
2T
1
⋅
⎛
⎜
⎜
⎞
γ1−
γ
⋅=TBFind TB
()
:=
(c) TB325 K⋅:= By Eq. (2),
34
⎜
Solve virial eqn. for final V.
Guess: V2
RT⋅
P2
:=
Given P2V2
⋅
RT⋅1B
V2
+C
V22
+=V2Find V2
()
:= V2241.33 cm3
mol
=
Eliminate P from Eq. (1.3) by the virial equation:
(b) Eliminate dV from Eq. (1.3) by the virial equation in P:
dV R T⋅1−
P2C’+
⎛
⎜
⎝
⎞
⎠
⋅dP⋅=WR−T⋅
P1
P2
P
1−
PC’ P⋅+
⎛
⎜
⎝
⎞
⎠
⌠
⎮
⎮
⌡
d⋅:=
3.30 B 242.5−cm3
mol
⋅:= C 25200 cm6
mol2
⋅:= T 373.15 K⋅:=
P11 bar⋅:= P255 bar⋅:=
B’ B
RT⋅
:= B’ 7.817−10 3−
×1
bar
=
C’ CB
2
−
R2T2
⋅
:= C’ 3.492−10 5−
×1
bar2
=
(a) Solve virial eqn. for initial V.
Guess: V1
RT⋅
P1
:=
Given P1V1
⋅
RT⋅1B
V1
+C
V12
+=V1Find V1
()
:= V130780 cm3
mol
=
35
⎜
(b) B00.083 0.422
Tr1.6
−:= B00.304−=
B10.139 0.172
Tr4.2
−:= B12.262 10 3−
×=
(c) For Redlich/Kwong EOS:
σ1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1
αTr() T
r0.5−
:= Table 3.1 qT
r
()
Ψα Tr
()
⋅
ΩTr
⋅
:= Eq. (3.54)
βTrPr
,
()
ΩPr
⋅
Tr
:= Eq. (3.53)
Note: The answers to (a) & (b) differ because the relations between the two
sets of parameters are exact only for infinite series.
3.32 Tc282.3 K⋅:= T 298.15 K⋅:= Tr
T
Tc
:= Tr1.056=
Pc50.4 bar⋅:= P 12 bar⋅:= Pr
P
Pc
:= Pr0.238=
ω0.087:= (guess)
(a) B 140−cm3
mol
⋅:= C 7200 cm6
mol2
⋅:= VRT⋅
P
:= V 2066 cm3
mol
=
Given PV⋅
RT⋅1B
V
+C
V2
+=
36
Calculate Z Guess: Z 0.9:=
Given Eq. (3.52)
Z1βTrPr
,
()
+qT
r
()
βTrPr
,
()
⋅ZβTrPr
,
()
−
Zεβ TrPr
,
()
⋅+
()
Zσβ TrPr
,
()
⋅+
()
⋅
⋅−=
(e) For Peng/Robinson EOS:
σ12+:= ε 12−:= Ω 0.07779:= Ψ 0.45724:= Table 3.1
Table 3.1
αTr ω,
()
1 0.37464 1.54226ω+ 0.26992ω2
−
()
1T
r
1
2
−
⎛
⎜
⎝
⎞
⎠
⋅+
⎡
⎢
⎣
⎤
⎥
⎦
2
:=
qT
r
()
Ψα Trω,
()
⋅
ΩTr
⋅
:= Eq. (3.54) βTrPr
,
()
ΩPr
⋅
Tr
:= Eq. (3.53)
Calculate Z Guess: Z 0.9:=
Given Eq. (3.52)
Z1βTrPr
,
()
+qT
r
()
βTrPr
,
()
⋅ZβTrPr
,
()
−
Zεβ TrPr
,
()
⋅+
()
Zσβ TrPr
,
()
⋅+
()
⋅
⋅−=
(d) For SRK EOS:
σ1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1
Table 3.1
αTrω,
()
1 0.480 1.574ω+ 0.176ω2
−
()
1T
r
1
2
−
⎛
⎜
⎝
⎞
⎠
⋅+
⎡
⎢
⎣
⎤
⎥
⎦
2
:=
qT
r
()
Ψα Trω,
()
⋅
ΩTr
⋅
:= Eq. (3.54) βTrPr
,
()
ΩPr
⋅
Tr
:= Eq. (3.53)
37
V 1791 cm3
mol
=
Given PV⋅
RT⋅1B
V
+C
V2
+=
(b) B00.083 0.422
Tr1.6
−:= B00.302−=
B10.139 0.172
Tr4.2
−:= B13.517 10 3−
×=
(c) For Redlich/Kwong EOS:
σ1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1
Calculate Z Guess: Z 0.9:=
Given Eq. (3.52)
Z1βTrPr
,
()
+qT
r
()
βTrPr
,
()
⋅ZβTrPr
,
()
−
Zεβ TrPr
,
()
⋅+
()
Zσβ TrPr
,
()
⋅+
()
⋅
⋅−=
3.33 Tc305.3 K⋅:= T 323.15 K⋅:= Tr
T
Tc
:= Tr1.058=
Pc48.72 bar⋅:= P 15 bar⋅:= Pr
P
Pc
:= Pr0.308=
ω0.100:= (guess)
(a) B 156.7−cm3
mol
⋅:= C 9650 cm6
mol2
⋅:= VRT⋅
P
:=
38
Table 3.1
αTrω,
()
1 0.480 1.574ω+ 0.176ω2
−
()
1T
r
1
2
−
⎛
⎜
⎝
⎞
⎠
⋅+
⎡
⎢
⎣
⎤
⎥
⎦
2
:=
qT
r
()
Ψα Trω,
()
⋅
ΩTr
⋅
:= Eq. (3.54) βTrPr
,
()
ΩPr
⋅
Tr
:= Eq. (3.53)
Calculate Z Guess: Z 0.9:=
Given Eq. (3.52)
Z1βTrPr
,
()
+qT
r
()
βTrPr
,
()
⋅ZβTrPr
,
()
−
Zεβ TrPr
,
()
⋅+
()
Zσβ TrPr
,
()
⋅+
()
⋅
⋅−=
(e) For Peng/Robinson EOS:
σ12+:= ε 12−:= Ω 0.07779:= Ψ 0.45724:= Table 3.1
αTr() T
r0.5−
:= Table 3.1 qT
r
()
Ψα Tr
()
⋅
ΩTr
⋅
:= Eq. (3.54)
βTrPr
,
()
ΩPr
⋅
Tr
:= Eq. (3.53)
Calculate Z Guess: Z 0.9:=
Given Eq. (3.52)
Z1βTrPr
,
()
+qT
r
()
βTrPr
,
()
⋅ZβTrPr
,
()
−
Zεβ TrPr
,
()
⋅+
()
Zσβ TrPr
,
()
⋅+
()
⋅
⋅−=
(d) For SRK EOS:
σ1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1
39
Pc37.6 bar⋅:= P 15 bar⋅:= Pr
P
Pc
:= Pr0.399=
ω0.286:=
(guess)
(a) B 194−cm3
mol
⋅:= C 15300 cm6
mol2
⋅:= VRT⋅
P
:= V 1930 cm3
mol
=
Given PV⋅
RT⋅1B
V
+C
V2
+=
(b) B00.083 0.422
Tr1.6
−:= B00.283−=
Table 3.1
αTr ω,
()
1 0.37464 1.54226ω+ 0.26992ω2
−
()
1T
r
1
2
−
⎛
⎜
⎝
⎞
⎠
⋅+
⎡
⎢
⎣
⎤
⎥
⎦
2
:=
qT
r
()
Ψα Trω,
()
⋅
ΩTr
⋅
:= Eq. (3.54) βTrPr
,
()
ΩPr
⋅
Tr
:= Eq. (3.53)
Calculate Z Guess: Z 0.9:=
Given Eq. (3.52)
Z1βTrPr
,
()
+qT
r
()
βTrPr
,
()
⋅ZβTrPr
,
()
−
Zεβ TrPr
,
()
⋅+
()
Zσβ TrPr
,
()
⋅+
()
⋅
⋅−=
3.34 Tc318.7 K⋅:= T 348.15 K⋅:= Tr
T
Tc
:= Tr1.092=
40