CMEN 522 Advanced Thermodynamics
Homework Assignment # 3
1. Exercise E3.3 (p. 148). Also, as part (d), calculate ∆S for the process given in part (c).
Answer
(a) From the relationship given on p. 130, we write
µ
¶
µ
¶
∂p
∂Vm
∂T Vm ∂T p
µ
¶ µ
¶
∂p
∂Vm
3
= R+T
2
∂T Vm ∂T p
Cp,m =CV,m + T
(1)
From the given equation of state, p = (RT − a)/Vm and, therefore,
µ
∂p
∂T
¶
=
Vm
R
.
Vm
Also from the equation of state, Vm = (RT − a)/p and so
µ
∂Vm
∂T
¶
=
Vm
R
.
p
Substituting into Eq. (1) above, we get
R2 T
3
Cp,m = R +
.
2
pVm
Since the equation of state can be written as pVm + a = RT , we obtain pVm = RT − a. Therefore,
R2 T
3
Cp,m = R +
2
RT − a
R
3
.
= R+
2
1 − a/(RT )
(b) The thermodynamic equation of state for isothermal internal energy change is [Eq. (3.25)]:
µ
∂Um
∂Vm
¶
=T
¶
=
T
µ
∂p
∂T
¶
− p.
Vm
From the equation of state given,
µ
∂Um
∂Vm
T
RT
a
−p=
.
Vm
Vm
Therefore,
a
dVm ,
Vm Ã
!
Vm,2
.
∆Um =a ln
Vm,1
dUm =
(c) Let us first convert the parameter a to more convenient units so that we can use R = 0.083145 L bar
K−1 mol−1 (1 dm3 = 1 L).
3
−1
a = 0.304 dm MPa mol
µ
1 bar
×
0.100 MPa
1
¶
= 3.04 L bar mol−1 .
From the equation of state, we calculate the molar volumes corresponding to the initial and final states:
Vm,1 =(RT − a)/p1
=(0.083145 L bar K−1 mol−1 × 300 K − 3.04 L bar mol−1 )/10.0 bar
=2.19 L mol−1
and Vm,2 = (RT − a)/p2 = 21.90 L mol−1 . Now,
w=−
VZ
m,2
Vm,1
pdVm = −(RT − a)
=−(RT − a) ln
Ã
Vm,2
Vm,1
−1
=−21.90 L bar mol
=−5042.7 J mol−1 .
∆Um =3.04 L bar mol
−1
=699.98 J mol−1 .
!
VZ
m,2
Vm,1
dVm
Vm
µ
¶
µ
¶
21.90
8.3145 J K−1 mol−1
× ln
×
2.19
0.083145 L bar K−1 mol−1
21.90
8.3145 J K−1 mol−1
× ln
×
2.19
0.083145 L bar K−1 mol−1
The heat q is calculated using the first law:
q=∆Um − w
=5742.7 J mol−1 .
The enthalpy change is most conveniently calculated using the equation of state
µ
∂Hm
∂p
¶
T
= Vm − T
µ
∂Vm
∂T
¶
.
p
From the equation of state, we get
RT
dp
RT − a
dp −
dp = −a
p µ ¶ p
p
p2
= 699.98 J mol−1 .
∆Hm =−a ln
p1
dHm =
(d) Since the temperature is constant during the process, the entropy change can be calculated directly from
the q calculated above, as
5742.7 J mol−1
q
=
T
300 K
=19.14 J K−1 mol−1 .
∆S=
2. Exercise E3.7 (p. 149).
Answer
We make use of the fundamental equation in calorimetry: heat lost (by the copper block) = −heat gained
(by the water). We denote the masses of the copper block and water as mCu and mw , the respective
2
specific heats as sCu and sw , the initial and final temperatures of the copper block as TiCu and TfCu , and
that of water as Tiw and Tfw . Then, we can write
mCu sCu (TfCu − TiCu ) = −mw sw (Tfw − Tiw ).
The only unknown is the final temperature of the water. Therefore, we solve for it:
³
´
50 g × 0.389 J K−1 g−1 × (323 − 373) K = −100 g × 4.18 J K−1 g−1 × Tfw − 298
Tfw
K.
− 298 = 2.32 K
Tfw = 300.32 K
Now, the entropy changes for each component is calculated as
dS =
dqrev
msdT
=
.
T
T
Therefore, for copper, we get
µ
¶
323
∆SCu = 50 g × 0.389 J K g × ln
= −2.80 J K−1 .
373
µ
¶
300.32
= 3.24 J K−1 .
∆Sw = 100 g × 4.18 J K−1 g−1 × ln
298
−1
−1
The net entropy change in the isolated system is, therefore, ∆SCu + ∆Sw = 0.44 J K−1 . Since no heat
transfer occurs between the system (isolated!) and the surroundings, this is also equal to ∆Suniv , which
confirms that the transfer of heat from a hot body to a cooler body is a spontaneous process.
3. A reversible Joule cycle consists of the following four steps: (i) isobaric expansion, (ii) adiabatic expansion, (iii) isobaric compression, (iv) adiabatic compression. Helium gas with the equation of state
pVm = RT + Bp (where B > 0) is carried through a Joule cycle.
(a) Sketch the paths taken by the system in the following sets of coordinates: (i) (p, V ), (ii) (U, V ), and (iii)
(T, S). [For the sketches, only the general shapes and slopes of the lines are important, not the actual
values of the variables at the end of each step.]
(b) If B = 0.015 L mol−1 , Th = 500 K, and Tc = 298 K, calculate w, q, ∆U , ∆H, and ∆S for each step
and for the total cycle.
(c) What is the efficiency of this engine? Can it be expressed as a function of the two temperatures only?
Answer
(a) See the last page for the sketches.
(b) The question does not provide the initial and final pressures (or volumes) and, therefore, numerical results
cannot be obtained. However, let us derive the expressions required. The Joule cycle involves (i) an
isobaric expansion (by constant pressure heating) from (ph , Vh , Th ) to (ph , V1 , T1 ), (ii) adiabatic expansion
to (pc , Vc , Tc ), (iii) isobaric compression (by constant pressure cooling) to (pc , V2 , T2 ), and finally (iv)
adiabatic compression back to (ph , Vh , Th ). The following expressions are useful for our analysis:
For an isobaric process :
dq = dH = Cp dT.
Cp
dS =
dT.
T
3
dU = CV dT = dw + dq
µ ¶
R
dw = −pdV = −p
dT = −RdT
p
dq − dU = −dw = (Cp − Cv )dT
Therefore, Cp − Cv = R.
Thus, assuming CV,m = (3/2)R, we can evaluate w, q, ∆U , ∆H, and ∆S for the isobaric expansion
and compression processes, provided the temperatures T1 and T2 are known.
For an adiabatic process :
dq = 0
dS = 0.
dU = CV dT
dH = Cp dT
dw = dU
Once again, we can evaluate w, q, ∆U , ∆H, and ∆S for the adiabatic expansion and compression
processes, provided the temperatures T1 and T2 are known.
(c) The efficiency of a heat engine is defined as the ratio of the net work to the amount of heat absorbed in
the expansion process:
w1 + w2 + w3 + w4
q1 + q2
−R(Th − T1 ) − CV (Tc − T1 ) − R(T2 − Tc ) − CV (Th − T2 )
.
=
Cp (T1 − Th ) + 0
²=
Without adequate information about the temperatures T1 and T2 , it is not possible to verify if the efficiency
can be expressed as a function of two temperatures only.
4. The model Cp,m = α + βT + γT 2 is applied to the data given in P4.3 (p. 199) in the range 14.98 ≤ T
≤ 50.20 K. The following parameters are found to be satisfactory for the range 14.98 ≤ T ≤ 50.20 K:
α = −18.6002, β = 1.95645, and γ = −0.0155003, in appropriate units.
(a) Find the parameter a and the characteristic temperature T ∗ that will ensure the smooth continuation of
this curve to absolute zero with the Debye equation, according to which Cp,m = aT 3 . Plot the graph of
aT 3 in the range 0 − T ∗ and α + βT + γT 2 from 14.98-50.20 K to verify that the function is smooth and
continuous.
(b) Calculate H50.20 K −H0 K by integrating the Debye equation up to T ∗ and the expression given above
for higher temperatures.
(c) Also, using the same assumption, calculate S50.20 K −S0 K .
Answer
(a) In order to smoothly continue the Cp,m curve to 0 K, we impose conditions of continuity in the function
as well as its derivative across the temperature T ∗ at which the match is made.
aT 3 =α + βT + γT 2
2
3aT =β + 2γT
Dividing Eq. (2) by Eq. (3), we get
α + βT + γT 2
T
=
3
β + 2γT
4
(2)
(3)
which rearranges to
1 2 2
γT + βT + α = 0.
3
3
Substituting the constants given, the roots are 15.1726 K and 237.268 K. Since the second root lies
outside the domain of the data, the temperature T ∗ is 15.1726 K. Substituting for T ∗ in Eq. (3). we get
a = 0.0021518 J mol−1 K−4 . The plot is given below.
HW 3.4 (a)
50
a = 0.0021518
T* = 15.1726 K
–1
Cp,m (J K–1 mol )
40
30
20
10
0
0
10
20
30
40
50
60
T (K)
(b) The enthalpy change H50.20 K −H0 K is calculated as the integral of Cp,m dT over the temperature range.
We write
Z T∗
H50.20 K − H0 K =
3
aT dT +
0
Z 50.20 K
T∗
(α + βT + γT 2 )dT
´
a
β
γ³
= (15.17)4 + α(50.20 − 15.17) + (50.202 − 15.172 ) +
50.203 − 15.173
4
2
3
=981.38 J mol−1
(c) The absolute entropy at 50.20 K is evaluated as the integral of (Cp,m /T )dT over the temperature range.
Z T∗
aT 2 dT +
Z 50.20 K
α
+ β + γT )dT
T¶
´
a
50.20
γ³
= (15.17)3 + α ln
+ β(50.20 − 15.17) +
50.202 − 15.172
3
15.17
2
=31.0325 J K−1 mol−1
S50.20 K − S0 K =
0
∗
Tµ
(
5