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Homework Set #3 solutions
Chem 544
TA: Yuanxi Fu
Problem 1
In class we showed that dSV =dqV,n…/T=dU/T at constant volume.
a. Now that you know that enthalpy is the Legendre transform H=U+PV, show similarly
that dS =dqP,n…/T=dH/T at constant pressure instead of constant volume.
b. Since enthalpy is heat flow at constant pressure, write down a formula for the heat
capacity Cp(T) at constant pressure in terms of an enthalpy derivative. Then write down a
formula for dH in terms of Cp(T).
c. In lecture, we derived the Clausius equation from the Gibbs-Duhem equation (lower
case quantities are molar entropy and volume):
dP sg − sℓ
=
dT vg − vℓ
Vaporization from a liquid to its gaseous vapor occurs at constant temperature and
pressure. Integrate the formula in a. at constant T to replace sg-sl by the constant Δhvap,
the molar enthalpy of vaporization. Next make the approximation vg-vl ≈vg because the
gas volume is much greater than the liquid volume, and use the ideal gas law to express
vg in terms of P, R and T. Write down the resulting Clausius-Clapeyron equation for
dP/dT.
d. The vapor pressure of water at 0 °C is 4.58 torr (760 torr = 1 atm). Assuming that the
heat of vaporization of water is 595 calories/gram (1 calorie = 4.18 Joules) and
temperature-independent, use the Clausius-Clapeyron equation to estimate the boiling
point of water. [Careful: cal/g ≠ cal/mole!]
Solutions
a. We know that: dH = dU + PdV + VdP = TdS-PdV+µdn+…+PdV+VdP =
TdS+VdP+µdn+…At constant pressure, dH/T = dS = dq,n,../T
b.
c.
!"
!" !
!"
dT
= 𝐶! (𝑇) ⇒ 𝑑𝐻 = 𝐶! 𝑇 𝑑𝑇
! !!
= !! !!! =
!
!
∆!!"# /!
!!
=
∆!!"# /!
!"/!
=
∆!!"# ∙!
!! !
1 d.
!"
!
=
!"
!!
∙
∆!!"#
!
⇒
!!"# !"
!!!! !
𝑙𝑛
=
∆!!"#
!
!!"#$$#%& !"
!
!!
𝑃!"# ∆ℎ!"# 1
1
=
(
−
)
𝑃!!!
𝑅
273 𝑇!"#$#%&
Heat of vaporization: 595 calories/gram, which equals:
595
𝑐𝑎𝑙𝑜𝑟𝑖𝑒𝑠
𝑗𝑜𝑢𝑙𝑒
𝑔𝑟𝑎𝑚
∙ 4.18
∙ 18
= 44767.8 𝐽/𝑚𝑜𝑙
𝑔𝑟𝑎𝑚
𝑐𝑎𝑙𝑜𝑟𝑖𝑒
𝑚𝑜𝑙
𝐽
44767.8 𝑚𝑜𝑙 1
760
1
𝑙𝑛
=
(
− ) 𝐽
4.58 8.314 273 𝑇!
𝑚𝑜𝑙 ∙ 𝐾
Therefore, we can estimate the boiling point to be Tb = 368.5 K = 95.5°C
Problem 2
Work out the problem of spins in a magnetic field B at the end of “Survey” chapter 3:
a. The energy of the system ranges from 0 (all N spins down, or szj=-1/2) to U=NB (all N
spins up, or szj=+1/2). With energy as a vertical axis, plot a couple of the lowest and
highest energy levels of this system. What is the degeneracy W of the ground state, first
excited state, highest state, and next-highest state?
b. Explain briefly why the formula for W(U) given in the text is the correct general
formula for this system. Calculate S=kBln[W(U)].
c. Now sketch what a plot of S(U) looks like from U=0 to U=NB. Is there a problem with
monotonicity?
d. Take the derivative ∂S/∂U = 1/T(U) and solve for energy as a function of T. Does the
energy go to NB when T approaches ∞? What is the highest value of U that can be
achieved by heating? Since we can never heat the system above T=∞, under equilibrium
conditions it can never enter a regime where S does not monotonically increase with U.
e. What is U if the temperature were just slightly below 0 K? This state, called an
“inverted population” cannot be reached in equilibrium in a closed system: negative
temperatures correspond to inversion populations. Inversions can be prepared however in
open systems, since the postulates of stat mech (and laws of thermo) do not hold for open
systems.
Solutions
2 a.
b. When a single spin is excited, the energy change difference will be (1/2-(-1/2))*B
= B. Therefore, when the system has an energy of U, there will be U/B spins at
the excited state. This is an “N choose U/B” problem. So the number of
microstate will be W(U) = N!/(N-U/B)!/(U/B)! Now we can calculate S by
evoking the Stirling expansion.
𝑁!
)
𝑈
𝑈
𝑁−𝐵 ! 𝐵 !
𝑈
𝑈
𝑈
≈ 𝑘! (𝑁𝑙𝑛 𝑁 − 𝑁 − 𝑁 −
ln 𝑁 −
+ 𝑁−
𝐵
𝐵
𝐵
𝑈
𝑈
𝑈
−
𝑙𝑛
+
)
𝐵
𝐵
𝐵
𝑈
𝑈
𝑈
𝑈
= 𝑘! (𝑁𝑙𝑛𝑁 − 𝑁 −
ln 𝑁 −
−
ln
)
𝐵
𝐵
𝐵
𝐵
𝑆 = 𝑘! ln 𝑊 𝑈
= 𝑘! ln (
= 𝑘! (𝑁𝑙𝑛𝑁 − 𝑁𝑙𝑛 𝑁 −
𝑈
𝑈
𝑁𝐵
+
ln
−1 )
𝐵
𝐵
𝑈
c. Here is the plot of S as a function of U. The monotonicity breaks down when
energy U is larger than NB/2
3 d. Take the formula we derived in (a)
𝑈
𝑈
𝑁𝐵
+
ln
−1 )
𝐵
𝐵
𝑈
𝑁𝐵
𝑁
1 𝜕𝑆
1
𝑁𝐵
𝑈 − 𝑈!
𝐵
=
= 𝑘! (
+ ln
−1 +
)
𝑈
𝑇 𝜕𝑈
𝐵
𝑈
𝐵 𝑁𝐵 − 1
𝑁−𝐵
𝑈
𝑘!
𝑁𝐵
𝑁𝐵
=
ln
−1 ⇒𝑈 =
𝐵
𝑈
1 + 𝑒 !/!"
When T approaches infinity, U = NB/2. This is the highest value of the energy the
system can reach by heating.
𝑆 = 𝑘! (𝑁𝑙𝑛𝑁 − 𝑁𝑙𝑛 𝑁 −
e. When T is slightly below zero, U → NB
Problem 3
For the problem of RNA folding in Survey Chapter 4, start with the formula for U and
a. show that the heat capacity at constant volume can be written as rUrF(e/T)2/R.
b. Pick e=10 kJ/mole and WF=1, and plot cV in kJ/mole/K against T in Kelvin over an
appropriate temperature range to show its features. Make the plots for WU=5 and WU=50.
Solutions
a. According to the class notes, the partition function of the system
!
𝑍 = 𝑊! + 𝑊! 𝑒 !!"
EF = 0 and EU = ε
𝑈 = 𝐸! 𝜌! + 𝐸! 𝜌! =
𝜕𝑈
𝐶! =
=
𝜕𝑇
𝜕
𝑊! 𝜀𝑒 !!/!"
𝑊! + 𝑊! 𝑒 !!/!"
𝑊! 𝜀
!
𝑊! 𝑒 !"
+ 𝑊!
𝜕𝑇
1 𝜀
=
𝑅 𝑇
!
= 𝑊! 𝜀 ∙
𝑊! 𝑒 !"
!
𝑊! 𝑒 !" + 𝑊!
!
!
𝑊! 𝑒 !!"
𝑊! + 𝑊! 𝑒
!
!
!"
𝑊!
𝑊! + 𝑊! 𝑒
b. Given, ε = 10 kJ/mole, WF = 1, R = 8.314J/mole/K
For Wu = 5
𝜀
𝑅𝑇 !
4 !
!
!"
!
=
1 𝜀
𝑅 𝑇
!
𝜌! 𝜌!
𝐶! =
1
10𝑘𝐽/𝑚𝑜𝑙𝑒
∙
!!
8.314 ∙ 10 𝑘𝐽
𝑇
𝑚𝑜𝑙𝑒 ∙ 𝐾
!
5∙𝑒
!
!.!∙!"! !
!
(1 + 5 ∙ 𝑒
!
For WF = 50
𝐶! =
1
10𝑘𝐽/𝑚𝑜𝑙𝑒
∙
!!
8.314 ∙ 10 𝑘𝐽
𝑇
𝑚𝑜𝑙𝑒 ∙ 𝐾
!
50 ∙ 𝑒
!
!.!∙!"! !
!
)!
!.!∙!"! !
!
(1 + 50 ∙ 𝑒 !
!.!∙!"! !
!
)!
Plots of Cv from 0 to 1000K
The heat capacity first increases with temperature, then decreases. For Wu = 5, Cv
peaks at ~ 390K. For Wu = 50, Cv peaks at ~ 260K. The two curves crosses at T
≈ 425K, bellow which, Cv(Wu=5) < Cv(Wu = 50)
Problem 4
Two identical closed systems “i” = 1,2 have fundamental relation Si = C(NiViUi)1/3, where
C is a constant.
a. Is this entropy properly extensive?
b. What is the fundamental relation for the composite system in terms of Ni, Vi, Ui, i=1,2,
with all constraints still in place? [Hint: entropy is an extensive quantity.]
c. Given that the two systems are identical, what is the fundamental relation of the closed
composite system in terms of N, V, and U (N=N1+N2, etc.) once the two systems are
combined and all internal restraints are relaxed?
d. Is the entropy in c. greater than, or equal to, the entropy in b.? Which should it be,
based on common sense? This is called the “Gibbs paradox.” Comment on how it might
be resolved if the systems truly are composed of identical particles.
5 Solutions
a. S(λN, λV, λU) = C((λN)(λV)(λU))1/3 = λC(NVU) = λS. Therefore, the entropy is
extensive.
b. Seq(N1,N2,V1,V2,U1,U2) = C[N1V1U1]1/3 + C[N2V2U2]1/3
c. Seq(N,V,U) = C[NVU]1/3 = C[(2N1)(2V1)(2U1)]1/3 = C[8N1V1U1]1/3
=2C[N1V1U1]1/3 (because the systems are identical)
d. The entropy in (c) equals that in (b), i.e. the sum of entropy of two identical
systems equals the entropy of the composite system. Gibbs paradox does appear
in this specific case, because we assumed that the subsystems are identical.
Problem 5
Instead of Legendre transforming U= U(S,V,n) to A(T,V,n), start with S(U,V,n), and go
through an analogous derivation to derive
a. The Legendre transform of S with respect to U, S[U](T,V,n).
b. Its differential dS[U].
Legendre transforms of the entropy are important in statistical mechanics and are called
Massieu functions. S[U](T,V,n) is the entropy analog of the Helmholtz free energy
A(T,V,n). It is a fundamental relation that contains all thermodynamic information about
the system.
Solutions
a. S =
PV µN
A
U PV µN
' ∂S $
∴ S [U ] = S − %
−
=−
+
−
" U=
T
T
T
T
T
T
& ∂U #V ,n
You have to write S in terms of new variables: [T,V,N]
" 1
b. dS[U] = d $−
# T
%
(1+ ( 1
+
(1+ P
µ
A' = −Ad * - + * (− )• dA - = −Ad * - + dV − dn
&
)T , ) T
,T
)T , T
T
𝑃𝑉 − 𝜇𝑛 𝑃
𝜇
𝑑𝑆 𝑈 =
+ 𝑑𝑉 − 𝑑𝑛
!
𝑇
𝑇
𝑇
Problem 6
10 g of NaCl and 15 g of sugar (C12H22O11) are dissolved in 50 grams of pure H2O. The
volume of the resultant simple system is 55 cm3. What are the mole numbers of the three
6 components? The mole fractions? The molar volume (i.e. the volume where the system
contains a grand total of 1 mole of particles, treating NaCl as a unit)?
Solutions
NaCl (58.44 g/mol) : 10g ≅ 0.1711 moles
C12H22O4 (342.3 g/mol) : 15g ≅ 0.0438 moles
H2O (18.015 g/mol) : 50g ≅ 2.7754 moles
Total moles = 2.9903
Mole fractions :
χNaCl = 0.1711/2.9903 = 0.0572
χsugar = 0.0146
χH2O = 0.9281
Molar volume : 55 cm3 / 2.9903 moles = 18.43 cm3 / mole
Problem 7
For a binary mixture, dG=-SdT+VdP+Σµidni where the sum over i goes from 1 to 2.
Show that at equilibrium (dG=0) and constant T and P, dµ1 = -(χ2/χ1) dµ2, where χi are
the mole fractions.
Solutions
From Gibbs-Duhem relation,
#χ &
0 = − SdT + VdP + n1 dµ1 + n2 dµ 2 , dT = dP = 0 , ⇒ d µ1 = − % 2 ( d µ2 $ χ1 '
Problem 8
By integrating F=–PdV, compute the work done in the slow isothermal compression from
V1 to V2 of: (a) an ideal gas P=nRT/V; (b) a van der Waals gas P=nRT/((V-nb) – a(n/V)2)
Discuss the constants (a) and (b) in terms of the differences in 8(a) and 8(b) and their
microscopic meaning (assume a>0, b>0).
Solutions
T = constant and n = constant (assume n = 1 mole for simplicity)
(a) Ideal gas: P1 =
V
RT
dV
⇒ dW1 = − P1 dV = dW1 = − RT
⇒ W1 = − RT ln 2
V
V
V1
7 (b) vdW:
P2=
&V − b # & 1 1 #
RT
a
dV
dV
!! − a$$ − !!
− 2 ⇒ dW2 = − RT
+ a 2 ⇒ W2 = − RT ln$$ 2
V −b V
V −b
V
% V1 − b " % V2 V1 "
# V − b V1 & # 1 1 &
ΔW = W2 − W1 = −RT ln % 2
( − a% − (
$ V2 V1 − b ' $ V2 V1 '
Consider first the “a” term – if the gas is compressed, it makes a negative contribution
to the difference, meaning the work required in compression of a vdW gas is less than
that of an ideal, and the work required in expansion is greater. Attractive forces make a
gas more difficult to expand and easier to contract.
In the case of the “b” term,
𝑏
1−
𝑉! − 𝑏 𝑉!
𝑉!
−𝑅𝑇𝑙𝑛
= −𝑅𝑇𝑙𝑛(
)
𝑏
𝑉! 𝑉! − 𝑏
1−𝑉
!
When V2<V1, this quantity is positive. Therefore, W2>W1 if v2<v1. This can be seen as
the effects of the van der Waals terms counteract each other.
8

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