Thermo and intro to Stat Mech
2016
Homework assignment 5, Solution
Reading assignment: Chapter 13 and 14.
Problem 1:
Revisit problem 2 in Homework assignment 4 and carry out calculations also of the
entropy change, ∆S.
(a) One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K
from a pressure of 10 atm to a pressure of 2 atm. What is the value of ∆S for this
process?
Solution
The change in the entropy of the gas (the system here) an isothermal, reversible
process is ∆σ = Q/τ = 4014/300 J/K = 13.4 J/K. (But, note that the total entropy,
entropy of the system plus the bath, remains constant since the process is reversible).
(b) Now assume the gas expands by the same amount again isothermally but now
irreversibly against 2 atm pressure (instead of reversible expansion) and calculate ∆S.
Compare with the values obtained in part (a).
Solution
Since the entropy is a state function and the initial and final states are the same as
in (a), the change in entropy is the same as in (a), i.e. ∆σ = 13.4 J/K.
(c) Compare now the isothermal expansion in (a) to an adiabatic, reversible expansion
from a pressure of 10 atm to a pressure of 2 atm. What is the value of ∆S for this
process?
Solution
For an adiabatic process, Q = 0 and there is no change in the entropy of the system.
Also, since the process is reversible, there is no change in the total entropy either.
Problem 2:
An ideal, monatomic gas undergoes the following reversible process: It is first compressed isothermally from state A to state B. It is then expanded adiabatically to state
C. Finally, it is heated at constant volume to its original state, A.
(a) Each state can be characterized by the volume, pressure, temperature and entropy
of the gas, (V, P, τ, σ). Sketch the cyclic process in the (P, V ) plane, the (P, τ ) plane,
in the (V, τ ) plane and in the (σ, τ ) plane.
Solution
1
P"
B"
B"
P"
A"
A"
C"
C"
V"
V"
C"
A"
τ"
σ"
B"
A"
B"
C"
τ"
τ"
(b) Calculate the heat flow, work done and change in internal energy in each one of the
three steps in the cyclic process.
Solution
A→B (isothermal compression):
The change in internal energy is zero since the
RV
temperature remains constant, ∆UAB = 0. The work done is WAB = −τA VAB dV /V =
−τA log (VB /VA ) = τA log (VA /VB ) > 0. The heat flow is QAB = UAB − WAB = −WAB =
τA log (VB /VA ) < 0
B→C (adiabatic expansion): No heat flow, so QBC = 0 and ∆UBC = WBC . First,
find out how much the gas cools down during the expansion. The relationship between
3/2
3/2
temperature and volume is V1 τ1 = V2 τ2 , so
τC = τB
VB
VC
2/3
2/3
= τA
VB
VC
2/3
since τA = τB .
∆UBC
3
3
= (τC − τA ) = τA
2
2
VB
VC
!
−1
= WBC
C→A (constant volume heating): Since the volume is fixed, there is no work done,
WCA = 0 and ∆UCA = QCA . The change in internal energy can be obtained from the
temperature change as
2/3 !
3
3
VB
∆UCA = (τA − τC ) = τA 1 −
= QCA .
2
2
VC
Clearly, ∆UCA = −∆UBC so the total change in internal energy is ∆U = ∆UAB +
∆UBC + ∆UCA = 0 as it must be in a cyclic process (the final state is the same as the
2
initial state, so there is no net change in the energy or any other state function). Note:
Heat and work are not state functions and there is a nonzero sum of the work and heat
flow in the cyclic process.
(c) Calculate the change in entropy of the gas in each one of the three steps and the
total change in the entropy.
A→B (isothermal compression): ∆σAB = Q/τA = log (VB /VA ) < 0.
B→C (adiabatic expansion): ∆σBC = 0.
C→A (constant volume heating):
Z τA
Z τA CV
τA
3 2
∂σ
dτ =
dτ = CV log
= · log (VA /VB )
∆σCA =
∂τ V
τ
τC
2 3
τC
τC
So, clearly, ∆σ = σAB + σBC + σCA = 0 as it should be in a cyclic process.
Problem 3:
In absorption refrigerators, the energy driving the process is supplied as heat from a
gas flame at a high temperature τhh > τh , rather than as work. Cabin refrigerators may
be of this type and use gas as fuel. Assume here reversible operation.
(a) Draw an energy flow diagram similar to figures 13.5 in B&B for such a refrigerator.
Note that no net work is involved, but energy and entropy flows between the three parts
with temperature τhh > τh > τl .
Solution
Flame"
Refrigerator"
τhh"
τl"
Qhh"
Ql"
W"
Qh"
Room"
τh"
(b) Calculate the ratio Ql /Qhh where Ql is the heat extracted from the cold body and
Qhh is the heat extracted from the energy source.
Solution
If the process is reversible, the total change in entropy is zero
0 = ∆σ =
Qhh Qh Ql
−
+
τhh
τh
τl
3
There is no net change in total energy, ∆U , since the process is cyclic and there is no net
work, W = 0, so
Q = Qhh − Qh + Ql = 0 .
By obtaining an expression for Qh from this equation and inserting it into the equation
derived from 0 = ∆σ one obtains
Ql
=
Qhh
1
τh
1
τl
−
−
1
τhh
1
τh
=
(τhh − τh )
τl
·
.
τhh
(τh − τl )
Problem 4:
Consider the entropy of mixing of more than two components.
(a) Give an expression for the number of arrangements of A and B atoms on a lattice
where some of the lattice sites can be left empty.
Solution
Let the total number of sites be N , the number of A atoms be NA and the number
of B atoms be NB . The number of ways of assigning the A atoms to the lattice is NNA .
−NA
There are then N −NA sites left. The number of ways of assigning the B atoms is NN
.
B
The number of ways of assigning both the A and B atoms on the lattice is the product
of the two
N
N − NA
N !(N − NA )!
Ω(N, NA , NB ) =
·
=
NA
NB
(N − NA )!NA !(N − NA − NB )!NB !
N!
NA !NB !NC !
where NC is the number of empty sites, NC = N − NA − NB .
=
(b) Now assume the empty sites in part (a) are occupied by atoms of type C and give
an expression for the entropy of mixing in terms of the molefractions xA , xB and xC .
Solution
The mole fractions are defined as xA = nA /N , xB = nB /N and xC = nC /N . The
entropy of mixing is the entropy difference between the situation when all three types of
atoms are mixed together as compared with the situation when each one of the components is separate and pure (with only one possible arrangement).
∆S = kB ln Ω(N, nA , nB ) − kB ln 1 = ln
N!
.
NA !NB !NC !
Using Sterling approximation, log N ! ≈ N log N − N gives
∆S = kB (N ln N − N − NA ln NA + NA − NB ln NB + NB − NC ln NC + NC )
= kB (N ln N − NA ln NA − NB ln NB − NC ln NC )
4
Introducing xt = Nt /N for each of the three components gives
∆S = −kB N (xA ln xA + xB ln xB + xC ln xC )
which is clearly positive since each of the three logarithms is negative and there is a minus
sign in front. So, the entropy must increase upon mixing.
(c) Generalize your results from (b) to arbitrary number, n, of components.
Solution
By inspecting the expression in the results of part (b) one can generalize it to the
mixing of n components
n
X
∆S = −N kB
xt log xt .
t=1
Note the similarity to the Gibbs expression for the entropy
∆S = −N kB
n
X
i=1
5
Pi log Pi .