Bab 1 Solutions

Chapter 1

The properties of gases

Exercises 1.2(b) (a)Could 25 g of argon gas in a vessel of volume 1.5L exert a pressure of 2.0 bar at 30ºC if it behaved as a perfect gas? If not, what pressure would it exert?(b) what pressure would it exert if it behaved as a van der Waals gas? Solution: (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 

25g  0.626mol 39.95gmol  1 ( 0.626mol )  ( 8.31  102 Lbark -1mol -1 )  (30  273k)  10. 5bar 1.5L

not 2.0 bar. (b) The van der Waals equation is

p

RT a  2 Vm  b Vm

so p 

(8.31  102 LbarK 1 )  ( 30  273)K  10.4bar (1.5L / 0.626mol) 2

1

1.5(b) A sample of hydrogen gas was found to have a pressure of 125 kPa when the temperature was 23ºC, what can its pressure be expected to be when the temperature is 11ºC? Solution:The relation between pressure and temperature at constant volume can be derived from the perfect gas law

pV  nRT so

p  T and

pi p f  Ti T f

The final pressure, then, ought to be

pf 

piT f Ti



(125kPa)  (11  273)K  120kPa (23  273)K

2

1.7(b) The following data have been obtained for oxygen gas at 273.15K. Calculate the best value of the gas constant R from them and the best value of the molar mass of O2. P/atm -1

Vm/Lmol -1

ρ/(gL )

0.750 000

0.500 000

0.250 000

29.8649

44.8090

89.6384

1.07144

0.714110

0.356975

Solution: All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pVm / T will give the best value of R. The molar mass is obtained from pV  nRT＝ Which upon rearrangement gives M＝

m RT M

m RT RT RT ρ  V p p p

The best value of M is obtained from an extrapolation of 

p

versus p to p= 0；the intercept is M/RT

Draw up the following table

( pVm / T ) /(L atm K -1 mol -1 )

p / atm

( / p ) /(g L-1 atm -1 ) 0.750 000 0.500 000 0.250 000 From Fig. 1.1(a)

0.082 0014 0.082 0227 0.082 0414

1.428 59 1.428 22 1.427 90

 pVm   0.082 0615 L atm K -1 mol -1    T  p 0    1.42755g L 1 atm -1  p  p 0

From Fig. 1.1(b) 

 M  RT    ( 0.082 0615L atm K -1 mol -1 )  ( 273.15K)  (1.42755g L 1 atm -1 )  p  p 0

 31.9987g mol 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. THE PROPERTIES OF GASES

3

1.8(b) At 100ºC and 120 Torr, the mass density of phosphorus vapour is 0.6388 Kgm-3 . What is the molecular formula of phosphorus under these conditions? Solution:The mass density  is related to the molar volume Vm by

Vm＝

M 

Where M is the molar mass. Putting this relation into the perfect gas law yields

pVm  RT so

pM  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 62.364 LTorr K -1 mol -1  [( 100  273)K]  (0.6388gL-1 )   124gmol 1 p 120Torr

The number of atoms per molecule is

124gmol 1  4.00 31.0gmol 1

4

1.10(b) A gas mixture consists of 320 mg of methane, 175 mg of argon, and 225 mg of neon. The partial pressure of neon at 300K is 66.5 Torr. Calculate(a) the volume and (b) the total pressure of the mixture. Solution: (a) The volume occupied by each gas is the same, since each completely fills the container. Thus solving for V from eqn 14 we have (assuming a perfect gas)

V 

nJ RT pJ

n Ne 

0.225g 20.18gmol-1

 1.115  10 2 mol, V 

pNe  66.5Torr,

T  300K

(1.115  10 2 mol)  ( 62.36LTorr K -1 mol -1 )  ( 300K) ＝3.137 L  3.14 L 66.5Torr

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

nCH 4 

0.320 g 0.175 g  1.995  10 -2 mol nAr   4.38  10  3 mol 1 1 16.04 g mol 39.95 g mol

n  (1.995  0.438  1.115)  10  2 mol  3.548  10  2 mol nJ RT ( 3.548  10  2 mol)  ( 62.36LTorr K -1 mol-1 )  ( 300K) p [1]   212 Torr V 3.137L

5

1.13(b) Determine the ratios of (a) the mean speeds, (b)the mean kinetic energies of He atoms and Hg atoms at 25ºC. Solution:(a) The mean speed of a gas molecule is

 8 RT  c    M 

1

2

c ( He )  M ( Hg )  so   c ( Hg )  M ( He ) 

1

2

 200.59     4.003 

1

(b) The mean kinetic energy of a gas molecule is

 3 RT  c    M  So

1

2

 7.079 1 m c2 ,where c is the root mean square speed 2

2

1 m c2 is independent of mass, and the ratio of mean kinetic energies of He and Hg is 1. 2

6

1.14(b) The best laboratory vacuum pump can generate a vacuum of about 1 nTorr. At 25ºC and assuming that air consists of N2 molecules with a collision diameter of 395 pm, calculate (a) the mean speed of the molecules, (b) the mean free path, (c) the collision frequency in the gas. Solution: (a) The mean speed can be calculated from the formula derived in Example 1.6 1

 8 （8.314JK -1 mol -1 )  ( 298K)   8 RT  2  c   = c   3 -1  M     ( 28.02  10 kg mol  (b) The mean free path is calculated from  

With   d    ( 3.95  10 2

10

kT 1

2 2 p

1

2

 4.75  10  2 ms 1

[33]

m) 2  4.90  10 19 m 2

Then,



（1.381  10 23 JK -1 )  ( 298K)  4  104 m 5 1  1 atm   1.013  10 Pa   2 2  (4.90  10 19 m 2 )  (1  10 9 Torr)      1 atm  760 Torr   

(c) The collision frequency could be calculated from eqn 31, but is most easily obtained from eqn 32, since

 and c have already been calculated z 

c 4.75c   1  10  2 s 1 4  4.46  10 m

Thus there are 100s between collisions, which is a very long time compared to the usual timescale apparatus used to generate the very low pressure.

7

1.16(b) At an altitude of 15 km the temperature is 217 K and the pressure 12.1 kPa. What is the mean free path of N2 molecules? (σ=0.43nm2) Solution: The mean free path is



kT 1

2 2 p

（1.381  10 23 JK 1 )  (217K)

＝ 2

1

2

0.43  (10

9



1

m)  (12.1  10 Pa atm ) 2

3

 4.1  10 7 m

1.17(b) How many collisions per second does an N2 molecule make at an altitude of 15 km?(See Exercise 1.16b for data.) 1

 16  2 Solution: Obtain data from Exercise 1.17(a) is z    p  mkT  Substituting   0.43nm , p  12.1  10 Pa , m  ( 28.02u), andT  217K 2

3

we

z

obtain

4  (0.43  10 18 m 2 )  (12.1  10 3 Pa)

  ( 28.02)  (1.6605  10

 27





k g)  1.381  10  23 J K 1  ( 217 K )



1

2

=9.9×108s-1

8

1.21(b) Estimate the critical constants of a gas with van der Waals parameters a=1.32 atmL2mol-2 and b=0.0436Lmol-1. Solution: The critical constants of a van der Waals gas are

Vc  3b  3( 0.0436 Lmol1 )  0.131 Lmol1 pc 

a 1.32 atm L2 mol-2   25.7 atm 27b 2 27( 0.08206L atm K 1 )2

8a 8(1.32atmL2 mol 2 ) and Tc    109 K 27 Rb 27(0.08206L atm K 1 )  (0.0436 L mol1 )

1.22(b) A gas at 350K and 12 atm has a molar volume 12 per cent larger than calculated from the perfect gas law. Calculate (a) the compression factor under these conditions and (b) the molar volume of the gas. Which are dominating in the sample, the attractive or the repulsive forces? Solution: The compression factor is

Z

pVm Vm  RT Vm . perfect

(a) BeacuseVm  Vm . perfect  0.12Vm . perfect （0.12）Vm . perfect，we have Z  1.12 Repulsiveforces dominate. (b) The molar volumeis  RT  V  (1.12)Vm . perfect  (1.12)     p   (0.08206L atm K -1 mol-1 )  (350 K)    2.7 L mol1 V  (1.12)   12 atm  

9

1.24(b) The density of water vapour at 1.00 bar and 383 K is 0.5678 kg m-3.(a) Determine the molar volume Vm of water and the compression factor Z, from these data. (b) Calculate Z from the van der Waals equation with a=5.536L2atm mol-2 and b=0.03049L mol-1. Solution: (a)

Vm 

M 18.015 g mol1   31.728 L mol1 ρ 0.5678 g mol1

pVm (1.00 bar)  (31.728 L mol1 )   0.9963 RT (0.083145L bar K 1 mol1 )  (383 K) RT a (b) Using p   2 and subsitituting into the expressionfor Z above we get Vm  b Vm Z

Z 

Vm a Vm  b Vm RT 31.728 L mol-1 5.536L2 atm mol 2   0.9954 31.728 L mol-1  0.03049 L mol-1 31.728 L mol-1  0.03049L mol-1

Comment. Both values of Z are very close to the perfect gas value of 1.000, indicating that water vapour is essentially perfect at 1.00 bar pressure.

10

1.25(b) At 300 K and 20 atm, the compression factor of a gas is 0.86. Calculate (a) the volume occupied by 3.2 mmol of the gas under these conditions and (b) an approximate value of the second virrial coefficient B at 300 K. Solution:

The molar volumeis obtained by solving Z  Vm 

pVm 34, for Vm，which yields RT

ZRT 0.08206 L atm K 1 )  (300 K)   1.059 L mol1 p 20 atm

(a) Then,V  nVm  (8.2  10 -3 mol)  (1.059 L mol-1 )  8.7  10 -3 L  8.7 mL (b) An approximate value of B can be obtainedfrom eqn 36 by truncationof the seriesexpansion  pV  B  Vm  m  1   Vm  Z-1  RT 

 (1.059 L mol-1 )  0.86  1  0.15 L mol-1

11

1.27(b) The critical constants of ethane are pc=45.6 atm, Vc=148 cm3 mol-1, and Tc=305.4 K. Calculate the van der Waals parameters of the gas and estimate the radius of the molecules. Solution:

The critical volumeof a van der Waals gas is Vc  3b





1 1 so b  Vc  148 cm 3 mol1  49.3cm 3 mol1  0.0493 L mol1 3 3 By interpreting b as the cxcluded volumeof a mole of spherical molecules,we can obtain an estimate of molecularsize.The centresof spherical particles are excludedfrom a sphere whose radius is the diameterof those spherical particles (i.e.,twice their radius); that volumetimesthe Avogadro constant is the molar excluded volumeb  4π 2r 3  1  3b   so r   b  N A   4πN   3 2 A     1  3(49.3 cm 3 mol1 )   r   2  4ππ(6.02 10 23 mol1 ) 

1

3

1

3

 1.347  10 8 cm  1.94  10 -10 m

The critical pressureis pc 

a 27b 2



so a  27 pc b 2  2748.20 atm  0.0493 L mol1



2

 3.16 L2 atm mol 2

But this problem is overdetermined. We have anotherpiece of information 8a 27 Rb According to the constantswe have already determined, Tc should be

Tc 





8 3.16 L2 atm mol 2  231 K 27 0.08206 L atm K 1 mol1 (0.0493 L mol1 ) However, the reportedTc is 305.4 K, suggestingour computed a/b is about 25 per cent lower than it

Tc 





should be.

12

1.29(b) Suggest the pressure and temperature at which 1.0 mol of (a) H2S, (b)CO2, (c) Ar will be in states that correspond to 1.0 mol N2 at 1.0 atm and 25ºC. Solution:

Statesthat have the same reduced pressure,temperature, and volumeare said to correspond. The reduced pressureand temperature for N2 at 1.0 atm and 25C are pr 

p 1.0 atm T (25  273) K   0.030 and Tr    2.36 pc 33.54 atm Tc 126.3 K

The corresponding statesare (a) For H 2S p  pr pc  (0.030)  (88.3 atm)  2.6 atm T  TrTc  (2.36)  (373.2 K)  881 K (Critical constantsof H 2S obtained from handbook of chem istry and physics.) (b)For CO 2 p  pr pc  (0.030)  (72.85 atm)  2.2 atm T  TrTc  (2.36)  (304.2K)  718 K (c) For Ar p  pr pc  (0.030)  (48.00atm) 1.4 atm T  TrTc  (2.36)  (150.72K)  356K

13

1.30(b) A certain gas obeys the van der Waals equation with a=0.76 m6 Pa mol-2. Its volume is found to be 4.00×10-4 m3 mol-1 at 288 K and 4.0Mpa. From this information calculate the van der Waals constant b. What is the compression factor for this gas at the prevailing temperature and pressure? Solution:

The van der Waals equationis p

RT a  2 Vm  b Vm

which can be solvedfor b b  Vm 

RT  4.00  10 4 m 3 mol1 a p 2 Vm

The compression factor is Z



 



pVm 4.0  106 Pa  4.00  10 4 m 3 mol 1   0.67 RT 8.3145 J K 1 mol1  288 K 





14

Problems Numerical problems 1.4 A meterological balloon had a radius of 1.0 m when released at sea level at 20°C and expanded to a radius of 3.0 m when it had risen to its maximum altitude where the temperature was -20°C. What is the pressure inside the balloon at that altitude? Solution:

pV  nRT[12] implies that, with n constant,

p fVf Tf



piVi Ti

Solving for p f , the pressureat its maximumaltitude, yields p f  Substituting Vi 

   

 4 ri 3 pf   3 3  4 r  3 f

Vi T f   pi V f Ti

4 3 4 3 ri and V f  r f 3 3

3  T   f  p   ri   T f  p i i  T r  Ti i f    3

 1.0m   253 K  2      1.0 atm  3.2  10 atm  3.0m   293 K 

15

1.10 A vessel of volume 22.4 L contains 2.0 mol H2, and 1.0 mol N2 at 273.15 K initially. All the H2 reacted with sufficient N2 to form NH3. Calculate the partial pressure and the total pressure of the final mixture.4 Solution:

We assumethat no H2 remainsafter the reaction has gone to completion. The balanced equationis N 2  3H 2  2NH 3 We can draw up the following table

N2

H2

NH3

Total

Initial amount

n

n'

0

Final amount

n

0

2 n' 3

n  n' 1 n  n' 3

Specifically

0.33mol

0

1.33 mol

1.66 mol

Mole fraction

0.20

0

0.80

1.00

p

1 n' 3





 8.206  10 2 L atm K 1 mol 1  273.15 K   nRT   1.66 atm  1.66mol   V 22.4 L  

p(H 2 )  x(H 2 ) p  0

p(N 2 )  x(N 2 ) p  0.20  1.66 atm  0.33 atm

p(NH 3 )  x(NH 3 ) p  0.80  1.66 atm  0.33 atm

16

1.15 Calculate the molar volume of chlorine gas at 350 K and 2.30 atm using (a) the perfect gas law and (b) the van der Waals equation. Use the answer to (a) to calculate a first approximation to the correction term for attraction and then use successive approximations to obtain a numerical answer for part(b). Solution:

RT (8.206  10 2 L atm K 1 mol1 )(350 K)   12.5 L mol1 p 2.30 atm RT a RT (b)From p   2 39b, we obtain Vm   brearrange39b Vm  b Vm   a p 2  Vm   Then, with a and b from Table 1.6 (a)Vm 

Vm 

(8.206  10  2 L atm K 1 mol1 )(350 K)  5.622  10  2 L mol1 2 2  6.579L atm mol   (2.30 atm)   1 2 (12.5 L mol )  





28.7 2L mol1  5.622  10  2 L mol1  12.3 L mol1 2.34 Substitution of 12.3 L mol1 into the denominator of the first expressionagain resultsin 





Vm  12.3 L mol1 , so the cycle of approximation may be terminated.

17

Chapter 2 The First Law：the concepts Exercises 2.4 (b) A sample consisting of 2.00 mol He is expanded isothermally at 22℃ from 22.8 L to 31.7 L (a) reversibly, (b) against a constant external pressure equal to the final pressure of the gas, and (c) freely(against zero external pressure). For the three processes calculate q,w,△U,and △H. Solution：For a perfect gas at constant temperature

U  0 so q  -w For a perfect gas at constant temperature, H is also zero

dH  d(U  pV ) we have already noted that U does not change at constant temperature; nor does pV if the gas obeys Boyle’s law. These apply to all three cases below. (a) Isothermal reversible expansion

w   nRTln

Vf Vi

 -(2.00 mol)  (8.3145J K -1 mol-1 )  (22  273) K  ln

31.7 L  1.62  10 3 22.8 L

q   w  1.62  10 3 J (b) Expansion against a constant external pressure

w   pex V Where pex

in this case can be computed from the perfect gas law

pV  nRT (2.00 mol)  (8.3145J K -1 mol-1 )  (22  273) K  (1000Lm  3 )  1.55  105 Pa 31.7 L 5 - (1.55  10 Pa)  (31.7 - 22.8)L and w   1.38  10 3 J 3 1000 L m 3 q   w  1.38  10 J so p 

(c)

Free expansion is expansion against no force, so w  0 , and q  -w  0 as well.

18

2.6(b) A sample of argon of mass 6.56 g occupies 18.5 L at 305 K. (a) Calculate the work done when the gas expands isothermally against a constant external pressure of 7.7 kPa until its volume has increased by 2.5 L. (b) Calculate the work that would be done if the same expansion occurred reversibly. Solution :

aw   pex V   7.7  10



Pa  2.5L  19 J 1000 L m  3

(b) w  -nRTln

3

Vf Vi

  2.5  18.5 L 6.56 g  -   8.3145J K -1mol-1  305 K   ln -1  18.5 L  39.95 g mol   -52.8 J





19

2.8(b) A sample of 2.00 mol CH3OH(g) is condensed isothermally and reversibly to liquid at 64℃. The standard enthalpy of vaporization of methanol at 64℃ is 35.3 kJ mol-1. Find q,w,△U,and △H for this process.



Solution: q  ΔH  n( Δvap H )  2.00 mol   35.3K J mol Θ

1

  70.6 kJ

Because the condensation also occurs at constant pressure, the work is

w    pex dV   pV The change in volume from a gas to a condensed phase is approximately equal in magnitude to the volume of the gas





w   p（- Vvapour） nRT  2.00mol  8.3145 kJ K -1mol-1  64  273K  5.60  103 J ΔU  q  w  (-70.6  5 .60 )kJ  -65.0kJ

20

2.11(b) The constant-pressure heat capacity of a sample of a perfect gas was found to vary with temperature according to the expression Cp,m/(JK-1)=20.17+0.4001(T/K). Calculate q,w,△U,and △H for 1.00 mol when the temperature of 1.00 mol of gas is raised from 0℃ to 100℃ (a) at constant pressure (b) at constant volume. Solution ：

(a)At constant pressure q   CpdT  

100 273K

0  273K

20.17  ( 0.4001 )T / K dT J K 1





1 373K    20.17T  0.4001  T 2 / K  J K 1 273K 2   1    20.17  373  273  0.4001  3732  2732  J  14.9  10 3 J  H 2   1 w   pΔΔ   nRΔR   1.00 mol  8.3145JＫ mol1  100 K   831J ΔU  q  w  14.9 - 0.831 k J  14.1 k J (b) ΔU and ΔH depend only on temperature in perfect gases.Thus, ΔH  14.9k J and ΔU  14.1k J

 





as above. At constant volume,w  0 and ΔU  q , so q  14 .1k J

21

2.13(b) A sample of nitrogen of mass 3.12 g at 23.0℃ is allowed to expand reversibly and adiabatically from 400 mL to 2.00 L. What is the work done by the gas? Solution: Reversible adiabatic work is

w  CV T  nC p ,m  R   T f  Ti  Where the temperatures are related by [solution to Exercise2.12b]

V Tf  Ti  i V  f

1c

    C p,m  R (29.125  8.3145) J K 1 mol1 C where c  v,m    2.503 R R 8.3145 J K 1 mol1

 400  10  3 L   So Tf  23.0  273.15 K     2.00 L 

1 2.503

 156 K

 3.12 g  and w     29.125  8.3145 J K 1 mol1  156  296 K  325 J -1   28.0 g mol 

22

2.15(b) Calculate the final pressure of a sample of water vapour of mass 1.4 g that expands reversibly and adiabatically from an initial temperature of 300 K and volume 1.0 L to a final volume of 3.0 L. Take γ=1.3. Solution: For reversible adiabatic expansion



p f v f  piVi



V so p f  pi  i V  f

   



We need pi , which we can obtain from the perfect gas law

pV  nRT

so

p

nRT V

 1.4 g     0.08206 L atm K 1 mol1  300 K  1  18 g mol  pi    1.9 atm 1.0 L



 1.0 L  p f  1.9 atm     3.0 L 



1.3

 0.46 atm

23

2.19(b) When 2.0 mol CO2 is heated at a constant pressure of 1.25 atm, its temperature increases from 250 K to 277 K. Given that the molar heat capacity of CO2 at constant pressure is 37.11 JK-1 mol-1, calculate q,△U,and △H . Solution:





ΔH  q p  C p ΔT  nC p,m ΔT  2.0 mol  37.11 J K 1mol1  277  250 K  2.0  10 3 J mol1 ΔH  ΔU  Δ pV   ΔU  nRΔR



so

ΔU  ΔH  nRΔR



ΔU  2.0  10 3 J mol1  2.0 mol  8.3145 J K 1 mol1  277  250 K  1.6  10 3 J mol1

24

2.21 (b) A sample consisting of 2.5 mol of perfect gas at 220 K and 200 kPa is compressed reversibly and adiabatically until the temperature reaches 255 K. Given that its molar constant-volume heat capacity is 27.6 JK-1 mol-1, calculate q,w,△U,and △H,and the final pressure and volume. Solution: For adiabatic compression, q=0 and





w  CV T  2.5 mol  27.6 K J 1 mol1  255  220 K  2.4  10 3 J ΔU  q  w  2.4  10 3 J ΔH  ΔU  Δ pV   ΔU  nRΔR





 2.4  10 3 J  2.5 mol  8.3145 J K 1 mol-1  255  220 K  3.1  10 3 J The initial and final states are related by c

T  V f T f  V i Ti so V f  V i  i  T   f C 27.6 J K -1 mol1 where c  V,m   3.32 R 8.314 J K 1 mol1 nRTi 2.5 mol  8.3145 J K 1 mol1  220 K Vi    0.0229 m 3 3 pi 200  10 Pa c



c



3.32

 220 K  3 V f  0.0229 m     0.014 m  14 L 255 K   nRTf 2.5 mol  8.3145 J K 1 mol1  255 K pf    3.8  105 Pa 3 Vf 0.014 m 3

25

2.24(b) Consider a system consisting of 3.0 mol O2 (assumed to be a perfect gas) at 25℃ confined to a cylinder of cross-section 22 cm2 at 820 kPa. The gas is allowed to expand adiabatically and irreversibly against a constant pressure of 1.0 atm. Calculate q,w,△U,and △H and △T when the piston has moved 15 cm. Solution: In an adiabatic process, q=0. Work against a constant external pressure is



w   pex V   110  10

3

 15 cm  22 cm 2  Pa    36 J (100 cm m 1 ) 3

ΔU  q  w  36 J

w  CV ΔT  nC p,m -RΔT

ΔT 

so

w

nC p,m -R

 36 J  0.57 K 3.0 mol  29.355  8.3145J K -1mol-1 H  ΔU  Δ pV   ΔU  nRΔR



 36 J  3.0 mol  (8.3145 J K -1 mol-1 )   0.57 K   50 J

26

2.27(b) The standard enthalpy of formation of phenol -165.0 kJ mol-1. Calculate its standard enthalpy of combustion. Solution: The reaction is

C6 H 5 OH  7O2  6CO2  3H 2 O

 c H   6 f H  CO2  3 f H  H 2 O   f H  C6 H 5 OH  7  f H  O 2   6 393.51  3 285.83   165.0  70K J mol1  3053.6 kJ mol1

27

Θ

2.29(b) From the following data, determine Δf H for diborane, B2H6(g), at 298 K:

1B 2 H 6 g   3O2 g   B 2O 3 s  3H 2Og  22Bs  3 O 2 g   B 2O 3 s 2 3H 2 g   1 O 2 g   H 2Og  2 Solution: We need  f H



 r H θ  1941 kJ mol1  r H θ  2368 kJ mol1  r H θ  241.8 kJ mol1

for the reaction

4 2Bs  3H2 g   B 2H 6 g  reaction(4)  reaction(2)  3  reaction(3) - reaction(1) Thus,  f H    r H  reaction(2) 3   r H  reaction(3)  r H  reaction(1)   2368  3   241.8   1941K J mol1  1152K J mol1

28

2.32(b) When 2.25 mg of anthracene, C14H10(s), was burned in a bomb calorimeter the temperature rose by 1.35 K. Calculate the calorimeter constant. By how much will the temperature rise when 135 mg of phenol, C6H5OH(s), is burned in the calorimeter under the same conditions? Solution: For anthracene the reaction is

C14 H10 s 

33 O 2 g   14CO2 g   5H 2Ol  2 5 ΔcU Θ  Δc H Θ  ng RT 26 ng  - mol 2 Θ 1 Δc H  7163 k J mol ( Hand book of chem istry and physics )  5  ΔcU Θ  7163 k J mol1   -  8.3  10  3 k J K -1 mol1  298K  (assumeT  298 K)  2  1  -7157 k J mol  2.25  10  3 g    7157 k J mol1 q  qV  nΔcU Θ   1   172.23 g mol   0.0935k J



C

q T



0.0935 k J  0.0693 k J K 1  69.3 J K 1 1.35 K

When phenol is used the reaction is C 6 H 5 OHs  Δc H Θ  3054 k J mol1 Table2.5



15 O 2 g   6CO2 g   3H 2 Ol  2

3 ng  - mol 2

ΔcU Θ  Δc H Θ  ng RT ,









 3   3054 k J mol1     8.314  10  3 k J K -1 mol1  298K   2  3050 k J mol1  135  10  3 g   3050 k J mol1  4.375 k J q   1  94.12 g mol   q 4.375 k J ΔT    63.1 K C 0.0693 k J K 1 Comment. In this case ΔcU Θ and Δc H Θ differedby  0.1 per cent. Thus, to within 3 significant





figures,it would not have matteredif we had used Δc H Θ insteadof ΔcU Θ , but for very precise work it would.

29

2.35(b) Given that the standard enthalpy combustion of graphite is -393.51 kJ mol-1 and that of diamond is -395.41kJ mol-1, calculate the enthalpy of the graphitediamond transition. Solution: The difference of the equations is Cgr   Cd 

Δtrans H Θ   393.51   395.41kJ mol 1  1.90 kJ mol 1

30

2.39(b) Use standard enthalpies of formation to calculate the standard enthalpies of the following reactions:

aCyclopropaneg  propeneg bHCl aq  NaOHaq  NaCl aq  H 2Ol  Solution:

( a ) Δr H Θ  Δf H Θ  propene , g   Δf H Θ cycloropane , g   20.42  53.30k J mol1  32.88 k J mol1

b  The met ionic reaction is obtainedfrom H  aq   Cl  aq   Na  aq   OH  aq   Na  aq   Cl  aq   H 2 Ol  and is H  aq   OH  aq   H 2 Ol  Δr H Θ  Δf H Θ H 2 O, l   Δf H Θ H  , aq   Δf H Θ OH  , aq    285.83  0    229.99 k J mol1  55.84 k J mol1

31

2.40(b) Given the reactions(1) and (2) below, determine (a) Δr H

Θ

Θ

and ΔrU for reaction (3), (b)

Δf H Θ for both HI(g) and H2O(g) all at 298 K. Assume all gases are perfect.

1H 2 g   I 2 g   2HI g  22H 2 g   O 2 g   2H 2Og  34HI g   O 2 g   I 2 s  2H 2Og 

 r H θ  52.96 kJ mol 1  r H θ  483.64 kJ mol 1

Solution:

reaction3  reaction2  2reaction1 ( a ) Δr H Θ 3  Δr H Θ 2  2 Δr H Θ 1



 483.64k J mol1  2 52.96k J mol1



 -589.56k Jmol1 ΔrU Θ  Δr H Θ  ng RT





 -589.56k Jmol1   3  8.314J K -1 mol1  298 K  1

1

 -589.56k Jmol  7.43k J mol  582.13k J mol1

b Δf H Θ HI   1 52.96k J mol1   26.48k J mol1

2 1 Δf H Θ H 2O    483.64k J mol1  241.82k J mol1 2





32

Θ 2.44(b) Calculate Δr H and Δr U

Θ

Θ

at 298 K and Δr H at 348 K for the hydrogenation of ethyne

(acetylene)to ethane (ethylene) from the enthalpy of combustion and heat capacity data in Tables 2.5 and 2.6. Assume the heat capacities to be constant over the temperature range involved. Solution: The hydrogenation reaction is

1C 2 H 2 g   H 2 g   C 2 H 4 g 

Δ r H Θ Tc   ?

The reactions and accompanying data which are to be combined in order to yield reaction (1) and Are

2H 2 g   1 O 2 g   H 2Ol 

Δ C H Θ 2  285.83 k J mol1 2 3C 2 H 4 g   3O2 g   2H 2Ol   2CO2 g  Δ C H Θ 3  1411k J mol1

4C 2 H 2 g   5 O 2 g   H 2Ol   2CO 2 g 

Δ C H Θ 4  1300k J mol1 2 reaction(1)  reaction(2) - reaction(3)  reaction(4) Hence,

aΔ r H Θ T   Δ r H Θ 2  Δ c H Θ 3  Δ c H Θ 4   285.83   1411   1300k J mol1  175k J mol1

ΔrU Θ T   Δr H Θ T   Δng RT 26 Δ





g

 -1

 - 175k J mol1  2.48k J mol1  173k J mol1

bΔ r H 348 K   Δ r H 298K   Δ r C p 348K  298K Example2.7 Δ r C p   v J C p ,m J 47  C p ,m C 2 H 4 , g   C p ,m C 2 H 2 , g   C p ,m H 2 , g  Θ

Θ

J

 43.56  43.93  28.82  10  3 k J K -1 mol1  29.19  10  3 k J K -1 mol1



 



Δ r H Θ 348K    175k J mol1  29.19  10  3 k J K -1 mol1  50K  1

 176k J mol

33

Problems 2.2 An average human produces about 10MJ of heat each day through metabolic activity. If a human body were an isolated system of mass 65 kg with the heat capacity of water, what temperature rise would the body experience? Human bodies are actually open systems, and the main mechanism of heat loss is through the evaporation of water. What mass of water should be evaporated each day to maintain constant temperature? Solution: Good approximate answers can be obtained from the data for the heat capacity and molar heat of vaporization of water at 25℃.[Table 2.6 and 2.3]

C p ,m H 2O   75.3J K 1 mol1

 vap H  H 2 O  44.0k J mol1

nH 2 O 

65 k g  3.6  10 3 mol 0.018 k g mol1 From ΔH  nC p,m ΔT , we obtain ΔT 

ΔH 1.0  104 k J   37 K nC p,m 3.6  10 3 mol  0.0753 k J K -1 mol1



 



m  vap H  M M  ΔH 0.018k g mol1  1.0  104 k J m   4.09k g  vap H  44.0k Jmol1

From ΔH  n vap H  



 



Comment: This estimate would correspond to about 30 glasses of water per day, which is much higher than the average consumption. The discrepancy may be a result of our assumption that evaporation of water is the main mechanism of heat loss.

34

Chapter 3 The First Law: the machinery Exersises 3.10(b) A vapour at 22 atm and 5℃ was allowed to expand adiabatically to a final pressure of 1.00 atm; the temperature fell by 10 K. Calculate the Joule-Thomson coefficient, µ, at 5℃, assuming it remains constant over this temperature range. Solution: The Joule-Thomson coefficient  is the ratio of temperature change to pressure change under conditions of isenthalpic expansion. So

 T  ΔT  10 K μ     0.48 K atm1   Δp 1.00  22 atm  p  H

35

3.11(b) For a van der Waals gas, πT  a

Vm

2

. Calculate U m for the isothermal reversible expansion of

argon from an initial volume of 1.00 L to 22.1 L at 298 K. What are the values of q and w? Solution:

 U   dU m  dU m    dT   m  dVm  T Vm  Vm  dT  0 in an isothermalprocess, so

U m  U m T,Vm 

 U  a dU m   m  dVm  2 dVm Vm  Vm  T Vm2 Vm2 22.1 L mol -1 dV a a m U m   dU m   dV  a  2 m -1 2 2  Vm1 m Vm1 m 1.00 L mol Vm Vm Vm

22.1L mol -1 1.00 L mol -1

a a 21.1a    0.95475 L mol-1 -1 -1 -1 22.1L mol 1.00 L mol 22.1L mol 2 2 a  1.345 atm L mol 



   1.01325  10

U m  0.95475 mol L1  1.345 atm L2 mol 2



 0.2841 atm L2 mol 2

 1m 3  1.01325  10 Pa atm   3  10 L  3 -1  130P am mol  130 J mol1 RT a w   pex dVm and p   2 Vm  b Vm



5

1



5



Pa atm1



   

for a van der Waals gas

 RT  a dVm   2 dVm   q  ΔU m so w     Vm  Vm  b  Thus q  

22.1 L mol -1

1.00 L mol

-1

 RT  mol - 1  dVm   RT lnVm  b  122.00.1 LL mol -1  Vm  b   22.1  3.22  10  2    8.314 J K -1 mol-1  298 K   ln 2   1.00  3.22  10 





 7.7469 k J mol-1 w  7747 J mol-1  130 J mol-1  7617 J mol-1  7.62 k J mol-1

36

3.12(b)

The

volume

of

a

certain



liquid

V  V ' 0.77  3.7  10 4 T/K   1.52  10 6 T/K 

2

varies

with

temperature

as



Where V' is its volume at 298 K. Calculate its expansion coefficient,  , at 310 K Solution: The expansion coefficient is

α

1 V





V' 3.7  10 4 K 1  2  1.52  10 6 T K 2  V     V  T  p





V' 3.7  10 4 K 1  2  1.52  10 6 T/K  K 1

T/K   1.52  10 T/K 2  3.7  104 K 1  2  1.52  106 310 K 1  1.27  103 K 1  2 0.77  3.7  10 4 310  1.52  10 6 310 



V' 0.77  3.7  10

4

6

37