Solucionario Atkins, Paula - 9ed

 

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Part 1: Equilibrium

 

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1

Th The e pr prop oper erti ties es of ga gase ses s

Solutions to exercises Discussion questions partial al press pressure ure of a gas in a mixture mixture of gases is th thee pressure pressure the gas w would ould ex exert ert if it occup occupied ied E1.1(b)   The parti alone the same container as the mixture at the same temperature. It is a limiting law because it holds exactly only under conditions where the gases have no effect upon each other. This can only be true in the limit of zero pressure where the molecules of the gas are very far apart. Hence, Dalton’s law holds exactly only for a mixture of perfect gases; for real gases, the law is only an approximation.

E1.2(b)   The critic critical al const constants ants repr represen esentt the state of a system at which the dis distincti tinction on between between the liquid and vapour phases disappears. We usually describe this situation by saying that above the critical temperature the liquid phase cannot be produced by the application of pressure alone. The liquid and vapour phases can no longer coexist, though fluids in the so-called supercritical region have both liquid and vapour characterist characteristics. ics. (See Box 6.1 for a more thoroug thorough h disc discussio ussion n of the super supercriti critical cal state.) E1.3(b)   The van de derr Waals Waals equatio equation n is a cubic equa equation tion in the vo volume, lume, V  . Any cubic equation has certain properties, one of which is that there are some values of the coefficients of the variable where the number of real roots passes from three to one. In fact, any equation of state of odd degree higher than 1 can in principle account for critical behavior because for equations of odd degree in  V   there are necessarily some values of temperature and pressure for which the number of real roots of   V  passes from  n (odd) to 1. That is, the multiple values of   V  V  converge from  n  to 1 as  T    →   T c . This mathematical result is consistent with passing from a two phase region (more than one volume for a given  T   and  p ) to a one phase region (only one  V  for a given  T   and  p  and this corresponds to the observed experimental result as the critical point is reached.

Numerical exercises E1.4(b)   Boyle’ Boyle’ss law applies. pV   = co cons nsta tant nt pf   =

pi V i V ff  

=

so   pf V f   = pi V i

(104 kPa kPa ) × (200 2000 0 cm3 ) (250 250 cm3 )

= 

832kPa

E1.5(b)   (a)   The perfect gas law is pV   = nRT 

implying that the pressure would be p=

nRT  V 

All quantities on the right are given to us except n, which can be computed from the given mass of Ar. n=

so p =

25 g 39.95gmol−1

 =

0.626 mol

(0.626 mol) × (8.31 × 10−2 LbarK−1 mol−1 ) × (30 + 273K)

1.5 L

not 2.0 bar. bar.

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= 

10.5 bar bar

 

4

INSTRUCTOR S MANUAL ’

(b)   The van der Waals Waals equation is RT 

p=

V m − b

−

a 2 V m

(8.31 × 10−2 LbarK−1 mol−1 ) × (30 + 273) K

so p =

(1.5 L/0.626 mol) − 3.20 × 10−2 Lmol−1 −

013 bar bar at atm m−1 ) (1.337L2 atm mol−2 ) × (1.013  

¯ (1.5 L/0.626mol )2

bar =   10.4 bar

Boyle’s law applies. E1.6(b)   (a)   Boyle’s pV   = co cons nsta tant nt

and pi   =

pf V ff   V i

=

so   pf V f   = pi V i (1.48 × 103 Torr) × (2.14dm3 ) 3

(2.14 + 1.80) dm (b)   The original pressure in bar is 2

pi   = (8.04 × 10 Torr) ×



  1atm 760 Torr

  ×

 

2 =   8.04 × 10 Torr

1.013 bar 1atm



1.07 bar =   1.07

Charles’ss law applies. E1.7(b)   Charles’ V i

V   ∝ T    so

and T f   =

V f T i

T i

=

V f  T ff  

(150 150 cm3 ) × (35 + 273) K

92.4 K   =   92.4 500 500 cm3 between etween press pressure ure and te temperature mperature at constant v volume olume can be deriv derived ed from th thee perfect E1.8(b)   The relation b gas law =

V i

pV   = nRT   

so   p ∝ T    and

pi

=

T i

pf  T f 

The final pressure, then, ought to be pf   =

pi T f  T i

=

(125 kPa) × (11 + 273) K (23 + 273) K

kPa =   120 kPa

E1.9(b)   Accor According ding to the perfect perfect gas law law,, one can compute the am amount ount of gas from pressur pressure, e, temperat temperature, ure, and volume. Once this is done, the mass of the gas can be computed from the amount and the molar mass using pV   = nRT 

so n =

pV  RT 

=

atm) × (1.013 × 105 Paatm−1 ) × (4.00 × 103 m3 ) (1.00 atm −1

(8.3145JK

−1

mol

) × (20 + 273) K

=

1.66 × 105 mol

and m = (1.66 × 105 mol) × (16.04gmol−1 ) = 2.67 × 106 g =   2.67 × 103 kg

E1.10(b)   All gases are perfect perfect in the limit of zero pressure pressure.. Therefore the ext extrapol rapolated ated val value ue of  pV   pV m /T   will give the best value of  R  R .

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THE PROPERTIES OF GASES

 

The molar mass is obtained from  pV   = nRT   =

5

m

RT  M  RT  m RT  which upon rearrangement gives  M   = =ρ p V  p

The best value of  M   M  is obtained from an extrapolation of  ρ /p  versus  p  to  p   =  0; the intercept is M/RT . Draw up the following table (pV m /T )/( )/(L atmK−1 mol−1 ) 0.082 0014 0.082 0227 0.082 0414

p/ atm   0.750 000 0.500 000 0.250 000

From Fig. 1.1(a),

From Fig. 1.1(b),

   pV m T 

−1

=   0.0820615LatmK

mol−1

p =0

ρ

p

(ρ (ρ/p /p))/(g L−1 atm−1 ) 1.428 59 1.428 22 1.427 90

=

1.4275 42755 5 g L−1 atm−1

p =0

8.20615

8.206

8.204

8.202     m

8.200 0

0.25

0.50

 

0.75 0.

1.0

Figure 1.1(a)

1.4288 1.4286 1.4284 1.4282 1.4280 1.4278 1.4276

1.42755

1.4274 0

0.25

0.50

0.75

1.0

Figure 1.1(b)

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6

 ρ

M   = RT 

p

INSTRUCTOR S MANUAL ’

−1

=   (0.0820615Latmmol

K−1 ) × (273.15 K) × (1.4275 42755 5 g L−1 atm−1 )

p =0

=

  31.9987 9987 g mo moll−1

The value obtained for R  deviates from the accepted value by 0.005 per cent. The error results from the fact that only three data points are available and that a linear extrapolation was employed. The molar mass, however, agrees exactly with the accepted value, probably because of compensating plotting errors.

E1.11(b)   The mass density ρ  is related to the molar volume  V m  by V m   =

M  ρ

where  M  is  is the molar mass. Putting this relation into the perfect gas law yields pV m   = RT    so

pM  ρ

= RT 

Rearranging this result gives an expression for  M ; once we know the molar mass, we can divide by the molar mass of phosphorus atoms to determine the number of atoms per gas molecule M   =

RT ρ

=

(62.364LTorrK−1 mol−1 ) × [(100 + 273) K] × (0.6388gL−1 )

31.0gmol−1

 =

124 124 g mo moll−1 .

120 Torr

p The number of atoms per molecule is

124 124 g mo moll−1

=

4.00

suggesting sugge sting a formu formula la of P4 perfectt gas equation to compute the amount; then con convert vert to mas mass. s. E1.12(b)   Use the perfec pV   = nRT    so   n =

pV  RT 

We need the partial pressure of water, which is 53 per cent of the equilibrium vapour pressure at the given temperature and standard pressure. p = (0.53) × (2.69 × 103 Pa) = 1.43¯ × 103 Pa (1.43 × 103 Pa) × (250m3 )

2

so n = (8.3145JK−1 mol−1 ) × (23 + 273) K = 1.45 × 10 mol or  m = (1.45 × 102 mol) × (18.0gmol−1 ) = 2.61 × 103 g =   2.61kg completely  fills lls the container. Thus E1.13(b)   (a)   The volume occupied by each gas is the same, since each completely fi solving for  V   from eqn 14 we have (assuming a perfect gas gas)) V   =

nJ RT  pJ

nNe   = =

V   =

0.225 225 g 20.18gmol−1 1.115 × 10−2 mol,

pNe   = 66.5 Torr,

(1.115 × 10−2 mol) × (62.36LTorrK−1 mol−1 ) × (300 300 K)

66.5 Torr

 

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T   = 300 300 K =

3.137L =   3. 3.14 14 L

THE PROPERTIES OF GASES

 

7

(b)   The total pressure is determined determined from the total amount of g gas, as, n = nCH4   + nAr + nNe . nCH4   =

0.320 320 g −1

 =

16.04gmol

1.995 × 10−2 mol   nAr   =

0.175g 39.95gmol−1

 =

4.38 × 10−3 mol

n = (1.995 + 0.438 + 1.115) × 10−2 mol = 3.548 × 10−2 mol p=

nRT  V 

[1] =

(3.548 × 10−2 mol) × (62.36LTorrK−1 mol−1 ) × (300K)

3.137L

=   212 Torr

E1.14(b)   This is similar to Exercise 1 1.14(a) .14(a) with the eexception xception that the density is fi is first rst calculated. M   = ρ ρ = M   =

RT  p

[Exercise 1.11(a)]

33.5 mg 250 250 mL

=

0.1340gL−1 ,

p = 152 Torr,

T   = 298 298 K

(0.1340gL−1 ) × (62.36LTorrK−1 mol−1 ) × (298K)

152 Torr

−1

=   16.4gmol

exercis ercisee is similar to Exerc Exercise ise 1.15( 1.15(a) a) in that it uses the de defi finition of absolute zero as that E1.15(b)   This ex temperature at which the volume of a sample of gas would become zero if the substance remained a gas at low temperatures. The solution uses the experimental fact that the volume is a linear function of the Celsius temperature. Thus V   = V 0 + αV 0 θ   = V 0 + bθ, b = αV 0 At absolute zero,  V   = 0, or 0 = 20.00 L + 0.0741 0741 L◦ C−1 × θ (abs. zero) θ (abs. zero) = −

20.00 L 0.0741 0741 L◦ C−1

◦

  = −270

C

which is close to the accepted value of −273◦ C.

E1.16(b)   (a)   p =

nRT 

V  moll n = 1.0 mo  =

T  (i) 273.15K;   (ii) 50  500 0K 414 L;   (ii) 15  150 0 cm3 V   = (i) 22.414

(i)   p =

(1.0mol) × (8.206 × 10−2 LatmK−1 mol−1 ) × (273.15 K)

22.414L

1.0 atm =   1.0 (ii)   p =

(1.0mol) × (8.206 × 10−2 LatmK−1 mol−1 ) × (500 500 K)

270 atm   (2 =   270

0.150L significant signifi cant fi  figures gures)

(b)   From Table Table (1.6) for H2 S a   = 4.484L2 atm mol−1 an 2 nRT  p= − V   − nb V 2

b = 4.34 × 10−2 Lmol−1

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8

INSTRUCTOR S MANUAL ’

(1.0mol) × (8.206 × 10−2 LatmK−1 mol−1 ) × (273.15 K)

(i)   p =

−

22.414 414 L − (1.0 mo moll) × (4.34 × 10−2 Lmol−1 ) (4.484L2 atm mol−1 ) × (1.0mol)2 (22.414 414 L)2

0.99 atm =   0.99 (1.0mol) × (8.206 × 10−2 LatmK−1 mol−1 ) × (500K)

(ii)   p =

−

0.150 150 L − (1.0 mo moll) × (4.34 × 10−2 Lmol−1 ) −1 2 (4.484L atm mol ) × (1.0 mo moll)2

(0.150 150 L)2 significant cant fi  figures gures). = 185.6atm ≈   190atm   (2 signifi

E1.17(b)   The critical constants of a van der Waals gas are 0436 L mo moll−1 ) =   0.131Lmol−1 V c   = 3b = 3(0.0436 a

pc   =

27b2

and T c   =

 =

8a

1.32atmL2 mol−2 27(0.0436 0436 L mo moll−1 )2

=

27Rb

25.7 atm   =   25.7

8(1.32atmL2 mol−2 ) 27(0.08206LatmK−1 mol−1 ) × (0.0436 0436 L mo moll−1 )

=   109K

E1.18(b)   The compression factor is Z =

pV m RT 

=

V m V m,perfect

(a)   Because  V m   = V m,perfect + 0.12 V m,perfect   = (1.12)V m,perfect , we have  Z   =   1.12

Repulsivee forces dominate. Repulsiv (b)   The molar volume is V   = (1.12)V m,perfect   = (1.12) × V   = (1.12) ×

E1.19(b)   (a)   V mo   =

RT  p

=

=



  RT  p

−1

(0.08206LatmK

−1

mol

) × (350K)

12 atm atm



−1

=   2.7Lmol

(8.314JK−1 mol−1 ) × (298.15 K) (200 bar) × (105 Pabar−1 )

1.24 × 10−4 m3 mol−1 =   0.124Lmol−1

(b)   The van der Waals Waals equation is a cubic equation in  V m . The most direct way of obtaining the molar volume would be to solve the cubic analytically analytically.. However However,, this approach is cumbersome, so we proceed as in Example 1.6. The van der Waals Waals equation is rearranged to the cubic form V m3 −



b+

RT  p

  V m2 +

a

p

V m −

ab p

=

3



0 or   x − b +

with x   = V m /(Lmol−1 ).

 

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RT  p

  x2 +

a

p

x−

ab p

=

0

THE PROPERTIES OF GASES

 

9

The coef ficients in the equation are evaluated as

b+

RT 

−2

= (3.183 × 10

p

Lmol−1 ) +

(8.206 × 10−2 LatmK−1 mol−1 ) × (298.15 K)

013 at atm m bar bar−1 ) (200 bar) × (1.013

−2

= (3.183 × 10

a p ab p

−1

+ 0.1208) Lmol

1.360L2 atm mol−2

=

(200 bar) × (1.013 013 atm atm ba barr =

−1

=

)

−1

=

0.1526 1526 L mo moll

6.71 × 10−3 (Lmol−1 )2

(1.360L2 atmmol−2 ) × (3.183 × 10−2 Lmol−1 ) −1

(200 bar) × (1.013 013 atm atm bar bar

=

)

2.137 × 10−4 (Lmol−1 )3

Thus, the equation to be solved is  x 3 − 0.1526x 2 + (6.71 × 10−3 )x − (2.137 × 10−4 ) = 0. Calculators and computer software for the solution of polynomials are readily available. available. In this case we we fi  find nd or   V m   =   0.112 112 L mo moll−1

x   = 0.112

The difference is about 15 per cent. M 

E1.20(b)   (a)   V m   = Z =

=

ρ RT 

Z  = =

−1

0.5678gL

pV m

(b)   Using p =

18.015 015 g mo moll−1

=

(1.00 ba barr) × (31.728 728 L mol mol−1 )

V m − b

V m − b

=   0.9963

(0.083145LbarK−1 mol−1 ) × (383 383 K)

RT 

V m

−1

728 L mol mol   =   31.728

−

−

a V m2

and substituting into the expression for  Z  above we get

a V m RT 

31.728Lmol−1 31.728 728 L mol mol−1 − 0.03049Lmol−1 5.464 464 L2 atm mol−2

−

−1

(31.728Lmol =   0.9954

−1

×

)

(0.08206LatmK

−1

mol

×

)

(383K)

Comment. Both values of  Z  Z  are very close to the perfect gas value of 1 .000, indicating that water vapour is essentially perfect at 1.00 bar pressure. pressure.

E1.21(b)   The molar volume is ob obtained tained by solving Z   =

V m   =

ZRT  p

=

pV m RT 

[1.20b], for  V m , which yields

(0.86) × (0.0820 08206 6 L atm atm K−1 mol−1 ) × (300 300 K)

20 atm atm

=

1.059Lmol−1

(a)   Then, V   = nV m   = (8.2 × 10−3 mol) × (1.059 059 L mol mol−1 ) = 8.7 × 10−3 L =   8. 8.7 7 mL

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10

INSTRUCTOR S MANUAL ’

(b)   An approximate value of  B  B  can be obtained from eqn 1.22 by truncation of the series expansion after the second term,  B/V   B /V m , in the series. Then, B   =   V m



pV m RT 



−1

= V m × (Z − 1)

059 L mo moll−1 ) × (0.86 − 1) = −0.15Lmol−1 =   (1.059

E1.22(b)   (a)   Mole fractions fractions are nN

xN   =

ntotal

=

2.5mol (2.5 + 1.5) mol

=   0.63

Similarly,  xH   =   0.37 (c)   According to the perfect gas gas law ptotal V   = ntotal RT 

so ptotal   = =

ntotal RT  V  moll) × (0.0820 08206 6 L atm atm mo moll−1 K−1 ) × (273.15 K) (4.0 mo

4.0 0 atm =   4.

22.4 L (b)   The partial pressures are

atm) =   2. 2.5 5 atm pN   = xN ptot   = (0.63) × (4.0 atm and pH   = (0.37) × (4.0 atm atm) =   1.5 1.5 atm

E1.23(b)   The critical volume of a van van der Waals gas is V c   = 3b

so b =   13 V c   =   31 (148 148 cm3 mol−1 ) = 49.3 cm3 mol−1 =   0.0493 0493 L mo moll−1 By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular molecular size. The centres of spher spherical ical particl particles es are excluded excluded from a sphere sphere whose whose radius radius is the diameter of those spherical particles (i.e., twice their radius); that volume times the Avogadro constant is the molar excluded volume b



b = N A

r =

1 2



4π( 2r )3 3



  so   r   =

  3b

2

4π N A

1/3

  3(49.3 cm3 mol−1 ) 4π( 6.022 × 1023

−1

mol

1/3

  

1

=

)

1.94 × 10−8 cm =   1.94 × 10−10 m

The critical pressure is pc   =

a

27b2

so a   = 27pc b2 = 27(48.20 atm atm) × (0.0493 0493 L mo moll−1 )2 =   3.16 L2 atm mol−2

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THE PROPERTIES OF GASES

 

11

But this problem is overdetermined. W Wee have another piece of information T c   =

8a 27Rb

According to the constants we have already determined, T c  should be T c   =

8(3.16 L2 atm mol−2 ) 27(0.08206LatmK−1 mol−1 ) × (0.0493 0493 L mo moll−1 )

=

231 231 K

However, the reported T c is 305.4 K, suggesting our computed a/b  is about 25 per cent lower than it should be. dZ vanishes. According to the E1.24(b)   (a)   The Boyle temperature is the temperature at which lim V m →∞ d(1/V m ) van der Waals equation pV m

Z  =

so

RT 

dZ

=

=

d(1/V m )



 a   RT  V m −b   − V m2

RT 

      d Z

V m2 b

=

− b)2

V m

=

d(1/V m )

  dZ dV m

2 = −V m

V m

  dV m

×

dV m



V m − b

 

2 = −V m

a

 −

−

a V m RT 

  −V m 1 a  + + 2 RT  V m − b V m (V m − b)2



RT  (V m In the limit of large molar volume, we have

dZ

lim

V m →∞

and T   =

d(1/V m ) a

Rb

=

=b−

a RT 

=

0

a

so

RT 

=b

(4.484L2 atm mol−2 )

0434 L mo moll−1 ) (0.0820 08206 6 L atm atm K−1 mol−1 ) × (0.0434

1259 K =   1259

(b)   By interpreting  b  as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e. twice their radius); the Avogadro constant constant times the volume is the molar excluded volume  b b = N A

r =

1 2





4π( 2r 3 ) 3



  so   r   =

  3b

2

4π N A

1/3

  3(0.043 0434 4 dm3 mol−1 ) 4π( 6.022 × 1023

−1

mol

1/3

  

1

=

)

1.286 × 10−9 dm = 1.29 × 10−10 m =   0.129 0.129 nm

have ve the sa same me reduced pres pressure, sure, temperature, and volume aare re said to correspond. correspond. The E1.25(b)   States that ha ◦ reduced pressure and temperature for N2 at 1.0 atm aand nd 25 C are pr   =

p pc

=

1.0atm 33.54 atm atm

=

0.030

and   T r   =

T  T c

=

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(25 + 273) K

126.3 K

=

2.36

 

12

INSTRUCTOR S MANUAL ’

The corresponding states are (a)  For H2 S

2.6 6 atm p = pr pc   = (0.030) × (88.3atm) =   2. T   = T r T c   = (2.36) × (373.2 K) =   881K

(Critical constants of H2 S obtained from  Handbook of Chemistry and Physics.) (b)  For CO2

atm) =   2. 2.2 2 atm p = pr pc   = (0.030) × (72.85 atm T   = T r T c   = (2.36) × (304.2 K) =   718K (c)  For Ar p = pr pc   = (0.030) × (48.00 atm atm) =   1. 1.4 4 atm T   = T r T c   = (2.36) × (150.72 K) =   356K

van an der Waals equation is E1.26(b)   The v a

RT 

2 p = V m − b − V m

which can be solved for  b b   =   V m −

RT  p +   a2

=

−4

4.00 × 10

3

−1

m mol

−

4.0

V m

=

(8.3145JK−1 mol−1 ) × (288K) × 106

Pa +

  1.3 × 10−4 m3 mol−1



  0.76 m6 Pamol−2 (4.00×10−4 m 3 mol−1 )2



The compression factor is Z =

pV m RT 

=

(4.0 × 106 Pa) × (4.00 × 10−4 m3 mol−1 ) (8.3145JK−1 mol−1 ) × (288K)

=   0.67

Solutions to problems

Solutions to numerical problems P1.2

 

Identifying  pex in the equation  p = pex + ρgh  [1.4] as the pressure at the top of the straw and  p  as the atmospheric pressure on the liquid, the pressure difference is p − pex   = ρgh   =   (1.0 × 103 kg m−3 ) × (9.8 1 m s−2 ) × (0.15 m) =

P1.4

 

  1.5 × 103 Pa   (= 1.5 × 10−2 atm)

pV   = nRT   [1.12] implies that, with  n  constant,

pf V f  T f 

=

pi V i T i

Solving for  pf , the pressure at its maximum altitude, yields  pf   =

www.elsolucionario.net  

V i V ff  

×

T f  T i

× pi

THE PROPERTIES OF GASES

 

13

Substituting V i   =   34 π ri3  and V f   =   43 π rf 3 pf   =



(4/3)π ri3 (4/3)π rf 3



×

T f  T i

× pi   =

3

     ri

×

rf 

T f  T i

× pi

3

=

P1.6

 

1.0 m 3.0 m

×

253K 293K

× (1.0atm) =   3.2 × 10−2 atm

The value of absolute zero can be expressed in terms of   α α  by using the requirement that the volume of a perfect gas becomes zero at the absolute zero of temperature. Hence 0 = V 0 [1 + αθ (abs. zero)] Then θ (abs. zero) = −

1 α

All gases become perfect in the limit of zero pressure, so the best value of   α α and, hence, θ (abs. zero) is obtained by extrapolating α to zero pressure. This is done in Fig. 1.2. Using the extrapolated value, α   = 3.6637 × 10−3◦ C−1 , or 1

θ (abs. zero) = −

◦

− ◦

3.6637 × 10

3

−

C

  = −272.95

C

1

which is close to the accepted value of −273.15◦ C.

3.672

3.670

3.668

3.666

3.664

3.662

P1.7

 

0

200

400  p / Torr

 

800

600

Figure 1.2

The mass of displaced gas is ρ V , where  V   is the volume of the bulb aand nd  ρ  is the density of the gas. The balance condition for the two gases is  m(bulb) = ρV (bulb), m(bulb) = ρ  V (bulb) which implies that  ρ   = ρ  . Because [Problem 1.5]  ρ   =

pM  RT 

the balance condition is  pM   = p M  p which implies that  M  =    × M  p This relation is valid in the limit of zero pressure (for a gas behaving perfectly).

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14

INSTRUCTOR S MANUAL ’

In experiment 1,  p = 423.22 Torr, p  = 327.10 Torr; hence M  =

423.22 Torr 327.10 Torr

014 g mo moll × 70.014

−1

=

90.59gmol−1

In experiment 2,  p = 427.22 Torr, p  = 293.22 Torr; hence M  =

427.22 Torr 293.22 Torr

014 g mo moll × 70.014

−1

=

102.0gmol−1

In a proper proper serie seriess of expe experimen riments ts one shou should ld reduce the pres pressure sure (e.g. by adjusting the balanced weight). Experiment 2 is closer to zero pressure than experiment 1; it may be safe to conclude that 102 g mo moll−1 . The molec molecules ules CH2 FCF3   or CHF CHF2 CHF2   have  M   ≈ 102 102 g mo moll−1 . M   ≈ 102

P1.9

 

We assume that no H2  remains after the reaction has gone to completion. The balanced equation is N2 + 3H2   → 2NH3 We can draw up the following table N2 Initial amount

 

 

H2

n

p=



 

0

0 0 0

moll) × = (1.66 mo

V 

NH3

n

Final amount   n −   31 n Specifi Speci fically 0.33 m mo ol Mole fraction 0.20

nRT 

 



n + n

 

n +   31 n 1.66 m mo ol 1.00

2 3 n

 

Total

1.33 m mo ol 0.80

(8.206 × 10−2 L a tmK−1 mol−1 ) × (273.15 K)

22.4 L



atm m =   1.66 at

p(H2 ) = x( H2 )p =   0 p(N2 ) = x( N2 )p = (0.20 × (1.66 atm atm)) =   0.33 at atm m p(NH3 ) = x( NH3 )p = (0.80) × (1.66 atm atm) =   1.33 at atm m

P1.10

 

(a)   V m   =

RT 

(b)   From p =

p

(8.206 × 10−2 L a tmK−1 mol−1 ) × (350 350 K)

=

2.30 atm atm

RT  V m − b

−

a V m2

[1.25b], we obtain V m   =

p

Then, with a  and  b  from Table 1.6 V m   ≈

(8.206 × 10−2 L a tmK−1 mol−1 ) × (350K)

atm) + (2.30 atm ≈

RT 



28.72Lmol−1 2.34



6.26 260 0 L2 atmmol −2 (12.5Lmol −1 )2 −2

+ (5.42 × 10



 

 a + 2 V m

−1

=   12.5 L mo l



  + b [rearrange

−2

+ (5.42 × 10

1.25b]

L mol mol−1 )

L mol mol−1 ) ≈   12.3 L mo l−1 .

Substitution of 12.3 L mo l−1 into the denominator of the fi the  first rst expression again results in  V m   = −1 12.3 L mo l , so the cycle of approximation may be terminated.

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THE PROPERTIES OF GASES

P1.13

 

 

15

(a)   Since  B  (T B ) = 0 at the Boyle temperature (section 1.3b):   B  (T B ) = a + b e−c/ T B2 = 0

Solving for  T B  :   T B   =

 

   

  −c

ln

 = −a

(b)   Perfect Gas Equation:   V m (p ,T) =

bar , 298.15 K) = V m (50 bar V m (50 bar bar , 373.15 K) =

ln

b

1131 K2 ) −(1131 −1

1993 3 bar −(−0.199

 =

)

501.0 K

(0.200 2002 2 bar−1 )



R T  p

0.08314 083145 5 L bar K−1 mol−1 (298.15 K) 50 bar bar 0.08314 083145 5 L bar K−1 mol−1 (373.15 K) 50 bar bar R T 

Virial irial Equ Equati ation on (eq (eqn n 1.2 1.21 1 to first order) order):: V m (p ,T) =

p

=

0.496 L mo moll−1

=

0.621 L mo moll−1

 (1 + B  (T)p ) = V pperfect erfect (1 + B (T)p )

 c



B (T ) = a + b e

− 2 T  B

−1



B (298.15 K) = −0.199 1993 3 bar

−1

2002 2 bar + 0.200

e

2002 2 bar−1 e 1993 3 bar−1 + 0.200 B  (373.15 K) = −0.199 bar , 298.15 K) = 0.496Lmol−1 V m (50 bar V m (50 bar bar , 373.15 K) = 0.621Lmol−1

−

  1131K2 (298.15 K)2

−

  1131K2 1131K2 (373.15 K)2

−1

= −0.00163 bar

−1

000720 0 bar = −0.00072

 

 

1 − 0.001 00163 63 bar−1 50 bar bar = 0.456 456 L mo moll−1

1 − 0.00072 000720 0 bar−1 50 ba barr = 0.599 599 L mol mol−1

The perfect gas law predicts predicts a molar volume that is 9% too large large at 298 K and 4% too large at 373 K. The negative value of the second virial coef ficient at both temperatures indicates the dominance of  very weak intermolecular attractive forces over repulsive forces.

P1.15

 

From Table 1.6  T c   = 1/2

    2a

2

  2a

×

3

 1

12

           2

×

3

12pc b

8

=

R

vmol   =

vmol   =

3b3

8



b

N A

4π 3

 3 4π

=

1 3

 V c

×

N A

(3)

× (6.022 × 1023

r3 1/3

−29

× (8.86 × 10

3



m )

R

8.206 × 10−2 L a tmK−1 mol−1

160 × 10−6 m3 mol−1

=

12bpc

(40 at atm m) × (160 × 10−3 L mol mol−1 )

×

3

 

. Thus

R

=   0.28nm

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×

2aR

pc V c

×

3

=

r =

, pc   =

3bR

1/2

   

may be solved for from the expression for pc  and yields

3bR

T c   =

1/2

  

mol−1 )

=



=   210K

8.86 × 10−29 m3

 

16

P1.16

 

V c   = 2b,

T c   =

a

4bR

INSTRUCTOR S MANUAL ’

[Table 1.6]

Hence, with V c  and  T c from Table 1.5, b =   21 V c   =   21   × (118.8 cm3 mol−1 ) =   59.4 cm3 mol−1 a   =   4bRT c   = 2RT c V c −2

=   (2) × (8.206 × 10 =

L a tmK−1 mol−1 ) × (289.75 K) × (118.8 × 10−3 L mol mol−1 )

  5.649 649 L2 atm atm mol mol−2

Hence p = =

RT  V m − b

a/RT T V m e−a/R =

nRT  V   − nb

na/RT RT V  e−na/

(1.0 mo moll) × (8.206 × 10−2 L a tmK−1 mol−1 ) × (298K)

mol−1 ) (1.0 L) − (1.0 mo moll) × (59.4 × 10−3 L mol



× exp

moll) × (5.649 649 L2 at atm m mol mol−2 ) −(1.0 mo

 

298 K) × (1.0 L2 at atm m mol mol−1 ) (8.206 × 10−2 L a tmK−1 mol−1 ) × (298

=  26 .0atm × e

−0.231



atm =   21 atm

Solutions to theoretical problems P1.18

 

This expansion expansion has already been given in th thee solutions solutions to Ex Exercise ercise 1.24(a)  and Problem 1.17; the result is p=

RT  V m

 

1+ b−

a RT 



 1 V m

Compare this expansion with  p =

and hence fi hence  find nd   B   = b −

a RT 

+



b2

+···

2 V m

RT  V m



1+

B V m

+

C V m2



+···

  [1.22]

and   C   = b2

Since  C   = 120 1200 0 cm6 mol−2 , b = C 1/2 =   34.6 cm3 mol−1 273 L atm atm mo moll−1 ) × (34.6 + 21.7) cm3 mol−1 a   =   RT(b − B) = (8.206 × 10−2 ) × (273 −1

=   (22.40Latmmol

P1.22

 

) × (56.3 × 10−3 L mol mol−1 ) =   1.26 L2 at atm m mol mol−2

For a re real al g gas as we may use the v virial irial eexpansion xpansion in terms of  p  p [1.21] p=

nRT  V 

(1 + B  p + · · ·) = ρ

which rearranges to

p ρ

=

RT  M 

+

RT  M 

(1 + B  p + · · ·)

RT B  M 

p +···

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THE PROPERTIES OF GASES

 

p

Therefore, the limiting slope of a plot of 

17

B  RT 

against   p  is

ρ

M 

. From Fig. 1.2 in the  Student’s

Solutions Manual, the limiting slope is B  RT  M 

(4.41 − 5.27) × 104 m2 s−2

=

× 104 Pa

(10.132 − 1.223) RT 

From Fig. 1.2,

M 

=

−2

5.39 × 104 m2 s−2 ; hence

9.7 × 10−2 kg−1 m3



B =−

kg−1 m3

= −9.7 × 10

5.39 × 104

m2 s−2

−6

  = −1.80 × 10

Pa−1

B  = (−1.80 × 10−6 Pa−1 ) × (1.0133 × 105 Pa at atm m−1 ) = −0.182 atm−1 B   = RT B  [Problem 1.21] −2

L a tmK−1 mol−1 ) × (298K) × (−0.182 atm−1 )

= (8.206 × 10

−1

= −4.4 L mo l

P1.23

 

    ∂V m

Write  V m   = f (T,p ); then dV m   =

∂ T 

∂ V m

dT   +

∂p

p

dp

T 

Restricting the variations of   T  T   and p  to those which leave  V m  constant, that is dV m   = 0, we obtain

                ∂ V m

=−

∂T 

∂V m ∂p

p

∂p

×

∂ T 

T 

  ∂p

=−

∂V m

V m

∂p

−1

− ∂T 

∂p

×

∂ T 

T 

=

V m

From the equation of state   ∂p

=−

∂ V m

T 

RT  V m2

∂p

−3

− 2(a + bT)V m

R

=

∂ T 

V m

V m

+

b

2 V m

Substituting

   ∂ V m

=−

∂T 

−

P 

 R   b V m + V m2

 2 (a +bT ) RT  − V m2 V m3

From the equation of state

Then

P1.25

 

  ∂ V m ∂ T 

V m2 m

RT  V m +

p

(a + bT )

R +  V b



=



V m





2 p −   RT  V  m

   R+

 = +

  =p−



 =

 2 (a +bT ) RT  V m + V m2

RT 

m

2p



  RT  − V  m

=

RV m + b

2pV m − RT 

o ,  where  V m   = the molar volume of a perfect gas V mo From the given equation of state

Z =

V m   = b +

RT  p

=b

o + V m

  then   Z   =

o b + V m

V mo

=

For V m   = 10b,  10 b = b + V mo  or  V mo   = 9b then Z   =

10b 9b

=

10 9

=

1.11

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

V m

R +  V b



 b V m

1+

b o V m

  ∂p ∂V m

V m

T 

 

18

P1.27

 

INSTRUCTOR S MANUAL ’

The two two mas masses ses re represent present the ssame ame v volume olume o off gas under identical conditions, and therefore, therefore, the same number of molecules (Avogadro’ (Avogadro’s principle) and moles,  n . Thus, the masses can be expressed as nM N   = 2.2990 2990 g

for for ‘  ‘chemical chemical nitrogen’ nitrogen’ and 3102 g nAr M Ar Ar + nN M N   = n[xAr M A Arr + (1 − xAr )M N ] = 2.3102 for for ‘  ‘atmospheric atmospheric nitrogen’ nitrogen’. Dividing the latter expression by the former yields xAr M A Arr

2.3102

+ (1 − xAr ) =

M N

so   xAr

2.2990



M A Arr M N



−1

=

2.3102 2.2990

−1

2.3102 2.3102 2.2990   − 1 2.2990   − 1 = =   0.011 M Ar 39.95gmol −1 Ar 1 −  − 1 M N 28.013gmol−1

and xAr   =

Comment. This value for the mole fraction of argon in air is close to the modern value.

P1.29

 

Z  = =

              

pV m

T c

=

×

RT  T   V r   8T r T r

But V r   =

3V r − 1 RT c V 

T r

8

3

3

r

V r

  T r

T r

V   −



  1

r

V r  − 1/8

−

−1

−

1/8

V r  − 1/8

−

=

RT c

V 

V r



=

  [1.29] V r2 8 pc V  pc V 

8T r

= V r

RT c

3

−

V r

=

pc V m

×

pc

V c = pc V c ×

Therefore Z   =

Z =

 p

−

pr V r T r

[1.20b, 1.28]

8



3 RT c [1.27] = 3 V r 3

    

8V r 2 3

27

64(V r )2 27

64T r (V r )2

27

(2)

64T r V r

To derive the alternative form, solve eqn 1 for V r , substitute the result into eqn 2, and simplify into polynomial form. V r  = ZT r pr Z =

=

=

ZT r /pr ZT r  1 pr − 8

−

8ZT r 8ZT r − pr

27 64T r −

    pr

ZT r

27pr 64ZT r2

512T r3  Z2 − 27pr × (8T r Z − pr ) 64T r2  × (8ZT r − pr )Z

64T r2  (8ZT r − pr )Z2 = 512T r3  Z 2 − 216T r pr Z + 27pr2





512T r3  Z3 − 64T r2  pr + 512T r3 Z2 + 216T r pr Z − 27pr2  = 0

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THE PROPERTIES OF GASES

3

Z −

 



 p r

8T r



+1

Z2 +

27pr

27pr2

64T r

512T r3

Z− 2

=

19

0

(3)

At T r   = 1.2 and pr   = 3 eqn 3 predicts that  Z  is the root of  3

Z −



27(3) 27(3)2   3 3 +1 Z + Z−  = 0 2 8(1.2) 64(1.2) 512(1.2)3



Z3 − 1.3125Z 2 + 0.8789Z − 0.2747 = 0

The real root is   Z   = 0.611 and this predictio prediction n is independen independentt of the spec speciific gas. Figure 1.27 indicates that the experimental result for the listed gases is closer to 0.55.

Solutions to applications P1.31

 

Refer to Fig. 1.3.

h

Air (environment)

Ground

Figure 1.3

The buoyant force on the cylinder is F buoy   =   F bottom − F ttop op =   A(pbottom − ptop )

according to the barometric formula. ptop   = pbottom e−Mgh/RT 

where M  is  is the mol molar ar mas masss of the en envir vironm onment ent (air). (air). Sin Since ce h is small, small, the expone exponenti ntial al can be ex expan panded ded 1 the  first-order rst-order term only in a Taylor series around  h   =  0 e−x = 1 − x + x 2 + · · · . Keeping the fi 2! yields



ptop   = pbottom



1−

Mgh Mg h RT 



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