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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):

dPdT

=sg − sℓvg − 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!(𝑇) ⇒ 𝑑𝐻 = 𝐶! 𝑇 𝑑𝑇

c. !"dT= !!!!!

!!!!!= ∆!!"#/!

!!= ∆!!"#/!

!"/!= ∆!!"#∙!

!!!

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d. !"!= !"

!!∙ ∆!!"#

!⇒ !"

!= ∆!!"#

!!"!!

!!"#$$#%&!

!!"#!!!!

𝑙𝑛𝑃!"#𝑃!!!

=∆ℎ!"#𝑅 (

1273−

1𝑇!"#$#%&

)

Heat of vaporization: 595 calories/gram, which equals:

595𝑐𝑎𝑙𝑜𝑟𝑖𝑒𝑠𝑔𝑟𝑎𝑚 ∙ 4.18

𝑗𝑜𝑢𝑙𝑒𝑐𝑎𝑙𝑜𝑟𝑖𝑒 ∙ 18

𝑔𝑟𝑎𝑚𝑚𝑜𝑙 = 44767.8  𝐽/𝑚𝑜𝑙

𝑙𝑛7604.58 =

44767.8 𝐽𝑚𝑜𝑙

8.314   𝐽𝑚𝑜𝑙 ∙ 𝐾

(1273−

1𝑇!)  

 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

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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.

𝑆 = 𝑘! ln 𝑊 𝑈 = 𝑘!ln  (𝑁!

𝑁 − 𝑈𝐵 ! 𝑈𝐵 !)

≈ 𝑘!(𝑁𝑙𝑛 𝑁 − 𝑁 − 𝑁 −𝑈𝐵 ln 𝑁 −

𝑈𝐵 + 𝑁 −

𝑈𝐵

−𝑈𝐵 𝑙𝑛

𝑈𝐵 +

𝑈𝐵 )

= 𝑘!(𝑁𝑙𝑛𝑁 − 𝑁 −𝑈𝐵 ln 𝑁 −

𝑈𝐵 −

𝑈𝐵 ln

𝑈𝐵 )

= 𝑘!(𝑁𝑙𝑛𝑁 − 𝑁𝑙𝑛 𝑁 −𝑈𝐵 +

𝑈𝐵 ln

𝑁𝐵𝑈 − 1 )

c. Here is the plot of S as a function of U. The monotonicity breaks down when

energy U is larger than NB/2

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d. Take the formula we derived in (a)

𝑆 = 𝑘!(𝑁𝑙𝑛𝑁 − 𝑁𝑙𝑛 𝑁 −𝑈𝐵 +

𝑈𝐵 ln

𝑁𝐵𝑈 − 1 )

1𝑇 =

𝜕𝑆𝜕𝑈 = 𝑘!(

𝑁𝐵

𝑁 − 𝑈𝐵+1𝐵 ln

𝑁𝐵𝑈 − 1 +

𝑈𝐵

−𝑁𝐵𝑈!𝑁𝐵𝑈 − 1

)

=𝑘!𝐵 ln

𝑁𝐵𝑈 − 1 ⇒ 𝑈 =

𝑁𝐵1+ 𝑒!/!"  

 When T approaches infinity, U = NB/2. This is the highest value of the energy the system can reach by heating.

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! =𝑊!𝜀𝑒!!/!"

𝑊! +𝑊!𝑒!!/!"  

𝐶! =𝜕𝑈𝜕𝑇 =

𝜕 𝑊!𝜀𝑊!𝑒

!!" +𝑊!𝜕𝑇 =𝑊!𝜀 ∙

𝑊!𝑒!!"

𝜀𝑅𝑇!

𝑊!𝑒!!" +𝑊!

!

=1𝑅

𝜀𝑇

! 𝑊!𝑒!!!"

𝑊! +𝑊!𝑒! !!"

𝑊!

𝑊! +𝑊!𝑒! !!"=1𝑅

𝜀𝑇

!𝜌!𝜌!

b. Given, ε = 10 kJ/mole, WF = 1, R = 8.314J/mole/K For Wu = 5

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𝐶! =1

8.314 ∙ 10!!𝑘𝐽𝑚𝑜𝑙𝑒 ∙ 𝐾

∙10𝑘𝐽/𝑚𝑜𝑙𝑒

𝑇

! 5 ∙ 𝑒!!.!∙!"!!

!

(1+ 5 ∙ 𝑒!!.!∙!"!!

! )!  

For WF = 50

𝐶! =1

8.314 ∙ 10!!𝑘𝐽𝑚𝑜𝑙𝑒 ∙ 𝐾

∙10𝑘𝐽/𝑚𝑜𝑙𝑒

𝑇

! 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.

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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. TN

TPV

TUS µ

−+= TA

TN

TPVU

USSUS

nV

−=−="#

$%&

'∂

∂−=∴

µ

,

][

You have to write S in terms of new variables: [T,V,N]

b. dS[U]= d −1TA

"

#$%

&'= −Ad 1

T(

)*

+

,-+ (− 1

T)•dA

(

)*

+

,-T

= −Ad 1T(

)*

+

,-+

PTdV − µ

Tdn

𝑑𝑆 𝑈 =

𝑃𝑉 − 𝜇𝑛𝑇! +

𝑃𝑇 𝑑𝑉 −

𝜇𝑇 𝑑𝑛

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

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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,

  ⇒==+++−= , 0 , 0 2211 dPdTdndnVdPSdT µµ dµ1 = −

χ2

χ1

#

$%

&

'( dµ2  

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: 1

211111 ln

VVRTW

VdVRTdWdVPdW

VRTP −=⇒−==−=⇒=

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(b) vdW:

!!"

#$$%

&−−!!

"

#$$%

&

−

−−=⇒+

−−=⇒−

−=

121

222222

11lnVV

abVbVRTW

VdVa

bVdVRTdW

Va

bVRTP

ΔW =W2 −W1 = −RT lnV2 − bV2

V1V1 − b

#

$%

&

'(− a

1V2−1V1

#

$%

&

'(

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,

−𝑅𝑇𝑙𝑛𝑉! − 𝑏𝑉!

𝑉!𝑉! − 𝑏

=  −𝑅𝑇𝑙𝑛(1− 𝑏

𝑉!1− 𝑏

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.