GROUP (22)
1
SHEET (3)
2nd Law of
Thermodynamics
2
Question No. (1)
The initial state of one mole of an ideal gas is P = 10 atm
and T = 300K. Calculate the entropy change in the gas for
(a) An isothermal decrease in the pressure to 1 atm;
(b) A reversible adiabatic decrease in the pressure to 1
atm;
(c) A constant volume decrease in the pressure to 1 atm.
The Answer of Question No. (1)
a) Isothermal
∆S = NRln
V2
P1
= NRln
V1
P2
10
= 1 ∗ 8.314 ∗ ln
1
∆S = 19.14 J/K
b) Reversible Adiabatic
∆S = 0
c) Constant Volume
T2
P2
∆S = NCv ln = NCv ln
T1
P1
3
3
1
= 1 ∗ ∗ 8.314 ∗ ln
2
10
= − 28.71 J/K
4
Question No. (2)
One mole of an ideal gas is subjected to the following
sequence of steps:
a.Starting at 25 ˚C and 1 atm, the gas expands freely into
a vacuum to double its volume.
b.The gas is next heated to 125 ˚C at constant volume.
c. The gas is reversibly expanded at constant temperature
until its volume is again doubled.
d.The gas is finally reversibly cooled to 25 ˚C at constant
pressure.
Calculate ∆U, ∆H, q, W, and ∆S in the gas.
The Answer of Question No. (2)
a) Expands freely into a vacuum to double its volume
RT1 0.082 ∗ 298
V1 =
=
= 24.436 L
P1
1
V2 = 2 V1 = 48.872 L
V2
P1
=
V1
P2
1
=2
P2
5
P2 = 0.5 atm
T2 = T1 = 298 K
∆U = W = q = ∆H = 0
V2
∆S = NR ln = 1 ∗ 8.314 ln 2
V1
= 5.763 J/K
b) Heated to 125 ˚C at constant volume
T3 = 398 K
W=0
∆U = NCv ∆T
3
= 1 ∗ ∗ 8.314 ∗ (398 − 298)
2
= 1247.1 J
q = ∆U = 1247.1 J
∆H = NCp ∆T
5
= 1 ∗ ∗ 8.314 ∗ (398 − 298)
2
= 2078.5 J
6
∆S = NCv ln
T2
T1
3
398
= 1 ∗ ∗ 8.314 ∗ ln
2
298
= 3.6086 J/K
c) Reversibly expanded at constant temperature until its
volume is again doubled
V4 = 2V3 = 2 ∗ 48.872 = 97.744 L
∆U = 0
W = NRT ln
V3
V4
1
= 1 ∗ 8.314 ∗ 398 ∗ ln
2
= − 2293.6 J
q = − W = 2293.6 J
∆H = 0
V4
∆S = NR ln = 1 ∗ 8.314 ∗ ln 2
V3
= 5.763 J/K
d) Reversibly cooled to 25 ˚C at constant pressure
W = P(V4 − V5 ) = NR(T4 − T5 )
= 1 ∗ 8.314 ∗ (398 − 298)
= 831.4 J
7
∆U = NCv ∆T
3
= 1 ∗ ∗ 8.314 ∗ (298 − 398)
2
= − 1247.1 J
q = ∆U − W
= −1247.1 − 831.4
= − 2078.5 J
∆H = qp = − 2078.5 J
T5
∆S = NCp ln
T4
5
298
= 1 ∗ ∗ 8.314 ∗ ln
2
398
= − 6.014 J/K
Process ∆U (J)
W (J)
q (J)
∆H (J) ∆S (J/K)
a)
0
0
0
0
5.763
b)
1247.1
0
1247.1
2078.5
3.6086
c)
0
0
5.763
d)
-1247.1
Total
0
-2293.6 2293.6
831.4
-2078.5 -2078.5 -6.014
-1462.2 1462.2
8
0
9.1206
Question No. (3)
Calculate the final temperature and the entropy produced
when 1500 grams of lead at 100 ˚C is placed in 100 grams
of adiabatically contained water, the initial temperature of
which is 25 ˚C. Given that: Cp(H2O) =18 cal/(deg.mole)
and Cp(Pb) =6.38 cal/(deg.g.atom). The molecular and
atomic weights of H2O and Pb are 18 and 207 grams,
respectively.
The Answer of Question No. (3)
∑ dq = 0
NCp ∆T)water + NCp ∆T)lead = 0
Nw Cp(w) (Tf − Tw ) = NPb Cp(Pb) (TPb − Tf )
100
1500
∗ 18.03 ∗ (Tf − 25) =
∗ 6.38 ∗ (100 − Tf )
18
207
∴ Tf = 48.6846 ℃
∆Stotal = ∆Swater + ∆Slead
∆Swater = Nw Cp(w) ln
9
Tf
Tw
=
100
321.6846
∗ 4.18 ∗ 18.03 ∗ ln
18
298
= 32.02 J/K
∆Slead
Tf
= NPb Cp(Pb) ln
TPb
1500
321.6846
=
∗ 4.18 ∗ 6.38 ∗ ln
207
373
= − 28.6 J/K
∆Stotal = 32.02 − 28.6
= 3.42 J/K
10
Question No. (4)
From 298 K up to its melting temperature of 1048 K, the
constant pressure molar heat capacity of RbF is given as:
Cp = 7.97+ 9.2 x 10-3 T + 1.21 x 105 T-2 cal/ (deg.mole)
and from the melting temperature to 1200 K, the constant
pressure molar heat capacity of liquid RbF is given as:
Cp = -11.3 + 0.833 x 10-3 T + 350.7 x 105 T-2 cal/ (deg.mole)
At its melting temperature the molar heat of fusion of RbF
is 6300 cal/mole. Calculate the increase in the entropy of
1 mole of RbF when it is heated from 300 K to 1200 K.
The Answer of Question No. (4)
∆S1 : Heating solid RbF from 300 K to 1048 K.
∆S2 : Melting RbF at 1048 K.
∆S3 : Heating liquid RbF from 1048 K to 1200 K.
Tm
∆S1 = ∫
Ti
Tm NC
δq
p(s)
=∫
dT
T
T
Ti
1048
=∫
(7.97T −1 + 9.2 ∗ 10−3 + 1.21 ∗ 105 T −3 ) dT
300
11
1048
5
1.21
∗
10
= [7.97 ln T + 9.2 ∗ 10−3 T +
T −2 ]
−2
300
∆S1 = 17.468 J/K
qm
∆S2 =
T
6300
=
= 6.01145 J/K
1048
Tf
Tf NC
δq
p(l)
∆S3 = ∫
=∫
dT
T
T
Tm
Tm
1200
=∫
1048
(
−11.3
+ 0.833 ∗ 10−3 + 350.7 ∗ 105 T −3 ) dT
T
1200
5
350.7
∗
10
= [−11.3 ln T + 0.833 ∗ 10−3 T +
T −2 ]
−2
1048
= 2.3846 J/K
∆Stotal = ∆S1 + ∆S2 + ∆S3
= 17.468 + 6.01145 + 2.3846
= 25.864 J/K
12
Question No. (5)
Two moles of an ideal gas are contained adiabatically at
30 atm pressure and 298 K. The pressure is suddenly
released to 10 atm, and the gas undergoes in an
irreversible adiabatic expansion as a result of which 500
cal of work are performed. Show that the final temperature
of the gas after the irreversible expansion is greater than
that which the gas would attain if the expansion from 30 to
10 atm had been conducted reversible. Calculate the
entropy produced as a result of the irreversible expansion.
Cv for the gas equals to 1.5R.
The Answer of Question No. (5)
If the process is reversible from 1
2
P2 𝛾−1
T2 𝛾
( )
=( )
P1
T1
5
−1
3
10
( )
30
5
3
T2
=(
)
298
T2 = 192 K
13
For the irreversible process from 1
3
∆U = W
NCv (T3 − T1 ) = W
3
2 ∗ ∗ 8.314 ∗ (T3 − 298) = −4.18 ∗ 500
2
T3 = 214.2 K
∴ T3 > T2
To calculate ∆Sirr , choose any reversible path from state
1 to state 3. Consider the reversible path 1
2
3
From 1
2 adiabatic
∆S = 0
From 2
3 constant pressure
∆S = NCp ln
T3
5
214.2
= 2 ∗ ∗ 8.314 ∗ ln
T2
2
192
= 4.548 J/K
∴ ∆Sirr = 4.548 J/K
14
Question No. (6)
At a pressure of 1 atm the equilibrium melting temperature
of lead is 600 K, and at this temperature the latent heat of
fusion of lead is 1150 cal/mole. If 1 mole of supercooled
liquid lead spontaneously freezes at 590 K and 1 atm
pressure, calculate the entropy produced. The constant
pressure molar heat capacity of liquid lead, as a function
of temperature at 1 atm pressure, is given as
Cp (Pb)l = 7.75 – 0.74 x 10-3 T cal/deg
And the corresponding expansion for solid lead is given as
Cp (Pb)s = 5.63 + 2.33 x 10-3 T cal/deg
The Answer of Question No. (6)
The freezing process happened spontaneously
(irreversibly), so to calculate ∆Sirr consider the following
reversible steps:
1. Heating supercooled liquid lead from 590 K to 600 K.
2. Solidifying the liquid lead at 600 K.
3. Cooling solid lead from 600 K to 590 K.
600
∆S1 = ∫
590
600 NC
δq
p(Pb(l))
=∫
dT
T
T
590
15
600
(7.75T −1 − 0.74 ∗ 10−3 ) dT
=∫
590
= [7.75 ln T − 0.74 ∗ 10−3 T]600
590
= 0.12285 cal/K
−qm
∆S2 =
T
−1150
=
600
= − 1.91667 cal/K
590
∆S3 = ∫
600
590 NC
δq
p(Pb(s))
=∫
dT
T
T
600
590
=∫
(5.63T −1 + 2.33 ∗ 10−3 ) dT
600
= [5.63 ln T + 2.33 ∗ 10−3 T]590
600
= − 0.1179 cal/K
∆Sirr = ∆S1 + ∆S2 + ∆S3
= 0.12285 − 1.91667 − 0.1179
= − 1.9117 cal/K
16
Question No. (7)
A sample of 1 mole Ar is expanded isothermally at 0 ˚C
from 22.4 lit to 44.8 lit; if the process is carried out: a)
reversibly, b) against constant external pressure equal to
the final pressure of the gas, or C) freely. Calculate ∆S (in
J/(deg.mole)) for each case.
The Answer of Question No. (7)
a) Isothermal
V2
∆S = NR ln
V1
44.8
= 1 ∗ 8.314 ∗ ln
22.4
= 5.763 J/K
b) Constant pressure
T2
V2
∆S = NCp ln = NCp ln
T1
V1
5
44.8
= 1 ∗ ∗ 8.314 ∗ ln
2
22.4
= 14.407 J/K
17
c) Freely: it is irreversible process. Consider the reversible
isothermal path (during free expansion temperature
does not change (ideal gas))
V2
∆S = NR ln
V1
44.8
= 1 ∗ 8.314 ∗ ln
22.4
= 5.763 J/K
18
Question No. (8)
A sample of 1 mole H2O(g) is condensed isothermally and
reversibly to liquid water at 100 ˚C. The standard enthalpy
of vaporization of water at 100 ˚C is 40.656 KJ/mole. Find
∆S in (KJ/(deg.mole)) for this process.
The Answer of Question No. (8)
−∆H
∆S =
T
=
−40.656
373
= − 0.109 KJ/K
19
Question No. (9)
The heat capacity of one mole sample of a perfect gas
was found to vary with temperature according to the
expression Cp(J/(deg.mole) = 20.17+0.3665 T. Calculate
∆S in (KJ/(deg.mole)) when the temperature is raised from
25 ˚C to 200 ˚C if the process is carried out: a) at constant
pressure, or b) at constant volume.
The Answer of Question No. (9)
a) Constant pressure
T2 NC
p
∆S = ∫
dT
T
T1
473
=∫
(20.17T −1 + 0.3665) dT
298
= [20.17 ln T + 0.3665T]473
298
= 73.456 J/K
= 0.073456 KJ/K
b) Constant volume
Cp − Cv = R
Cv = Cp − R
20
Cv = 20.17 + 0.3665T − 8.314
= 11.856 + 0.3665T
T2
NCv
∆S = ∫
dT
T
T1
473
=∫
(11.856T −1 + 0.3665) dT
298
= [11.856 ln T + 0.3665T]473
298
= 69.615 J/K
= 0.069615 KJ/K
21
Question No. (10)
Two moles of N2 expands from 10 lit, 25 ˚C to 20 lit final
state. Find ∆S obtained if the process is isothermal,
assuming that Cv= (5/2) R, and Cp= (7/2) R, and the gas
behaves ideally.
The Answer of Question No. (10)
V2
∆S = NR ln
V1
= 2 ∗ 8.314 ∗ ln
= 11.525 J/K
22
20
10
Question No. (11)
For each step, and for the net cycle shown in the next
page, if 1 mole of a perfect gas performed the cycle and
assuming that all processes to be reversible; find: a) P, V,
T at points 1, 2, 3, and 4 and b) ∆S for every step and for
the net cycle.
P (atm.)
1 10 atm
↗ Isothermal
2
→ adiabatic
4
0
0.5
3
1
1.5
2
2.5
V (lit)
The Answer of Question No. (11)
At point 1
V1 = 0.5 L
P1 = 10 atm
T1 =
P1 V1 0.5 ∗ 10
=
= 60.976 K
NR
0.082
23
At point 2
V2 = 2 L
T2 = T1 = 60.976 K
NRT2
P2 =
V2
1 ∗ 0.082 ∗ 60.976
=
= 2.5 atm
2
At point 4
V4 = 1 L
V4 𝛾
P1
( ) =( )
V1
P4
5
3
1
10
( ) =( )
0.5
P4
P4 = 3.15 atm
P4 V4 1 ∗ 3.15
T4 =
=
= 38.41 K
NR
0.082
At point 3
V3 = 2 L
P3 = P4 = 3.15 atm
P3 V3 2 ∗ 3.15
T3 =
=
= 76.82 K
NR
0.082
24
Process from 1
2 (Isothermal)
∆S = NR ln
V2
V1
2
= 1 ∗ 8.314 ∗ ln
0.5
= 11.525 J/K
Process from 2
3 (Constant volume)
T3
∆S = NCv ln
T2
3
76.82
= 1 ∗ ∗ 8.314 ∗ ln
2
60.967
= 2.88 J/K
Process from 3
4 (Constant pressure)
T4
∆S = NCp ln
T3
5
38.42
= 1 ∗ ∗ 8.314 ∗ ln
2
76.82
= − 14.4 J/K
Process from 4
1 (Adiabatic)
∆S = 0
25
For the net cycle
∆Scycle = 0
Point
P (atm)
V (L)
T (K)
1
10
0.5
60.976
2
2.5
2
60.976
3
3.15
2
76.82
4
3.15
1
38.41
Process
1
2
2
3
3
4
4
1
Total
∆S (J/K)
11.525
2.88
-14.4
0
0
26