Student Solutions Manual for Physical Chemistry PDF

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i ! ,. 1 j 1 j ! 1 Student Solutions Manual for Physical Chemistry David W. Ball Cleveland State University THOMSON * BROOKS/COLE Australia- Canada> Mexico· Singapore> Spain· United Kingdom. United States CHAPTER 1. GASES AND THE ZEROTH LAW OF THERMODYNAMICS 1.1. The drawing is left to the st...

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i ! ,.

1 j

1 j

!

Student Solutions Manual

1

for

Physical Chemistry

David W. Ball Cleveland State University

THOMSON

*

BROOKS/COLE

Australia- Canada> Mexico· Singapore> Spain· United Kingdom. United States

CHAPTER 1. GASES AND THE ZEROTH LAW OF THERMODYNAMICS 1.1. The drawing is left to the student. The calorimeter, water bath, and associated equipment

(thermometers, ignition system, and so forth) are the system, while the surroundings are

everything outside the apparatus.

(b) 4SOC + 273.15 = 318 K

(c) 1.055 atm x 1.0~~~bar x 100;~: Pa = 1.Q~~~~)Os Pa (d) 1233 mmHgx ltorr x 1atm x 1.01325 bar 1mmHg 760 torr 1atm

= 1.644 bar

l cm"

(e) 125mLx-- = 125cm 3 'lmL (f) 4.2 K-273.15 =-269.0°C

1bar (g) 25750Pax = 0.2575 bar

. 100,000 Pa .

1.5. In terms of the zeroth law ofthennodynamics, heat will flow from the (hot) burner or flame on the stove into the (cold) water, which gets hotter. Then heat will move from the hot water into the (colder) egg.

~

0.0250L. =!J)4 xl 0- 4 ~• (33.0+273.l5)K

.he temperature.) tfthe volume is going to be 66.9 cnr' =

For this sample orgas under these conditions, F(p)

....~. (Note the conversion to. .:' 0.0669 L: 0.0669 L . .

//' lOC -4 .04 x 10 ~,

1.9. There are many possi 8.314

~:~'~43 K .

. \.0 4,-x II) ~ L {c::...

hL/

{

.>'f7;/ '.

0 versions. Using

J x 1cal = 1.987 cal

mol.K 4.184J mol.K ,-

=_

. -"

~. _i

..

--~'_

.

the fact that 1 cal = 4.184

J~'

_---­II_

\;

.. .....

1.11. Calculations using STP and SATP use different numerical values of R because the sets of conditions are defmed using different units. It's still the same R, but it's expressed in different units of pressure, atm for STP and bar for SATP.

1

lb 1.13. The partial pressure ofNz = 0.80x 14:71b =11.8. m. m. 7 . I pressure 0 f 0. z = 0 .20 x14. 2 9 -. lb. . lb =. The partia . m. m. 1.15. (a) This is an equation for a straight line with slope = 5, so at x simply 5.

d (b). The slope of this function is given by its first derivative: - (3x 2 dx 5, the slope is 6(5) - 5 = 25. At x = 10, the slope is 6(10) - 5 = 55. (c) The slope of this function is given by its first derivative:

· s1ope IS

- -7 ;

25

1.17. (a)

at x = 10, thee slone s ope iIS

(BVJ Bp

T,n

= ~(nRTJ = -

p,n

(d)

tJ

2T

(e) . (.Bp) BT r =~(nRT)=nR er v van

TB =

At x

= 5, the

»

p

.

3.592 ea~ . .. mol. = 1026 K (0.04267 Llmol{0.08205 Latm) \. molK 1.360 I!a~

mol

= 521K

(0.03183L1mol{0.08205 Latm)

\. molK

2

.

(c)

=~(nRT)=RT

(Bp)

r, =..!!.-, and using data from Table 1.6, we have: bR

. for CO2 : Tn =

:2).

.

Jv. . (~(dTJ dV· dp vJ or (!!-(dT) dp dV

2.. Using / /

. 5x + 2) = 6x - 5. At x = ....

~ (:) = -(

1.19.

p

-

nRpan. (b) (BV) = RT r,» = ~(nRTJ an p p

Bp p

BT) =~(PV)=L (BV BV nR nR

7 100

- -

= 5 and x = 10, the slope is

T,V

an v

v

1.390ea~ forN z: Tn =

mol = 433 K (0,03913 Llmo/ 0.08205 Latm)

'\ molK

.

1.23. The C term is or

e.

;2 .

In order for the term to be unitless, C should have units of (volumej',

The C' term is C'pz, and in order for this term to have the same units as p V (which would

be Latm), C' would need units of ~. (The unit bar could also be substituted foratm ifbar

atm

units are used for pressure.)

1.25. Gases that have lower Boyle temperatures will be most ideal (at least at high temperatures). Therefore, they should be ordered as He, Hz, Ne, N z, Oz, AI, CI-Lt, and COz..

rn Let us assume standard conditions of temperature and pressure, so T= 273.15 K andp = \~ atm.

Also, let us assume a molar volume of22.412 L = 2.2412 have for hydrogen:

=1 + B = 1+ 15 cm3:mo~ RT V 2.2412xl0 em Imol compressibility. For HzO, we have: pV

X

104 cnr'. Therefore, we

=1.00067 , which is a,0.067% increase in the

,--~

(O,~ J~.

.t. 0

0 ";;;

3/mol·

=

pV = 1 + -= B ·1'· . th e + -1126cm 4 3 =09497 '. , whi Ch'IS a 50°/ . 10 d ecrease In RT V 2.2412xl0 em Imol

compressibility with respect to an ideal gas.

-

~Y comparing th~ tw~ expressions from the~\

~ +(0 - -!!:-) 1 +f( b) + ~ .. and Z = 1 +'~ RT V V v'

C

C + ..,\ V2 ~ it seems straightforward to suggest that, at the first"approximation, C = bZ• . Additional terms involving V 2 may occuririlater terms of the first expression, necessitating additional corrections to this approximation for C. f

1.31. In terms ofp, V, and T, we can also write the following two expressions using the cyclic rule: = (av) er p

C~fl ap )

(

av

T

and

(avJ

ap

= T

(:1 (aT) avo

There are other constructions possible that would

p

be reciprocals of these relationships or the one giveri in Figure 1.11.

3

1.33, Since the expansion coefficient is defined as

"

-.!-(8V) , a will have units of. V M . p

1 volume 1 --- ----- = , so it will have units ofK- 1, Similarly, the isothermal volume temperature temperature

is defime d as - -1 (av) its of , so K. WI'II h ave units 0 compresslibili I ity IS V

EJp

T

1 . volume volume pressure

=

I pressure

,

or atm-I or bar'] , 1.35. For an ideal gas,

.

K

= _-.!-(av) = _.!-~(nRT) =...!.. nRT , Since nRT = V, this last

V ap

T

V EJp P

V p2

P

ion b ecomes 1V = -I lor c. an Iidea1 gas, Th e-"expresston ,T expressIOn -a.IS eva 1uated as

Vp p P TTl (av) T a (nRT) T-nR, For an Ideal , gas, the Ideal , gas law can be -a p =-p V -er =pV et -p- =pV P ' -nR =-, V b . .IS rearranged to give so we su ' stitute to get th at thiIS Iast expression p

:v , ~

T

p

which = ~

Thus, the two sides of the equation ultin.Iately yield the same expression and

so are equal.

",

- RT

1.37. For an Ideal gas, V = - . Therefore, the expression for density becomes, substituting for p the molar volume, d = temperature is

(ad) aT

(:a. =-~

p.«

M" = pM . The derivative of this expression with respect to RTf p RT = - P~ . Using the definition of V,. this can be rewritten as RT

4

CHAPTER 2. THE 'FIRST LAW OF THERMODYNAMICS

=F· s = IFIlsl cosO work = 30 N . 30 m· cos 0° = 900 Nrn = 900 J. work = 30 N· 30 m· cos 45° = 900·0.7071 Nrn =

2.1. work (a) (b) 2.3.

.

W

.

= -Pext8V = --c(2.33 atm)(450.mL - 50.mL)x

640 J. 1L

1000mL

.' 101.32 J . = 94.4 1. lL·atm·

= 0.932 L· atm x

2.5. (a) The work would be less because the external pressure is less. (b) The work would be greater because the external pressure is greater. (c) No work would be performed because the external pressure is (effectively) zero. (d) The work would be greater if the process were irreversible. o

(kli7~'" £,rGtf\.!',/l'N-~I-" It-I lifz T-(,/·t~ lk't;l#z.u-

I'

.

2.7. These three compounds experience hydrogen bonding between their molecules. Because it requires more energy to overcome the effects of this hydrogen bonding, these compounds have hi=:ecific heat capacities than other, similar-mass molecules. ,

I '?!l1'l,./.q~

5-1 ~~~ ";~~PlI"1

""'\11-{-1 °t~-""l1~fW:;'~?1

LVl~'/1~. (572 mol)(21.0 J/mol·K)(140.0°C - 20.0°C) = 1.44x106 J

I1U= q + w = 1.44x106 J + 0 = 1.44x106 J.

~ +:

" J

VI'

.23. w=-nRTln-=~(0.505mol)(8.314

17;

.

" 0 10L ' )(5.0+273.15K)1h '=+2690J mol-K 1.0L

q = -1270 J (given) I1U= q + w = +2689 J -1270 J = +1420 J Mi = I1U + 11(pV)Since the process occurs at constant temperature, Boyle's lawapplies and 11(pV)= O. Therefore, Mi= +l420 J. "

." 0!j' ,

In terms ofpressure and volume: dU

(au)

=

Up ,dp +

(au) av

dV . p

(aH) ,dp + (aH) ­

For enthalpy: dH = -

ap

II

av

dV. p

~ In order for each term to have units of J/mol·K for each term, the first term has units

\1bR!I-K; the second term has units J/mol·K2 ; and the third term has units J·K/mol.

6

Q(OHJ ==(O(U + PV») ==~ +(OPV) ==O+(OPV) ==(OPV) ==(ORT) ~ op T Op T(~ )T OPT OPT OPT OpT . v~~·

which is zero at constant

~ 2 .3 1

temperature~Therefore, (~H)

. op

is zero . T

This derivation is given explicitly in the text in section 2.7 .

. ForRe:

T == -2a == Rb

ForH2: 2a T == - == Rb

2(0.244 L2 • arm/mol")

.

. '. == 224 K (compared to 202 K from text) (0.08205 L· atm/mol- K)(0.0266 Llmol) ,

. ~~,)~bl.. . : Because the pressur~ change isn.'t !90 dras~ic, our ~swer to exercise 2.34 is probably hin a few degrees ofbemg correcr--tf a truly isenthalpic process can be arranged.

~ 2.~

e.

, . 2(0.03508 arm/mol") 36 I K (compared to 40 K fr'om text)" ==. (0~08205 L· atm/mol- K)(0.0237 Llmol)

~ "

"P'(i6\

,

.

-:

..

..

~.41?1 b~,

1M;, :t\~" .

2.37. Because strictly speakmg, heat capacities are extensiveproperties; they depend on the am t of matter in the system.. Thus, the form in equation 2.37 is the most general expression / t re "" the two ~s. ..

~

L~al

2 . First, calculate pressure and assume that this is the exte (a good approximation; since t

co stant throughout ~-~

== nRT == (0.145mol)(0.08205L·atmlmol·K)(273.15K) == 0.650 P V 5.00L

~~:~ression

Now we can calculate work as -Pext~V: .

.

w == -(0.650 atm)(3.92L -5.00 L)x

101321·

== +71.1 J

l Lvatm.

To determine ~U, we need to calculate q first. We need the final temperature, which can be determined by Charles' Iaw (since pressure is constant):

V; == Vf 1'; Tf

5.00L _ 3.92L ,273.15K

Tr '>

T ==214K f

_____

Thismeansthat~T==(214-273.l5)==-59'K.

Using q = n·c·~T: q =(O.l41\rr.~r;z.u.J.2.#m~gJ(-59 K) == -178 J. ~u =

q + w == -178 J + 71.1 J = -107 .)

2.41. Starting with - R In V (I v; == C v lnT (I T,: both logarithm terms can be evaluated similarly. Since In(a) -In(b) = In (alb), this The logarithm on the left is evaluated as -R(ln(Vr) -In

(Vi»'

7

simplifies to -R·ln(VtlVi), which is the left side of equation 2.44. The right side gets evaluated and simplified similarly. R Performing a similar substitution as in

r

C =

C

p -

y

c,

7 5 -R--R

= 2

'2 ~R

2

2 -R

2 = -",-, 2/2 =2 f or ann 1id ea1diiatomic . gas. =-5-R 5/2 5

2

2.4 . If the melting process occurs at standard pressure, then Mf = qp = 333.5 J'{from Table To correct for the volume change, we need to calculate the volumes of both water and ice at 0 C: 1 , 1mL ' 1.00016 mL for water: 1gram , 0.99984g 1mL for ice: 1gram x = 1.0907mL Therefore, fl.V = 1.00016 --: 1.0907 = -0.0905 mL 0.9168g . ~ Using the equation fl.U = Mf + fI.(p,)1) = Mf + pfl.V (for constant pressure): JMO:::-Dl)~6ptf) 1L' )(10132J) AU = 333.5 J + (1 atm)(-0.090,mL), m1. . = 333.5 - 0.0090 J ~ 333.5 J . , 1000 1Lvatm

x=

This shows that fl.U and Mf can be very close, ifnot v phase processes. ~

y

e same, for many condensed.

m~ water bums because steam gives up a considerable amount

2.47. Steam bums hurt ofheat as the heat of vaporization.

' e h eat of fusioh ~1\\.. . water can b e transferre ~~d (at least m in part part) to th e given up by th e fr eezmg 2.49. Th citrus fruit, keeping them warmer and (hopefully) keeping the fruit itself from freezing. Arx"H = LfI. [H(prods)-LfI. [H(rcts)

=(2mol)(26.5kJ/mol)- 0- 0 = 53.0kJ

aRe03 (s) -7 Na (s) + Yz H2 (g) + C (s) + 3/202 (g)] 2 Na (s) + C (8) + 3/202 (g) -7 Na2C03 (s) C (s) + 02 (g) -7 C02 (g) H 2 (g) + Yz O2 (g) -7 H 20 (1)

-2xfl.rH= +1901.62 kJ fl.rH= -1130.77 kJ fl.rH= -393.51 kJ ArB = -285.83 kJ

.I

r--:

This yields the overall reaction (you can verify that), and the overall fl.rxJl is the sum of the values on the rightfl.rxnH = 91.51 kJ. 2.55. Since process is constant-volume, qv = AU = -31,723 J. .

----

>

8

w ~ 0 since the process is in a constant-volume calorimeter. To determine MI, we need to know

the balanced chemical equation for the combustion ofbe~d:

.CJI,COOH(s) +

9

~+ 3H,OOj

For every mole of benzoic acid combust~ere isa Ch~~.sofq - 15/2) ~es of g~ 1.20 grams of benzoic acid are ~gx

(

l

~--'-

~

(~~ mo~ofbenz01c aCI . Therefore, the

!-~~.Q~ .~,

..............­

:a~~:~~;n rmmbe:F-clinol~s~)X(~0.5) .= .-0,0~492Il1;()10f gas.__ :qsing the

= AU +.6(pV) =.J1U + t:.(nRT) = t:.U + (.6ri)RT, we can determine the Mi of the process:

Ml= -31,723 J + (-0.00492 mol)(8.314 J/mol·K)(24..6 +273.151~:

Superhea\el steam has the advantage of a higher temperature, so (hopefully) there will he a higher effici~~cyfor a heat engine using superheated steam.

i

ne definition of a p~~~~ m~tion machine is a machine that creates more energy than it ses. Ifit were to do so in an iso1atea'ystem, it would violate the first law of thermodynamics. .erpetual motions of this sort are not known (and probably never will be).

~ll definiti~~s o~ e~~~~y are appl~~JP real-g~~ as well ~ ideal gases. ~~~::r~l~~~~~n IS m{{ependent oft]}elype of~~tl involved m a p~~ss. ~ A better statement of the secondlaw of thermodynamics includes the conditions under _ _~.~:ch the second .law is ~W7t~ly apyVf,able: for an isolated system, a ~...