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Unformatted text preview: 7 SECOND LAW OF
THERMODYNAMICS:
SOME MORE
STATISTICAL
MECHANICS COMMENTS The combined ﬁrst and second lav.r statement and the use of H, A, G, and S
are examined in this chapter. In many cases, ideal gases are involved, so
that the relationships you have to use are not very complicated. The main
difﬁculty will be in seeing what to do rather than how to do it. A good deal
of attention is paid to irreversible processes, and it is important to remember
that S is a state function, so that as does not depend on path. On the other
hand, to calculate d3 from the data, it may be necessary to make use of the
relationship as = cmf'l", which means that you will want to formulate a
set of reversible stops that take you from the given initial to the given ﬁnal
condition. The criteria [or equilibrium must he kept in mind. Two of the more useful
ones are that sis is zero for a reversible process occurring in an isolated system and that dG is zero for a reversible process at constant T and P.
Also, the Carnot cycle is used a good deal, as are applications of it to heal engines and heat pumps.
Eince E, A, H, G, and S are state functions, they each have a total differ ential, usually expressed in terms of those variables most convenient for the particular quantity. The total diﬁ‘erentials, in turn, give rise to a number of Int] Second law ofthennpd‘yanmies; some more statistical mechanics ltll crossdiﬁenential or Euler relationships [see Appendix], and there is now
considerable scope for derivations ofpartial diﬁ‘erential equations. A number of problems are of this type. . _
Turning next to statistical mechanics, the second law relationshlp between entropy and heat capacity can be combined with Eq. (57] to express entropy
in terms of Q, the partition function. If the third law of thermodynamics
holds, that is, if SmC = t}, then absolute entropy values can be obtained.
The free energy can also be evaluated from {3, so that the entire set of thermo—
dynamic quantities can be obtained by means of the statistical mechanical
approach. Remember, though, that the applications here are of the simplest type. The ability to obtain thermodynamic quantities from Q is no panacea ;
the analytical evaluation of Q is usually a very tricky matter, and is often accomplished only by slipping in various distinctly simplifying assumptions. EQUATIONS AND CONCEPTS c = H — rs {v.1} _____‘__=_,_E_:_E]" A=E—TS as = res — saw so (g)? = r (g—EL = Fe no}
tiH = TdS + Vtilp (g); = T [65%]: = V [ll—3}
as = —ser= PdV (:44)? = —s (ng = —P p4}
JG = —StllT + VtiP (2E)? = —S [g—glr = F [1—5] Entropy in Terms of 9 d5 = dg,,,ﬂT or d8 = Cdln'l' where C = heat capacity. Thus d3 =
ICPd In T for a constantpressure process and d3 = [Cpln T for a constant
volume process [both should be reversible]. Ideal Gas
dS=CudlnT+Rdan {as}
as = ICEc‘ In T for a constantvolume process [it1"} —i_t 101 MM” physical ohensisrry ss= I R d in V: R in I71 I'oraeonstantteinperature process [1—3]
I as=lc,me+RInEE loranyprocess [as]
I Note that the change need not be reversible here, since an irreversible change
from V, and T, to V3 and T2 can always be replaced by two reversible steps
consisting of an isothermal change from V1 to 1": followed by an isoehorie
change from T1 to T1, and the overall :15 is again given by the above
equation. An alternative form is :13 = [CF st in T — R 1a Fifi”... If the ideal
gas is monatomie, we have Ta V1 T; F:
1532 zgiitlnT—1+RlnyI iRlnTl Rlnpl ['i'lll]
Reversible Adiahat :13 is zero for a reversible adiabatic change. For an ideal gas it follows that T: V3
0 ln— = Rln (“r1n
” T1 V1
1‘ P
c ln—‘iRlni {112
" T, p, i Carnot Cycle, Heat Engines, and Heat Pumps The two basic equations for a Carnot cycle operating between an upper
temperature T; and a lower temperature T1 are W I {h 'l' 92 {7‘13}
I'll—11"“ 21— = n [i,e., es, + as; = '1] {7'14}
1 In the ease of a heat engine, we want or in terms of 4:1,, so eliminate q, : T1 T1  Tl
_ _ = —l5
[T1)il'z '1" iii ‘l'z( T1) {7 1’
In the ease of a refrigerator, we want a. in terms of w, so eliminate q; :
T2 T1
h" = — — = T
‘3‘: {TJQI 0" 4 "(T1 _ T2) I: lﬁ} Second low ofrhermmtvnnmirs; same we srertrrimi mechanics 103 The quantity in parentheses is negative, but a: is negative and q, is positive,
so the signs come out all right. Eulerﬂelan'mslnps'
If dz = M dx + N dy is a total differential, then:
as} , [a] “17,
a}: I 3x J.
Thus
(‘2— ill= is $1 W
Statistical Thermodynamics Since d3 = dqﬂJT it follows that at constant volume
3:3,, +IC,dlnT [ii19] where 3,, is the entropy at (PK. 0n replacing C, by the statistical mechanical
expression of Eq. [ZST}, we obtain T
1 31an Mailing (1% l — e — sr=— irso] 3 3“ Kl “1‘36th later? «and integration by parts leads to
s — 3,3 = are??? + sin 12 a Rln a, (ran
or E as
s—su=—+alnn—alng, {r y T ﬁbere Q0 1s the partition function at [PK One concludes that So and R In {.230
not unrelated constants, but that as a separate condition 3.3— — R in go,
._ at is, that the entire entropy at 1]” K is given by the probability of the lowest Illl Undsrﬂcnd'iag physi cal chemistry energy state. It then follows that s = %+ a in o oza At O‘K. all the molecules are in the lowest possible energy state, and ﬂu
is just g0, the statistical weight of the ground state. In the case of a perfect
crystalline solid, there are no alternative structures or arrangements possible
at WK, so that in this case go = l and SD is therefore zero. This last amounts
to a statement of the Third Law of Thermodynamics. Combination of EC]. [Tl23] with the deﬁnitions of A and G [Eq. {3" ll] gives .4 = R in o [T44]
o = rt Ino + Pa (125}
SaelnauTeo'ode Equation A. very important application of wave mechanics to statistical mechanics
is the calculation of the absolute entropy of an ideal gas. Equation {519}
gives Q for a monatomic ideal gas (in, translation only}. but it is now neces
sary to add a slight complexity to make the treatment adequately rigorous. It
is assumed that translational energies are independent, that is, that the partic
ular energy that one molecule has does not affect the chance of some other
molecule having some other particular energy ; then for one mole of molecu [as Q... = as... o—ssi The idea is essentially the same as that involved in obtaining Eu. {55}. The
energy per molecule is _ 2sin Q _
Entilcc _ k? a?! I? 2?] and the energy per mole is then s in Q
_ 2 let '
Em... k?" or n as} but Bin :2."  a In l2” — N a In Q, and since kN n R, the ﬁnal result for E
is the same as given in Eq. [541 In the case of translation. there is an added aspect which, while it cancels
out and does not affect the evaluation of E, does affect that of S. In con— Second hm of thermodynamics: some WW statistics! masonic: Ills sidering a set of N molecules, Eq. ("i26] implies N distinguishable molecules,
and consequently '12.... as given by En. [125] overeounts the probabilities by
the number of ways ol‘permuting N molecules—molecules which are, in fact, not distinguishable. As a consequence. the correct expression for QM is 1 I.
QIUI = Ell—lg” l? 29} For translation, then 3f] N
c... = %[ 2m” V] new _h1—
Since N l is a very large number, it can be approximated by Sterling‘s
formula1 In N l = N in N — N, so that Eqs. til—3U] and (”i23} become E  In
T I: l 31'2
2runk V] — lclNlnN — N] + Nltinl] h: Sll'll'll
ﬁts in the case of Eq. {123}, Eq. [Tl23]: is used in the more correct form s = ~Eg+ it mom {132} In the case of an ideal gas, HIT = 311;; and Eq. [1le i can be put in the form 3R 21::ka “3 1 RT
sum = 3— + a + a in“ s1 ] [ﬁllFl] o—ssi Further manipulation gives SR 21: 3": l 5:: T5’2M3E) $34
3"“; = T + R Inn?) (Flu: } + R in P I: :i where M is now the molecular weight. For the ideal gas at one atmosphere
pressure the ﬁnal result is 3.....tca1rdesmolo= R Intrti’Mt’ti — an can which is the famous Sacker—Tetrode equation [units must be watched
carefully in deriving {Tl35] from [T34}]. i __ _ ___________.t I06 Uridc rattanding physical chemistry The same derivation as the above can be applied to the case of a two—
dimensionai gas. However, Q is now given by Eq. [513]. so that the expression
corresponding to Eq. [Tl33} is On inserting numerical values for the various constants,
Sm,“ [iidim} = R lnlMTc] + 65.8 {calfdegmcle} {13?} where o denotes the area available per molecule. The complication of distinguishahiiity, that is. with N !, appears only in
the case of the translational entropy; rotational and vibrational entropies
per mole are just N times those per molecule. Thus insertion of expression
[512] For [2..." and Eqs. (5—23] and [123] gives SM 2 Rﬂn (3“,, + l} {Tl33} Second tear of thermodynamics .' me armors statistica! mechanics It]? V FIGURE 7] [PC and an upper temperature 1",, and after adjustment, operated between PROBLEMS 0°C and STD, calculate [a] the wattsihour needed {alter adjustment] to
' convert llrg water at IT‘C to 1kg ice at 0°C and {b} the original upper 1. ill] min] A refrigerator operates at 50% of ideal efﬁciency, that is, the
ideal work is 50% of the actual work. ll it operates between 0°C and 2512,
calculate the work to freeze 1 kg of ice [heat of fusion Eﬂcalfg] and the heat
discharged at 25°C. 1. {ti min] Show what percent T. is of T2 for a heat engine whose ideal
efﬁciency is 10%. 3. [til min] Figure T—l shows a carnot cycle as often depicted in texts. The
same cycle may, alternatively, be given as a plot of 3 versus T, as is done in
Fig. 'i—Z. {s} indicate which are the corresponding steps in the two ﬁgures. Do this
by labeling the arrows in Fig. 72 with A, B, {3, and D. _ [b] An ideal heat engine operating between T, 2 23°F. and T2 produces Lilli} cal of work per cycle. The entropy changes that the working ﬂuid goes
through are shown in Fig. 1—2. Calculate all and q; for one cycle, and T2. 4. [15 minlﬁi homeowner is iortu nate enough to possess an ideally operating refrigerator; nonetheless he feels it is using too much power and calls in a
repair man, who moves the machine back from the wall to have better air circulation around the hot coils. The homeowner now ﬁnds his power
consumption to be halved. Assuming that the machine operated between temperature TE. [1 ca] = 4.21 or Wisec, beat of fusion of water is Bﬂealig} 5. [13 min} a reversible heat engine absorbs heat g; at WK, per cycle,
and evolves heat qr, at SOUR. Its work output is used to run a hoist, and,
owing to friction in the pulleys, mm of w is converted into heat at SEEK. S, calfdeg T1 T TI
FIGURE ‘in IN Understanding physfcuf chemistry For the system engine plus pulleys, the total heat evolved is 12,000 cal per
cycle. {a} Calculate in, ex, and w. {b} Calculate as per cycle, for the system
engine plus pulleys. .r 6. {12 min} A homeowner has the idea of using an extra refrigerator to cool
his living room during the summer. He therefore sets up the machine in The
middle of the room, leaving the refrigerator door open to get the beneﬁt of
its cooling coils. Room temperature is 25°C, and it may be assumed that the
refrigerator is operating between 25°C and 0°C. The machine ordinarily is
capable of freezing I kg of icefh with these operating temperatures. Calculate
the temperature change in the living room. that is, the new temperature,
after 1 h of operation of the refrigerator. Assume ideal operation, and that the heat capacity of the room is 100 kcalfdeg. The heat of fusion of water is
30 cal;f g. T. {14 min} a farmhouse has a heat pump for heating purposes; on a par
ticular day it is operating between 25°C and 0°C. Since no electricity is
available, the heat pump is operated by a heat engine leg, a gasoline motor}
operating between I,0tIl°C and 25°C. Both machines operate ideally {fortun—
ate farmerl]. Calculate the performance factor for the system. that is, the ratio of the
amount ofheat delivered to the house at 25°C to the amount of heat prod need
at 1,000°C by the burning gasoline. The motor is located outside the house. 8. [l2 min} A. heat engine operating between 130th1: and 25°C produces , work that is entirely used to run a refrigerating machine, operating between
0°C and 25°C. Calculate the ratio of the heat absorbed by the engine to that
absorbed by the refrigerator, assuming ideal operation for both. 9. [12 min} A." ideally operating refrigerator works between 0°C and T°C.
It will freeze 1 kg of ice per hour [that is, remove 30 kcal of heat per hour
from water at 0°C]. At the same time, the total heat output of the refrigerator
to the room is 100 kcalfh. Calculate the value of T°C. M0. [12 min} Calculate {or give with explanation} dH and as when a Ikg
bar of copper at 100°C is placed in 2kg of water at 0°C in an insulated container maintained at 1 atm pressure. Heat capacities. Cu, 0.] ealfdeg g;
H100], l calfdeg— g. . 11. [2 min} Calculate no for the process of Problem 2. Actually there is  insufﬁcient information, so explain what additional information is needed,
and set up the so nations to show clearly how you would make the calculations. 1 12. [12 min} Cine mole of an ideal monatomic gas is taken from the state
{22.4 liter. 223°K. 5' = 20cah‘deg} to the state f2atm, 303°Kjl. Calculate
dE, did, as, and no for this change. \ Second l'mv of thermodynamics . some more stemmed! minerair: 109 13. [10 min} Give a process for which {a} dE = 0, [b] Mal = QWA = 0,
{d} AG = 0, and [e] 115 = 0. State all necessary conditions or restrictions
clearly. ,/ 14. [10 min] Cine mole of an ideal, monatomic gas initially at STP expands
isothermally and irreversibly to 44.3 liter, under conditions such that
w = 1m cal. Calculate 11.5 and d6. 15. [20 min} One mole of a perfect monatomic gas initially at volume
V, = 5 liter, pressure P., and temperature T1 = 293°K experiences the
following reversible changes: [A]: lsothermat compression to one half the
volume, the new volume and pressure being V1 2 iVl, and P2. {B} Cooling
at constant volume, until the pressure is returned to the original value of P. ,
the ﬁnal temperature being T2. These changes are shown schematically in
Fig. 2—3. Notice that process C, reduction in volume at constant pressure P, ,
is equivalent to the sum of steps a and E. [a] Calculate P1, Pz, and T3. Also q, w, 5E, dH, £3, and so for A. and B
separately. [b] Are the magnitudes [without regard for sign] ofﬁE, q, and w for step C
greater than, less than, or equal to the values of these quantities for the sum
of steps a. and E? IE. {12 min} One mole of an ideal monatomic gas initially at 10 atm pres
sure and 0°C is allowed to expand against a constant external pressure of
1.0 atm. Conditions are such that the ﬁnal volume is 10 times the initial volume; the ﬂue] gas pressure equals the external pressure.
{a} Calculate the initial and ﬁnal volume and the ﬁnal temperature. {b} Calculate o, w, dB, dH, as, and so for the process. V. libe‘t'
FIGURE 23 1 ll] Utn'fs'rsta aiding physical chemistry ('13. {19 min] One mole ol‘ an ideal monatomic gas is contained in a piston
and cylinder at 300° K and Ill atm. The pressure is suddenly released to 2 atm,
so that the gas expands adiabatically at a constant pressure of 2 atm.
[a] Explain whether as for this expansion should be positive, negative, or zero. lb] Calculate dE and as. To speed your calculation, 233 cal of work are
done by the gas in this expansion. 13. [3 min] An isolated system comprises 1 mole of ideal gas, a heat reservoir
at T1 , a machine, and a source of energy for the machine. The gas, at T, , ﬁrst
expands freely to twice its initial volume ; it is then compressed by the machine [at 3",} hack to the initial volume: the gas again is allowed to expand freely to
twice the initial volume 1What is the minimum as for the system at the end ol‘
these three steps? v”lit. [13 min] Cine mole of He is heated from 203°C to 4Uil°C at a constant
pressure of 1 atm. Given that the absolute entropy of He at 200°C is 32.3
cah'deg—mole', and assuming that He is a perfect gas, calculate dG, dH, and as for this process: so comes out negative; does this mean that the process
is spontaneous? Explain. 2]t {to ruin] One mole of an ideal monatomic gas undergoes an irreversible
adiabatic process in which the gas ends up at STP and for which ad is .5 calfdeg
and w is 3th} cal. The third law entropy for the gasat ST? is 45 calfdeg‘mole. Calculate dB and so for the process and also the initial state of the gas {its
initial P, V, and T]. 21. [12 min] An ideal gas expands isothermally at 25°C, but somewhat
irreversibly, producing LOW cal of work. The entropy change is 10 calecg. Calculate the degree ol‘ irreversibility, that is, the ratio w,,,,,,.,lw,,, where w," is the reversible work for the isothermal expansion of the same gas to the
same ﬁnal volume. 22. {l 2 min] One mole olan ideal monatomic gas initially at STP undergoes
an irreversible isothermal expansion to 13.5 atm pressure, thereby doing
It'll} cal of work. Calculate or explain the value of ﬁll, dH, d5, :5}, and as.
Calculate also the reversible work for this isothermal change of state. 2.3. [12 min]: It has been possible to superccc] small drops of water to —4tl°C.
Such drops are unstable, of course, and eventually nucleation occurs and ice
crystals form. Assume that such a drop is thermally isolated so that, when
spontaneous ice formation occurs, it does so adiabatically [andjsabaricallyy Cplis lcalyldeg for water {I} and {licah‘degg for water {a}, and the beat of
theme of water is ill} callr g at {1°C. Calculate the ﬁnal temperature ol‘ the drop after the spontaneous process has occurred, and dH and £13 for the process,
in calfg. Second low cfrhemwc’Jvmmics: some store statistical mechanics Ill 24. [9 min] Calculate dH and do? for the process
CH3OHﬂ, 64°C, 1 atm} = CH3DHfg, 64°C, 0.5 aim] [The normal boiling point of CH 30H is 64" C, and the heat of vaporization is
2613 callg.) ﬁll 15. {IE min} One gram of water enclosed in a vial is placed in an evacuated
ﬂask maintained at 25°C. By means ol'a lever, the vial is broken so that the
water is free to vaporize and, when equilibrium is reached, one—half the water
has vaporized. The vapor pressure of water at 25°C is 24 mm Hg, and the
enthalpy of vaporization is 590 calig. Calculate o. w, an, no, and tie? for this
process. ER {15 min] Cine mole ol'liquid water is allowed to expand into an evacuated
.frﬂaslt of volume such that the ﬁnal pressure is (1.3 atm. The bulb containing ® the liquid and the ﬂask are thermostated so that a constant temperature of
100°C is maintained. It is found that ll,t][lll cal ol'heat are absorbed when this process occurs. Calculate or explain what the values are for Irv, ﬁll, an, as,
and so. Neglect PF quantities for liquid water. 2?. {til min] Cine mole of water vapor at lllll°C is compressed to 2 atm. It
is sufﬁciently dust free so that it supersaturatcs; alter a while, however, con
densation occurs and goea to completion, since T and P are kept constant.
The process is thus Hlﬂlg. lIZITC, 2 atm} = Hzﬂll, 100°C, 2 atm}. Data: heat
capacities : CF for the vapor is "l calfdegmole, and is 13 calfdegmole for the
liquid. Heat of vaporization under these conditions: 11 kcalfmole. It is per
missible to assume the vapor to be ideal and liquid water to be incompressible
at a molar volume of 13 cc. Calculate tilH. no, and 113 for the process. 28. [18 min} 100 g of ice at [1°C is dropped into an insulated beaker con
taining 15D g of water at 100°C. Calculate 53 for the process that then occurs.
{You may take the heat of fusion olice to be 30 caLl'g, and the heat capacities
of water and of ice to be 1 and {3.5 calfdeg—g, respectively.) 29'. [12 mind in} Derive [dedeg = PET. (Hint: Make use of the total dimer
ential for E expressed as a function old and V.) {b} Verify the above equation
for an ideal gas; that is, evaluate [ESdelg directly. {3}. {9 min] Derive from the ﬁrst and second laws of thermodynamics and
related deﬁnitions: {dTy‘dPJs = [t3 wasp. 31. {3 min} Show that {dHldFls = V, using the ﬁrst and second laws of
thermodynamics and related deﬁnitions. 1 i1 Umfersmiadr'ng physical chemistry 32. {ll} min} Derive from the ﬁrst and second laws of thermodynamics and
related deﬁnitions '. ﬁlAf'l'Zlde = — Elia. 33. {‘l min] Derive from the ﬁrst and second laws and related deﬁnitions:
(salami. = —P'_ 34. [15 min] Derive from the ﬁrst and second laws and related deﬁnitions:
lal’lﬂTlp = ll3'3l+'3‘Flr~ 35. {12 mini Given the process:
0.20 mole 0; {g 0.2 atm] + £13 mole I‘llz lg. [LE atm] = mixture [g] all at 25°C, which is carried out by having the oxygen and nitrogen initially in
separate flasks, and then by opening the stopcock connecting the two ﬂasks.
{a} Calculate the ﬁnal P. {b} Calculate n, w, dE, as, and so for the process.
{c} Calculate the reversible q and w for the isothermal process returning the
mixture to its initial state. 315. [12 min} The element krypton exists in nature as a mixture of about
1D% Kr“, 'lﬂ‘iFE Kr“, and 20% Kt“. Calculate :13 and of} at 25°C for the FIGCE'SS I 1 mole Kr [natural mlxt are] 1 atm. 25°C 0.7 mole KrEm = Ill mole Krszll atm, 25°C] + mixture of 0.2 mole Kr86
at i atmI 25°C 3?. [[2 min} {a} Explain why, according to the Eackur—Tetrode equation,
the entropy of an ideal gas depends on the molecular weight of the gas. It is
not sufﬁcient to copy a derivation! What is wanted is an explanation of what
is going on physically that brings about this eﬁ‘ect. {b} Explain whether the
Sacker—Tetrode equation does or does not predict that the heat capacity of a
gas should depend on its molecular weight. Here, the best approach will be
through a derivation rather than physical arguments. ® [lll min] The rotational partition function for a diatomic molecule
{ignoring a nuclear spin factor} is Enzlk'l'fhz, where .lr is the moment of inertia.
Calculate the difference in rotational entropies per mole for diatomic mole
cules AA and BB. Both gases are at 25°C, but AA has twice the molecular weight and twice the moment of inertia of EB. 5'th low of ilternmdynnmics; some more statistical mechanics 113 39. til] min} Show how the entropy of an ideal monatomic gas should vary
with temperature, according to the Sackueretrode equation. Obtain the same result by another means. 4]. {lﬂ min} Give a qualitative, but deﬁnite explanation of whether the
rotational entropy for CH4 gas at STP should be greaterI the same as, or less
than the translational entropy. 41. {13 min} For a diatomic molecule. Em. = lirﬂ'lt'ljlll2 [as noted in Prob~
lein 33}. {a} Estimate {to within about Ellie} the value of QM?!“ for H1.
lb] Estimate the temperature at which Sm. might come out negative on
application of Eq. {Tl—23}. {cl Why is the situation described in [b] nonsense
and how does such a situation appear to be possible? @ [14 min} A. vapor adsorbed on a liquid surface can behave very much
ll e an ideal twodimensional gas. Equation [Tl3?] gives the entropy for such
a gas in terms ole, the area available per molecule. Derive the expression for
the translational entropy of a twodimensional gas in a standard state
deﬁned as one in which the molecules are the same average distance apart as
for an ideal three—dimensional gas at STP. 43. [13 min} An alternative and somewhat more general form of Eq. [54] is E” = £3 + gar + arise: ohyar where E" is the energy in a chosen standard state, E3, the energy of the lowest
state {zeropoint energy} The equipartition value for translational energy is assumed. and QM and om, have been lumped together as QM." = QMQHb.
Given the above approach, derive with the use of other equations of this chapter the expression
ro° — Egirr= —§1nM — sin r— a In pm + v.27 for an ideal gas at 1 atm pressure. [The quantity {5" — EEHT is known as the
free energy function] {T his may look a hit forbidding. As a hintT a good place to start is with the dening equations for G and H.) 44. {a min] Qvib and its contribution to the heat capacity of gases was fairly
well emphasized in Chapter 5, yet has hardly been mentioned in the present
chapter. Why might this be? {A serious answer, please.) 45. [ill min} Derive Eq. til33]. [Hint . focus on the temperature dependence
or gentl 46. Suppose that the molecules of a certain substance can exist either in a
ground energy state [of energy taken to be acre] or an excited state of energy Ill Understanding physical chemistry the expression for the partition a; there are no other excited states. Write
mole and the entropy for the function. Calculate the average energv per
case of inﬁnite temperature. ANSWERS ‘— [TII'IITIMI 5 —gﬂ1ﬂﬂﬂ X 1. Here, we want at in terms of til: w = q,
LTl'Je ZSfETS = —T,35ft c l. The actual work will be twice this: —
heat discharged at 5°C will be $111,111} plus MED!) = 9457110. I. 1titlewantthcecpiationw = o1 — quiflglfivjq: = D.l,then (l — Tully} = fitl, or '1".sz = 119 so “I", is 90% of 1",.
he higher temperature, so it must card 3. Step {at is isothermal and at t
72; the other labels follow in counter respond to the step marked X in Fig.
clockwise order. ' [b] For the cycle quill; + opt}; = fl = as, + as}. From Fig. 4—2,
as, = —20 and as, = so, see, = —so T. and q, = so T1. The work,
at = o, + q; = Ltlfrllcal, so —2fl'>< 213 + EDT] =1oze and! T2 = SITE;
cl = #63150 cal. _
4. [a] Use the equation w = q, — tTﬁTﬂq. = 813,013!) x [l — MERE] =
43,300 cal {per h]. In Mace. 3,311) x 4.2,“3,ﬁ{}fl = “Ll W. Us] The original work must have been —l’f,ﬁfl{l cal for the same job;
hence l — Tsz, =  [lollflfﬂﬂslﬂﬂ = 43.21 so TZHT, = 1.22 and T3 =
333°K or ETC [inspection of the formula for or would have told you that to
double w. the temperature difference must be doubled, and hence be 60" instead of 312)“).
5. {a} First, 111w + hm = llﬂﬂfl cal. Also, w = q, — {Tlf'nlql =
itucle of :11 and argue [ﬁllﬂﬂﬂflllqll — Elqil. [It is easier to use the magn
intuitively that it is then to he added to 0.1 w.) Then ﬂ.21q,l + lshl = 123ml
—‘l'll,flftll cal. Next, q: = —[T2tTllq, — 490D.“ and lfhl 7 lﬂﬂflll Ul‘ q1 =
mnlqi = 3'3,” cal. Finally, at = q, [bl Since the engine is reversible. d5
also receive the frictional heat, which is lllflll cal from part [a], at which produces an entropyr of 2,0flllf3ﬂﬂl = as? calfdeg per cycle.
= —se.ooox 25ers = 7".35Dcaltperh}. (a, = will) and q: = — 87,351] cal.) The homeowrier is evidentlg.r in for a surprise,
as the net effect is that the work of running the refrigerator appears as heat! The room heats up by $.33] calfh or 135W“? = 0.0135 degfh.
T. For the heat pump, w = q. + :31 Ttt'Tzl: while for the heat + q; 3 21],.“ cal.
—— fl, per cycle, for it. Its surroundings
3ﬂﬂ°K, = ‘i‘ell "' Second law It!“ thermodynamics .' some more statistical mr'hn‘m'cs llﬁ Tmnf._w+= q“. + it”: = rfztl — T._.FT’1J. Since is" = —w, it follows that
_q1 , _ 1f?2}=o1{l — TH'TEJ, and hence that the desired ratio of
q,,tq, — tr, — TllﬂttTl — The; = ass x asses x iris = ELI. B. F '  ‘ w' = or the! engine, as — qr. + q: .= qgfﬂ — "ELT1. For the refrigerator
15],.13513 —qu[—"~:Iﬂl;l a Til“T, Smce w = a”, it follows that q3{1.fﬂ} —’
”+11: q, .5 — 1. T3 and hence that qzt'q'l = {25f2'l3HL2T3JW5} = 9. Sinceth + T = '
= 341°K ﬂ; asideen“ ﬂ. 1t follows that T2 = —[q3,tql}:r, = {IDDIEEIET} Lush Since the system is adiabatic at constant pressure. 5H = I). d5 however
i e sum of the entrop}.r changes in the copper and the water. Firsl. the ﬁnal temperature is givenh 1.0m x ()1 _ _
For the copper, 3"  Illflfl tl — 2,0012! x 1 x t,orr = 4.7TDC. as = c,1n mi", = Lona x at x 2.3 log silvers
= —23.T calfdeg. For the water, as = acne Intsivwsissi = 2,ﬂl]l]1nl{1 + nets}
e acne x oars = 36calfdeg. The not do? is then —23.? + 36 = T calfdeg. 11. Since the initial and ﬂu
al states are not at the same to
ﬁg? must be assembled as follows. For the copper, $621211er
1‘: 1:3  T131)“; for the water, d6 = ﬁHHzU — [T233 — T.5.}H a. déunet
31¢ Emilia? s w1fllCeancell The additional information needed is then
. rep}r o u and H10 at some particular tem er
_ at , f
wh1ch the values at the desired temperatures can he calculgted ilisrfngrgi: given heat capacities, that is. T:
32 = 3. +I c,s1nr
,. L [2. First. complete the identiﬁcation of initial and ﬁnal states Initial: V = 21 4 liter Final '
. : V = .4
T = 2T3°K T = $34231 So P = 1 atm P = 2 atm
S = m CEl_.'ltleg—mnle 1111 Understanding physical chemistry Since the gas is ideal, 13E = 13,111" I: {331:2} x 31] = 91112511; 2111' = [392‘ —— {51232} x 311 = 1511 cat. .
For as, the general equation could be used, but since as also 15 wanted, it is better to write out explicitly the two steps:
[1} 22.4 liter, 2333K, 1 arm I: 24.3 liter, 303°K, 1 atm as, = c, In stir, = {says} 111 1.11 = 11.53 calideglso s,.,,. = 211.53} 5H, —['1'}_S] — TIS1J= 151] — [3113 x 211.53 — 273 x 2D}
—tirl)11ca1 as, [2] 24.31itcr. 3ﬂ3°K, 1 atm = 12.4 liter, 3932K, 2 atm 11133 = Rln 2".sz m 1.93 x 2.31ogi = —1.3'1ca1,ideg
1361 = RTln P112}. = 41? cal Overall,then..i3.S = [1.53 + [r1321 = —[1.34ca1,‘dcg,n(} = —600 + 41":r =
—133ca1. . 13. {a} as = [I [or any isochoric adiabatic process to, = 111. [111} Similarly,
13H = d for any isobaric adiabatic process lqp = [11. is] Since 114 = — 3 11'1" —
P5114 it will be zero for any reversible constant T and 1’ process: {11} Since
1116 = —SdT + Firth21.6 will be zero for any reversible constant T and P
process. {c} Since 115 = em,“ 1", 1113 will be zero for any reversible adiabatic process. 14. [Since the gas is ideal and the process is isothermal, 1311‘ = 11, so
a = w— = 190cm.) Actually, all that is needed' ts the knowledge that the ﬁnal
state is 2'13“K and 44.3 liter [so P = [1.5 atm}. Hence 111,3 = R In Iﬁﬂﬁ 2 1.93 x 2.31og2 .——1.3'1'clli'deg,andr11.G=RT x In Pp? = 1.93 a 213 x
2..31cg[15=' r314cal. 15. ta] From the ideal gas law, P. = RSI"1 ,W, = 4.89 arm. after the iso—
thermal compression to halt" the volume, pressure must be doubled, so
P2 = 9.13 arm. To halve this pressure by cooling, ‘1"2 = ”hi2 = 149°K. Stepa: Since itis isothermal, as m 319' = 11. c = w = RT x in 1311’. =
l.93 x 293 x 2.31og§= —41J'3caL 13.3 = R111 V1.31”. = —l.3&ellyl1eg and
so = 111111133111"l = 41131:“. SrepB:S1nce1t1s1soc11or1c w = 11.1; = 11152 213,312“ = {311,121 x {—149}
—442 cal. dH= deT = {Emmi—149] = —'1'411eal.dS = (3,111 T311".
1311,1212. 31 logly : —2.ﬂﬁee1,‘deg. Since 1316:1311 — {T232— T131], it can‘
not be obtained without a knowledge of the absolute entropies and these
are not given. 1 SeeJed Few ofihernmdymmirs; some more sioiisitcni' niceﬂanks 111' 1b} EtE is independent of path and must be the same. Since the area under
step C in the P—V plot o1" Fig. '11—3 is less than under steps A plus E, w must be
smaller in magnitude for step (3. Since $13 = q — w, 1; must likewise be smaller. 16. [a] 1’. is 2.24 liters; hence 1’2 is 22.4 liters. Since the ﬁnal pressure equals
the external pressure of 1 atm, the ﬁnal temperature must be 0°C. The overall
process is then isothermal; hence 151E = SH = 11. lb} The work done is P1111” = latm x [22.4 — 2.241 = 211.2liter—atrn or
493 cal. Since 11.13 is zero. a = w = 493 cal. .133 = Rln V3114 = 1.93 x
2.3 log II} = 4.55 calfdeg and ME = RT In szP. {P = PM} = — 1.93 x
2'13 x 2.3103 [1.1 = —l,24'rl] cal. 17. {a} This is an irreversible process for which 1; = 11', therefore AS :2 {1.
[hi Since 1} = I}. dE = C,_.13T= —w or {3M2} x 11.1" = —233 and diT=
—9?“, so '1"; = 2113“K. and therefore V1 2 3.32 liters Since SS = C11 in Tsz.
— R111 PIIPI. 1313 = {5112,1212} x logi2ﬂ3f3ﬂ‘ﬂ]  1.93 x 2.3 log 2,1111} 2
—1.92 + 3.19 = 1.21r calfdeg. 13. Consider ﬁrst the gas only: step 1 plus step 2 returns it to the initial state.so d3. = 11. Step 3tas does step liinyolves d5 = R In 2 = 1.31:1 calideg.
For the rest of the system, in step 2, w = R2111 P3311”. = 1.93 x T. x 2.31og2 = 1.363". =1 4, since the gas suffers no energy change. Therefore
1.362] cal are delivered to the heat reservoir, and it gains 1.3!}{Tlli‘T11 entropy or 1.36 calideg. The minimum total as is then 2 x 1.315 cali'deg. 19. as = (3,1 T1 = [5.31121 )1: 2.3 Ioglﬁ'13f4231 = 1.25 calfdeg. Then
‘3; = 32.3 + = 34.1 calfdeg; 11.111 = (3,211": tSRi‘2l2t1ti = 9911 cal
Then 11113 = 13H — {T232 — T;S.] = 9911 — [6‘13 1: 34.1 — 433 x 32.31:
i§.5611 cal. The process need not be spontaneous; if as is negative for a
process at constant Tend P, then it is spontaneous. 211. By the ﬁrst law, 1'3E 1:: 11  w = —300; hence 13'1": —3i11],i'[3Ri2] —
—11'.11"‘, and T1 = 2'13 + 101 = 374°1t. By the second law equation as =
13,, Iangf'ﬂl + Rlnﬂéﬂ’ﬂ, which applies to any change of state, reversible
cr not,R1n[V3,iV1} = 5 — [3111121 lnlj22311324} = 5.94. Then Iogleﬂr’ﬂ = 1.311.
and V. = 22.431211 = 1.12 liter. P. is then (1.1132 x 32433.12 = 17.311111. 21113 = .dH — MTS], with 1311' =1 —1111[SR,1'21=: —500. Also. S. = 45
— 5 = 411, so 1313 == —5Cl3 — {2'13 x 45  3'14 x 4111= 2,1711ea1. 21. In this case, the amount of gas is not speciﬁed, so 11.3 = 1111 lntVzli‘l/l], where 11 is the number cl'ntoles. w," is just nRTIntVﬁV.) or w," = ’1'" 11.3 =
asst]. w,,,,,,iw,,, = incomes:1 = ass. 22. The process is isothermal. so SE = {land 13H = 11. Also,tb1e:n q = w =
teessi. as is given by —sin 111.111. = F.n 11.5.11 = 1.37 cali'deg; this is
also 1;"va and w," is therefore 223 x 1.3? = .373 ea]. so 2 did — This =
11 — 373 = —3?3 cal. ”3 Us demos ding physicat chemistry 23. It is ﬁrst necessary to decide whether the drop ends up entirely as ice
{at T g 0°C} or whether it only partly freezes {and so is at 9°C}. A little
reﬂection suggests that the ED calt'g liberated if it all froze should be more
than enough to raise the temperature to 0°C, and that the drop will therefore
not entirely solidify. To check this, set up the two reversible steps and work
out the heat balance [clearly q? and also so are zero here}: lg [l] HIDEI, —4ﬂ°C] = Hglel,ﬂ°Ci
[2] xHZCHI, 13°C] = xHIDIs, 13°C] 511', = 40 caly'g
tilHz = —Eli Since dH. + tilllg = ﬂ, .1: must be [1.5 g. The fact that a; is a physically possible answer conﬁrms the guess as to
which happens. Then 5.3. = C, In UT. 2 1 x 2.3 og[2"l3f233] = 0.159
calfdeg. 5.32 = qt?" = —4ﬂ,l2l3 = 43.1415. d3 net is then 0.013 caltdeg. 14. Set up a reversible path as follows: [ll CHSDHIEI, 64°C, 1 atm] = Cﬂyﬂﬂfg, 64°C, 1 atm}
dH. = 2613 x 32 = E,3lﬂcal,lrnole ﬁGl = ﬂ .
t2} CHaﬂﬂfg. 64°C, 1 atm} = priming, see, as :1th
as, = o ' 2162 = RTln szPl 2 1.93 x 33? x gtgiﬂgé = —4ﬁ_tleal lCIverall then, diH = 8,311} calltnole, so = ~4atiﬂcaltrnole.
25. The process is evidently tg Hzﬂfl. 25°C {and 24 mm Hg] = g; Hzﬂfg, assc. 24mm Hg} Then q = 590,51 = 295 cal, w = (3 [no external work is done}. We assume
q is our. so dE = 295 cal. The corresponding reversible process would simply
involve placing the vial in a piston and cylinder at 25°C, and slowly expa riding
until halfthe liquid vaporized,so that for this prooessq = on = or, + FEW:
295 + [”36] moles x RT = 312ca] {neglecting the volume of liquid]. AH is
then 312 cal. 312 is the om, so :13 = 3  2,"293 = LII5 calfdeg. .933 is zero since
we have a reversible constant T and P process. Answers based on dH = 295
cal were equally accepted, in which case q would be 223 cal. 26. No external werk is done, so in = {l and q = q” = 513 = ll,llﬂﬂ' cal. dH = all + MPH = H.090 + P3P} = ll,[l{lI3 + RT = 11,24tleal. To get
the other quantities, set up ﬂ‘lf: reversible steps: i ll Second tow of thermodynmtcs: some more statistical mechanics “'9‘ [1] HIGH, lﬂﬂ“C} = Hyﬂig. 190°C. 1 atm} $31 = stT = ll.?4ﬂt3?3 = 31.5 calfcleg
so, = o [2} Hlotg. intro. I atm} = Hzﬂfg, 1oo°c,o.s atm} as, = sin Igor, = 1.93 x 2.31 legions] = 2.3Tca1fdeg
4562 = RTinng'IPl = —886C3.l Then 5.5 total is 33.9 calldeg and 5.6 total is —886 cal. 2'3. The process is at constant P. so is = {fi = diff = — 11,9013 cal. New
set up equivalent reversible steps: {1} HIDtg. lﬂﬂ°C,2atm} = Hzﬁfg. lﬂtWC, 1 atm}
as, = o _
£1.31 = Rln VLW. = 1.3? calfcleg
til.6. = RTln log“P1 = —509 cal
[2] Hlﬂg. IWC, l atml = HIGH, IWC, l atm}
{3] Hgﬂfl, IWC, l atm} = Hzﬂfl, lﬂﬂ’CJatm} Since the liquid is incompressible and FL is small, oilI3, as,“ and as, are
small and may be neglected. For process [2}, then, til13 must be — 11,909 = £12.: hence 5.31 = —11,IZt313,nr
333 = — 29.5 calldeg. at}: is zero, as we have a constant Tand P reversible process. lOver—all. then. as = —ll,l]l]ll ca], 215 = —28.1 caltdeg, and
so = —5l]9 cal. 28. It is ﬁrst necessary to set up a heat balance to determine the ﬁnal
temperature. Preliminary inspection suggests that all of the ice will melt,
so the overall process can be written as the sum of the following three: longtsmc = lﬂﬂg[,ﬂ°C] q,=1oo x so
longtime} = lﬂﬂgﬂ. so} a, = toot
lﬁﬂgfl, 100°C} = 159g“. no] q, = —1sotioo — it Since em, = t], it then follows that 3,1203 + 1139:: litltll] — 1543:, or
t = 23°C or 3Ul°K.The corresponding entropy changes are :13. = E,t:tllt}t2'l3
= 29.3, as2 = c,1nrr=,tr,] = tan moonsts] = 9.2, as, = isoinrsoiorsi
= —32.1. The overall. as is then 29.3 + 9.? — 32.1 = £3 caleeg. 29. [a] Following instructions {i.e.. the hint], write dE = {dﬁtﬁﬁjy rts + [Ill Understanding physical chemtsrry [dEfdl’lsdli The desired partial derivative follows, on dividing by if!” at
constant E: tetra E .E
esvevg W, 01‘
(E a _tﬂEiElfls
av ,, testes}; The two partial derivatives can be evaluated by comparing with the ﬁrst law
equat1on d5” = TdS — PriV, that is, [dEirZ‘SJV .= T and {tiEfﬂ If}5 = —P.
Substitution then yields the equation to be derived. {ht For an ideal gas, E is a ftmetion of T only, so the partial derivative
becomes tests VJT. Since d5 = R rt in V = R tit/IV for an isothermal process
with an ideal gas, tleti'V = {fiSthlr = RIF = PIT QED. 31]. We have a derivative by P with S constant equal to one by S with P constant, so this is clearly a Euler or crossdiﬂerentiation relationship. We look for something equal to Tris + l’dF, and a little reﬂection yields that =
no + VdPtfrom of: = Tris — PdVand H = E + PVJL 31. These types of relationships generally come from one of the versions
of the combined ﬁrst and second laws Try clH = TriS + Veil”; dividing by
ti? w1th S constant yields the desired partial derivative. 32. It is necessary to recognize that the partial derivative must be at con . stant Kata/fryer: tirrttearerty — All": but to: = —ser— P av so {age}??? = —s and tremor: —srr — and = —ttrr=ttrs + A} = 33. Try various versions of the combined ﬁrst and second laws. in fact. the
one for da suggests itself: rial = —SdT, — P :tV [from all: = res — 11th and A a E — TE}. Division by ill” with Teonstant yields the desired partial
dlﬁemntial. 34. We have a differential by Twith P constant equal to one by P with T
constant, which suggests a Euler or crossdifferential relationship. We there
fore lcolt for a relationship involving Val? — Sci?“ and a little reﬂection yields rt'G = —SdT+ VdP [from d5: Tat? PdV and G = H — TS
H=PE+PV1 ’ 35. [a] The volume of 0.2 mole at 13.2 atm is the same as that of [1.3 mole at
[1.3 atm. so the two ﬂasks must have been equal in volume. We end up with {l} rSnole of gas in twice the volume of one ﬂaslt, so the ﬁnal pressure must he
. atm. Second low oj'rhermodynumies; some more .il'ttl'iiﬂicm’ mechanics 111 [ht The ﬁnal partial pressures must he [14 for N2 and ill for 02. Then
for the 03, d3 = [1.2K 1n tall/1, and for the H2, dS= DER In Vﬂ‘l’, or
as total is R in WV. = LEE x 1.3 log 2 = 1.321r ealfdeg. Similarly. so total
is RTln PZIP, = Lilli x 293 x 2.3logitP1fP, = {r for each gas} or so =
dill? cal. From the nature of the operation no work is performed, so
or = I}. The gases are ideal so their energies do not change with volume:
hence dnE = 0. Finally, it follows that o = t}. [e] Since an is zero [the gases being ideal]. q = w. Also, em = Th3 or
q = w = 293[—1.3‘ll = —4tl9eal. 36. Since the Kr“ and Kr“ are not separated, their mixture may he
treated as a single species, and the process can be rewritten as _lrnole{ltl% A,9ﬂ% lit] = eta + (193 as is then —dtS = —R{tl.l l'n Ill + I19 In 5.9] = 9.642 or :53 = "9.641 eat!
tleg. Since MT is zero, d6 = — res = ”‘2 cal. 37. [a] Referring hack to Eq. It'sl3} the wave mechanical expression for
the soparation of translational energy states contains the mass m of the
particle in the denominator. This means that the larger the molecular weight,
the more closely spaced are these states, and the lower the temperature at which the system can gain energy. {bl The Sackur—Tetrode equation predicts
that the translational heat capacity should not depend on the molecular weight of the gas. The equation can be written 3 = [constant] + [ERR] in T,
and since CF = [33rd in T},, it then follows directly that C, = SEQ. 33. Since 3 = Et'T + R in Q, SM — Sea for rotation will reduce to
it In we)“, the rotational energy being expected to he at the same equi
partition value in both cases. The ratio of Q‘s further reduces to hating or 2,
so the answer is R In 2 or lﬁﬂeatfdegntote. 39. According to the Sackur—Tetrode equation 5n... = Rln T5” plus
'ton'tperature—independent quantities. Then dedT=tﬁitt2HltTt As the
alternative route. C, is 5R3? by the equipartition principle, and d3 =
£5”de in T or 3",,“ = C." In T plus constants; again, dedT = [SREZHIITZL It}. The entropies are obtained through the equation d3 = (3,,ti'ln ”I; so
the lower the temperature at which equipartition heat capacity is reached,
the larger will be this integral. in view of the very close spacing of translational
energy levels as compared with rotational, the translational entropy should
he [and is] much larger than the rotational. 4]. [a] it will be about {Ito x 10“)“ x ill—all. and inserting values for
the other constants, er'T is about 0114. {b} From Eq. {2—23}, 3 = EXT + [12 Unricrsin'ndirig physieui chemistry R ln {3; the maximum possible HIT is the equipartition value of ERIE or R.
3 must be negative, then. ii In {I} e 1__ or {3 we ifs, or if T is less than about
10°K. to} The situation in {b} is nonsensical sinoe S as an absolute entropy
must be greater than zero. The difﬁculty is evidently that the given expression
for Q, being based on an integration, is not accurate at such a low temperature.
From Eq. [5—21) diam between the states with .i = l and .l’ = ﬂ is itzi’Sirii,
so the expression for Q itself {by the integration procedure} turns out to be
just itdesm. At Iﬂ°K, k1" is thus about onethird of dam, and it is to be
expected that the integration approximation would be poor. 42. For a gas at STP, the volume per molecule is 224mm x Iii” or
3.? x 10'” cclitnole, oorTeSponding to an average distance apart of
3.3 x lIllI'T cm, and an area er 2: 1.! x HF 13 cm2. 3"“, itwmdimensional}
then becomes Rn{1.1 x 1U"3]+ R IniMT] + 65.3 = F. IniMT} + 6.6. 43. First, by deﬁnition c“ = H” — rs“ = s” + PV—— rs“ = E” + RT
— T3”. The translational part of S” is given by the Sackur—Tetrode equation,
and the rest by $3,. = EﬂulT+ Rln 12,“, = RTtiln Q1anT+ Rln QM. On
substitution into the deﬁning equation for G“, G“ = as + sar+ RT’tiln o,,,,,ttr+ sr— sat"Instr — garin r
+ 2.3]T— arsstnolnysr— RTanim On cancelling terms, and rearranging,
{cs — Eights": —%RlnM  its in r_ Ringim+{2.31+§R + it]
The constants total to 12?, to give the desired result. 4d. Qua can easily make an appreciable contribution to the heat capacity
of a gas around room temperature, but does not add much to the entropy of the gas As a reason for the latter, entropy development can be regarded in
terms of the integral C,ti' In T, and the vibrational contribution begins to
appear only at a relatively high temperature. 45. From Eqs. {i231 and {54} s,,,= ENJT+Ranm = ars1ngwysr+ Ring”, Since on, may as written at. equal to air", it follows that a la ogre?" = or,
so that 3",, is simply R + R [1'] gm. Setond imp ofiﬂwmodymks; some more statistical mechanics 123 «iii. Q in this case is simply i + expi—EiltT] and s is ll] + 9—HT!"
1 + rs” its inﬁnite temperature expi—sfitT} approaches unity, so s becomes er? or
E = Neil and S = EfT+ R In E = ii'lnl. E: ...
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 Fall '09
 Goux
 Physical chemistry, pH

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