termodinamika Chapter 3
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Chapter 3
Problems
Problem 3.1:
Express the volume expansivity β and isothermal compressibility κ as functions of density ρ and its partial
derivatives. The isothermal compressibility coefficient () of water at 50 oC and 1 bar is 44.18∗10−6
bar-1. To what pressure must water be compressed at 50 oC to change its density by 1%? Assume that is
independent of P.
Given Data:
Volume expansivity=β=
1 ∂V
V ∂T
( )
P
Or
β=
1 dV
V dT
( ) →(1)
P
Isothermal Compressibilty=κ=
−1 ∂ V
V ∂P
( )
Or
T
−1 dV
→ ( 2)
V dP T
Temperature=T =50
κ=
Pressure=P1=1 ¯¿
¿¯−1
κ=44.18∗10−6 ¿
( )
(a)
We know that
ρ=
1
1
V
V=
1
ρ
C
Density of water =ρ1=1
P2=?
Solution:
0
kg
m3
ρ2=( 1+1 )
kg
m3
ρ2=1.01
kg
m3
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Put in (1) & (2)
β=ρ
( dTd 1ρ )
β=
P
−ρ dρ
ρ2 dT
( )
β=
P
−1 dρ
ρ dT
( ) Proved
P
Now,
κ=−ρ
( dPd 1ρ )
κ=
T
ρ dρ
ρ2 dP
( )
κ=
T
1 dρ
ρ dP
( ) Proved
T
(b)
As
κ=
1 dρ
ρ dP
( )
κdP=
T
dρ
ρ
Integrating on both sides
P2
ρ2
κ ∫ dP=∫
P1
ρ1
dρ
ρ
P2
ρ2
1
1
κ|P|P =|ln ρ|ρ
κ ( P2 −P1 )=( ln ρ2−ln ρ1 )
κ ( P2 −P1 )=ln
ρ2
ρ1
Putting values
¯¿ ( P2−1 )= ln1.01
1
−6
44.18∗10
¿
¯¿
44.18∗10−6
P2−1=0.00995∗¿
P2=( 225.22+1 ) ¯¿
P2=226.22 ¯¿
Answer
Problem 3.2:
Generally, volume expansivity β and isothermal compressibility κ depend on T and P. Prove that
( ∂∂ Pβ ) =−( ∂T∂κ )
T
P
Solution:
We know that
Volume expansivity=β=
Since
2
β
is very small
1 ∂V
V ∂T
( )
P
August 20,
2013
PROBLEMS
ZAID YAHYA
V=
1 ∂V
∂ β ∂T
( ) → ( 1)
11-CH-74
Isothermal Compressibilty=κ=
P
−1 ∂ V
V ∂P
( )
T
Since κ is very small
V=
−1 ∂ V
∂κ ∂ P
1 ∂V
∂β ∂T
∂V
( ) =−1
(
∂κ ∂ P)
( )
T
P
1 1
∂ β ∂T
1
∂V
( ) ∂V =−1
(
∂κ ∂P )
T
P
T
( ∂T∂ κ ) =( ∂∂ Pβ )
−
P
T
( ∂∂ Pβ ) =−( ∂T∂κ ) Proved
T
P
Problem 3.3:
The Tait equation for liquids is written for an isotherm as:
AP
V =V 0 1−
B+ P
Where V is specific or molar volume, Vo is the hypothetical molar or specific volume at P = 0 and A & B
are positive constant. Find an expression for the isothermal compressibility consistent with this equation.
(
)
Solution:
We Know That,
Isothermal Compressibilty=κ=
−1 ∂ V
V ∂P
( ) → ( 1)
T
Given that
(
V =V 0 1−
AP
B+ P
)
Where
V0 = Hypothetical molar/specific volume at zero pressure, so it is constant
V = Molar/specific volume
Now,
AP
V =V o−
V
B+ P o
V −V o=
−AP
V
B+ P o
V −V o −AP
=
Vo
B +P
Differentiate w.r.t Pressure
1 ∂
−∂ AP
V −V o ) =
(
V o ∂P
∂ P B+ P
(
3
)
[
A ( B+ P )− AP ( 1 )
1 ∂V
−0 =−
V o ∂P
( B+ P )2
(
)
]
August 20,
2013
PROBLEMS
ZAID YAHYA
−1 ∂V
AB+ AP− AP
=
Vo ∂ P
( B+ P )2
( )
11-CH-74
−1 ∂V
AB
=
V o ∂ P ( B+ P )2
( )
Since, Temperature is constant
Therefore,
−1 ∂V
AB
=
V o ∂ P T ( B+ P )2
Or, From (1)
( )
κ=
AB
Proved
( B+P )2
Problem 3.4:
For liquid water the isothermal compressibility is given by:
c
κ=
V ( P+ b )
Where c & b are functions of temperature only if 1 kg of water is compressed isothermally & reversibly
from 1 bar to 500 bars at 60 oC, how much work is required?
At 60 oC, b=2700 bars and c = 0.125 cm3 g-1
Given Data:
Isothermal compressibility=κ=
b=2700 bars
c=0.125 cm3 / g
c
Mass of water=m=1 kg
V ( P+b )
P2=500 bars
Temperature=T =60 0C
Work=W =?
Solution:
We know that
W =−∫ PdV →(1)
κ=
c
→(2)
V ( P+ b )
Also
κ=
−1 dV
V dP
( ) → ( 3)
T
Comparing (2) & (3)
4
Pressure=P1=1 ¯
¿
August 20,
2013
PROBLEMS
ZAID YAHYA
−1 dV
c
=
V dP V ( P+ b )
−dV =
11-CH-74
c dP
P+b
Put in (1)
W =−∫ −P
P2
c dP
P+b
P2
W =c ∫ dP−b c ∫
P1
P1
P2
P2
P
W =c ∫
dP
P+ b
P
P+ b−b
W =c ∫
dP
P+ b
P
1
1
dP
P+b
P2
P2
1
1
[
P2
1
1
P+ b
b
W =c ∫
dP−c ∫
dP
P+ b
P+b
P
P
1
W =c|P|P −bc|ln(P+b)|P
P2
W =c ( P2−P 1) −bc [ ln ( P2+ b ) −ln ( P1 +b ) ]
W =c ( P2−P 1) −bc ln
P2 +b
P1 +b
]
Putting values
3
3
W =0.125
3
cm (
¯ ¯¿ 0.125 cm ∗ln 500+ 2700
∗ 500−1 )−2700
g
g
1+2700
cm3∗¯¿
∗1 m3
g
∗101325 N
3
3
100 cm
∗J
2
1.01325 ¯¿ m
Nm
W =5.16 ¿
W =0.516
cm ∗¯¿
g
3
¯
cm ∗¿
−57.216 ¿
g
W =62.375 ¿
3
cm ∗¯¿
g
W =5.16 ¿
J
Answer
g
Problem 3.5:
Calculate the reversible work done in compressing 1 ft 3 of mercury at a constant temperature of 32F from
1(atm) to 3,000(atm). The isothermal compressibility of mercury at 32F is:
κ/(atm)-1 = 3.9 x 10-6 - 0.1 x10-9P(atm)
Given Data:
Work done=W =?
Volume=V =1 ft
Pressure=P2=3000 atm
Where
5
3
Temperature=T =32 F
κ /atm−1=3.9∗10−6−0.1∗10−9 P(atm)
Pressure=P1=1 atm
August 20,
2013
PROBLEMS
Term,
ZAID YAHYA
3.9*10-6
has
unit
atm-1
of
11-CH-74
&
0.1*10-9
has
units
of
atm-2
Solution:
We know that, work done for a reversible process is
W =−∫ PdV →(1)
Also
κ=
−1 dV
V dP
( )
dV =−κVdP
T
Put in (1),
P2
W =−∫ P (−κVdp )
W =V ∫ κ P dP
P1
P2
P2
W =V ∫ ( 3.9∗10−6−0.1∗10−9 P ) P dP
P1
3000
−6
W =3.9∗10 V
∫
P dP−0.1∗10 V
−6
P1
∫
P2 dP
1
3000
||
P2
W =3.9∗10 V
2
W=
P1
3000
−9
1
−6
P2
W =V ∫ 3.9∗10−6 P dP−V ∫ 0.1∗10−9 P2 dP
1
3000
||
P3
−0 .1∗10 V
3
−9
3
1
−11
3
1.95∗10 ∗1 ft (
3.333∗10 ∗1 ft (
30002−12 ) atm2−
∗ 30003−13 ) atm3
2
atm
atm
W = (17.55−0.8991 ) atm∗ft 3
W =16.65 atm∗ft 3 Answer
Problem 3.6:
Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state
at 1 bar during which the temperature change from 0 oC to 20oC. Determine ΔVt, W, Q, and ΔUt. The
properties for liquid carbon tetrachloride at 1 bar & 0 oC may be assumed independent of temperature: β =
1.2 x 10-3 K-1 Cp = 0.84 kJ kg-1 K-1, ρ = 1590 kg m-3
Given Data:
Mass=m=5 kg
6
Pressure=P=1 ¯¿
Temperature=T 1=0
0
C
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
T 1 =273.15 K
Temperature=T 2=20
T 2 =293.15 K
T 2 =( 20+273.15 ) K
−3
β=1.2∗10 K
0
C
C P=0.84
−1
kJ
kg∗K
ρ=1590
kg
m3
t
Q=?
∆ U =?
Solution:
As
1
ρ
V=
V 1=
1
ρ1
V 1=
1 m3
1590 kg
Also,
we know that
Volume expansivity=β=
1 dV
V dT
( )
βdT =
P
1
dV
V
Integrating on both sides,
T2
V2
β ∫ dT =∫
T1
V1
dV
V
T2
V2
1
1
β|T |T =|lnV |V
β ( T 2−T 1 )=( ln V 2−ln V 1 )
β ( T 2−T 1 )=ln
V2
V1
Putting values
V ∗1590 kg
1.2∗10−3
∗( 293.15−273.15 ) K =ln 2
K
m3
e
0.024
=
V 2∗1590 kg
m3
1.024
∗m3
1590
V 2=
kg
3
V 2=0.000644
m
kg
Now,
∆ V =V 2−V 1
1 m3
∆ V = 0.000644−
1590 kg
(
)
∆ V =15.28∗10−6
m3
kg
Now, for total volume,
t
∆ V =∆ V ∗m
∆ V t =( 15.28∗10−6∗5 )
Now,
We know that for a reversible process,
7
m3
∗kg
kg
t
−5
3
t
∆ V =?
∆ V =7.638∗10 m Answer
W =?
August 20,
2013
PROBLEMS
ZAID YAHYA
Work done=W =−P ∆ V t
W =−1 ¯¿ 7.638∗10
−5
m3∗101325 N
∗J
1.01325 ¯¿ m2
∗1 kJ
Nm
1000 J
11-CH-74
W =−7.638∗10−3 kJ Answer
Now,
For a reversible process at constant pressure,we have
Q=∆ H
Q=m C P ∆ T
Q=5 kg∗0.84
kJ
∗( 293.15−273.15 ) K
kg∗K
Q=84 kJ Answer
Now,
According to first law of thermodynamics,
∆ U t =( 84−7.368∗10−3 ) kJ
∆ U t =Q+W
∆ U t =83.99 kJ Answer
Problem 3.7:
A substance for which k is a constant undergoes an isothermal, mechanically reversible process from
initial
state
(P1,
V1)
to
(P2,
V2),
where
V
is
a
molar
volume.
a) Starting with the definition of k, show that the path of the process is described by
V = A ( T ) exp(−κP)
b) Determine an exact expression which gives the isothermal work done on 1 mol of this constant-k
substance.
Solution:
(a)
We know that
Isothermal compressibilty =κ=
−1 dV
V dP
( )
T
dV
=−κdP
V
Integrating on both sides,
dV
=−κ ∫ dP
V
lnV =−κP+lnA ( T )
∫
Where ln A (T) is constant of integration & A depends on T only
lnV −lnAT =−κP
ln
Taking anti log on both sides,
V
=e−κP
AT
8
V = A ( T ) e−κP
V
=−κP
A (T )
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Or
V = A ( T ) exp (−κP ) Proved
(b)
Work done=W=?
For a mechanically reversible process, we have,
dW =−PdV → ( 1 )
Using,
d ( PV ) =PdV +VdP
−PdV =VdP−d ( PV )
Put in (1)
dW =VdP−d ( PV ) → ( 2 )
We know that
Isothermal compressibilty =κ=
−1 dV
V dP
( )
T
−dV
=VdP
κ
Put in (2)
dW =
−dV
−d ( PV )
κ
Integrating on both sides,
−1
∫ dW = κ ∫ dV −∫ d ( PV )
W=
−1
Δ V −Δ ( PV )
κ
Since volume changes from V1 to V2 & pressure changes from P1 to P2 ,
Therefore,
W=
−1
( V 2−V 1 ) −( P2 V 2−P1 V 1 )
κ
W=
( V 1−V 2)
κ
+ P1 V 1−P2 V 2 Proved
Problem 3.8:
One mole of an ideal gas with C V = 5/2 R, CP = 7/2 R expands from P 1 = 8 bars & T1= 600 K to P2 = 1 bar
by each of the following path:
a) Constant volume
b) Constant temperature
c) Adiabatically
Assuming mechanical reversibility, calculate W, Q, ∆ U, and ∆ H for each of the three processes.
Sketch each path in a single PV diagram.
9
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Given Data:
5
CV = R
2
7
C P= R
2
P1=8 ¯
¿
T 1 =600 K
P2=1 ¯
¿
W =?
Q=?
∆ U =?
∆ H=?
Solution:
(a)
According to first law of thermodynamics,
∆ U =Q+W →(1)
For a constant volume process,
∆ U =CV ∆T
W =0
Put in (1)
Q=∆U =C V ∆ T
Q=∆U =C V ( T 2 −T 1 ) →(2)
For T2 , We know that for an ideal gas
T1 T 2
=
P1 P2
T2=
T1
∗P
P1 2
Put in (2),
8 ¯¿∗1 ¯¿
600 K
T2= ¿
5
Q=∆U = R ( 75−600 ) K
2
T 2 =75 K
Q=∆U =
−5
J
∗8.314
∗525 K
2
mol∗K
Q=∆U =−10.912
Q=∆U =−10912
kJ
Answer
mol
Also
For a mechanically reversible process we have,
∆ H=C P ∆ T
7
∆ H= R ( T 2−T 1 )
2
7
J
∆ H= ∗8.314
∗( 75−600 ) K
2
mol∗K
kJ
∆ H=−15.277
Answer
mol
∆ H=−15277
(b)
For a constant temperature process,
∆ U =0
∆ H=0
We know that at constant temperature, work done is
W =R T 1 ln
W =8.314
10
J
mol
J
1
∗600 K∗ln
mol∗K
8
P2
P1
W =−10373
Now, according to first law of thermodynamics,
J
mol
W =−10.373
kJ
Answer
mol
J
mol
August 20,
2013
PROBLEMS
∆ U =Q+W
ZAID YAHYA
0=Q+ W
11-CH-74
Q=−W
Or
Q=10.373
kJ
Answer
mol
(c)
We know that for an adiabatic process
Q=0
Now, according to first law of thermodynamics,
∆ U =Q+W
∆ U =W →(1)
∆ U =CV ∆T
Put in (1)
W =∆ U=C V ∆ T
W =∆ U=C V ( T 2−T 1 ) →(2)
For T2 , We know that for an adiabatic process
T1 P
( 1−γ )
γ
1
=T 2 P
( 1−γ )
γ
2
T 2 =T 1
P1
P2
( )
( 1−γ )
γ
T 2 =600 K
8
1
()
( 1−1.4 )
1.4
T 2 =331.23 K
Put in (2)
5
W =∆ U= R∗( 331.23−600 ) K
2
W =∆ U=
−5
J
∗8.314
∗268.77 K
2
mol∗K
W =∆ U=5.5864
W =∆ U=−5586.4
J
mol
J
Answer
mol
For a mechanically reversible adiabatic process we have
7
∆ H= R ( T 2−T 1 )
2
J
∆ H=−7.821
Answer
mol
∆ H=C P ∆ T
7
J
∆ H= ∗8.314
∗( 331.23−600 ) K
2
mol∗K
Problem 3.9:
An ideal gas initially at 600k and 10 bar undergoes a four-step mechanically reversible cycle in a closed
system. In step 12, pressure decreases isothermally to 3 bars; in step 23, pressure decreases at constant
volume to 2 bars; in step 34, volume decreases at constant pressure; and in step 41, the gas returns
adiabatically to its initial state. Take CP = (7/2) R and CV = (5/2) R.
a) Sketch the cycle on a PV diagram.
b) Determine (where unknown) both T and P for states 1, 2, 3, and 4.
c) Calculate Q, W, ∆ U, and ∆ H for each step of the cycle.
11
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Given Data:
Initial Temperature=T 1=600 K
Initial Pressure=P1=10 ¯
¿
7
C P= R
2
5
CV = R
2
Solution:
(b)
Step 12, an Isothermal process,
Since
For an isothermal process, temperature is constant
Therefore,
P2=3 ¯¿
T 2 =T 1=600 K
We know that, for an ideal gas
P2 V 2 =R T 2
V 2=
RT 2
P2
¯¿∗1.01325 ¯¿ m2
∗N∗m
101325 N
mol∗K∗3
J
8.314∗J∗600 K
V 2=
¿
V 2=0.0166
m3
mol
Step 23, an Isochoric process,
Since
For an isochoric process, Volume is constant
Therefore,
3
V 3=V 2=0.0166
m
mol
P3=2 ¯¿
We know that, for an ideal gas
P3 V 3 =RT 3
T3=
P3 V 3
R
2 ¯¿ 0.0166 m 3∗mol∗K
∗J
mol∗8.314 J
∗101325 N
N∗m
T3=
1.01325 ¯¿ m2
Step 34, an Isobaric process,
Since
For an isobaric process, pressure is constant
Therefore,
12
T 3 =400 K
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
P4 =P 3=2 ¯¿
For T4 , we will use an adiabatic relation of temperature and pressure
As
T 4 P4
=
T1
P1
( )
R
CP
T 4=T 1
P4
P1
( )
R
CP
2
T 4=600 K∗
10
( )
2∗R
7R
T 4=378.83 K
We know that, for an ideal gas
P4 V 4 =RT 4
V 4=
¯¿∗1.01325 ¯¿ m2
∗Nm
101325 N
mol∗K∗2
J
8.314 J∗378.83 K
V 4=
¿
RT4
P4
V 4 =0.0157
m3
mol
Step 41, an adiabatic process,
Since
Gas returns to its initial state adiabatically
Therefore,
T 1 =600 K
P1=10 ¯¿
We know that, for an ideal gas
P1 V 1=R T 1
V 1=
RT 1
P1
¯¿∗1.01325 ¯¿ m2
∗Nm
101325 N
mol∗K∗10
J
8.314 J∗600 K
V 1=
¿
(c)
Step 12, an Isothermal process,
Since
For an isothermal process, temperature is constant
Therefore
∆ U 12 =0
13
∆ H 12=0
For an isothermal process, we have
V 1=4.988∗10−3
3
m
mol
August 20,
2013
PROBLEMS
Q=−R T 1 ln
P2
P1
ZAID YAHYA
Q=−8.314
J
3
∗600 K∗ln
mol∗K
10
Q=6006
11-CH-74
J
mol
Q=6.006∗103
J
Answer
mol
According to first law of thermodynamics
∆ U 12=Q12 +W 12
0=Q12+W 12
W 12=−Q12
W 12=−6.006∗103
J
Answer
mol
Step 23, an Isochoric process,
Since
For an isochoric process, Volume is constant
Therefore,
W 23=0
At constant volume we have
Q23=∆ U 23=CV ∆ T
Q23=∆ U 23=CV ( T 3−T 2 )
5
∗8.314 J
2
Q23=∆ U 23=
∗(−200 ) K
mol∗K
5
Q23=∆ U 23= R ( 400−600 ) K
2
Q23=∆ U 23=−4157
J
mol
Q23=∆ U 23=−4.157∗10 3
J
Answer
mol
We know that
∆ H 23=C P ∆ T
7
∆ H 23= R ( 400−600 ) K
2
∆ H 23=C P ( T 3−T 2 )
∆ H 23=−5820
J
mol
∆ H 23=−5.82∗103
7
J
∆ H 23= ∗8.314
∗(−200 ) K
2
mol∗K
J
Answer
mol
Step 34, an Isobaric process,
Since
For an isobaric process, pressure is constant
Therefore, at constant pressure we have,
Q34=∆ H 34 =C P ∆ T
7
Q34=∆ H 34 = R ( T 4−T 3 )
2
7
J
Q34=∆ H 34 = ∗8.314
∗( 378.83−400 ) K
2
mol∗K
J
Q34=∆ H 34 =−616
Answer
mol
For an Isobaric process we have
W 34=−R ∆ T
14
W 34=−R ( T 4−T 3 )
W 34=−8.314
J
∗( 378.83−400 ) K
mol∗K
W 34=176
J
Answer
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
We know that,
∆ U 34 =CV ∆T
5
∆ U 34 = R ( T 4 −T 3 )
2
5
J
∆ U 34 = ∗8.314
∗( 378.83−400 ) K
2
mol∗K
J
∆ U 34 =−440
Answer
mol
Step 41, an adiabatic process,
Since
For an adiabatic process there is no exchange of heat
Therefore,
Q41=0
We know that,
5
∆ U 41= R ( T 1−T 4 )
2
5
J
∆ U 41= ∗8.314
∗( 600−378.83 ) K
2
mol∗K
J
∆ U 41=4.597∗103
Answer
mol
∆ U 41=C V ∆ T
∆ U 41=4597
J
mol
We know that
∆ H 41=C P ∆T
7
∆ H 41= R ( T 1−T 4 )
2
7
J
∆ H 41= ∗8.314
∗( 600−378.83 ) K
2
mol∗K
3 J
∆ H 41=6.4358∗10
Answer
mol
∆ H 41=6435.8
J
mol
According to first law of thermodynamics
∆ U 41=Q 41+W 41
∆ U 41=W 41
W 41=4.597∗103
J
Answer
mol
Problem 3.10:
An ideal gas, CP= (5/2) R and CV= (3/2) R is changed from P1 = 1bar and V t1 = 12m3 to P2 = 12 bar and
V t2 = 1 m3 by the following mechanically reversible processes:
a) Isothermal compression
b) Adiabatic compression followed by cooling at constant pressure.
c) Adiabatic compression followed by cooling at constant volume.
d) Heating at constant volume followed by cooling at constant pressure.
e) Cooling at constant pressure followed by heating at constant volume.
Calculate Q, W, change in U, and change in H for each of these processes, and sketch the paths of all
processes on a single PV diagram.
Given Data:
15
August 20,
2013
PROBLEMS
5
C P= R
2
t
V 2=1 m
3
ZAID YAHYA
3
Initial pressure=P1=1 ¯
¿
CV = R
2
Q=?
W =?
∆ H=? ∆ U =?
t
3
11-CH-74
V 1=12 m
Final pressure=P 2=12 ¯¿
12
∗101325 N
1
∗J
1.01325 ¯¿ m2
∗1 kJ
Nm
3
¯
Q=−1 ¿ 12 m ∗ln
1000 J
Q=−2981.88 kJ Answer
Solution:
Since
Temperature=constant
Therefore, for all parts of the problem,
∆ H=0
∆ U =0
(a)
Isothermal compression,
For an isothermal process, we have
Q=−R T 1 ln
P2
P1
Since
For an ideal gas, we have
P1 V 1=R T 1
Therefore,
Q=−P1 V 1 ln
P2
P1
According to first law of thermodynamics
∆ U =Q+W
0=Q+ W
W =−Q
W 12=2981.88 kJ Answer
(b)
Adiabatic compression followed by cooling at constant pressure
Since
For an adiabatic process, there is no exchange of heat
Therefore,
Q=0 Answer
The process completes in two steps
First step, an adiabatic compression to final pressure P 2 , intermediate volume can be given as
16
August 20,
2013
PROBLEMS
'
' γ
V =V 1
P2 ( V ) =P 1 V 1
ZAID YAHYA
P1
P2
( )
11-CH-74
1
γ
For mono atomic gas, we have
γ =1.67
'
3
1
12
( )
V =12 m ∗
1
1.67
V ' =2.71 m3
We know that,
W 1=
P2 V ' −P 1 V 1
γ−1
( 12∗2.71−1∗12 ) ¯¿ m3
∗101325 N
1.67−1
∗J
1.01325 ¯¿ m2
∗1 kJ
Nm
W 1=
1000 J
W 1=3063 kJ →(1)
Second step, cooling at constant pressure P2
We know that, for a mechanically reversible process
m3∗101325 N
∗J
1.01325 ¯¿ m2
∗1kJ
Nm
¯
W 2=−12 ( 1−2.71 )
1000 J
W 2=−P2 ( V 2−V ' )
W 2=2052 kJ →( 2)
Now
W =W 1+ W 2
W = ( 3063+ 2052 ) kJ
W =5115 kJ Answer
(c)
Adiabatic compression followed by cooling at constant volume
Since
For an adiabatic process, there is no exchange of heat
Therefore,
Q=0 Answer
First step, an adiabatic compression to volume V2 , intermediate pressure can be given as
'
γ
P V 2 =P1 V 1
P' =P1
V1
V2
( )
For mono atomic gas, we have
17
γ
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
γ =1.67
12¯ 1.67
P =1
1
'
( )
P' =63.42 ¯¿
We know that,
P' V 2−P 1 V 1
W 1=
γ−1
( 63.42∗1−1∗12 ) ¯¿ m3
∗101325 N
1.67−1
∗J
1.01325 ¯¿ m 2
∗1 kJ
Nm
W 1=
1000 J
W 1=7674.76 kJ
Second step, cooling at constant Volume,
Therefore, No work will be done
W 2=0
Now
W =W 1+ W 2
W = (7674.76+ 0 ) kJ
W =7674.76 kJ Answer
(d)
Heating at constant volume followed by cooling at constant pressure
The process completes in two steps
Step 1, Heating at constant volume to P2
Therefore no work will be done
W 1=0
Step 2, Cooling at constant pressure P2 To V2
We know that, for a mechanically reversible process
3
W 2=−P2 ∆ V
W 2=−P2 ( V 2−V 1 )
m ∗101325 N
∗J
2
1.01325 ¯¿ m
∗1 kJ
Nm
¯ )
W 2=−12 ( 1−12
1000 J
Now
W =W 1+ W 2
W = ( 0+13200 ) kJ
W =13200 kJ Answer
According to first law of thermodynamics
∆ U =Q+W
(e)
18
0=Q+ W
Q=−W
Q=−13200 kJ Answer
W 2=13200 kJ
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Cooling at constant pressure followed by heating at constant volume
The process completes in two steps
Step 1, Cooling at constant Pressure P1 to V2
Therefore, for a mechanically reversible process
W 1=−P1 ( V 2−V 1 )
W 1=−P1 ∆ V
m3∗101325 N
∗J
1.01325 ¯¿ m2
∗1 kJ
Nm
¯ )
W 1=−1 ( 1−12
1000 J
W 1=1100 kJ
Step 1, Heating at constant Volume V2 to pressure P2
Therefore no work will be done
W 2=0
Now
W =W 1+ W 2
W = (1100 +0 ) kJ
W =1100 kJ Answer
According to first law of thermodynamics
∆ U =Q+W
0=Q+ W
Q=−W
Q=−1100 kJ Answer
Problem 3.11:
dT
characterizes the local variation of temperature with elevation in the
dz
earth's atmosphere. Atmospheric pressure varies with elevation according to the hydrostatic formula,
dP
=−M ρg
dz
The environmental lapse rate
Where M is a molar mass, ρ is molar density and g is the local acceleration of gravity. Assume that the
atmosphere is an ideal gas, with T related to P by the polytropic formula equation (3.35 c). Develop an
expression for the environmental lapse rate in relation to M, g, R, and δ.
Solution:
Given that
dP
=−M ρg → ( 1 )
dz
The polytropic relation is
TP
19
1−δ
δ
=Constant
August 20,
2013
PROBLEMS
ZAID YAHYA
Or
TP
1−δ
δ
=T o P
1−δ
δ
o
Where
To =Temperature at sea level, so it is constant
Po = Pressure at sea level, so it is constant
T
P
δ−1
δ
T
To
( )
=
To
P
δ
δ−1
δ−1
δ
o
P
=
Po
T
P
=
T o Po
( )
δ−1
δ
T
P=Po
To
( )
δ
δ −1
→(a)
P=
Po
To
δ
δ−1
δ
∗T δ −1
Differentiate w.r.t to Temperature on both sides
Po
δ
δ −1
Po
∗δ
dP T o
=
∗T
dT
δ−1
1
δ−1
δ
δ−1
∗δ
1
To
δ −1
dP=
∗T
dT →(2)
δ−1
We know that, for an ideal gas
ρ=
P
RT
Where
R=Specific gas constant=R ' /M
Put (a) in above equation
ρ=
1
T
∗Po
RT
To
( )
Put in (1)
dP
g∗1
T
=−M
Po
dz
RT
To
( )
Put (2) in above
20
δ
δ −1
dP=−M
g∗1
T
Po
RT
To
( )
δ
δ −1
∗dz
δ
δ −1
11-CH-74
August 20,
2013
PROBLEMS
Po
δ
δ−1
ZAID YAHYA
−δ −1
∗M g
δ
∗Po
δ
R
∗T o δ−1
Po
∗δ
1
To
g∗1
T
∗T δ −1 dT =−M
Po
δ −1
RT
To
( )
δ
δ−1
∗dz
To
To
dT
=
dz
−δ −1
∗M g
δ
∗Po
δ
R
∗T o δ−1
Po
dT
=
dz
11-CH-74
δ
δ−1
δ
δ−1
δ
∗T δ−1
1
T∗T δ−1
δ
∗T δ−1
δ
T δ−1
−δ
∗M g
dT δ−1
=
Proved
dz
R
Problem 3.12:
An evacuated tank is filled with gas from a constant pressure line. Develop an expression relating the
temperature of the gas in the tank to temperature T’ of the gas in line. Assume that gas is ideal with
constant heat capacities, and ignore heat transfer between the gas and the tank. Mass and energy balances
for this problem are treated in Ex. 2.13.
Solution:
Choose the tank as the control volume. There is no work, no heat transfer & kinetic & potential energy changes are
assumed negligible.
Therefore, applying energy balance
d ( mU )tank
+∆ ( Hm ) =0
dt
d ( mU )tank
+ H '' m '' −H ' m ' =0
dt
Since
Tank is filled with gas from an entrance line, but no gas is being escaped out,
Therefore,
d ( mU )tank
+0− H ' m' =0
dt
21
d ( mU )tank
−H ' m' =0→( 1)
dt
August 20,
2013
PROBLEMS
ZAID YAHYA
Where prime (‘) denotes the entrance stream
Applying mass balance
m' =
d mtank
→ ( 2)
dt
Combining equation (1) & (2)
d ( mU )tank
d mtank
−H '
=0
dt
dt
1
d ( mU )tank−H ' d mtank }=0
{
dt
'
d ( mU )tank =H d mtank
Integrating on both sides
m2
m2
∫ d ( mU )tank=H ' ∫ d mtank
m1
∆ ( mU )tank =H ' ( m2−m1 )
m1
'
m2 U 2 −m1 U 1=H ( m2−m1 )
Because mass in the tank initially is zero, therefore
m1=0
m2 U 2 =H ' m2
'
U 2=H →(3)
We know that
U=C V T
U 2=C V T 2 → ( a )
Also
'
'
H =C P T → ( b )
Put (a) & (b) in (3)
CV T =C P T '
T=
CP '
T
CV
Since heat capacities are constant, therefore
γ=
CP
CV
T =γ T ' Proved
Problem 3.14:
22
11-CH-74
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
A tank of 0.1-m3 volume contains air at 25 oC and 101.33 kPa. The tank is connected to a compressed-air
line which supplies air at the constant conditions of 45oC and 1,500 kPa. A valve in the line is cracked so
that air flows slowly into the tank until the pressure equals the line pressure. If the process occurs slowly
enough that the temperature in the tank remains at 25 oC, how much heat is lost from the tank? Assume
air to be an ideal gas for which CP = (7/2) R and CV = (5/2) R
Given Data:
Volume=V =0.1 m
3
T 1 =25 o. C =298 K
7
C P= R
2
Heat lost=Q=?
P1=101.33 kPa
T 2 =45 o.C =318 K
5
CV = R
2
Solution:
According to first law of thermodynamics
∆ U =Q+W →(1)
Since
∆ H=∆ U +∆ ( PV )
∆ U =∆ H−∆ ( PV )
∆ U =∆ H−P ∆ V −V ∆ P→ ( a )
Also, we know that
W =−P ∆ V → ( b )
Put (a) & (b) in (1)
∆ H−P ∆V −V ∆ P=Q−P ∆ V
Also, we have
∆ H=nC P ∆T
∆ H=nC P ( T 2−T 1 )
Put in (2)
n C P ( T 2−T 1 ) −V ∆ P=Q →(3)
For “n”
We know that for an ideal gas,
PV =nRT
Initial number of moles of gas can be obtained as,
23
∆ H−V ∆ P=Q → ( 2 )
P2=1500 kPa
August 20,
2013
P1 V =n1 R T 1
PROBLEMS
n1=
ZAID YAHYA
11-CH-74
P1 V
RT1
The final number of moles of gas at temperature T1 are
P2 V =n2 R T 1
n2=
P2 V
RT1
Now, Applying molar balance
n=n1−n2
n=
P1 V P2 V
−
R T1 R T1
n=
( P1−P 2) V
RT1
Put in (3)
( P 1−P2 ) V
R T1
( P1−P2 ) V
C ( T 2−T 1 )−V ∆ P=Q
P
( P1−P2 ) V
T1
2
RT1
2
∗7
R∗( T 2−T 1 )−V ∆ P=Q
∗7
∗( T 2 −T 1 ) −V ( P 2−P1 )=Q
( 101.33−1500 ) kPa∗0.1 m3
∗7
298 K
3
∗( 318−298 ) K −0.1 m ( 1500−101.33 ) kPa=Q
2
kPa∗1 kN
∗1 kJ
2
3 1 kPa∗m
Q=−172.717 m
1 kNm
Q=−172.717 kJ Answer
Problem 3.17:
A rigid, no conducting tank with a volume of 4 m 3 is divided into two unequal parts by a thin membrane.
One side of the membrane, representing 1/3 of the tank, contains nitrogen gas at 6 bars and 100 oC, and
the other side, representing 2/3 of the tank, is evacuated. The membrane ruptures and the gas fills the
tank.
a) What is the final, temperature of the gas? How much work is done? Is the process reversible?
b) Describe a reversible process by which the gas can be returned to its initial state, How much work
is done
Assume nitrogen is an ideal gas for which CP = (7/2) R & CV = (5/2) R
Given Data:
24
August 20,
2013
PROBLEMS
ZAID YAHYA
3
Volume of the tank=V 1=4 m
V 3=
V 1∗2 8 3
= m
3
3
V 2=
V 1∗1 4 3
= m
3
3
Pressure=P2=6 ¯¿
11-CH-74
o
Temperature=T 1=100 . C
Solution:
(a)
Finaltemperature=T 2=?
According to first law of thermodynamics
∆ U =Q+W
Since
No work is done & no heat is transferred
Therefore
Q=W =0
mC V ∆ T =0
∆ U =0
∆ T =0
T 2 −T 1=0
T 2 =T 1
T 2 =100℃ Answer
No, process is not reversible
(b)
Since
Therefore, the process is isothermal
For an isothermal process we have
W =−R T 2 ln
V2
V1
As, for an ideal gas
P2 V 2 =R T 2
W =−P2 V 2 ln
V2
V1
4
4
m3∗101325 N
kJ
W =−6 ¯¿ m3 ln
W =8.788 ¯¿
∗1
2
3
3∗4
1000 Nm
1.01325 ¯¿ m
W =878.8 kJ Answer
Problem 3.18:
An ideal gas initially at 30 0C and 100 kPa undergoes the following cyclic processes in a closed system:
a) In mechanically reversible processes, it is first compressed adiabatically to 500 kPa then cooled at
a constant pressure of 500 kPa to 30 0C and finally expanded isothermally to its original state
25
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
b) The cycle traverses exactly the same changes of state but each step is irreversible with an efficiency
of 80% compared with the corresponding mechanically reversible process NOTE: the initial step
can no longer be adiabatic
Find Q W ∆ U and ∆ H for each step of the process and for the cycle Take C p = (7/2) R and CV =
(5/2) R
Given Data:
T 1 =30 0.C
T 1 =303.15 K
P1=100 kPa
Q=?
W =?
∆ U =?
∆ H=?
7
C P= R
2
5
CV = R
2
lution:
(a)
P2=500 kPa
1) Adiabatic Compression from point 1 to point 2
Q12=0
Now, from first law of thermodynamics,
∆ U 12=Q12 +W 12
∆ U 12=W 12
W 12=∆ U 12=C V ∆ T 12
5
W 12=∆ U 12= R ( T 2−T 1 ) → ( 1 )
2
For ‘T2’
We know that
T 2 P2
=
T 1 P1
( )
γ −1
γ
T 2 =T 1
P2
P1
( )
γ −1
γ
T 2 =303.15 K
500
100
( )
1.4−1
1.4
T 2 =480.13 K
Put in (1)
5
J
kJ
kJ
W 12=∆ U 12= ∗8.314
( 480.13−303.15 ) K∗1
W 12=∆ U 12=3.679
2
mol∗K
1000 J
mol
Also, we have
∆ H 12=C P ( T 2−T 1)
7
J
kJ
∆ H 12= ∗8.314
( 480.13−303.15 ) K∗1
2
mol∗K
1000 J
2) Cooling at constant pressure from point 2 to point 3
Therefore at constant pressure we have,
Q23 =∆ H 23=C P ∆T 23
26
7
Q23=∆ H 23= R ( T 3−T 2 )
2
∆ H 12=5.15
kJ
mol
So
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Here
T 3 =303.15 K
7
J
kJ
kJ
Q23=∆ H 23= ∗8.314
( 303.15−480.13 ) K∗1
Q23=∆ H 23=−5.15
2
mol∗K
1000 J
mol
Also, we have
∆ U 23 =C V ( T 3−T 2)
5
J
kJ
∆ U 23= ∗8.314
( 303.15−480.13 ) K∗1
2
mol∗K
1000 J
kJ
∆ U 23=−3.679
mol
Now, from first law of thermodynamics,
∆ U 23=Q23 +W 23
W 23=∆U 23−Q23
W 23=−3.679+5.15
W 23=1.471
3) Isothermal expansion from point 3 to point 1
Since for an isothermal process temperature remains constant
Therefore,
∆ U 31 =∆ H 31=0
Here
P3=P2=500 kPa
For an Isothermal process we have
500
∗1 kJ
P3
J
100
W 31=−RT 3 ln W 31=−8.314
∗303.15 K∗ln
P1
mol∗K
1000 J
W 31=−4.056
kJ
mol
According to first law of thermodynamics
∆ U 31 =Q31 +W 31
0=Q31+W 31
Q31=−W 31
Q31=4.056
kJ
mol
For the complete cycle,
Q=Q12 +Q23+ Q31
27
Q=0−5.15+4.056
Q=−1.094
kJ
Answer
mol
kJ
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
W =W 12+W 23 +W 31
W =3.679+ 1.471−4.056
W =1.094
∆ H=∆ H 12+ ∆ H 23+ ∆ H 31
∆ U =∆ U 12 +∆ U 23 +∆ U 31
11-CH-74
kJ
Answer
mol
∆ H=5.15−5.15+0
∆ U =3.679−3.679+0
kJ
Answer
mol
kJ
∆ U =0
Answer
mol
∆ H=0
(b)
If each step that is 80% accomplishes the same change of state then values of
in part (a) but values of Q & W will change.
1. Adiabatic Compression from point 1 to point 2
W 12=
W 12
0.8
W 12=
3.679
0.8
W 12=4.598
∆U
&
∆H
will remain same as
kJ
mol
According to first law of thermodynamics
∆ U 12 =Q12 +W 12
3.679
kJ
kJ
=Q12+ 4.598
mol
mol
Q12=3.679
kJ
kJ
−4.598
mol
mol
Q12=−0.92
kJ
mol
2. Cooling at constant pressure from point 2 to point 3
W 23=
W 23
0.8
W 23=
1.471
0.8
W 23=1.839
kJ
mol
According to first law of thermodynamics
∆ U 23 =Q23 +W 23
−3.679
kJ
kJ
=Q23 +1.839
mol
mol
Q23=−3.679
kJ
kJ
−1.839
mol
mol
kJ
∗0.8
mol
W 31=3.245
3. Isothermal expansion from point 3 to point 1
Since initial step can no longer be adiabatic , therefore
W 31=W 31∗0.8
W 31=−4.056
According to first law of thermodynamics
∆ U 31 =Q31 +W 31
28
Q31=−W 31+0
kJ
mol
Q23=−5.518
kJ
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
Q31=3.245
11-CH-74
kJ
mol
For the complete cycle,
Q=Q12 +Q 23+Q31
W =W 12 +W 23+W 31
Q=−3.193
Q=−0.92−5.518+ 3.245
W =4.598+1.839−3.245
W =3.192
kJ
Answer
mol
kJ
Answer
mol
Problem 3.19:
One cubic meter of an ideal gas at 600 K and 1,000 kPa expands to five times its initial volume as follows:
a) By a mechanically reversible, isothermal process
b) By a mechanically reversible adiabatic process
c) By adiabatic irreversible process in which expansion is against a restraining pressure of 100 kPa
For each case calculate the final temperature, pressure and the work done by the gas, Cp=21 J mol-1K-1.
Given Data:
V 1=1 m3
T 1 =600 K
P1=1000 kPa
V 2=5 V 1
V 2=5 m3
C P=21
J
mol K
W =?
Solution:
We know that,
C P−C V =R
CV =C P −R
CV =( 21−8.314 )
J
mol∗K
CV =12.686
As
γ=
CP
CV
γ =1.6554
(a)
Since, for an isothermal process
Temperature remains constant, therefore
T 2 =T 1=600 K Answer
29
For an ideal gas we have
J
mol∗K
CV =?
T 2 =?
P2=?
August 20,
2013
PROBLEMS
P1 V 1 P 2 V 2
=
T1
T2
ZAID YAHYA
P1 V 1
∗T 2
T1
P 2=
V2
11-CH-74
1000 kPa∗1 m3
∗600 K
600 K
P 2=
5 m3
P2=200 kPa Answer
We know that, for an isothermal process
W =−R T 1 ln
V2
V1
Since
P1 V 1=R T 1
Therefore,
5
∗N
1
∗J
Pa∗m2
3
W =−1000 kPa∗1m ln
Nm
V2
W =−P1 V 1 ln
V1
W =−1609.43 kJ Answer
(b)
We know that, for an adiabatic process
P1 V 1γ =P2 V 2γ
V1
V2
γ
( )
P2 V 2
T2=
∗T
P1 V 1 1
N
∗J
( 69.65∗5−1000∗1 ) kPa∗m Pa∗m2
W=
1.6554−1
Nm
3
Pr=100 kPa
Since, for an adiabatic process
Q=0
30
P2=69.65 kPa Answer
69.65 kPa∗5 m3
T2=
∗600 K
1000 kPa∗1 m3
(c)
1.6554
T 2 =208.95 K Answer
For an adiabatic process work done is
P V −P1 V 1
W= 2 2
γ −1
1
5
()
P2=1000 kPa∗
For an ideal gas we have
P1 V 1 P2 V 2
=
T1
T2
P2=P1
According to first law of thermodynamics
W =−994.43 kJ Answer
August 20,
2013
PROBLEMS
ZAID YAHYA
∆ U =Q+W
∆ U =W
kPa∗m3∗N
∗J
Pa∗m2
∆ U =W =−100 ( 5−1 )
Nm
∆ U =W =−Pr dV
∆ U =−400 kJ
T2=
n CV ∆T =−400 kJ
n CV ( T 2−T 1 )=−400 kJ
−400 kJ
+T 1 → ( 1 )
n CV
n=
P1 V 1
RT1
1000 kPa∗1 m3∗mol∗K
∗kN
8.314 J∗600 K
∗kJ
kPa∗m2
n=
kNm
n=0.2005 mol
Put in (1)
−400 kJ∗mol∗K
∗1000 J
0.2005 mol∗12.686 J
T2=
+600 K
1 kJ
∆ U =W =−Pr ( V 2−V 1)
For an ideal gas we have,
P1 V 1=nR T 1
11-CH-74
T 2 =−157.26 K +600 K
T 2 =442.74 K Answer
For an ideal gas we have
P1 V 1 P2 V 2
=
T1
T2
P1 V 1
∗T 2
T1
P2=
V2
1000 kPa∗1 m3
∗442.74 K
600 K
P2=
5 m3
P2=147.58 kPa Answe r
Problem 3.20:
One mole of air, initially at 150 0C and 8 bars undergoes the following mechanically reversible changes. It
expands isothermally to a pressure such that when it is cooled at constant volume to 50 0C its final
pressure is 3 bars. Assuming air is an ideal gas for which C P = (7/2) R and CV = (5/2) R, calculate W, Q,
∆ U , and ∆ H
Given Data:
Mole of air=n=1mol
Initial Temperature=T 1=150 0.C =423.15 K
0
Finaltemperature =T 3=50 .C =323.15 K
Final pressure=P 3=3 ¯
¿
Initial pressure=P1=8 ¯¿
7
C P= R
2
5
CV = R
2
Solution:
Since process is reversible
Two different steps are used in this case to reach final state of the air.
31
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Step 12:
For step 12 temperatures is constant,
T 1 =T 2
Therefore
∆ U 12 =∆ H 12=0
For an isothermal process we have
W 12=R T 1 ln
V1
V2
As
V 2=V 3
W 12=R T 1 ln
V1
→(1)
V3
We know that
P1 V 1 P3 V 3
=
T1
T3
V 1 P3 ¿T 1
=
V 3 T 3∗P1
8.314 J∗423.15 K
∗1 kJ
mol∗K
3∗423.15
W 12=
∗ln
1000 J
8∗323.15
kJ
W 12=−2.502
mol
P ¿T
W 12=R T 1 ln 1 3
T 1∗P 3
According to first law of thermodynamics
∆ U 12=Q12 +W 12
0=Q12 +W 12
Q12=−W 12
Q12=2.502
kJ
mol
Step 23:
For step 23 volume is constant,
Therefore,
W 23=0
According to first law of thermodynamics
∆ U 23=Q23 +W 23
∆ U 23 =Q23 +0
Q 23=∆ U 23
Q23=∆ U 23 =CV ∆ T
Q23=∆ U 23=CV ( T 3−T 2 )
5
Q23=∆ U 23= R ( 323.15−423.15 ) K
2
J
∗1 kJ
5
mol∗K
kJ
Q23=∆ U 23= ∗8.314
(−100 ) K Q23=∆U 23 =−2.0785
2
1000 J
mol
32
W e know that
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
J
∗1 kJ
7
mol∗K
∆ H 23= ∗8.314
( 423.15−323.15 ) K
2
1000 J
kJ
∆ H 23=2.91
mol
∆ H 23 =C P ∆ T
∆ H 23 =C P ( T 3−T 2 )
For the complete cycle,
Work=W =W 12 +W 23
Q=Q12 +Q23
W = (−2.502+0 )
Q=( 2.502−2.0785 )
∆ U =∆ U 12 +∆ U 23
∆ H=∆ H 12+ ∆ H 23
kJ
mol
kJ
mol
W =−2.502
Q=0.424
kJ
Answe r
mol
kJ
Answer
mol
∆ U =( 0−2.0785 )
kJ
mol
∆ U =−2.0785
∆ H= ( 0−2.91 )
kJ
mol
∆ H=−2.91
kJ
Answe r
mol
kJ
Answe r
mol
Problem 3.21:
An ideal gas flows through a horizontal tube at steady state. No heat is added and no shaft work is done.
The cross-sectional area of the tube changes with length, and this causes the velocity to change. Derive an
equation relating the temperature to the velocity of the gas. If nitrogen at 150 0C flows past one section of
the tube with a velocity of 2.5 m/s, what is the temperature at another section where its velocity is 50 m/s?
Let CP = (7/2) R
Given Data:
0
Temperature=T 1=150 . C=423.15 K
Velocity=u 1=2.5
m
sec
Molecualr weight of Nitrogen=28
Solution:
Applying energy balance for steady state flow process
∆ H+
∆ u2
+ g ∆ z=Q+W S
2
Since
∆ z=W S=Q=0
33
T 2 =?
g
mol
u2=50
m
sec
7
C P= R
2
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Therefore,
∆ u2
∆H+
=0
2
−∆u 2
CP ∆ T =
2
2
−u −u1
C P ( T 2 −T 1 ) = 2
2
2
−u22−u12
T2=
+ T1
2C P
−( 50 2−2.52 )∗2∗m 2∗mol∗K
∗28 g Nitrogen
2∗7∗8.314 J∗sec 2
∗J
1 mol Nitrogen
T 2 =−1.199 K + 423.15 K
∗N∗sec 2
N∗m
∗1 kg
kg∗m
T2=
+ 423.15 K
1000 g
T 2 =421.95 K
Answe r
T 2 =( 421.95−273.15 ) 0.C
T 2 =148.8 0.C
Problem 3.22:
One mole of an ideal gas, initially at 30 0C and 1 bar, is changed to 130 0C and 10 bars by three different
mechanically reversible processes:
a) The gas is first heated at constant volume until its temperature is 130 0C; then it is compressed
isothermally until its pressure is 10 bar
b) The gas is first heated at constant pressure until its temperature is 130 0C; then it is compressed
isothermally to 10 bar
c) The gas is first compressed isothermally to 10 bar; then it is heated at constant pressure to 130 0C
Calculate Q, W, ∆ U ∧∆ H in each case. Take CP = (7/2) R and CV = (5/2) R. alternatively, take CP =
(5/2) R and CV = (3/2) R
Given Data:
0
T 1 =30 .C
T 1 =( 30+273.15 ) K
T 3 =403.15 K
P1=1 ¯
¿
T 1 =303.15 K
P3=10 ¯¿
Q=?
W =?
Solution:
7
C P= R
2
5
CV = R
2
Each part consist of two steps, 12 & 23
For the overall processes
∆ U =∆ U 12=∆U 23=CV ∆T
34
5
∆ U =∆ U 12=∆U 23= R ( T 3 −T 1 )
2
0
T 2 =130 .C
∆ U =?
T 3 =( 130+273.15 ) K
∆ H=?
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
5
J
K∗1 kJ
∆ U =∆ U 12=∆U 23= ∗8.314
( 403.15−303.15 )
2
mol∗K
1000 J
∆ U =∆ U 12=∆U 23=2.079
kJ
→(a) Answe r
mol
Now
∆ H=∆ H 12 =∆ H 23=C P ∆ T
7
∆ H=∆ H 12 =∆ H 23= R ( T 2−T 1 )
2
7
J
kJ
∆ H=∆ H 12=∆ H 23= ∗8.314
( 403.15−303.15 ) K∗1
2
mol∗K
1000 J
kJ
∆ H=∆ H 12=∆ H 23=2.91
→ ( b ) Answer
mol
(a)
Step 12:
For step “12” volume is constant
Therefore
W 12=0
Here
T 2 =T 3
According to first law of thermodynamics
∆ U 12 =Q12 +W 12
∆ U 12 =Q12
Q12=∆ U 12=C V ∆ T
Also we have
∆ H 12=2.91
kJ
[ ¿( b)]
mol
Step 23:
Since for step “23” process is isothermal
Therefore
∆ U 23 =∆ H 23=0
Here
35
Q12=∆ U 12=2.079
kJ
[¿ ( a ) ]
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
T 2 =T 3
Now, intermediate pressure can be calculated as
P 1 P2
=
T1 T 2
P 2=
P1
∗T
T1 2
1
¯¿
∗403.15 K
303.15 K
P2=¿
For an isothermal process we have
W 23=R T 2 ln
P3
P2
W 23=8.314
J
K∗1 kJ
10
∗403.15
∗ln
mol∗K
1000 J
1.329
W 23=6.764
P2=1.329 b ar
kJ
mol
According to first law of thermodynamics
∆ U 23=Q23 +W 23
0=Q23+W 23
Q23=−W 23
Q23=−6.764
kJ
mol
For the complete cycle,
Work=W =W 12 +W 23
Q=Q 12 +Q23
W = ( 0+6.764 )
Q=( 2.079−6.764 )
∆ U =∆ U 12 +∆ U 23
kJ
mol
kJ
mol
W =6.764
Q=−4.685
kJ
Answe r
mol
∆ U =( 2.079+0 )
kJ
mol
∆ U =2.079
∆ H= ( 2.91+0 )
kJ
mol
∆ H=2.91
∆ H=∆ H 12+ ∆ H 23
kJ
Answe r
mol
kJ
Answe r
mol
kJ
Answe r
mol
(b)
Step 12:
For step “12” volume is constant
Therefore, at constant pressure we have
Q12 ¿ ∆ H 12=2.91
kJ
mol
[ ¿(b) ]
Also,
∆ U 12 =2.079
36
kJ
mol
[ ¿( a) ]
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
According to first law of thermodynamics
∆ U 12=Q12 +W 12
W 12=∆ U 12−Q 12
W 12=( 2.079−2.91 )
kJ
mol
W 12=−0.831
Step 23:
Since for step “23” process is isothermal ( T = Constant)
Therefore
∆ U 23 =∆ H 23=0
Here
T 2 =T 3 ∧P 1=P2
For an isothermal process we have
W 23=R T 2 ln
P3
P2
W 23=8.314
J
K∗1 kJ
10
∗403.15
∗ln
mol∗K
1000 J
1
W 23=7.718
kJ
mol
According to first law of thermodynamics
∆ U 23=Q23 +W 23
0=Q23+W 23
Q23=−W 23
Q23=−7.718
kJ
mol
For the complete cycle,
Work=W =W 12 +W 23
Q=Q 12 +Q23
W = (−0.831+7.718 )
Q=( 2.91−7.718 )
∆ U =∆ U 12 +∆ U 23
∆ H=∆ H 12+ ∆ H 23
kJ
mol
kJ
mol
Q=−4.808
kJ
mol
∆ U =2.079
∆ H= ( 2.91+0 )
kJ
mol
∆ H=2.91
Since for step “12” process is isothermal ( T = Constant)
Therefore
∆ U 12 =∆ H 12=0
P2=P3
37
kJ
Answer
mol
kJ
Answe r
mol
∆ U =( 2.079+0 )
(c)
Here
W =6.887
kJ
Answe r
mol
kJ
Answe r
mol
kJ
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
For an isothermal process we have
W 12=R T 1 ln
P2
P1
W 12=8.314
W 12=5.8034
11-CH-74
J
K∗1 kJ
10
∗303.15
∗ln
mol∗K
1000 J
1
kJ
mol
According to first law of thermodynamics
∆ U 12=Q12 +W 12
0=Q12+W 12
Q12=−W 12
Q12=−5.8034
kJ
mol
Step 23:
For step “23” volume is constant
Therefore, at constant pressure we have
Q23=∆ H 23=2.91
kJ
[ ¿ (b)]
mol
Here
T 2 =T 3
Now
∆ U 23=2.079
kJ
[ ¿(a)]
mol
According to first law of thermodynamics
∆ U 23=Q23 +W 23
W 23=∆U 23−Q23
W 23=( 2.079−2.91 )
kJ
mol
W 23=−0.831
For the complete cycle,
Work=W =W 12 +W 23
Q=Q 12 +Q23
Q=(−5.8034+2.91 )
∆ U =∆ U 12 +∆ U 23
∆ H=∆ H 12+ ∆ H 23
38
W = (5.8034−0.831 )
kJ
mol
kJ
mol
W =4.972
Q=−2.894
kJ
Answer
mol
kJ
Answe r
mol
∆ U =( 0+2.079 )
kJ
mol
∆ U =2.079
∆ H= ( 0+2.91 )
kJ
mol
∆ H=2.91
kJ
Answe r
mol
kJ
Answe r
mol
kJ
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Solution:
5
C P= R
2
3
CV = R
2
Each part consist of two steps, 12 & 23
For the overall processes
∆ U =∆ U 12=∆U 23=CV ∆T
3
∆ U =∆ U 12=∆U 23= R ( T 3 −T 1 )
2
3
J
K∗1 kJ
∆ U =∆ U 12=∆U 23= ∗8.314
( 403.15−303.15 )
2
mol∗K
1000 J
∆ U =∆ U 12=∆U 23=1.247
kJ
→(a) Answe r
mol
Now
∆ H=∆ H 12 =∆ H 23=C P ∆ T
5
∆ H=∆ H 12=∆ H 23= R ( T 2−T 1 )
2
5
J
kJ
∆ H=∆ H 12=∆ H 23= ∗8.314
( 403.15−303.15 ) K∗1
2
mol∗K
1000 J
kJ
∆ H=∆ H 12=∆ H 23=2.079
→ ( b ) Answe r
mol
(a)
Step 12:
For step “12” volume is constant
Therefore
W 12=0
Here
T 2 =T 3
According to first law of thermodynamics
∆ U 12=Q12 +W 12
39
∆ U 12 =Q12
Q12=∆ U 12=C V ∆ T
Q12=∆ U 12=1.247
kJ
[ ¿ (a)]
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Also we have
∆ H 12=2.079
kJ
[ ¿ (b)]
mol
Step 23:
Since for step “23” process is isothermal
Therefore
∆ U 23 =∆ H 23=0
Here
T 2 =T 3
Now, intermediate pressure can be calculated as
P 1 P2
=
T1 T 2
P1
P2= ∗T 2
T1
1
¯¿
∗403.15 K
303.15 K
P2=¿
For an isothermal process we have
W 23=R T 2 ln
P3
P2
W 23=8.314
J
K∗1 kJ
10
∗403.15
∗ln
mol∗K
1000 J
1.329
W 23=6.764
P2=1.329 b ar
kJ
mol
According to first law of thermodynamics
∆ U 23=Q23 +W 23
0=Q23+W 23
Q 23 =−W 23
Q23=−6.764
kJ
mol
For the complete cycle,
Work=W =W 12 +W 23
Q=Q12 +Q23
40
W = ( 0+6.764 )
Q=( 1.247−6.764 )
kJ
mol
kJ
mol
W =6.764
Q=−5.516
kJ
Answe r
mol
kJ
Answe r
mol
∆ U =∆ U 12 +∆ U 23
∆ U =( 1.247+0 )
kJ
mol
∆ U =1.247
kJ
Answer
mol
∆ H=∆ H 12+ ∆ H 23
∆ H= ( 2.079+ 0 )
kJ
mol
∆ H=2.079
kJ
Answe r
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
(b)
Step 12:
For step “12” volume is constant
Therefore, at constant pressure we have
Q12 ¿ ∆ H 12 =2.079
kJ
mol
[ ¿(b) ]
Also,
∆ U 12=1.247
kJ
mol
[ ¿( a) ]
According to first law of thermodynamics
∆ U 12 =Q12 +W 12
W 12=∆ U 12−Q 12
W 12=( 1.247−2.079 )
kJ
mol
W 12=−0.832
Step 23:
Since for step “23” process is isothermal ( T = Constant)
Therefore
∆ U 23 =∆ H 23=0
Here
T 2 =T 3 ∧P 1=P2
For an isothermal process we have
W 23=R T 2 ln
P3
P2
W 23=8.314
J
K∗1 kJ
10
∗403.15
∗ln
mol∗K
1000 J
1
W 23=7.718
kJ
mol
According to first law of thermodynamics
∆ U 23=Q23 +W 23
0=Q23+W 23
Q23=−W 23
Q23=−7.718
kJ
mol
For the complete cycle,
Work=W =W 12 +W 23
Q=Q12 +Q23
41
W = (−0.832+7.718 )
Q=( 2.079−7.718 )
kJ
mol
kJ
mol
W =6.886
Q=−5.639
kJ
Answe r
mol
kJ
Answer
mol
kJ
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
∆ U =∆ U 12 +∆ U 23
∆ U =( 1.247+0 )
kJ
mol
∆ U =1.247
kJ
Answer
mol
∆ H=∆ H 12+ ∆ H 23
∆ H= ( 2.079+ 0 )
kJ
mol
∆ H=2.079
kJ
Answe r
mol
(c)
Step 12:
Since for step “12” process is isothermal ( T = Constant)
Therefore
∆ U 12=∆ H 12=0
Here
P2=P3
For an isothermal process we have
W 12=R T 1 ln
P2
P1
W 12=8.314
W 12=5.8034
J
K∗1 kJ
10
∗303.15
∗ln
mol∗K
1000 J
1
kJ
mol
According to first law of thermodynamics
∆ U 12=Q12 +W 12
0=Q12+W 12
Q12=−W 12
Q12=−5.8034
Step 23:
For step “23” volume is constant
Therefore, at constant pressure we have
Q23=∆ H 23=2.079
kJ
[ ¿(b)]
mol
Here
T 2 =T 3
Now
∆ U 23=1.247
42
According to first law of thermodynamics
kJ
[ ¿(a) ]
mol
kJ
mol
August 20,
2013
PROBLEMS
∆ U 23=Q23 +W 23
ZAID YAHYA
W 23=∆U 23−Q 23
11-CH-74
W 23=( 1.247−2.079 )
kJ
mol
W 23=−0.832
kJ
mol
For the complete cycle,
Work=W =W 12 +W 23
Q=Q12 +Q23
W = (5.8034−0.832 )
Q=(−5.8034+2.079 )
kJ
mol
W =4.9714
kJ
mol
Q=−3.724
kJ
Answe r
mol
kJ
Answe r
mol
∆ U =∆ U 12 +∆ U 23
∆ U =( 0+1.247 )
kJ
mol
∆ U =1.247
kJ
Answer
mol
∆ H=∆ H 12+ ∆ H 23
∆ H= ( 0+2.079 )
kJ
mol
∆ H=2.079
kJ
Answe r
mol
Problem 3.23:
One mole of an ideal gas, initially at 30 ℃ and 1 bars, undergoes the following mechanically
reversible changes. It is compressed isothermally to point such that when it is heated at constant volume to
120 ℃ its final pressure is 12 bars. Calculate Q, W, ∆ U ∧∆ H for the process. Take CP = (7/2) R and
CV = (5/2) R.
Given Data:
T 1 =30℃
T 3 =393.15 K
T 1 =( 30+273.15 ) K
P3=12 ¯¿
Q=?
T 1 =303.15 K
W =?
P1=1 ¯
¿
∆ U =?
∆ H=?
Solution:
The process consist of two steps, 12 & 23
Step 12:
Since for step “12” process is isothermal ( T = Constant)
Therefore
∆ U 12 =∆ H 12=0
Now, intermediate pressure can be calculated as
43
T 3 =120 ℃
7
C P= R
2
T 3 =( 120+273.15 ) K
5
CV = R
2
August 20,
2013
PROBLEMS
P2
T2
ZAID YAHYA
P
¿ 3
T3
P
P2= 3 ∗T 2
T3
12
¯¿
∗303.15 K
393.15 K
P2=¿
P2=9.25 b ar
For an isothermal process we have
W 12=R T 1 ln
P2
P1
W 12=8.314
J
K∗1 kJ
9.25
∗303.15 .15
∗ln
mol∗K
1000 J
1
W 12=5.607
11-CH-74
kJ
mol
According to first law of thermodynamics
∆ U 12=Q12 +W 12
0=Q12+W 12
Q12=−W 12
Q12=−5.607
kJ
mol
Step 23:
Since for step “23” volume is constant
Therefore
W 23=0
According to first law of thermodynamics
∆ U 23=Q23 +W 23
∆ U 23=Q 23
Q23 =∆ U 23 =CV ∆ T
5
Q23=∆ U 23= R ( T 3−T 1 )
2
5
J
K∗1 kJ
kJ
Q23=∆ U 23= ∗8.314
( 393.15−303.15 )
Q23=∆ U 23=1.871
2
mol∗K
1000 J
mol
Now
∆ H 23=C P ∆ T
7
∆ H 23= R ( T 2−T 1 )
2
7
J
kJ
∆ H 23= ∗8.314
( 393.15−303.15 ) K∗1
2
mol∗K
1000 J
∆ H 23=2.619
For the complete cycle,
Work=W =W 12 +W 23
Q=Q 12 +Q23
44
W = (5.607 +0 )
Q=(−5.607+1.871 )
kJ
mol
kJ
mol
W =5.607
Q=−3.736
kJ
Answe r
mol
kJ
Answe r
mol
kJ
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
∆ U =∆ U 12 +∆ U 23
∆ U =( 0+1.871 )
kJ
mol
∆ U =1.871
kJ
Answe r
mol
∆ H=∆ H 12+ ∆ H 23
∆ H= ( 0+2.691 )
kJ
mol
∆ H=2.691
kJ
Answe r
mol
Problem 3.24:
A process consists of two steps: (1) One mole of air at T = 800 K and P = 4 bars are cooled at constant
volume to T = 350 K. (2) The air is then heated air constant pressure until its temperature reaches 800 K.
If this two step process is replaced by a single isothermal expansion of the air from 800 K and 4 bar to
some final pressure P, what is the value of P that makes the work of two step processes the same? Assume
mechanical reversibility and treat air as an ideal gas with CP = (7/2) R and CV = (5/2) T.
Given Data:
T 1 =800 K
P1=4 ¯
¿
T 2 =350 K
P=?
Solution:
For the first step volume is constant
Therefore,
W 12=0
For the work done is
W =W 23=−P 2 ∆ V →(1)
For one mole of an ideal gas we have,
P ∆ V =R ∆ T
P2 ∆ V =R ∆ T
Put in (1)
W =−R ∆ T
W =−R ( T 3−T 2 )
Since
T 3 =T 1
Therefore
W =−R ( T 1−T 2) →(2)
45
For an isothermal process we have
August 20,
2013
W =R T 1 ln
PROBLEMS
ZAID YAHYA
11-CH-74
P
→ (3)
P1
Compare (2) and (3)
P
−R ( T 1−T 2 )=R T 1 ln
P1
P
4 ¯¿
(350−800 ) K
=ln ¿
800 K
¯
P=2.279 Answer
T 2−T 1
P
=ln
T1
P1
P
T 2 −T 1=T 1 ln
P1
4 ¯¿ 0.5698=P
4 ¯¿
P
e−0.5625 = ¿
Problem 3.25:
A scheme for finding the internal volume V tB of the gas cylinder consists of the following steps. The
cylinder is filled with a gas to low pressure P 1, and connected through a small line and valve to an
evacuated reference tank of known volume V tA . The valve is opened, and the gas flows through the line
into the reference tank. After the system returns to its initial temperature, a sensitive pressure transducer
provides a valve for the pressure change ∆ P in the cylinder. Determine the cylinder volume V tB
from the following data:
a) V tA =256 cm3
b) ∆ P/ P1=−0.0639
Given Data:
t
V B =?
t
V A =256 cm
3
∆P
=−0.0639
P1
Solution:
∆P
=−0.0639
P1
P2 −P 1
=−0.0639
P1
P2
−1=−0.0639
P1
P2
=−0.0639+1
P1
P2
=0.9361 →(1)
P1
Assume that gas is ideal & P2 is th
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Chapter 3
Problems
Problem 3.1:
Express the volume expansivity β and isothermal compressibility κ as functions of density ρ and its partial
derivatives. The isothermal compressibility coefficient () of water at 50 oC and 1 bar is 44.18∗10−6
bar-1. To what pressure must water be compressed at 50 oC to change its density by 1%? Assume that is
independent of P.
Given Data:
Volume expansivity=β=
1 ∂V
V ∂T
( )
P
Or
β=
1 dV
V dT
( ) →(1)
P
Isothermal Compressibilty=κ=
−1 ∂ V
V ∂P
( )
Or
T
−1 dV
→ ( 2)
V dP T
Temperature=T =50
κ=
Pressure=P1=1 ¯¿
¿¯−1
κ=44.18∗10−6 ¿
( )
(a)
We know that
ρ=
1
1
V
V=
1
ρ
C
Density of water =ρ1=1
P2=?
Solution:
0
kg
m3
ρ2=( 1+1 )
kg
m3
ρ2=1.01
kg
m3
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Put in (1) & (2)
β=ρ
( dTd 1ρ )
β=
P
−ρ dρ
ρ2 dT
( )
β=
P
−1 dρ
ρ dT
( ) Proved
P
Now,
κ=−ρ
( dPd 1ρ )
κ=
T
ρ dρ
ρ2 dP
( )
κ=
T
1 dρ
ρ dP
( ) Proved
T
(b)
As
κ=
1 dρ
ρ dP
( )
κdP=
T
dρ
ρ
Integrating on both sides
P2
ρ2
κ ∫ dP=∫
P1
ρ1
dρ
ρ
P2
ρ2
1
1
κ|P|P =|ln ρ|ρ
κ ( P2 −P1 )=( ln ρ2−ln ρ1 )
κ ( P2 −P1 )=ln
ρ2
ρ1
Putting values
¯¿ ( P2−1 )= ln1.01
1
−6
44.18∗10
¿
¯¿
44.18∗10−6
P2−1=0.00995∗¿
P2=( 225.22+1 ) ¯¿
P2=226.22 ¯¿
Answer
Problem 3.2:
Generally, volume expansivity β and isothermal compressibility κ depend on T and P. Prove that
( ∂∂ Pβ ) =−( ∂T∂κ )
T
P
Solution:
We know that
Volume expansivity=β=
Since
2
β
is very small
1 ∂V
V ∂T
( )
P
August 20,
2013
PROBLEMS
ZAID YAHYA
V=
1 ∂V
∂ β ∂T
( ) → ( 1)
11-CH-74
Isothermal Compressibilty=κ=
P
−1 ∂ V
V ∂P
( )
T
Since κ is very small
V=
−1 ∂ V
∂κ ∂ P
1 ∂V
∂β ∂T
∂V
( ) =−1
(
∂κ ∂ P)
( )
T
P
1 1
∂ β ∂T
1
∂V
( ) ∂V =−1
(
∂κ ∂P )
T
P
T
( ∂T∂ κ ) =( ∂∂ Pβ )
−
P
T
( ∂∂ Pβ ) =−( ∂T∂κ ) Proved
T
P
Problem 3.3:
The Tait equation for liquids is written for an isotherm as:
AP
V =V 0 1−
B+ P
Where V is specific or molar volume, Vo is the hypothetical molar or specific volume at P = 0 and A & B
are positive constant. Find an expression for the isothermal compressibility consistent with this equation.
(
)
Solution:
We Know That,
Isothermal Compressibilty=κ=
−1 ∂ V
V ∂P
( ) → ( 1)
T
Given that
(
V =V 0 1−
AP
B+ P
)
Where
V0 = Hypothetical molar/specific volume at zero pressure, so it is constant
V = Molar/specific volume
Now,
AP
V =V o−
V
B+ P o
V −V o=
−AP
V
B+ P o
V −V o −AP
=
Vo
B +P
Differentiate w.r.t Pressure
1 ∂
−∂ AP
V −V o ) =
(
V o ∂P
∂ P B+ P
(
3
)
[
A ( B+ P )− AP ( 1 )
1 ∂V
−0 =−
V o ∂P
( B+ P )2
(
)
]
August 20,
2013
PROBLEMS
ZAID YAHYA
−1 ∂V
AB+ AP− AP
=
Vo ∂ P
( B+ P )2
( )
11-CH-74
−1 ∂V
AB
=
V o ∂ P ( B+ P )2
( )
Since, Temperature is constant
Therefore,
−1 ∂V
AB
=
V o ∂ P T ( B+ P )2
Or, From (1)
( )
κ=
AB
Proved
( B+P )2
Problem 3.4:
For liquid water the isothermal compressibility is given by:
c
κ=
V ( P+ b )
Where c & b are functions of temperature only if 1 kg of water is compressed isothermally & reversibly
from 1 bar to 500 bars at 60 oC, how much work is required?
At 60 oC, b=2700 bars and c = 0.125 cm3 g-1
Given Data:
Isothermal compressibility=κ=
b=2700 bars
c=0.125 cm3 / g
c
Mass of water=m=1 kg
V ( P+b )
P2=500 bars
Temperature=T =60 0C
Work=W =?
Solution:
We know that
W =−∫ PdV →(1)
κ=
c
→(2)
V ( P+ b )
Also
κ=
−1 dV
V dP
( ) → ( 3)
T
Comparing (2) & (3)
4
Pressure=P1=1 ¯
¿
August 20,
2013
PROBLEMS
ZAID YAHYA
−1 dV
c
=
V dP V ( P+ b )
−dV =
11-CH-74
c dP
P+b
Put in (1)
W =−∫ −P
P2
c dP
P+b
P2
W =c ∫ dP−b c ∫
P1
P1
P2
P2
P
W =c ∫
dP
P+ b
P
P+ b−b
W =c ∫
dP
P+ b
P
1
1
dP
P+b
P2
P2
1
1
[
P2
1
1
P+ b
b
W =c ∫
dP−c ∫
dP
P+ b
P+b
P
P
1
W =c|P|P −bc|ln(P+b)|P
P2
W =c ( P2−P 1) −bc [ ln ( P2+ b ) −ln ( P1 +b ) ]
W =c ( P2−P 1) −bc ln
P2 +b
P1 +b
]
Putting values
3
3
W =0.125
3
cm (
¯ ¯¿ 0.125 cm ∗ln 500+ 2700
∗ 500−1 )−2700
g
g
1+2700
cm3∗¯¿
∗1 m3
g
∗101325 N
3
3
100 cm
∗J
2
1.01325 ¯¿ m
Nm
W =5.16 ¿
W =0.516
cm ∗¯¿
g
3
¯
cm ∗¿
−57.216 ¿
g
W =62.375 ¿
3
cm ∗¯¿
g
W =5.16 ¿
J
Answer
g
Problem 3.5:
Calculate the reversible work done in compressing 1 ft 3 of mercury at a constant temperature of 32F from
1(atm) to 3,000(atm). The isothermal compressibility of mercury at 32F is:
κ/(atm)-1 = 3.9 x 10-6 - 0.1 x10-9P(atm)
Given Data:
Work done=W =?
Volume=V =1 ft
Pressure=P2=3000 atm
Where
5
3
Temperature=T =32 F
κ /atm−1=3.9∗10−6−0.1∗10−9 P(atm)
Pressure=P1=1 atm
August 20,
2013
PROBLEMS
Term,
ZAID YAHYA
3.9*10-6
has
unit
atm-1
of
11-CH-74
&
0.1*10-9
has
units
of
atm-2
Solution:
We know that, work done for a reversible process is
W =−∫ PdV →(1)
Also
κ=
−1 dV
V dP
( )
dV =−κVdP
T
Put in (1),
P2
W =−∫ P (−κVdp )
W =V ∫ κ P dP
P1
P2
P2
W =V ∫ ( 3.9∗10−6−0.1∗10−9 P ) P dP
P1
3000
−6
W =3.9∗10 V
∫
P dP−0.1∗10 V
−6
P1
∫
P2 dP
1
3000
||
P2
W =3.9∗10 V
2
W=
P1
3000
−9
1
−6
P2
W =V ∫ 3.9∗10−6 P dP−V ∫ 0.1∗10−9 P2 dP
1
3000
||
P3
−0 .1∗10 V
3
−9
3
1
−11
3
1.95∗10 ∗1 ft (
3.333∗10 ∗1 ft (
30002−12 ) atm2−
∗ 30003−13 ) atm3
2
atm
atm
W = (17.55−0.8991 ) atm∗ft 3
W =16.65 atm∗ft 3 Answer
Problem 3.6:
Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state
at 1 bar during which the temperature change from 0 oC to 20oC. Determine ΔVt, W, Q, and ΔUt. The
properties for liquid carbon tetrachloride at 1 bar & 0 oC may be assumed independent of temperature: β =
1.2 x 10-3 K-1 Cp = 0.84 kJ kg-1 K-1, ρ = 1590 kg m-3
Given Data:
Mass=m=5 kg
6
Pressure=P=1 ¯¿
Temperature=T 1=0
0
C
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
T 1 =273.15 K
Temperature=T 2=20
T 2 =293.15 K
T 2 =( 20+273.15 ) K
−3
β=1.2∗10 K
0
C
C P=0.84
−1
kJ
kg∗K
ρ=1590
kg
m3
t
Q=?
∆ U =?
Solution:
As
1
ρ
V=
V 1=
1
ρ1
V 1=
1 m3
1590 kg
Also,
we know that
Volume expansivity=β=
1 dV
V dT
( )
βdT =
P
1
dV
V
Integrating on both sides,
T2
V2
β ∫ dT =∫
T1
V1
dV
V
T2
V2
1
1
β|T |T =|lnV |V
β ( T 2−T 1 )=( ln V 2−ln V 1 )
β ( T 2−T 1 )=ln
V2
V1
Putting values
V ∗1590 kg
1.2∗10−3
∗( 293.15−273.15 ) K =ln 2
K
m3
e
0.024
=
V 2∗1590 kg
m3
1.024
∗m3
1590
V 2=
kg
3
V 2=0.000644
m
kg
Now,
∆ V =V 2−V 1
1 m3
∆ V = 0.000644−
1590 kg
(
)
∆ V =15.28∗10−6
m3
kg
Now, for total volume,
t
∆ V =∆ V ∗m
∆ V t =( 15.28∗10−6∗5 )
Now,
We know that for a reversible process,
7
m3
∗kg
kg
t
−5
3
t
∆ V =?
∆ V =7.638∗10 m Answer
W =?
August 20,
2013
PROBLEMS
ZAID YAHYA
Work done=W =−P ∆ V t
W =−1 ¯¿ 7.638∗10
−5
m3∗101325 N
∗J
1.01325 ¯¿ m2
∗1 kJ
Nm
1000 J
11-CH-74
W =−7.638∗10−3 kJ Answer
Now,
For a reversible process at constant pressure,we have
Q=∆ H
Q=m C P ∆ T
Q=5 kg∗0.84
kJ
∗( 293.15−273.15 ) K
kg∗K
Q=84 kJ Answer
Now,
According to first law of thermodynamics,
∆ U t =( 84−7.368∗10−3 ) kJ
∆ U t =Q+W
∆ U t =83.99 kJ Answer
Problem 3.7:
A substance for which k is a constant undergoes an isothermal, mechanically reversible process from
initial
state
(P1,
V1)
to
(P2,
V2),
where
V
is
a
molar
volume.
a) Starting with the definition of k, show that the path of the process is described by
V = A ( T ) exp(−κP)
b) Determine an exact expression which gives the isothermal work done on 1 mol of this constant-k
substance.
Solution:
(a)
We know that
Isothermal compressibilty =κ=
−1 dV
V dP
( )
T
dV
=−κdP
V
Integrating on both sides,
dV
=−κ ∫ dP
V
lnV =−κP+lnA ( T )
∫
Where ln A (T) is constant of integration & A depends on T only
lnV −lnAT =−κP
ln
Taking anti log on both sides,
V
=e−κP
AT
8
V = A ( T ) e−κP
V
=−κP
A (T )
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Or
V = A ( T ) exp (−κP ) Proved
(b)
Work done=W=?
For a mechanically reversible process, we have,
dW =−PdV → ( 1 )
Using,
d ( PV ) =PdV +VdP
−PdV =VdP−d ( PV )
Put in (1)
dW =VdP−d ( PV ) → ( 2 )
We know that
Isothermal compressibilty =κ=
−1 dV
V dP
( )
T
−dV
=VdP
κ
Put in (2)
dW =
−dV
−d ( PV )
κ
Integrating on both sides,
−1
∫ dW = κ ∫ dV −∫ d ( PV )
W=
−1
Δ V −Δ ( PV )
κ
Since volume changes from V1 to V2 & pressure changes from P1 to P2 ,
Therefore,
W=
−1
( V 2−V 1 ) −( P2 V 2−P1 V 1 )
κ
W=
( V 1−V 2)
κ
+ P1 V 1−P2 V 2 Proved
Problem 3.8:
One mole of an ideal gas with C V = 5/2 R, CP = 7/2 R expands from P 1 = 8 bars & T1= 600 K to P2 = 1 bar
by each of the following path:
a) Constant volume
b) Constant temperature
c) Adiabatically
Assuming mechanical reversibility, calculate W, Q, ∆ U, and ∆ H for each of the three processes.
Sketch each path in a single PV diagram.
9
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Given Data:
5
CV = R
2
7
C P= R
2
P1=8 ¯
¿
T 1 =600 K
P2=1 ¯
¿
W =?
Q=?
∆ U =?
∆ H=?
Solution:
(a)
According to first law of thermodynamics,
∆ U =Q+W →(1)
For a constant volume process,
∆ U =CV ∆T
W =0
Put in (1)
Q=∆U =C V ∆ T
Q=∆U =C V ( T 2 −T 1 ) →(2)
For T2 , We know that for an ideal gas
T1 T 2
=
P1 P2
T2=
T1
∗P
P1 2
Put in (2),
8 ¯¿∗1 ¯¿
600 K
T2= ¿
5
Q=∆U = R ( 75−600 ) K
2
T 2 =75 K
Q=∆U =
−5
J
∗8.314
∗525 K
2
mol∗K
Q=∆U =−10.912
Q=∆U =−10912
kJ
Answer
mol
Also
For a mechanically reversible process we have,
∆ H=C P ∆ T
7
∆ H= R ( T 2−T 1 )
2
7
J
∆ H= ∗8.314
∗( 75−600 ) K
2
mol∗K
kJ
∆ H=−15.277
Answer
mol
∆ H=−15277
(b)
For a constant temperature process,
∆ U =0
∆ H=0
We know that at constant temperature, work done is
W =R T 1 ln
W =8.314
10
J
mol
J
1
∗600 K∗ln
mol∗K
8
P2
P1
W =−10373
Now, according to first law of thermodynamics,
J
mol
W =−10.373
kJ
Answer
mol
J
mol
August 20,
2013
PROBLEMS
∆ U =Q+W
ZAID YAHYA
0=Q+ W
11-CH-74
Q=−W
Or
Q=10.373
kJ
Answer
mol
(c)
We know that for an adiabatic process
Q=0
Now, according to first law of thermodynamics,
∆ U =Q+W
∆ U =W →(1)
∆ U =CV ∆T
Put in (1)
W =∆ U=C V ∆ T
W =∆ U=C V ( T 2−T 1 ) →(2)
For T2 , We know that for an adiabatic process
T1 P
( 1−γ )
γ
1
=T 2 P
( 1−γ )
γ
2
T 2 =T 1
P1
P2
( )
( 1−γ )
γ
T 2 =600 K
8
1
()
( 1−1.4 )
1.4
T 2 =331.23 K
Put in (2)
5
W =∆ U= R∗( 331.23−600 ) K
2
W =∆ U=
−5
J
∗8.314
∗268.77 K
2
mol∗K
W =∆ U=5.5864
W =∆ U=−5586.4
J
mol
J
Answer
mol
For a mechanically reversible adiabatic process we have
7
∆ H= R ( T 2−T 1 )
2
J
∆ H=−7.821
Answer
mol
∆ H=C P ∆ T
7
J
∆ H= ∗8.314
∗( 331.23−600 ) K
2
mol∗K
Problem 3.9:
An ideal gas initially at 600k and 10 bar undergoes a four-step mechanically reversible cycle in a closed
system. In step 12, pressure decreases isothermally to 3 bars; in step 23, pressure decreases at constant
volume to 2 bars; in step 34, volume decreases at constant pressure; and in step 41, the gas returns
adiabatically to its initial state. Take CP = (7/2) R and CV = (5/2) R.
a) Sketch the cycle on a PV diagram.
b) Determine (where unknown) both T and P for states 1, 2, 3, and 4.
c) Calculate Q, W, ∆ U, and ∆ H for each step of the cycle.
11
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Given Data:
Initial Temperature=T 1=600 K
Initial Pressure=P1=10 ¯
¿
7
C P= R
2
5
CV = R
2
Solution:
(b)
Step 12, an Isothermal process,
Since
For an isothermal process, temperature is constant
Therefore,
P2=3 ¯¿
T 2 =T 1=600 K
We know that, for an ideal gas
P2 V 2 =R T 2
V 2=
RT 2
P2
¯¿∗1.01325 ¯¿ m2
∗N∗m
101325 N
mol∗K∗3
J
8.314∗J∗600 K
V 2=
¿
V 2=0.0166
m3
mol
Step 23, an Isochoric process,
Since
For an isochoric process, Volume is constant
Therefore,
3
V 3=V 2=0.0166
m
mol
P3=2 ¯¿
We know that, for an ideal gas
P3 V 3 =RT 3
T3=
P3 V 3
R
2 ¯¿ 0.0166 m 3∗mol∗K
∗J
mol∗8.314 J
∗101325 N
N∗m
T3=
1.01325 ¯¿ m2
Step 34, an Isobaric process,
Since
For an isobaric process, pressure is constant
Therefore,
12
T 3 =400 K
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
P4 =P 3=2 ¯¿
For T4 , we will use an adiabatic relation of temperature and pressure
As
T 4 P4
=
T1
P1
( )
R
CP
T 4=T 1
P4
P1
( )
R
CP
2
T 4=600 K∗
10
( )
2∗R
7R
T 4=378.83 K
We know that, for an ideal gas
P4 V 4 =RT 4
V 4=
¯¿∗1.01325 ¯¿ m2
∗Nm
101325 N
mol∗K∗2
J
8.314 J∗378.83 K
V 4=
¿
RT4
P4
V 4 =0.0157
m3
mol
Step 41, an adiabatic process,
Since
Gas returns to its initial state adiabatically
Therefore,
T 1 =600 K
P1=10 ¯¿
We know that, for an ideal gas
P1 V 1=R T 1
V 1=
RT 1
P1
¯¿∗1.01325 ¯¿ m2
∗Nm
101325 N
mol∗K∗10
J
8.314 J∗600 K
V 1=
¿
(c)
Step 12, an Isothermal process,
Since
For an isothermal process, temperature is constant
Therefore
∆ U 12 =0
13
∆ H 12=0
For an isothermal process, we have
V 1=4.988∗10−3
3
m
mol
August 20,
2013
PROBLEMS
Q=−R T 1 ln
P2
P1
ZAID YAHYA
Q=−8.314
J
3
∗600 K∗ln
mol∗K
10
Q=6006
11-CH-74
J
mol
Q=6.006∗103
J
Answer
mol
According to first law of thermodynamics
∆ U 12=Q12 +W 12
0=Q12+W 12
W 12=−Q12
W 12=−6.006∗103
J
Answer
mol
Step 23, an Isochoric process,
Since
For an isochoric process, Volume is constant
Therefore,
W 23=0
At constant volume we have
Q23=∆ U 23=CV ∆ T
Q23=∆ U 23=CV ( T 3−T 2 )
5
∗8.314 J
2
Q23=∆ U 23=
∗(−200 ) K
mol∗K
5
Q23=∆ U 23= R ( 400−600 ) K
2
Q23=∆ U 23=−4157
J
mol
Q23=∆ U 23=−4.157∗10 3
J
Answer
mol
We know that
∆ H 23=C P ∆ T
7
∆ H 23= R ( 400−600 ) K
2
∆ H 23=C P ( T 3−T 2 )
∆ H 23=−5820
J
mol
∆ H 23=−5.82∗103
7
J
∆ H 23= ∗8.314
∗(−200 ) K
2
mol∗K
J
Answer
mol
Step 34, an Isobaric process,
Since
For an isobaric process, pressure is constant
Therefore, at constant pressure we have,
Q34=∆ H 34 =C P ∆ T
7
Q34=∆ H 34 = R ( T 4−T 3 )
2
7
J
Q34=∆ H 34 = ∗8.314
∗( 378.83−400 ) K
2
mol∗K
J
Q34=∆ H 34 =−616
Answer
mol
For an Isobaric process we have
W 34=−R ∆ T
14
W 34=−R ( T 4−T 3 )
W 34=−8.314
J
∗( 378.83−400 ) K
mol∗K
W 34=176
J
Answer
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
We know that,
∆ U 34 =CV ∆T
5
∆ U 34 = R ( T 4 −T 3 )
2
5
J
∆ U 34 = ∗8.314
∗( 378.83−400 ) K
2
mol∗K
J
∆ U 34 =−440
Answer
mol
Step 41, an adiabatic process,
Since
For an adiabatic process there is no exchange of heat
Therefore,
Q41=0
We know that,
5
∆ U 41= R ( T 1−T 4 )
2
5
J
∆ U 41= ∗8.314
∗( 600−378.83 ) K
2
mol∗K
J
∆ U 41=4.597∗103
Answer
mol
∆ U 41=C V ∆ T
∆ U 41=4597
J
mol
We know that
∆ H 41=C P ∆T
7
∆ H 41= R ( T 1−T 4 )
2
7
J
∆ H 41= ∗8.314
∗( 600−378.83 ) K
2
mol∗K
3 J
∆ H 41=6.4358∗10
Answer
mol
∆ H 41=6435.8
J
mol
According to first law of thermodynamics
∆ U 41=Q 41+W 41
∆ U 41=W 41
W 41=4.597∗103
J
Answer
mol
Problem 3.10:
An ideal gas, CP= (5/2) R and CV= (3/2) R is changed from P1 = 1bar and V t1 = 12m3 to P2 = 12 bar and
V t2 = 1 m3 by the following mechanically reversible processes:
a) Isothermal compression
b) Adiabatic compression followed by cooling at constant pressure.
c) Adiabatic compression followed by cooling at constant volume.
d) Heating at constant volume followed by cooling at constant pressure.
e) Cooling at constant pressure followed by heating at constant volume.
Calculate Q, W, change in U, and change in H for each of these processes, and sketch the paths of all
processes on a single PV diagram.
Given Data:
15
August 20,
2013
PROBLEMS
5
C P= R
2
t
V 2=1 m
3
ZAID YAHYA
3
Initial pressure=P1=1 ¯
¿
CV = R
2
Q=?
W =?
∆ H=? ∆ U =?
t
3
11-CH-74
V 1=12 m
Final pressure=P 2=12 ¯¿
12
∗101325 N
1
∗J
1.01325 ¯¿ m2
∗1 kJ
Nm
3
¯
Q=−1 ¿ 12 m ∗ln
1000 J
Q=−2981.88 kJ Answer
Solution:
Since
Temperature=constant
Therefore, for all parts of the problem,
∆ H=0
∆ U =0
(a)
Isothermal compression,
For an isothermal process, we have
Q=−R T 1 ln
P2
P1
Since
For an ideal gas, we have
P1 V 1=R T 1
Therefore,
Q=−P1 V 1 ln
P2
P1
According to first law of thermodynamics
∆ U =Q+W
0=Q+ W
W =−Q
W 12=2981.88 kJ Answer
(b)
Adiabatic compression followed by cooling at constant pressure
Since
For an adiabatic process, there is no exchange of heat
Therefore,
Q=0 Answer
The process completes in two steps
First step, an adiabatic compression to final pressure P 2 , intermediate volume can be given as
16
August 20,
2013
PROBLEMS
'
' γ
V =V 1
P2 ( V ) =P 1 V 1
ZAID YAHYA
P1
P2
( )
11-CH-74
1
γ
For mono atomic gas, we have
γ =1.67
'
3
1
12
( )
V =12 m ∗
1
1.67
V ' =2.71 m3
We know that,
W 1=
P2 V ' −P 1 V 1
γ−1
( 12∗2.71−1∗12 ) ¯¿ m3
∗101325 N
1.67−1
∗J
1.01325 ¯¿ m2
∗1 kJ
Nm
W 1=
1000 J
W 1=3063 kJ →(1)
Second step, cooling at constant pressure P2
We know that, for a mechanically reversible process
m3∗101325 N
∗J
1.01325 ¯¿ m2
∗1kJ
Nm
¯
W 2=−12 ( 1−2.71 )
1000 J
W 2=−P2 ( V 2−V ' )
W 2=2052 kJ →( 2)
Now
W =W 1+ W 2
W = ( 3063+ 2052 ) kJ
W =5115 kJ Answer
(c)
Adiabatic compression followed by cooling at constant volume
Since
For an adiabatic process, there is no exchange of heat
Therefore,
Q=0 Answer
First step, an adiabatic compression to volume V2 , intermediate pressure can be given as
'
γ
P V 2 =P1 V 1
P' =P1
V1
V2
( )
For mono atomic gas, we have
17
γ
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
γ =1.67
12¯ 1.67
P =1
1
'
( )
P' =63.42 ¯¿
We know that,
P' V 2−P 1 V 1
W 1=
γ−1
( 63.42∗1−1∗12 ) ¯¿ m3
∗101325 N
1.67−1
∗J
1.01325 ¯¿ m 2
∗1 kJ
Nm
W 1=
1000 J
W 1=7674.76 kJ
Second step, cooling at constant Volume,
Therefore, No work will be done
W 2=0
Now
W =W 1+ W 2
W = (7674.76+ 0 ) kJ
W =7674.76 kJ Answer
(d)
Heating at constant volume followed by cooling at constant pressure
The process completes in two steps
Step 1, Heating at constant volume to P2
Therefore no work will be done
W 1=0
Step 2, Cooling at constant pressure P2 To V2
We know that, for a mechanically reversible process
3
W 2=−P2 ∆ V
W 2=−P2 ( V 2−V 1 )
m ∗101325 N
∗J
2
1.01325 ¯¿ m
∗1 kJ
Nm
¯ )
W 2=−12 ( 1−12
1000 J
Now
W =W 1+ W 2
W = ( 0+13200 ) kJ
W =13200 kJ Answer
According to first law of thermodynamics
∆ U =Q+W
(e)
18
0=Q+ W
Q=−W
Q=−13200 kJ Answer
W 2=13200 kJ
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Cooling at constant pressure followed by heating at constant volume
The process completes in two steps
Step 1, Cooling at constant Pressure P1 to V2
Therefore, for a mechanically reversible process
W 1=−P1 ( V 2−V 1 )
W 1=−P1 ∆ V
m3∗101325 N
∗J
1.01325 ¯¿ m2
∗1 kJ
Nm
¯ )
W 1=−1 ( 1−12
1000 J
W 1=1100 kJ
Step 1, Heating at constant Volume V2 to pressure P2
Therefore no work will be done
W 2=0
Now
W =W 1+ W 2
W = (1100 +0 ) kJ
W =1100 kJ Answer
According to first law of thermodynamics
∆ U =Q+W
0=Q+ W
Q=−W
Q=−1100 kJ Answer
Problem 3.11:
dT
characterizes the local variation of temperature with elevation in the
dz
earth's atmosphere. Atmospheric pressure varies with elevation according to the hydrostatic formula,
dP
=−M ρg
dz
The environmental lapse rate
Where M is a molar mass, ρ is molar density and g is the local acceleration of gravity. Assume that the
atmosphere is an ideal gas, with T related to P by the polytropic formula equation (3.35 c). Develop an
expression for the environmental lapse rate in relation to M, g, R, and δ.
Solution:
Given that
dP
=−M ρg → ( 1 )
dz
The polytropic relation is
TP
19
1−δ
δ
=Constant
August 20,
2013
PROBLEMS
ZAID YAHYA
Or
TP
1−δ
δ
=T o P
1−δ
δ
o
Where
To =Temperature at sea level, so it is constant
Po = Pressure at sea level, so it is constant
T
P
δ−1
δ
T
To
( )
=
To
P
δ
δ−1
δ−1
δ
o
P
=
Po
T
P
=
T o Po
( )
δ−1
δ
T
P=Po
To
( )
δ
δ −1
→(a)
P=
Po
To
δ
δ−1
δ
∗T δ −1
Differentiate w.r.t to Temperature on both sides
Po
δ
δ −1
Po
∗δ
dP T o
=
∗T
dT
δ−1
1
δ−1
δ
δ−1
∗δ
1
To
δ −1
dP=
∗T
dT →(2)
δ−1
We know that, for an ideal gas
ρ=
P
RT
Where
R=Specific gas constant=R ' /M
Put (a) in above equation
ρ=
1
T
∗Po
RT
To
( )
Put in (1)
dP
g∗1
T
=−M
Po
dz
RT
To
( )
Put (2) in above
20
δ
δ −1
dP=−M
g∗1
T
Po
RT
To
( )
δ
δ −1
∗dz
δ
δ −1
11-CH-74
August 20,
2013
PROBLEMS
Po
δ
δ−1
ZAID YAHYA
−δ −1
∗M g
δ
∗Po
δ
R
∗T o δ−1
Po
∗δ
1
To
g∗1
T
∗T δ −1 dT =−M
Po
δ −1
RT
To
( )
δ
δ−1
∗dz
To
To
dT
=
dz
−δ −1
∗M g
δ
∗Po
δ
R
∗T o δ−1
Po
dT
=
dz
11-CH-74
δ
δ−1
δ
δ−1
δ
∗T δ−1
1
T∗T δ−1
δ
∗T δ−1
δ
T δ−1
−δ
∗M g
dT δ−1
=
Proved
dz
R
Problem 3.12:
An evacuated tank is filled with gas from a constant pressure line. Develop an expression relating the
temperature of the gas in the tank to temperature T’ of the gas in line. Assume that gas is ideal with
constant heat capacities, and ignore heat transfer between the gas and the tank. Mass and energy balances
for this problem are treated in Ex. 2.13.
Solution:
Choose the tank as the control volume. There is no work, no heat transfer & kinetic & potential energy changes are
assumed negligible.
Therefore, applying energy balance
d ( mU )tank
+∆ ( Hm ) =0
dt
d ( mU )tank
+ H '' m '' −H ' m ' =0
dt
Since
Tank is filled with gas from an entrance line, but no gas is being escaped out,
Therefore,
d ( mU )tank
+0− H ' m' =0
dt
21
d ( mU )tank
−H ' m' =0→( 1)
dt
August 20,
2013
PROBLEMS
ZAID YAHYA
Where prime (‘) denotes the entrance stream
Applying mass balance
m' =
d mtank
→ ( 2)
dt
Combining equation (1) & (2)
d ( mU )tank
d mtank
−H '
=0
dt
dt
1
d ( mU )tank−H ' d mtank }=0
{
dt
'
d ( mU )tank =H d mtank
Integrating on both sides
m2
m2
∫ d ( mU )tank=H ' ∫ d mtank
m1
∆ ( mU )tank =H ' ( m2−m1 )
m1
'
m2 U 2 −m1 U 1=H ( m2−m1 )
Because mass in the tank initially is zero, therefore
m1=0
m2 U 2 =H ' m2
'
U 2=H →(3)
We know that
U=C V T
U 2=C V T 2 → ( a )
Also
'
'
H =C P T → ( b )
Put (a) & (b) in (3)
CV T =C P T '
T=
CP '
T
CV
Since heat capacities are constant, therefore
γ=
CP
CV
T =γ T ' Proved
Problem 3.14:
22
11-CH-74
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
A tank of 0.1-m3 volume contains air at 25 oC and 101.33 kPa. The tank is connected to a compressed-air
line which supplies air at the constant conditions of 45oC and 1,500 kPa. A valve in the line is cracked so
that air flows slowly into the tank until the pressure equals the line pressure. If the process occurs slowly
enough that the temperature in the tank remains at 25 oC, how much heat is lost from the tank? Assume
air to be an ideal gas for which CP = (7/2) R and CV = (5/2) R
Given Data:
Volume=V =0.1 m
3
T 1 =25 o. C =298 K
7
C P= R
2
Heat lost=Q=?
P1=101.33 kPa
T 2 =45 o.C =318 K
5
CV = R
2
Solution:
According to first law of thermodynamics
∆ U =Q+W →(1)
Since
∆ H=∆ U +∆ ( PV )
∆ U =∆ H−∆ ( PV )
∆ U =∆ H−P ∆ V −V ∆ P→ ( a )
Also, we know that
W =−P ∆ V → ( b )
Put (a) & (b) in (1)
∆ H−P ∆V −V ∆ P=Q−P ∆ V
Also, we have
∆ H=nC P ∆T
∆ H=nC P ( T 2−T 1 )
Put in (2)
n C P ( T 2−T 1 ) −V ∆ P=Q →(3)
For “n”
We know that for an ideal gas,
PV =nRT
Initial number of moles of gas can be obtained as,
23
∆ H−V ∆ P=Q → ( 2 )
P2=1500 kPa
August 20,
2013
P1 V =n1 R T 1
PROBLEMS
n1=
ZAID YAHYA
11-CH-74
P1 V
RT1
The final number of moles of gas at temperature T1 are
P2 V =n2 R T 1
n2=
P2 V
RT1
Now, Applying molar balance
n=n1−n2
n=
P1 V P2 V
−
R T1 R T1
n=
( P1−P 2) V
RT1
Put in (3)
( P 1−P2 ) V
R T1
( P1−P2 ) V
C ( T 2−T 1 )−V ∆ P=Q
P
( P1−P2 ) V
T1
2
RT1
2
∗7
R∗( T 2−T 1 )−V ∆ P=Q
∗7
∗( T 2 −T 1 ) −V ( P 2−P1 )=Q
( 101.33−1500 ) kPa∗0.1 m3
∗7
298 K
3
∗( 318−298 ) K −0.1 m ( 1500−101.33 ) kPa=Q
2
kPa∗1 kN
∗1 kJ
2
3 1 kPa∗m
Q=−172.717 m
1 kNm
Q=−172.717 kJ Answer
Problem 3.17:
A rigid, no conducting tank with a volume of 4 m 3 is divided into two unequal parts by a thin membrane.
One side of the membrane, representing 1/3 of the tank, contains nitrogen gas at 6 bars and 100 oC, and
the other side, representing 2/3 of the tank, is evacuated. The membrane ruptures and the gas fills the
tank.
a) What is the final, temperature of the gas? How much work is done? Is the process reversible?
b) Describe a reversible process by which the gas can be returned to its initial state, How much work
is done
Assume nitrogen is an ideal gas for which CP = (7/2) R & CV = (5/2) R
Given Data:
24
August 20,
2013
PROBLEMS
ZAID YAHYA
3
Volume of the tank=V 1=4 m
V 3=
V 1∗2 8 3
= m
3
3
V 2=
V 1∗1 4 3
= m
3
3
Pressure=P2=6 ¯¿
11-CH-74
o
Temperature=T 1=100 . C
Solution:
(a)
Finaltemperature=T 2=?
According to first law of thermodynamics
∆ U =Q+W
Since
No work is done & no heat is transferred
Therefore
Q=W =0
mC V ∆ T =0
∆ U =0
∆ T =0
T 2 −T 1=0
T 2 =T 1
T 2 =100℃ Answer
No, process is not reversible
(b)
Since
Therefore, the process is isothermal
For an isothermal process we have
W =−R T 2 ln
V2
V1
As, for an ideal gas
P2 V 2 =R T 2
W =−P2 V 2 ln
V2
V1
4
4
m3∗101325 N
kJ
W =−6 ¯¿ m3 ln
W =8.788 ¯¿
∗1
2
3
3∗4
1000 Nm
1.01325 ¯¿ m
W =878.8 kJ Answer
Problem 3.18:
An ideal gas initially at 30 0C and 100 kPa undergoes the following cyclic processes in a closed system:
a) In mechanically reversible processes, it is first compressed adiabatically to 500 kPa then cooled at
a constant pressure of 500 kPa to 30 0C and finally expanded isothermally to its original state
25
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
b) The cycle traverses exactly the same changes of state but each step is irreversible with an efficiency
of 80% compared with the corresponding mechanically reversible process NOTE: the initial step
can no longer be adiabatic
Find Q W ∆ U and ∆ H for each step of the process and for the cycle Take C p = (7/2) R and CV =
(5/2) R
Given Data:
T 1 =30 0.C
T 1 =303.15 K
P1=100 kPa
Q=?
W =?
∆ U =?
∆ H=?
7
C P= R
2
5
CV = R
2
lution:
(a)
P2=500 kPa
1) Adiabatic Compression from point 1 to point 2
Q12=0
Now, from first law of thermodynamics,
∆ U 12=Q12 +W 12
∆ U 12=W 12
W 12=∆ U 12=C V ∆ T 12
5
W 12=∆ U 12= R ( T 2−T 1 ) → ( 1 )
2
For ‘T2’
We know that
T 2 P2
=
T 1 P1
( )
γ −1
γ
T 2 =T 1
P2
P1
( )
γ −1
γ
T 2 =303.15 K
500
100
( )
1.4−1
1.4
T 2 =480.13 K
Put in (1)
5
J
kJ
kJ
W 12=∆ U 12= ∗8.314
( 480.13−303.15 ) K∗1
W 12=∆ U 12=3.679
2
mol∗K
1000 J
mol
Also, we have
∆ H 12=C P ( T 2−T 1)
7
J
kJ
∆ H 12= ∗8.314
( 480.13−303.15 ) K∗1
2
mol∗K
1000 J
2) Cooling at constant pressure from point 2 to point 3
Therefore at constant pressure we have,
Q23 =∆ H 23=C P ∆T 23
26
7
Q23=∆ H 23= R ( T 3−T 2 )
2
∆ H 12=5.15
kJ
mol
So
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Here
T 3 =303.15 K
7
J
kJ
kJ
Q23=∆ H 23= ∗8.314
( 303.15−480.13 ) K∗1
Q23=∆ H 23=−5.15
2
mol∗K
1000 J
mol
Also, we have
∆ U 23 =C V ( T 3−T 2)
5
J
kJ
∆ U 23= ∗8.314
( 303.15−480.13 ) K∗1
2
mol∗K
1000 J
kJ
∆ U 23=−3.679
mol
Now, from first law of thermodynamics,
∆ U 23=Q23 +W 23
W 23=∆U 23−Q23
W 23=−3.679+5.15
W 23=1.471
3) Isothermal expansion from point 3 to point 1
Since for an isothermal process temperature remains constant
Therefore,
∆ U 31 =∆ H 31=0
Here
P3=P2=500 kPa
For an Isothermal process we have
500
∗1 kJ
P3
J
100
W 31=−RT 3 ln W 31=−8.314
∗303.15 K∗ln
P1
mol∗K
1000 J
W 31=−4.056
kJ
mol
According to first law of thermodynamics
∆ U 31 =Q31 +W 31
0=Q31+W 31
Q31=−W 31
Q31=4.056
kJ
mol
For the complete cycle,
Q=Q12 +Q23+ Q31
27
Q=0−5.15+4.056
Q=−1.094
kJ
Answer
mol
kJ
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
W =W 12+W 23 +W 31
W =3.679+ 1.471−4.056
W =1.094
∆ H=∆ H 12+ ∆ H 23+ ∆ H 31
∆ U =∆ U 12 +∆ U 23 +∆ U 31
11-CH-74
kJ
Answer
mol
∆ H=5.15−5.15+0
∆ U =3.679−3.679+0
kJ
Answer
mol
kJ
∆ U =0
Answer
mol
∆ H=0
(b)
If each step that is 80% accomplishes the same change of state then values of
in part (a) but values of Q & W will change.
1. Adiabatic Compression from point 1 to point 2
W 12=
W 12
0.8
W 12=
3.679
0.8
W 12=4.598
∆U
&
∆H
will remain same as
kJ
mol
According to first law of thermodynamics
∆ U 12 =Q12 +W 12
3.679
kJ
kJ
=Q12+ 4.598
mol
mol
Q12=3.679
kJ
kJ
−4.598
mol
mol
Q12=−0.92
kJ
mol
2. Cooling at constant pressure from point 2 to point 3
W 23=
W 23
0.8
W 23=
1.471
0.8
W 23=1.839
kJ
mol
According to first law of thermodynamics
∆ U 23 =Q23 +W 23
−3.679
kJ
kJ
=Q23 +1.839
mol
mol
Q23=−3.679
kJ
kJ
−1.839
mol
mol
kJ
∗0.8
mol
W 31=3.245
3. Isothermal expansion from point 3 to point 1
Since initial step can no longer be adiabatic , therefore
W 31=W 31∗0.8
W 31=−4.056
According to first law of thermodynamics
∆ U 31 =Q31 +W 31
28
Q31=−W 31+0
kJ
mol
Q23=−5.518
kJ
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
Q31=3.245
11-CH-74
kJ
mol
For the complete cycle,
Q=Q12 +Q 23+Q31
W =W 12 +W 23+W 31
Q=−3.193
Q=−0.92−5.518+ 3.245
W =4.598+1.839−3.245
W =3.192
kJ
Answer
mol
kJ
Answer
mol
Problem 3.19:
One cubic meter of an ideal gas at 600 K and 1,000 kPa expands to five times its initial volume as follows:
a) By a mechanically reversible, isothermal process
b) By a mechanically reversible adiabatic process
c) By adiabatic irreversible process in which expansion is against a restraining pressure of 100 kPa
For each case calculate the final temperature, pressure and the work done by the gas, Cp=21 J mol-1K-1.
Given Data:
V 1=1 m3
T 1 =600 K
P1=1000 kPa
V 2=5 V 1
V 2=5 m3
C P=21
J
mol K
W =?
Solution:
We know that,
C P−C V =R
CV =C P −R
CV =( 21−8.314 )
J
mol∗K
CV =12.686
As
γ=
CP
CV
γ =1.6554
(a)
Since, for an isothermal process
Temperature remains constant, therefore
T 2 =T 1=600 K Answer
29
For an ideal gas we have
J
mol∗K
CV =?
T 2 =?
P2=?
August 20,
2013
PROBLEMS
P1 V 1 P 2 V 2
=
T1
T2
ZAID YAHYA
P1 V 1
∗T 2
T1
P 2=
V2
11-CH-74
1000 kPa∗1 m3
∗600 K
600 K
P 2=
5 m3
P2=200 kPa Answer
We know that, for an isothermal process
W =−R T 1 ln
V2
V1
Since
P1 V 1=R T 1
Therefore,
5
∗N
1
∗J
Pa∗m2
3
W =−1000 kPa∗1m ln
Nm
V2
W =−P1 V 1 ln
V1
W =−1609.43 kJ Answer
(b)
We know that, for an adiabatic process
P1 V 1γ =P2 V 2γ
V1
V2
γ
( )
P2 V 2
T2=
∗T
P1 V 1 1
N
∗J
( 69.65∗5−1000∗1 ) kPa∗m Pa∗m2
W=
1.6554−1
Nm
3
Pr=100 kPa
Since, for an adiabatic process
Q=0
30
P2=69.65 kPa Answer
69.65 kPa∗5 m3
T2=
∗600 K
1000 kPa∗1 m3
(c)
1.6554
T 2 =208.95 K Answer
For an adiabatic process work done is
P V −P1 V 1
W= 2 2
γ −1
1
5
()
P2=1000 kPa∗
For an ideal gas we have
P1 V 1 P2 V 2
=
T1
T2
P2=P1
According to first law of thermodynamics
W =−994.43 kJ Answer
August 20,
2013
PROBLEMS
ZAID YAHYA
∆ U =Q+W
∆ U =W
kPa∗m3∗N
∗J
Pa∗m2
∆ U =W =−100 ( 5−1 )
Nm
∆ U =W =−Pr dV
∆ U =−400 kJ
T2=
n CV ∆T =−400 kJ
n CV ( T 2−T 1 )=−400 kJ
−400 kJ
+T 1 → ( 1 )
n CV
n=
P1 V 1
RT1
1000 kPa∗1 m3∗mol∗K
∗kN
8.314 J∗600 K
∗kJ
kPa∗m2
n=
kNm
n=0.2005 mol
Put in (1)
−400 kJ∗mol∗K
∗1000 J
0.2005 mol∗12.686 J
T2=
+600 K
1 kJ
∆ U =W =−Pr ( V 2−V 1)
For an ideal gas we have,
P1 V 1=nR T 1
11-CH-74
T 2 =−157.26 K +600 K
T 2 =442.74 K Answer
For an ideal gas we have
P1 V 1 P2 V 2
=
T1
T2
P1 V 1
∗T 2
T1
P2=
V2
1000 kPa∗1 m3
∗442.74 K
600 K
P2=
5 m3
P2=147.58 kPa Answe r
Problem 3.20:
One mole of air, initially at 150 0C and 8 bars undergoes the following mechanically reversible changes. It
expands isothermally to a pressure such that when it is cooled at constant volume to 50 0C its final
pressure is 3 bars. Assuming air is an ideal gas for which C P = (7/2) R and CV = (5/2) R, calculate W, Q,
∆ U , and ∆ H
Given Data:
Mole of air=n=1mol
Initial Temperature=T 1=150 0.C =423.15 K
0
Finaltemperature =T 3=50 .C =323.15 K
Final pressure=P 3=3 ¯
¿
Initial pressure=P1=8 ¯¿
7
C P= R
2
5
CV = R
2
Solution:
Since process is reversible
Two different steps are used in this case to reach final state of the air.
31
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Step 12:
For step 12 temperatures is constant,
T 1 =T 2
Therefore
∆ U 12 =∆ H 12=0
For an isothermal process we have
W 12=R T 1 ln
V1
V2
As
V 2=V 3
W 12=R T 1 ln
V1
→(1)
V3
We know that
P1 V 1 P3 V 3
=
T1
T3
V 1 P3 ¿T 1
=
V 3 T 3∗P1
8.314 J∗423.15 K
∗1 kJ
mol∗K
3∗423.15
W 12=
∗ln
1000 J
8∗323.15
kJ
W 12=−2.502
mol
P ¿T
W 12=R T 1 ln 1 3
T 1∗P 3
According to first law of thermodynamics
∆ U 12=Q12 +W 12
0=Q12 +W 12
Q12=−W 12
Q12=2.502
kJ
mol
Step 23:
For step 23 volume is constant,
Therefore,
W 23=0
According to first law of thermodynamics
∆ U 23=Q23 +W 23
∆ U 23 =Q23 +0
Q 23=∆ U 23
Q23=∆ U 23 =CV ∆ T
Q23=∆ U 23=CV ( T 3−T 2 )
5
Q23=∆ U 23= R ( 323.15−423.15 ) K
2
J
∗1 kJ
5
mol∗K
kJ
Q23=∆ U 23= ∗8.314
(−100 ) K Q23=∆U 23 =−2.0785
2
1000 J
mol
32
W e know that
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
J
∗1 kJ
7
mol∗K
∆ H 23= ∗8.314
( 423.15−323.15 ) K
2
1000 J
kJ
∆ H 23=2.91
mol
∆ H 23 =C P ∆ T
∆ H 23 =C P ( T 3−T 2 )
For the complete cycle,
Work=W =W 12 +W 23
Q=Q12 +Q23
W = (−2.502+0 )
Q=( 2.502−2.0785 )
∆ U =∆ U 12 +∆ U 23
∆ H=∆ H 12+ ∆ H 23
kJ
mol
kJ
mol
W =−2.502
Q=0.424
kJ
Answe r
mol
kJ
Answer
mol
∆ U =( 0−2.0785 )
kJ
mol
∆ U =−2.0785
∆ H= ( 0−2.91 )
kJ
mol
∆ H=−2.91
kJ
Answe r
mol
kJ
Answe r
mol
Problem 3.21:
An ideal gas flows through a horizontal tube at steady state. No heat is added and no shaft work is done.
The cross-sectional area of the tube changes with length, and this causes the velocity to change. Derive an
equation relating the temperature to the velocity of the gas. If nitrogen at 150 0C flows past one section of
the tube with a velocity of 2.5 m/s, what is the temperature at another section where its velocity is 50 m/s?
Let CP = (7/2) R
Given Data:
0
Temperature=T 1=150 . C=423.15 K
Velocity=u 1=2.5
m
sec
Molecualr weight of Nitrogen=28
Solution:
Applying energy balance for steady state flow process
∆ H+
∆ u2
+ g ∆ z=Q+W S
2
Since
∆ z=W S=Q=0
33
T 2 =?
g
mol
u2=50
m
sec
7
C P= R
2
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Therefore,
∆ u2
∆H+
=0
2
−∆u 2
CP ∆ T =
2
2
−u −u1
C P ( T 2 −T 1 ) = 2
2
2
−u22−u12
T2=
+ T1
2C P
−( 50 2−2.52 )∗2∗m 2∗mol∗K
∗28 g Nitrogen
2∗7∗8.314 J∗sec 2
∗J
1 mol Nitrogen
T 2 =−1.199 K + 423.15 K
∗N∗sec 2
N∗m
∗1 kg
kg∗m
T2=
+ 423.15 K
1000 g
T 2 =421.95 K
Answe r
T 2 =( 421.95−273.15 ) 0.C
T 2 =148.8 0.C
Problem 3.22:
One mole of an ideal gas, initially at 30 0C and 1 bar, is changed to 130 0C and 10 bars by three different
mechanically reversible processes:
a) The gas is first heated at constant volume until its temperature is 130 0C; then it is compressed
isothermally until its pressure is 10 bar
b) The gas is first heated at constant pressure until its temperature is 130 0C; then it is compressed
isothermally to 10 bar
c) The gas is first compressed isothermally to 10 bar; then it is heated at constant pressure to 130 0C
Calculate Q, W, ∆ U ∧∆ H in each case. Take CP = (7/2) R and CV = (5/2) R. alternatively, take CP =
(5/2) R and CV = (3/2) R
Given Data:
0
T 1 =30 .C
T 1 =( 30+273.15 ) K
T 3 =403.15 K
P1=1 ¯
¿
T 1 =303.15 K
P3=10 ¯¿
Q=?
W =?
Solution:
7
C P= R
2
5
CV = R
2
Each part consist of two steps, 12 & 23
For the overall processes
∆ U =∆ U 12=∆U 23=CV ∆T
34
5
∆ U =∆ U 12=∆U 23= R ( T 3 −T 1 )
2
0
T 2 =130 .C
∆ U =?
T 3 =( 130+273.15 ) K
∆ H=?
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
5
J
K∗1 kJ
∆ U =∆ U 12=∆U 23= ∗8.314
( 403.15−303.15 )
2
mol∗K
1000 J
∆ U =∆ U 12=∆U 23=2.079
kJ
→(a) Answe r
mol
Now
∆ H=∆ H 12 =∆ H 23=C P ∆ T
7
∆ H=∆ H 12 =∆ H 23= R ( T 2−T 1 )
2
7
J
kJ
∆ H=∆ H 12=∆ H 23= ∗8.314
( 403.15−303.15 ) K∗1
2
mol∗K
1000 J
kJ
∆ H=∆ H 12=∆ H 23=2.91
→ ( b ) Answer
mol
(a)
Step 12:
For step “12” volume is constant
Therefore
W 12=0
Here
T 2 =T 3
According to first law of thermodynamics
∆ U 12 =Q12 +W 12
∆ U 12 =Q12
Q12=∆ U 12=C V ∆ T
Also we have
∆ H 12=2.91
kJ
[ ¿( b)]
mol
Step 23:
Since for step “23” process is isothermal
Therefore
∆ U 23 =∆ H 23=0
Here
35
Q12=∆ U 12=2.079
kJ
[¿ ( a ) ]
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
T 2 =T 3
Now, intermediate pressure can be calculated as
P 1 P2
=
T1 T 2
P 2=
P1
∗T
T1 2
1
¯¿
∗403.15 K
303.15 K
P2=¿
For an isothermal process we have
W 23=R T 2 ln
P3
P2
W 23=8.314
J
K∗1 kJ
10
∗403.15
∗ln
mol∗K
1000 J
1.329
W 23=6.764
P2=1.329 b ar
kJ
mol
According to first law of thermodynamics
∆ U 23=Q23 +W 23
0=Q23+W 23
Q23=−W 23
Q23=−6.764
kJ
mol
For the complete cycle,
Work=W =W 12 +W 23
Q=Q 12 +Q23
W = ( 0+6.764 )
Q=( 2.079−6.764 )
∆ U =∆ U 12 +∆ U 23
kJ
mol
kJ
mol
W =6.764
Q=−4.685
kJ
Answe r
mol
∆ U =( 2.079+0 )
kJ
mol
∆ U =2.079
∆ H= ( 2.91+0 )
kJ
mol
∆ H=2.91
∆ H=∆ H 12+ ∆ H 23
kJ
Answe r
mol
kJ
Answe r
mol
kJ
Answe r
mol
(b)
Step 12:
For step “12” volume is constant
Therefore, at constant pressure we have
Q12 ¿ ∆ H 12=2.91
kJ
mol
[ ¿(b) ]
Also,
∆ U 12 =2.079
36
kJ
mol
[ ¿( a) ]
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
According to first law of thermodynamics
∆ U 12=Q12 +W 12
W 12=∆ U 12−Q 12
W 12=( 2.079−2.91 )
kJ
mol
W 12=−0.831
Step 23:
Since for step “23” process is isothermal ( T = Constant)
Therefore
∆ U 23 =∆ H 23=0
Here
T 2 =T 3 ∧P 1=P2
For an isothermal process we have
W 23=R T 2 ln
P3
P2
W 23=8.314
J
K∗1 kJ
10
∗403.15
∗ln
mol∗K
1000 J
1
W 23=7.718
kJ
mol
According to first law of thermodynamics
∆ U 23=Q23 +W 23
0=Q23+W 23
Q23=−W 23
Q23=−7.718
kJ
mol
For the complete cycle,
Work=W =W 12 +W 23
Q=Q 12 +Q23
W = (−0.831+7.718 )
Q=( 2.91−7.718 )
∆ U =∆ U 12 +∆ U 23
∆ H=∆ H 12+ ∆ H 23
kJ
mol
kJ
mol
Q=−4.808
kJ
mol
∆ U =2.079
∆ H= ( 2.91+0 )
kJ
mol
∆ H=2.91
Since for step “12” process is isothermal ( T = Constant)
Therefore
∆ U 12 =∆ H 12=0
P2=P3
37
kJ
Answer
mol
kJ
Answe r
mol
∆ U =( 2.079+0 )
(c)
Here
W =6.887
kJ
Answe r
mol
kJ
Answe r
mol
kJ
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
For an isothermal process we have
W 12=R T 1 ln
P2
P1
W 12=8.314
W 12=5.8034
11-CH-74
J
K∗1 kJ
10
∗303.15
∗ln
mol∗K
1000 J
1
kJ
mol
According to first law of thermodynamics
∆ U 12=Q12 +W 12
0=Q12+W 12
Q12=−W 12
Q12=−5.8034
kJ
mol
Step 23:
For step “23” volume is constant
Therefore, at constant pressure we have
Q23=∆ H 23=2.91
kJ
[ ¿ (b)]
mol
Here
T 2 =T 3
Now
∆ U 23=2.079
kJ
[ ¿(a)]
mol
According to first law of thermodynamics
∆ U 23=Q23 +W 23
W 23=∆U 23−Q23
W 23=( 2.079−2.91 )
kJ
mol
W 23=−0.831
For the complete cycle,
Work=W =W 12 +W 23
Q=Q 12 +Q23
Q=(−5.8034+2.91 )
∆ U =∆ U 12 +∆ U 23
∆ H=∆ H 12+ ∆ H 23
38
W = (5.8034−0.831 )
kJ
mol
kJ
mol
W =4.972
Q=−2.894
kJ
Answer
mol
kJ
Answe r
mol
∆ U =( 0+2.079 )
kJ
mol
∆ U =2.079
∆ H= ( 0+2.91 )
kJ
mol
∆ H=2.91
kJ
Answe r
mol
kJ
Answe r
mol
kJ
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Solution:
5
C P= R
2
3
CV = R
2
Each part consist of two steps, 12 & 23
For the overall processes
∆ U =∆ U 12=∆U 23=CV ∆T
3
∆ U =∆ U 12=∆U 23= R ( T 3 −T 1 )
2
3
J
K∗1 kJ
∆ U =∆ U 12=∆U 23= ∗8.314
( 403.15−303.15 )
2
mol∗K
1000 J
∆ U =∆ U 12=∆U 23=1.247
kJ
→(a) Answe r
mol
Now
∆ H=∆ H 12 =∆ H 23=C P ∆ T
5
∆ H=∆ H 12=∆ H 23= R ( T 2−T 1 )
2
5
J
kJ
∆ H=∆ H 12=∆ H 23= ∗8.314
( 403.15−303.15 ) K∗1
2
mol∗K
1000 J
kJ
∆ H=∆ H 12=∆ H 23=2.079
→ ( b ) Answe r
mol
(a)
Step 12:
For step “12” volume is constant
Therefore
W 12=0
Here
T 2 =T 3
According to first law of thermodynamics
∆ U 12=Q12 +W 12
39
∆ U 12 =Q12
Q12=∆ U 12=C V ∆ T
Q12=∆ U 12=1.247
kJ
[ ¿ (a)]
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
Also we have
∆ H 12=2.079
kJ
[ ¿ (b)]
mol
Step 23:
Since for step “23” process is isothermal
Therefore
∆ U 23 =∆ H 23=0
Here
T 2 =T 3
Now, intermediate pressure can be calculated as
P 1 P2
=
T1 T 2
P1
P2= ∗T 2
T1
1
¯¿
∗403.15 K
303.15 K
P2=¿
For an isothermal process we have
W 23=R T 2 ln
P3
P2
W 23=8.314
J
K∗1 kJ
10
∗403.15
∗ln
mol∗K
1000 J
1.329
W 23=6.764
P2=1.329 b ar
kJ
mol
According to first law of thermodynamics
∆ U 23=Q23 +W 23
0=Q23+W 23
Q 23 =−W 23
Q23=−6.764
kJ
mol
For the complete cycle,
Work=W =W 12 +W 23
Q=Q12 +Q23
40
W = ( 0+6.764 )
Q=( 1.247−6.764 )
kJ
mol
kJ
mol
W =6.764
Q=−5.516
kJ
Answe r
mol
kJ
Answe r
mol
∆ U =∆ U 12 +∆ U 23
∆ U =( 1.247+0 )
kJ
mol
∆ U =1.247
kJ
Answer
mol
∆ H=∆ H 12+ ∆ H 23
∆ H= ( 2.079+ 0 )
kJ
mol
∆ H=2.079
kJ
Answe r
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
(b)
Step 12:
For step “12” volume is constant
Therefore, at constant pressure we have
Q12 ¿ ∆ H 12 =2.079
kJ
mol
[ ¿(b) ]
Also,
∆ U 12=1.247
kJ
mol
[ ¿( a) ]
According to first law of thermodynamics
∆ U 12 =Q12 +W 12
W 12=∆ U 12−Q 12
W 12=( 1.247−2.079 )
kJ
mol
W 12=−0.832
Step 23:
Since for step “23” process is isothermal ( T = Constant)
Therefore
∆ U 23 =∆ H 23=0
Here
T 2 =T 3 ∧P 1=P2
For an isothermal process we have
W 23=R T 2 ln
P3
P2
W 23=8.314
J
K∗1 kJ
10
∗403.15
∗ln
mol∗K
1000 J
1
W 23=7.718
kJ
mol
According to first law of thermodynamics
∆ U 23=Q23 +W 23
0=Q23+W 23
Q23=−W 23
Q23=−7.718
kJ
mol
For the complete cycle,
Work=W =W 12 +W 23
Q=Q12 +Q23
41
W = (−0.832+7.718 )
Q=( 2.079−7.718 )
kJ
mol
kJ
mol
W =6.886
Q=−5.639
kJ
Answe r
mol
kJ
Answer
mol
kJ
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
∆ U =∆ U 12 +∆ U 23
∆ U =( 1.247+0 )
kJ
mol
∆ U =1.247
kJ
Answer
mol
∆ H=∆ H 12+ ∆ H 23
∆ H= ( 2.079+ 0 )
kJ
mol
∆ H=2.079
kJ
Answe r
mol
(c)
Step 12:
Since for step “12” process is isothermal ( T = Constant)
Therefore
∆ U 12=∆ H 12=0
Here
P2=P3
For an isothermal process we have
W 12=R T 1 ln
P2
P1
W 12=8.314
W 12=5.8034
J
K∗1 kJ
10
∗303.15
∗ln
mol∗K
1000 J
1
kJ
mol
According to first law of thermodynamics
∆ U 12=Q12 +W 12
0=Q12+W 12
Q12=−W 12
Q12=−5.8034
Step 23:
For step “23” volume is constant
Therefore, at constant pressure we have
Q23=∆ H 23=2.079
kJ
[ ¿(b)]
mol
Here
T 2 =T 3
Now
∆ U 23=1.247
42
According to first law of thermodynamics
kJ
[ ¿(a) ]
mol
kJ
mol
August 20,
2013
PROBLEMS
∆ U 23=Q23 +W 23
ZAID YAHYA
W 23=∆U 23−Q 23
11-CH-74
W 23=( 1.247−2.079 )
kJ
mol
W 23=−0.832
kJ
mol
For the complete cycle,
Work=W =W 12 +W 23
Q=Q12 +Q23
W = (5.8034−0.832 )
Q=(−5.8034+2.079 )
kJ
mol
W =4.9714
kJ
mol
Q=−3.724
kJ
Answe r
mol
kJ
Answe r
mol
∆ U =∆ U 12 +∆ U 23
∆ U =( 0+1.247 )
kJ
mol
∆ U =1.247
kJ
Answer
mol
∆ H=∆ H 12+ ∆ H 23
∆ H= ( 0+2.079 )
kJ
mol
∆ H=2.079
kJ
Answe r
mol
Problem 3.23:
One mole of an ideal gas, initially at 30 ℃ and 1 bars, undergoes the following mechanically
reversible changes. It is compressed isothermally to point such that when it is heated at constant volume to
120 ℃ its final pressure is 12 bars. Calculate Q, W, ∆ U ∧∆ H for the process. Take CP = (7/2) R and
CV = (5/2) R.
Given Data:
T 1 =30℃
T 3 =393.15 K
T 1 =( 30+273.15 ) K
P3=12 ¯¿
Q=?
T 1 =303.15 K
W =?
P1=1 ¯
¿
∆ U =?
∆ H=?
Solution:
The process consist of two steps, 12 & 23
Step 12:
Since for step “12” process is isothermal ( T = Constant)
Therefore
∆ U 12 =∆ H 12=0
Now, intermediate pressure can be calculated as
43
T 3 =120 ℃
7
C P= R
2
T 3 =( 120+273.15 ) K
5
CV = R
2
August 20,
2013
PROBLEMS
P2
T2
ZAID YAHYA
P
¿ 3
T3
P
P2= 3 ∗T 2
T3
12
¯¿
∗303.15 K
393.15 K
P2=¿
P2=9.25 b ar
For an isothermal process we have
W 12=R T 1 ln
P2
P1
W 12=8.314
J
K∗1 kJ
9.25
∗303.15 .15
∗ln
mol∗K
1000 J
1
W 12=5.607
11-CH-74
kJ
mol
According to first law of thermodynamics
∆ U 12=Q12 +W 12
0=Q12+W 12
Q12=−W 12
Q12=−5.607
kJ
mol
Step 23:
Since for step “23” volume is constant
Therefore
W 23=0
According to first law of thermodynamics
∆ U 23=Q23 +W 23
∆ U 23=Q 23
Q23 =∆ U 23 =CV ∆ T
5
Q23=∆ U 23= R ( T 3−T 1 )
2
5
J
K∗1 kJ
kJ
Q23=∆ U 23= ∗8.314
( 393.15−303.15 )
Q23=∆ U 23=1.871
2
mol∗K
1000 J
mol
Now
∆ H 23=C P ∆ T
7
∆ H 23= R ( T 2−T 1 )
2
7
J
kJ
∆ H 23= ∗8.314
( 393.15−303.15 ) K∗1
2
mol∗K
1000 J
∆ H 23=2.619
For the complete cycle,
Work=W =W 12 +W 23
Q=Q 12 +Q23
44
W = (5.607 +0 )
Q=(−5.607+1.871 )
kJ
mol
kJ
mol
W =5.607
Q=−3.736
kJ
Answe r
mol
kJ
Answe r
mol
kJ
mol
August 20,
2013
PROBLEMS
ZAID YAHYA
11-CH-74
∆ U =∆ U 12 +∆ U 23
∆ U =( 0+1.871 )
kJ
mol
∆ U =1.871
kJ
Answe r
mol
∆ H=∆ H 12+ ∆ H 23
∆ H= ( 0+2.691 )
kJ
mol
∆ H=2.691
kJ
Answe r
mol
Problem 3.24:
A process consists of two steps: (1) One mole of air at T = 800 K and P = 4 bars are cooled at constant
volume to T = 350 K. (2) The air is then heated air constant pressure until its temperature reaches 800 K.
If this two step process is replaced by a single isothermal expansion of the air from 800 K and 4 bar to
some final pressure P, what is the value of P that makes the work of two step processes the same? Assume
mechanical reversibility and treat air as an ideal gas with CP = (7/2) R and CV = (5/2) T.
Given Data:
T 1 =800 K
P1=4 ¯
¿
T 2 =350 K
P=?
Solution:
For the first step volume is constant
Therefore,
W 12=0
For the work done is
W =W 23=−P 2 ∆ V →(1)
For one mole of an ideal gas we have,
P ∆ V =R ∆ T
P2 ∆ V =R ∆ T
Put in (1)
W =−R ∆ T
W =−R ( T 3−T 2 )
Since
T 3 =T 1
Therefore
W =−R ( T 1−T 2) →(2)
45
For an isothermal process we have
August 20,
2013
W =R T 1 ln
PROBLEMS
ZAID YAHYA
11-CH-74
P
→ (3)
P1
Compare (2) and (3)
P
−R ( T 1−T 2 )=R T 1 ln
P1
P
4 ¯¿
(350−800 ) K
=ln ¿
800 K
¯
P=2.279 Answer
T 2−T 1
P
=ln
T1
P1
P
T 2 −T 1=T 1 ln
P1
4 ¯¿ 0.5698=P
4 ¯¿
P
e−0.5625 = ¿
Problem 3.25:
A scheme for finding the internal volume V tB of the gas cylinder consists of the following steps. The
cylinder is filled with a gas to low pressure P 1, and connected through a small line and valve to an
evacuated reference tank of known volume V tA . The valve is opened, and the gas flows through the line
into the reference tank. After the system returns to its initial temperature, a sensitive pressure transducer
provides a valve for the pressure change ∆ P in the cylinder. Determine the cylinder volume V tB
from the following data:
a) V tA =256 cm3
b) ∆ P/ P1=−0.0639
Given Data:
t
V B =?
t
V A =256 cm
3
∆P
=−0.0639
P1
Solution:
∆P
=−0.0639
P1
P2 −P 1
=−0.0639
P1
P2
−1=−0.0639
P1
P2
=−0.0639+1
P1
P2
=0.9361 →(1)
P1
Assume that gas is ideal & P2 is th