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CHAPTER 1 6The Properties of Cases

PROB LEMS AN D SOLUTIONS

1 6-1. In an issue of the journal Science a few years ago, a research group discussed experiments inwhich they determined the structure of cesium iodide crystals at a pressure of 302 gigapascals(GPa), How many atmospheres and bars is this pressure?

2.98 x I 06 atm, 3.02 x 106 bar

16-2. In meteorology, pressures are expressed in units of millibars (mbar). Convert 985 mbar to torrand to atmsspheres.

739 torr, A.972 atm

16-3. Calculate the value of the pressure (in atm) exerted by a 33.9-foot column of water. Take thedensity of water to be 1.00 g.mL-1.

We first convert the height of the column to metric units: 33.9 ft = 19.33 m. Now

P : pgh: (1.00 kg,dm-3;198.067 dm.s-2)(103.3 dm)

: 1.013 x 104 kg.dm-r.s-2

: 1.013 x 105 Pa: 1.00 atm

16-4. At which temperature arp the Celsius and Farenheit temperature scales equal?

-40"

16-5. A travel guide says that to convert Celsius temperatures to Farenheit temperatures, double theCelsius temperature and add 30. Comment on this recipe.

481

482 Clrapter 15

This will provide a rough estimate of the temperature, decreasing in accuracy as tsmperature

increases. (of course, it is not valid for celsius temperatures below zero degrees.) At room

temperatures, it is accurate enough for ordinary purposes'

Actual 7 ("C) Actual 7 ("F) Travel 7 ('F)

0

10

zv

30

40

50

68

86

ta4

30

50

'70

90

110

1 G-6. Research in surface science is carried out using ultra-high vacuum chambers that can sustain

pressures as low as 10-r2 torr. How many molecules are there in a 1.00-cm3 volume inside such an

apparatus at2g8K? What is the corresponding molar volume 7 at this pressure and temperature?

We will assume ideal gas behavior, so

PVRT

(10-'2 torr)(1.00 cm3)

(16.1a)

(82.058 cm3 . atm. mol-' . K-' ) (760 torr' atm-' ) 1298 K) =17

5.38 x 10-20 mol : r

so there arc 3.24 x 104 molecules in the apparatus. The molar volume is

v 1.00 cm3tl -

_ : 1.86 x l0re cm3'mol-ln 5.38 x 10- mol

16-7. Use the following data for an unknown gas at 300 K to determine the molecular mass of the gas.

The line of best fit of a plot of P/p versus p will have an intercept of RT I M . Plotting, we flnd that

the intercept of this plot is 0.56558 bar'g-''dm3, and so M : 44.10 g'mol-r.

16-8. Recall from general chemistry that Dalton's law of partial pressures says that each gas in a

mixture of ideal gases acts as if the other gases were not present. Use this fact to show that the

partial pressure exerted by each gas is given by

/rt\P, = IJ- I P,^.., : y,P.^,^,-r

\Inr/ tota' /r toral

where P, is the partial pressure of the jth gas and ), is its mole fraction.

486 Chapter 16

1.2

i.0

0.8

0.6

0,4

^,,0

KKKKK

1 6*1 5. Use both the van der Waals and the Redlich-Kwong equations to calculate the molar volur:of CO at20A K and 1000 bar. Compare your result to the result you would get using the ideal-s-.equation. The experimental value is 0.04009 L.mol-r.

We can use the Newton-Raphson method (MathChapter G) to solve these cubic equations of s::We can express f (T) for the van der Waals equation as (Example 16-2)

and f '{V) as

r(v) :r' - (ul- #) v' + 3v - #

f' (v) = 3v' - z (r, + T), . ",s:F

3:i&.,3

;r:':

fr.*i.,+

For co, a:1.4734 dm6.bar.mol-2 and b:0.039523 dm3.mol-t lTable 16.3). Then, usr;:the Newton-Raphson method, we find that the van der Waals equation gives a result of T --0.04998 dm3'mol-r. Likewise, we can express 717) tor the Redlich-Kwong equation as(Equation 16.9)

r(v) :T' - Yv' - ( n' + Bf;r - -4-) v - AB/-' p \"' p iEF)'*Tvzp

and /'(V) as

f'(V)=3V'-tIV-(n'+BRr - A \P' \" ' P fiTF)For co, A = 17.208 dm6.bar.mol*z.Kt/2 and B : 0.02i394 dm3.mol-r (Table r6.4). App;..the Newton-Raphson method, we find that the Redlich-Kwong equation gives a result r-:7: 0.03866 dm3.mol-r. Finally, the ideal gas equation gives (Equation 16.1)

_ (0.083145 dms.bar.mol-t.K-1)(200 K) : 0.01663 dmj.mol_l1000 bar

The experimental value of 0.04009 dm3 .mol-r is closest to the result given by the Redlich-Kequation (the two values differ by about3To).

16-16. Compare the pressures given by (a) the ideal-gas equation, (b) the van der Waals equa:.(c) the Redlich-Kwong equation, and (d) the Peng-Robinson equarion for propane at 40C iand p - 10.62 mol'dm-3. The experimental value is 400 bar. Take cu = 9.6938 L2.mol-r :-f = 0.0563? L.mol-r for the Peng-Robinson equation.

'j'r:,..;

' .{i:

nftry -- P

Tlre Properties of Cases 487

-,.e molar volume corresponding to a density of 10.62 mol'dm-3 is 0.09416 dm3.mol-r.

:. The ideal gas equation gives a pressure of (Equation 1 6. 1 )

p : y- (0'083145 dm3 bar'mol-r'Kr)(400 K) : 3s3.2barV 0.09416 dm' .mol-'

b, The van der Waals equation gives a pressure of (Equation 16.5)

'= s!--2

v -b v"For propane, a = 9.3919 dm6,bar.mol-2 and D : 0.090494 dm3.mol-r (Table 16.3). Then

^ (0.083145dm3.bar.moI-r.K-')(400K) 9.3919dm6.bar.mol-2r=- 0.09416 dm3.mol-r - 0.090494 dm3.mol-r (0.09416 dmr.mol-r)2

: 8008 bar

c. The Redlich-Kwong equation gives a pressure of (Equation 16.7)

D_ RT A

' - v-BFor propane, A : 183.02 dm6'bar.mol-2.Kl/2 and B = 0.062723 dm3.mol-r (Table 16.4).Then

^ (0.083145dm3.bar,mol-r.-';i+OOf;'

0.09416 dm3.mol-r - A.06ziZ3 dm3.mol-r

1 83.02 dm6. bar. mo] 2 .Rt /2

(400 K)' /'z(0.094 1 6 dm3. mol-r ; 10.094 I 6 dm3. mol-r + 0.a627 23 dm3 . mol- I

): 438.4 bar

d. The Peng-Robinson equation gives a pressure of (Equation 16,8)

p'rD

- ^'l a._;-__

Forpropane, a :9.6938 dm6.bar'mol-z and fl :0.05632dm3,mo1-r. Then

- (0.083l45dm3.bar.mol-r.-'){+Oof;._- 0,09416 dm3.mol-r - 0.05632dm3.moI-r

9.6938 dm6'bar'mol-2(0.09416)(0.09416+ 0.05632) dm6'mol-2 + (0.05632)(0.09416 - 0.05632) dm6.mol-2

:284.2bar

The Redlich-Kwong equation of state gives a pressure closest to the experimentally observedpressure (the two values differ by about 107o).

16-17 . Use the van der Waals equation and the Redlich-Kwong equation to calculate the value of thepressure of one mole of ethane at 400.0 K confined to a volume of 83.26 cm3. The experimentalvalue is 400 bar.

{

Here, the moiar volume of ethane is 0.08326 dm3 .mol-l ,

CHAPTER 19The First Law of Thermodynamics

PROB LEMS AN D SOLUTIONS

19-1. Suppose that a 10-kg mass of iron at 20"C is dropped from a height of 100 meters. What is thekinetic energy of the mass just before it hits the ground? What is its speed? What would be the finaltemperature of the mass if all its kinetic energy at impact is transformed into internal energy? Takethe molar heat capacity of iron to be d" :25.1 J.mol-r.K-r and the gravitational accelerationconstant to be 9.80 m.s-t.

Just before the mass hits the ground, all of the potential energy that the mass originally had will beconverted into kinetic energy. So

PE : msh: (10 kg)(9.80 m.s-2)(t00 m) - 9.8 kJ : KE

Since kinetic energy can be expressed as ,nuz f 2, the speed of the mass just before hitting the groundis

/zKE\tlz I zo.B kD l'/2',:\;) =L+ffi1 :44ms-r

For a solid, the difference between eu and E" is small, so we can write A,lJ : ue ,LT(Equation 19.39). Then

LT: 9.8 kJ--2.2K

Hp

flll

ii

i

II

f;liil

The final temperature of the iron mass is rhen22.2"C.

19-2. Consider an ideal gas that occupies 2.50 dm3 at a pressure of 3.00 bar. If the gas is compressedisothermally at a constant external pressure, P.^,, so that the final volume is 0.500 dm3, calculatethe smallest value {*, can have. Calculate the work involved using this value of P"*,.

( ,::to^+-) ,rt.t r.mot-''K-')\55.85 B.mol-' 7 '--'

Since the gas is ideal, we can write

p" = P,!,

- (3.00 bar)(2.50 dm3) : 15.0 bar. v, 0.500 dm,

The smallest possible value of {*, is Pr. The work done in this case is (Equation 19.1)

*=) : 3ooo r

585

585 Chapter 19

19-3. A one-mole sample of CO,(S) occupies 2.00 dmr at a temperature of 300 K. If the gas is

compressed isothermally at a constant external pressure, P..,, so that the final volume is 0.750 dm3,

calculate the smallest value {^, can have, assuming that COr(g) satisfies the van der Waals equationof state under these conditions. Calculate the work involved using this value of {*,.

The smallest value {*, can have is P' where { is the final pressure of the gas. We can use the van

der Waals equation (Equation 16.5) and the constants given in Table 16.3 to find P,:

RT" ap -

z _-- vt-b V;

_ (0.083145 dm3.bar.mol-'.K-')(300 K) _ 3.6551 dm6.bar.mol-2

0.750 dm3.mol-r - 0.042816 drn3.mol-r (0.750 dm3.mol-r)2

= 28.8 bar

The work involved is (Equation 19.1)

'ur : * P LV :-(2S.8 x 105 Pa)i-1.25 x 10-3 m3) : 3.60kJ

19-4. Calculate the work involved when one mole of an ideal gas is compressed reversibly from1.00 bar to 5.00 bar at a constant temDerature of 300 K.

r"llf$i

;i

4-{{55{€

Using the ideal gas equation, we find that

nRTV,: -{ and

j

$

rn)\P,))(300 K) In0.2 : 4.01 kJ

(f)(8.31s

Equation 19.2 to find the work

/ v^\w: _rtRT tn ( # I\ "t/

: (-1 mol)(8.315 J.mol-

tt nRTy^:-zp'2

We can therefore wfite Vr/ V, : P, I Pr. Now we substitute into Equation 19.2 to find

1'):-lo*,ov--lt#r,: -nRT ln * -rtRT ln

J.mol-r.K-r

19*5. Caiculatetheworkinvolvedwhenonemoleofanidealgasisexpandedreversiblyfrom20.0dmrto 40.0 dm3 at a constant temperature of 300 K.

We can integrate involved:

' .K-') (3oo K) ln 2 : - I .73 kJ

- (-1 mol)

19-6. Calculate the mlnlmumisothermallv at 300 K from

amount of work required to compress 5.00 moles of an ideal gas

a volume of 100 dm3 to 40.0 dm3.

594 Chapter 19

and, finally,

19-18. Show that

o rrtTr,+nllc, - o rrGr+n\lc y,l rl - ,2rz

for an adiabatic expansion of an ideai gas. Show that this formula reduces to Equation 19.23 for a

monatomic gas.

For an ideal gas,

PtVt _ Tl

P,v' Tz

We can substitute this expression into the equation from Problem 19-15 to write

T. / P.\RIT,-:-l--!lTt \P'l

Taking the reciprocal gtves

and rearranging yields

PrV, - (\\^'''

PrV" \Vr)

P,v, - (L\o'',PrV, \%/

* _.(r+n72,,) - ..(r+nZ")P, %t '' = PzVz'

For a monatomic ideal gas, E, - ] R, so

P,Vllt = PrVltt (19.23,

19-1 9. Calculate the work involved when one mole of a monatomic ideal gas at298 K expandsreversibly and adiabatically finm a pressure of 10.00 bar to a pressure of 5.00 bar.

Because this process is adiabatic, 6q - 0. This means that

6w: dU = re ,dT

where eu is temperature-independent (since the gas is ideal). We can

Problem 19-17 to writeuse the equation from

/ P \RlI ,rr:rrl*l\r'l

For an ideal gas, C p:5R12, so

r,: (2e8r (#H)''' :226K

The First Law of ThermodYnamics

19-24, Liquid sodium is being considered as an engine coolant. How many grams of sodium are

needed to absorb 1.0 MJ of heat if the temperature of the sodium is not to increase by more than

lO'C. Take do : 30.8 J.K*r.mol-r for Na(l) and75.2J'K-r ,mol-r for HrO(l).

We must have a coolant which can absorb 1.0 x 106 J without changing its temperature by morethan l0 K. The smallest amount of sodium required will allow the temperature to change byexactly 10 K. We can consider this a constant-pressure process, because liquids are relativelyincompressible. Then, substituting LT :10 K into Equation 19.40, we flnd

LH :e ,l,r: 308 J.mol-r

Werequireonemoleof sodiumtoabsorb30SJof heat.Therefore,toabsorb l.0MJof heat,werequire

(r o x 10'D fg) f?ry) :746ks\308J/ \ I mol /

74.6kgof liquid sodium is needed.

19-25. A 25.0-9 sample of copper at 363 K is placed in 100.0 g of water at 293 K. The copper andwater quickly come to the same temperature by the process of heat transfer from copper to water.Calculate the final temperature of the water. The molar heat capacity of copper is24.5 J.K-t.mol-rand that of water is75.2 J'K-t'mol-t.

The heat lost by the copper is gained by the water. Since LH : nC ,LT (Equation 19.40), we can

let.r be the final temperature of the system and write the heat lost by the copper as

.K-l)(363 K - x)

and the heat gained by the water as

.K-')(x -293K)

Equating these two expressions gives

3495J - (9.628 J.K-')x: (418.0J.K-r)x - 1.224 x 10s J

1.259 x 105 K: 42'1.6x

295 K: x

The final temperature of the water is 295 K.

'19-26. A 10.0-kg sample of liquid water is used to cool an engine. Calculate the heat removed (injoules) from the engine when the temperature of the water is raised from 293 K to 373 K. Takee p :'75.2J'K-r .mol-r for Hro(l).

We can use Equation 19.40. where LT :373 K - 293K: 80 K. This eives

L.H : nC rLT = I9.D

597

) rto.tr'mol-r1-

t

g

mo

go.o

2.5

4663.5

) rtt., r.mol-l-l

00.0 g

52 g.molI

i8.0(t

o

"1

g'm

x

2

.0

l)

l0.018 ) rrt

rr mot-r

3340 kJ of heat is removed bv the water.

-l'K-')(80 K) = 3340 kJ

602 Chapter 1 9

19-34. Given the following data for sodium, plot ff 1fl - fltO) against I for sodium: melt-ing point,361 K; boiling point, 1156 K; A,u.Ho:2.60 kJ.mol-l; A,ooHo =9'1.4 kJ.mol-r;C" {r) : 28.2 1' mo|-t. x-' ; 2"(t) : 32.i J. mol-t . t<-' ; Zr(g) : z}.BJ. mol*' . K-' .

We can use an extended form of Equation 19.46:

Err> - n(o) : ['"'e ,{r)n, + ar,,,E * ['- e ,tt)ar +J o J r,,,,

Notice the very lalge jump between the liquid and gaseous phases.

AH\'ap

c r(gdr7T+l

r l,n

140

r00

20

t

I

F-l*

T/K

19-35. The A,H' values for the following equations are

2Fe(s) + lor(e) -+ Fero,(s) A,H": -206kJ'mol*'

3 Fe(s) + 2 Or(e) --+ Fe.,Oo(s) L,H" : -136 kJ.mol-l

Use these data to calculate the value of A,H for the reaction described by

4 FerO.,(s) * Fe(s) ---+ 3 Fe,Oo(s)

Set up the problem so that the summation of two reactions will

4[FerOr(s) + 2 Fe(s) + jOr(S)J

* 3[3 Fe(s) + 2 Oz(g) --+ FerOo(s)1

give the desired reaction:

L,H :4(206) kJ

L,H :3(-136) kJ

4 FerOr(s) * Fe(s)---+3 FerOu(s) a.Fl - 416 kJ

19-36. Given the following data,

;Hr(c) + j Fr(s) -+ HF(g) A.Ho : -213.3 kJ.mol-l

Hr(s) * ; or(e) -+ Hro(l) A,Ho : -285.8 kJ.mol-'

calculate the value of A,H for the reaction described by

zFr($ + 2 Hzo(l) ---+ a HF(g) + or(s)

Tlre First Law of Thermodynamics 603

Set up the ploblem so that the summation of two reactions will give the desiled reaction:

4ljH2(s) + jFr(s) --+ HF(g)l L,H :4(*273.3)kJ

+ 2[H,O(l) -.> Hr(g) + jo](c)l L,H - 2(285.8) kJ

2 F"(g) * 2 H,o(l)---+a HF(g) * O,(g) L,H = -521.6 ki

',9-37. The standard molar heats of combustion of the isorners rn-xylene and p-xylene are

-4553'9 kJ'mol-r and -4556.8 kJ.mol-r, respectively. Use these data, together with Hess'sLaw, to calculate the value of A.11'for the reaction described by

rn-xylene -> p-xylene

Because m-xylene and p-xylene are isomers, their combustion equations are stoichiometricallyequivalent. We can therefore write

rn-xylene -> combustionproducts L,H = -4553.9kJ* combustion products --+ p-xylene L,"H : +4556.8 kJ

rn-xylene----+p-xylene L.H : +2.9kJ

19-38. Given that A,H" : -2826.7 kJ for the combustion of 1 .00 mol of fructose at 298.15 K,

CuH,rOu(s) + 6 Or(e) -> 6 CO'(B) + 6 HrO(l)

and the ArH" data in Table 19.2, calculate the value of A,H' for fructose at 298.15 K.

We ate given A,F/' for the combustion of fructose in the statement of the problem. We use thevalues given in Table 19.2 for COr(S), HrO(l), and Or(g):

arH'[Co2(g)] : -393.509 kJ.mol-r arH'[H2O(I)] : -285.83 kJ.mol-'

A,H"[or(g)l :0

Now, by Hess's law, we write

L,H" :f nrA"lproductsl - I o,"'freaoanrs]

-2826.7 kJ.mol-i :6(-393.509 kJ.mol-r) + 6(-285.83 kJ'mol-r) - A.H"[fructose]

A.H'ffluctosel = 1249.3 kJ.mol -l

1 9-39. Use the A, H" data in Table 19 .2 to calculate the value of A" H" for the combustion leactionsdescribed by the equations:

a. cH3oH(l) + I or(s) ---+ cor(g) + 2 Hro(t)

b. N2H4(l) + 02(g) ---+ N2(8) + 2 H'O(l)

Compare the heat of combustion per gram of the fuels CH.OH(l) and N,Hn(1).

604 Chapter 19

We will need the following values from Table 19.2:

A,H"[CO,(B)] = -393.509 kJ.mol-' ArH'[HzO(l)] - -285.83 kJ'mol-r

Af 11.[N2H4(l)] : +50.6 kJ.mol-r Af H"[CH1OH(I)] : -239.1kJ'mol-'ArH"[Nr(g)J: o

a. Using Hess's law,

L,Ho - | n,A"lproductsl - I ort"lreactants]

: 2(*28s.83 kJ) + (-393'5 kJ) - (-239.1 kJ)

: ( -126.:kI ,) f +g\ = _zz.tkJ s_,\mol methanol / \32.OaZ g /

b. Again, by Hess's law,

L,H" : I lrH'[products] - I o,g"lreactants]

:2(-285.83 kJ) - (+50.6 kJ)

: ( -u?? =t=\_t) f J++) : _ re 4 kJ.g-,

\ molNrH, ) \32.046 g/

More energy per gram is produced by combusting methanol.

Ig-40. Using Table 19.2, calculate the heat required to vaporize 1.00 mol of CCln(l) at298 K

CCl,(l) -+ CClr(B)

We can subrracr ArH"[CCl4(1)] from ArH'[CCl4(g)] to find the heat required to vaporize CClu:

A"ooH : -102.9 kJ + 135.44 kJ : 32.5 kJ

19-41. Using the ArH" data in Table 19.2, calculate the values of A,H' for the following:

a' crHoG) + HrO0l ----+ CrHtoH(l)

b. CHo(e) 1+ Clr(S) ----+ CClo(l) + a HCl(g)

In each case, state whether the reaction is endothermic or exothermic.

a. Using Hess's law,

L,Ho : -27'7'69 kJ - (-285.83 kJ + 52'28kJ) - -44'14kJ

This reaction is exothermic.

b. Again, by Hess's law,

L,H" -- 4(-92.31kJ) - 135'44 kJ - (-74'81 kJ) : -429'87 kI

This reaction is also exothermic.

1942, Use the following data to calculate the value of A,,ooH'of water at298 K and compare

your answer to the one you obtain from Table 79.2: A,,orH" at 373 K=40'7 kJ'mol-t;

e rQ) : 75.2 J.mol-r.K-r; e rG) = 33-6J'mol-r'K-r.

608 Chapter 19

We can solve this polynomial using Simpson's rule or a numerical sofware package. Working inMathematica, we find that the final temperature will be 4040 K.

1947. Explain why the adiabatic flame temperature defined in the previous problem is also called the

maximum flame temperature.

The adiabatic flame temperature is the temperature of the system if all the energy released as

heat stays within the system. Since we are considering an isolated system, the adiabatic flametemperature is also the maximum temperature which the system can achieve.

19-48. How much energy as heat is required to raise the temperature of 2.00 moles of O2(g) from298 K to 1273 K at 1.00 bar? Take

?"tor{s)l/n :3.094+ (1.561 x l0-3 K-t)T - (4.65 x l0-7 9-2172

We can use Equation 19.M:

fr,A,H: I nC,dT,4

: (2.00mol)R ["" 1r.rro+ (1.561 x i0-3 K-t)r - g.6sx r0-7 l".-2)72]drJzss

:64.795 kJ.mol-r

1949. When one mole of an ideal gas is compressed adiabatically to one-half of its original volume.the temperature of the gas increases from 273 K to 433 K. Assuming that eu is independent oftemperature, calculate the value of Zu for this gas.

t

f-:.d$illg['yrt-

#ihdI

H;lii..'dl,:

# Equation 19.20 gives an expression for the reversible adiabatic expansion of an ideal gas:

e ,ar : -ffavIntegrating both sides and substituting the temperatures given, we find that

[9tar:-[!a,J T JVe ,nl= -Rh f

433Culnfr: -Rln2e-J - 1.59

19-50. Use the van der Waals equation to calculate the minimum work required to expand one moleof COr(g) isothermally from a volume of 0.100 dm3 to a volume of 100 dm3 at n3 K. Compareyour result with that which you calculate assuming ideal behavior.

Hu tA,l0: t,s,r6 tlq : l,4lq J130,41,4Y0 5}.^|d w{f}\ 1l^ri"

R= o,caa' ff#^rw^w=

?350I ^efua"L

+?*0 )Tncr*na* KL --

ry' 1 U Uv q4rr^A^e )WtAPE -- Pr (xo^) - PE bk /r) bl*'-""'f'"-d

@a

| = - 4q0,5 hd m, s-L : -4gO,S

@ | *.*lputrc ^K3WKlFor :{ tA* lEec tWC ' ot+ Coz,

%= -3q3 ^hf,lmot" (r^qgtre t"lc\*"r

tlx," ilra"*D

PE= rlKL=D

.tr?*Im

D,3lt+:34'P:LJ1" n"l

wo^lob@ \$40prt *1,v4 r tooo s"d- K 4 \14 -t =lW_ t),?d. lJg,L4 *lALO \00 W =,* * ^l

M,r^ 33 @ ^f6ff =u tdrf[

Q |-t Lo*.: Alt= cA*@* Fcru on .rrtul

Tae oo'/LLL__ nCv,, AT

A,nt ,nd,,*t* &u,\rwyn o"

', fuaPt6na*l

0"1\1t11= Vi \/

$o.{fu,wnatfu ,rn{a 14 AT = 6r.(

k*ry*^"f.h 1 eb,--,tej ro:

AltL= pf cur.oir=")1r= - k)c

t*q dw,,w.*n Tcon clta'r,rX<" U har.tdrogo \

rCI

P

(carnpuausn on lo brr-)

f ^ ttw attlnr-url "/Yxtavttcrwptpe^'

Firu,r\^o&ts %urnLs {g= - [/t R{V) JV,lV

L

F'u ^a,"il T, pUr= ng]

F^r Vi,V; *\lGLxYV

Vi = 0.1ttrUV'=4'1xtlu

(tat^W%51.*r)

Tt a* ctl co'tu , p,taA,\il^! , @ laerot'I,k2

dV -- - pfve-lf,) ' -pAV

Dt\r/"e- pex is \\e Uba'at- o-+ uh*,h

/b i.aAAInd- uu*,

t r,)c

tl)c= 440,q? bo^'l= 44,0q? \Alt. =e (blc Ar=o)

AlJe=Q=++rDe(.de= -(c

cb.= -14,ffi7T

Xp,org.n 5kp? 5or.,'o cta al,ar^s b^* Po= lO ba'L, Vr -- iSlfZL,d^& Vi- o'1ffi9? L,

frr = _

_^sl;

NU-e-= we+ teQ = hle r$e

Qe= -ll]ev

[r) e = -pexav = - (lo b*0 (.q tqq ? L- 0,1\997 Dtl)e=

' - 4\;011^=- 4,4095 Tt;= *4,4 oq,tf

Fon- bU, skpo

(e suh.Afr ,lr* Jrqr*r,*;)

$N'll-ro+ = AUgf NIIg= 01 *DJ=C[ldto+ -- 4A,01?T -4 $oqnr = j1,bt?.[T

Tot -- -j1,6?4f

trtru*mrl& {nM, p= S t" phrro6X},v$D u$Jv-!44}il^-^

^oJqqtLtds= -t N dVJqsuou V

h,RJ an" <-erv.o\a,,t l't^al copto ov* $ "t*sr*p,

Ltic= -npr \o' tnt?L I ey

Jqntr+r- V

wc-- - n Rr^0nV 1o

Awt+t-

l4.qwr,t-

[r)c = - Grn*)to,oze iar M^.1. rwt-t K ,)baf !^/t lggjz\r.'c-= l\1.1? L bon- = I r,4t? J

\< n867 LJ

(u, i-@ btcl-\u u ,trrtf^r.*,,\

41,t.= oI= llflsq: + ?c

$"= -lt,{8?T v

bo t\u. sorna, Icvr'v+= 4mf

arya'r'dry''tb, bru+r l4'r mr,vr''aA98r+l

A?/" : q er L^Je Ce srrl*,.rp dJ^d^a "4p*r*r*)D-

he= -nR1,Unl{etr?L\ = -f t/S"l\ DlqwL/

t" = +11,4?? f<t

4^ t"+'o ptLoue'ry"a ,l,t-ror---Os

Qrc+ = OT'(Awv =otr

runarirttpa"rra

@ &dlurtt fo*t^" P' 4--rio-.- f V-nb Tv,

\dcn-h-= -(v* {u> dyJV. I

rWt'tu!tte./rY\!6Aa ptU) 'L",& he aof t" p cy\ a-Qj SkX."r,5 Hr.cry*,

rruu/L = - ['tt nR-t dV u (totlL dV)qv V-nb Jqu TV'

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f 'otj- dt/)as VL

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c*hqanrr? 1tfrd

(ht) ^(tr),-

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+\ =tlJ

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+ -Iruft>t hL! - = (eS?J"tlrV

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Grfttz)ruegeg)tr)1.')f (Baf-098)r u*1* bLt- = (astJ"\\'V

-,,07i :fff;'7r:atW{=d)v },n\o? rrVbv t(*az).'dy =tort] "bk

-{\ ryuleg 11 19'g1t- = (sbtJ"HrV

=(grzl"Hrq

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lD \N" @- 3>oK'"

lDy'u x\rn,*,-pox -tq3els =Ft{3k\ \ = gew{ e3{tt" " lr\,r{"Nb" =b

lro

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fil'L'I4r" trath-oo '

d\s^, \nn f ;-fltl* rLa6/mo-a4q 6 T (nt",W-n,:,",*)

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A 50 K )r^rjr* ,nl twar^l,f'.e- al"a-* t h" |fi^ok_,u* O {&i/ rt t^rl ^- lo }-\ A [^)lr). l,l* cl*." a,^ {,1^._tl.,t'fu ao cLbn^J b,S "f