SHEET (3)
2nd Law of
Thermodynamics
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=NRlnV2V1=NRlnP1P2
=1*8.314*ln101
∆S=19.14 J/K
b) Reversible Adiabatic
∆S=0
c) Constant Volume
∆S=NCvlnT2T1=NCvlnP2P1
=1*32*8.314*ln110
=- 28.71 J/K
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
V1=RT1P1=0.082*2981=24.436 L
V2=2 V1=48.872 L
V2V1= P1P2
1P2=2
P2=0.5 atm
T2=T1=298 K
∆U=W=q=∆H=0
∆S=NRlnV2V1=1*8.314ln2
=5.763 J/K
b) Heated to 125 ˚C at constant volume
T3=398 K
W=0
∆U=NCv∆T
=1*32*8.314*398-298
=1247.1 J
q=∆U=1247.1 J
∆H=NCp∆T
=1*52*8.314*398-298
=2078.5 J
∆S=NCvlnT2T1
=1*32*8.314*ln398298
=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=NRTlnV3V4
=1*8.314*398*ln12
=- 2293.6 J
q=- W=2293.6 J
∆H=0
∆S=NRlnV4V3=1*8.314*ln2
=5.763 J/K
d) Reversibly cooled to 25 ˚C at constant pressure
W=P(V4-V5)=NRT4-T5
=1*8.314*398-298
=831.4 J
∆U=NCv∆T
=1*32*8.314*298-398
=- 1247.1 J
q=∆U-W
=-1247.1-831.4
=- 2078.5 J
∆H=qp=- 2078.5 J
∆S=NCplnT5T4
=1*52*8.314*ln298398
=- 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 / -2293.6 / 2293.6 / 0 / 5.763
d) / -1247.1 / 831.4 / -2078.5 / -2078.5 / -6.014
Total / 0 / -1462.2 / 1462.2 / 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
NwCp(w)Tf-Tw=NPbCpPb(TPb-Tf)
10018*18.03*Tf-25=1500207*6.38*100-Tf
∴ Tf=48.6846 ℃
∆Stotal=∆Swater+∆Slead
∆Swater=NwCpwlnTfTw
=10018*4.18*18.03*ln321.6846298
=32.02 J/K
∆Slead=NPbCpPblnTfTPb
=1500207*4.18*6.38*ln321.6846373
=- 28.6 J/K
∆Stotal=32.02-28.6
=3.42 J/K
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.
∆S1=TiTmδqT=TiTmNCpsTdT
=30010487.97T-1+9.2*10-3+1.21*105T-3 dT
=7.97lnT+9.2*10-3T+1.21*105-2T-23001048
∆S1=17.468 J/K
∆S2=qmT
=63001048=6.01145 J/K
∆S3=TmTfδqT=TmTfNCplT dT
=10481200-11.3T+0.833*10-3+350.7*105T-3dT
=-11.3lnT+0.833*10-3T+350.7*105-2T-210481200
=2.3846 J/K
∆Stotal=∆S1+∆S2+∆S3
=17.468+6.01145+2.3846
=25.864 J/K
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
P2P1γ-1=T2T1γ
103053-1=T229853
T2=192 K
For the irreversible process from 1 3
∆U=W
NCvT3-T1=W
2*32*8.314*T3-298=-4.18*500
T3=214.2 K
∴T3T2
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=NCplnT3T2=2*52*8.314*ln214.2192
=4.548 J/K
∴ ∆Sirr=4.548 J/K
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.
∆S1=590600δqT=590600NCp(Pbl)TdT
=5906007.75T-1-0.74*10-3dT
=7.75lnT-0.74*10-3T590600
=0.12285 cal/K
∆S2=-qmT
=-1150600
=- 1.91667 cal/K
∆S3=600590δqT=600590NCpPbsT dT
=6005905.63T-1+2.33*10-3dT
=5.63lnT+2.33*10-3T600590
=- 0.1179 cal/K
∆Sirr=∆S1+∆S2+∆S3
=0.12285-1.91667-0.1179
=- 1.9117 cal/K
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
∆S=NRlnV2V1
=1*8.314*ln44.822.4
=5.763 J/K
b) Constant pressure
∆S=NCplnT2T1=NCplnV2V1
=1*52*8.314*ln44.822.4
=14.407 J/K
c) Freely: it is irreversible process. Consider the reversible isothermal path (during free expansion temperature does not change (ideal gas))
∆S=NRlnV2V1
=1*8.314*ln44.822.4
=5.763 J/K
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)
∆S=-∆HT
=-40.656373
=- 0.109 KJ/K
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
∆S=T1T2NCpT dT
=29847320.17T-1+0.3665 dT
=20.17lnT+0.3665T298473
=73.456 J/K
=0.073456 KJ/K
b) Constant volume
Cp-Cv=R
Cv=Cp-R
Cv=20.17+0.3665T-8.314
=11.856+0.3665T
∆S=T1T2NCvT dT
=29847311.856T-1+0.3665 dT
=11.856lnT+0.3665T298473
=69.615 J/K
=0.069615 KJ/K
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)
∆S=NRlnV2V1
=2*8.314*ln2010
=11.525 J/K
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.
The Answer of Question No. (11)
At point 1
V1=0.5 L
P1=10 atm
T1=P1V1NR=0.5*100.082=60.976 K
At point 2
V2=2 L
T2=T1=60.976 K
P2=NRT2V2
=1*0.082*60.9762=2.5 atm
At point 4
V4=1 L
V4V1γ=P1P4
10.553=10P4
P4=3.15 atm
T4=P4V4NR=1*3.150.082=38.41 K
At point 3
V3=2 L
P3=P4=3.15 atm
T3=P3V3NR=2*3.150.082=76.82 K
Process from 1 2 (Isothermal)
∆S=NRlnV2V1
=1*8.314*ln20.5
=11.525 J/K
Process from 2 3 (Constant volume)
∆S=NCvlnT3T2
=1*32*8.314*ln76.8260.967
=2.88 J/K
Process from 3 4 (Constant pressure)
∆S=NCplnT4T3
=1*52*8.314*ln38.4276.82
=- 14.4 J/K
Process from 4 1 (Adiabatic)
∆S=0
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 / ∆S (J/K)
1 2 / 11.525
2 3 / 2.88
3 4 / -14.4
4 1 / 0
Total / 0
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