Chapter 1 The properties of gases

Exercises1.2(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=nRTV

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

of Ar.

n=25 g

39 .95 gmol−1=0 .62 { 6̄mol ¿

so p=

( 0.626mol )×(8 .31×10−2Lbark -1 mol-1)×(30+273k )1 .5 L

=10 . 5̄ bar

not 2.0 bar.

(b) The van der Waals equation is

p= RTV m−b

− aV

m2

so p=

( 8. 31×10−2 LbarK−1 )×(30+273) K

(1. 5 L/0 .62 6mol )2=10 . 4bar

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 α T and pi

Ti=

p f

T f

The final pressure, then, ought to be

p f=pi T f

T i

=(125 k Pa)×(11+273 )K

(23+273 )K=120 k Pa

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 0.750 000 0.500 000 0.250 000

Vm/Lmol-1 29.8649 44.8090 89.6384

ρ/(gL-1) 1.07144 0.714110 0.356975

Solution:

All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pV m/T

will give

the best value of R.

The molar mass is obtained from pV=nRT ＝ m

MRT

Which upon rearrangement gives M ＝ m

VRTp

=ρRTp

=ρRTp

The best value of M is obtained from an extrapolation of

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

Draw up the following table

p/atm ( pV m/T )/(L atm K -1 mol -1)

( ρ / p) /(g L-1 atm -1)

0.750 000 0.082 0014 1.428 59

0.500 000 0.082 0227 1.428 22

0.250 000 0.082 0414 1.427 90

From Fig. 1.1(a)

1-1-

0

mol K atm L 0615 082.0

p

m

T

pV

From Fig. 1.1(b) ( ρ

p )p=0

=1.42755 g L− 1 atm -1

M=RT (ρp )

p=0=(0 .0820615 L atm K -1 mol -1)×(273.15 K )×(1 . 42755 g L

− 1 atm -1)

=31 . 9987 g 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 ρ is related to the molar volume Vm by

V m ＝ Mρ

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

pV m=RT so pMρ

=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

p=

62 .364 LTorr K-1 mol-1×[(100+273 )K ]×(0 . 6388gL-1)120 Torr

=124 gmol−1

The number of atoms per molecule is

124gmol−1

31 .0gmol−1=4 . 00

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

nNe=0 .225g20 .18gmol-1

=1 .115×10−2 mol, pNe=66 .5 Torr, T=300K

V=(1.115×10−2 mol )×(62. 36LTorr K-1 mol-1)×(300 K )66 .5Torr

＝3 .137 L=3 .14 L

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

Torr 212L73.13

K)300()mol K LTorr36.62(mol)1083.54([1]p

mol1083.54mol10)511.1438.051.99(

mol1038.4mol g 39.95

g 0.175 mol1051.99

mol g 16.04

g 0.320

1-1-2

22

31

2-14

V

RTn

n

nn

J

ArCH

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

c=(8RTπM )

12

soc (He )c (Hg )

=(M (Hg )M (He ) )

12=(200 . 59

4 . 003 )1

2=7 . 079

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

12

mc2

,where c is the root mean square speed

c=( 3RTM )

12

So

12

mc2

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

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

c=( 8 RT

πM )1

2

=

c=( 8×（8. 314 JK-1 mol -1)×(298 K )π×(28 .02×10−3 kg mol-1 )

12=4 . 75×10−2 ms−1

(b) The mean free path is calculated from

λ= kT

21

2 σp

[ 33 ]

With σ=πd2=π×(3 . 95×10−10 m)2=4 .90×10−19 m2

Then,

λ=（1. 381×10−23JK -1)×(298 K )

21

2×(4 .90×10−19 m2 )×(1×10−9Torr )×( 1 atm760 Torr )×( 1 . 013×105 Pa

1 atm )=4×104 m

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

λ and c have already been calculated z= c

λ= 4 .75c

4 . 46×104 m=1×10−2 s−1

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

λ= kT

21

2σp

＝ （1 .381×10−23 JK−1)×(217K )

21

2 [0 .43×(10−9m )2]×(12.1×103 Pa atm−1)

=4 .1×10−7m

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.)

Solution: Obtain data from Exercise 1.17(a) is z=( 16

π mkT )1

2 σp

Substituting σ=0 . 43 nm2 , p=12 .1×103 Pa ,m=(28 .02 u ), andT=217 K

we obtain

z=4×(0 . 43×10−18m2 )×(12. 1×103 Pa )

[ π×(28. 02 )×(1 . 6605×10−27 kg )×(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

V c=3 b=3(0 .0436 Lmol−1 )=0 .131 Lmol−1

pc=a

27 b2=1 . 32 atm L2 mol -2

27 (0 .08206L atm K−1 )2=25. 7 atm

and T c=8 a27 Rb

=8 (1.32atmL2 mol−2)27 (0 .08206 L atm K−1 )×(0 .0436 L mol−1)

=109 K

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=pV m

RT=

V m

V m . perfect

(a ) Beacuse V m=V m . perfect+0 . 12V m . perfect=（0.12）V m . perfect ， we have Z=1.12 Repulsive forces dominate .(b ) The molar volume is

V=(1.12)V m . perfect=(1 .12 )×(RTp )

V=(1 .12)×((0 .08206 L atm K-1 mol-1 )×(350 K )12 atm )=2 .7 L mol−1

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)

V m=Mρ

=18 .015 g mol−1

0.5678 g mol−1=31 .72 8 L mol−1

Z=pVmRT

=(1 .00 bar)×(31 .72 8 L mol−1 )(0.083145 L bar K−1 mol−1 )×(383 K)

=0 . 9963

(b ) Using p=RTV m−b

−aV

m2

and subsitituting into the expression for Z above we get

Z=V m

V m−b -

aV m RT

=31.72 8 L mol-1

31.728 L mol-1−0 .03049 L mol-1−5.536L2 atm mol−2

31.72 8 L mol-1−0 .03049L mol-1=0 .995 4

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 volume is obtained by solving Z=pV m

RT[ 34 ] , for V m， which yields

V m=ZRTp

=0. 08206 L atm K−1)×(300 K )20 atm

=1 .059 L mol−1

(a ) Then, V =nV m=(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 obtained from eqn 36 by truncation of the series expansion

B=V m( pV m

RT−1)=V m× (Z-1 )

=(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 volume of a van der Waals gas isV c=3 b

so b=13

V c=13

(148 cm3 mol−1 )=49 .3cm3 mol−1=0. 0493 L mol−1

By interpreting b as the cxcluded 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 thediameter of those spherical particles ( i . e ., twice their radius ) ; that volume times the Avogadro constant is the molar excluded volume b

b=N A(4 π (2 r )3

3 ) so r=12 (3 b

4 πN A)1

3

r=12 (3( 49 .3 cm3 mol−1 )

4π(6 . 022×1023mol−1 ) )1

3=1.347×10−8cm=1 .94×10-10 m

The critical pressure is

pc=a

27 b2

so a=27 pc b2=27 ( 48 .20 atm )×( 0. 0493 L mol−1)2=3 .16 L2 atm mol−2

But this problem is overdetermined . We have another piece of information

T c=8a27 Rb

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

T c=8 ( 3. 16 L2atm mol−2 )27 ( 0 .08206 L atm K−1 mol−1)(0 . 0493 L mol−1)

=231 K

However, the reported T c is 305 . 4 K, suggesting our computed a/b is about 25 per cent lower than itshould 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:

States that have the same reduced pressure, temperature, and volume are said to correspond . The reduced pressure and temperature for N2 at 1 . 0 atm and 25 °C are

pr=ppc

=1. 0 atm33 .54 atm

=0. 030 and T r=TTc

=(25+273) K126 . 3 K

=2 . 36

The corresponding states are(a ) For H2 S p=pr pc=(0 .030 )×(88 .3 atm )=2 .6 atm

T=Tr T c=(2. 36 )×(373 . 2 K )=881 K(Critical constants of H2 S obtained from handbook of chemistry and physics . )(b )For CO2

p=pr pc=(0 . 030 )×(72 .85 atm )=2. 2 atm T=T r T c=(2. 36 )×(304 .2K )=718 K

(c ) For Ar p=pr pc=(0 . 030)×( 48. 00atm )=1 . 4 atm T=T r T c=(2. 36 )×(150 .72K )=3 56 K

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 equation is

p=RTV m−b

−aV

m2

which can be solved for b

b=V m−RT

p+aV

m2

=4 .00×10−4 m3 mol−1

The compression factor is

Z=pV m

RT=

( 4 . 0×106 Pa )×(4 . 00×10−4 m3 mol−1)( 8.3145 J K−1 mol−1 )¿ (288 K )

=0. 67

14

Problems

Numerical problems1.4 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 f V f

T f

=pi V i

T i

Solving for p f , the pressure at its maximum altitude, yields p f=V i

V f

×T f

T i

×pi

Substituting V i=43

πri3 and V f=

43

πrf 3

p f=((4

3)πri3

(4 3)πrf3 )×T f

T i

×pi=(ri

rf)3

×T f

T i

×pi

=(1.0m3 .0m )

3

×(253 K293 K )×(1 .0 atm )=3 .2×10−2 atm

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 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 H2 NH3 Total

Initial amount n n ' 0 n+n'Final amount

n−13

n '0 2

3n ' n+ 1

3n'

Specifically 0.33mol 0 1.33 mol 1.66 mol

Mole fraction 0.20 0 0.80 1.00

p=nRTV

=(1.66mol )×(( 8. 206×10−2L atm K−1mol−1)×(273 .15 K )22 . 4 L )=1 . 6 6 atm

p( H2 )=x ( H2 ) p=0

p( N2 )=x ( N2 ) p=(0 . 20×(1 . 66 atm ) )=0 .33 atm

p(NH3 )=x (NH3 ) 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:

(a )V m=RTp

=(8 . 206×10−2L atm K−1 mol−1 )(350 K )2.30 atm

=12 .5 L mol−1

(b )From p=RTV m−b

−aV

m2

[ 39b ] , we obtain V m=RT

( p+aV

m2 )+b [ rearrange39b ]

Then , with a and b from Table 1. 6

V m≈(8 .206×10−2 L atm K−1mol−1 )(350 K)

(2. 30 atm )+(6. 579L2 atm mol−2

(12. 5 L mol−1)2 )+(5 .622×10−2L mol−1)

¿28. 7 2L mol−1

2.34+(5 . 622×10−2L mol−1 )≈12. 3 L mol−1

Substitution of 12.3 L mol−1 into the denominator of the first expression again results in V m=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

ΔU=0 so q=-w

For a perfect gas at constant temperature, ΔH 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=−nRT lnV f

V i

=−(2.00 mol )×(8. 3145J K -1 mol-1 )×(22+273 ) K×ln31 .7 L22 . 8 L

=−1 .62×103

q=−w=1 .62×103 J

(b) Expansion against a constant external pressure

w=−pex ΔV

Where pex in this case can be computed from the perfect gas law

pV=nRT

so p=(2. 00 mol )×( 8. 3145J K -1 mol-1 )×(22+273 ) K31 .7 L

×(1000 Lm−3 )=1. 55×105 Pa

and w=−(1. 55×105Pa )×(31. 7-22. 8) L

1000 L m−3=−1 . 38×103J

q=−w=1 .38×103 J (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 ΔV =−(7 . 7×103Pa )×(2 .5L )1000 L m−3

=−19 J

(b )w=-nRT lnV f

V i

=−(6.56 g39 .95 g mol -1 )×(8 .3145 J K-1 mol -1 )×(305 K )×ln

(2 .5+18 . 5 ) L18 .5 L

=-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 HΘ)=(2. 00 mol )×(−35 .3K J mol−1)=−70 .6 kJ

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

w=−∫ pex dV=−pΔV

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 -1 mol-1 )× (64+273 ) K=5. 60×103 JΔU=q+w=(−70 . 6+5 . 60)kJ=−65 . 0 kJ

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=∫0+273K

100+273K [20.17+(0 .4001)T /K ]dT J K−1

¿[ (20 .17 )T +12

(0 .4001 )×(T 2 /K ) ] J K−1|273K373K

¿[ (20 .17 )×(373−273 )+12

(0 .4001 )×(3732−2732)] J=14 .9×103 J=ΔH

w=−pΔV =−nRΔT=−(1 .00 mol )×(8 .3145JＫ−1 mol−1)×(100 K )=−831 JΔU=q+w=(14 .9-0 .831 ) kJ=14 .1 kJ(b ) ΔU and ΔH depend only on temperature in perfect gases . Thus, ΔH=14 .9kJ and ΔU=14 .1kJ as above. At constant volume, w=0and ΔU=q , so q =+14 .1kJ

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 ΔT=n (Cp , m−R )×(T f −T i )

Where the temperatures are related by [solution to Exercise2.12b]

Tf =T i(V i

V f)1/c

where c=Cv,m

R=

C p,m−R

R=

(29 . 125−8.3145 ) J K−1 mol−1

8 . 3145 J K−1 mol−1=2 .503

So Tf= [ (23. 0+273 .15 ) K ]×(400×10−3 L2. 00 L )1/2. 503

=156 K

and w=(3 .12 g28 .0 g mol-1 )×(29 .125−8 .3145 ) J K−1mol−1×(156−296 ) K=−325 J

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 vfγ=pi V

iγ so pf =p i( V i

V f)

γ

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

pV=nRT so p=nRTV

pi=(1 . 4 g18 g mol−1 )×( 0.08206 L atm K−1 mol−1 )×(300 K )

1.0 L=1 . 9 atm

p f=(1. 9 atm )×(1.0 L3.0 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 .0×103 J mol−1

ΔH=ΔU +Δ ( pV )=ΔU+nRΔT so ΔU=ΔH−nRΔT

ΔU=2 .0×103 J mol−1−(2.0 mol )×(8 .3145 J K−1 mol−1)× (277−250 ) K=1 .6×103 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 ΔT=(2 .5 mol )×(27 . 6 K J−1 mol−1)×(255−220 ) K=2.4×103 J

ΔU=q+w=2 . 4×103 JΔH=ΔU +Δ ( pV )=ΔU+nRΔT

¿2 .4×103 J+ (2.5 mol )×( 8 .3145 J K−1 mol -1 )× (255−220 ) K=3 . 1×103 J

The initial and final states are related by

V f Tf c=V i T ic so V f =V i(T i

T f)

c

where c=CV,m

R=27 . 6 J K -1 mol−1

8. 314 J K−1 mol−1=3. 32

V i=nRT i

pi

=2. 5 mol×8 .3145 J K−1 mol−1¿220 K200×103 Pa

=0 . 0229 m3

V f =(0 . 0229 m3 )¿(220 K255 K )

3 . 32

=0 . 014 m3 =14 L

pf =nRT f

V f

=2 .5 mol×8 . 3145 J K−1 mol−1 ¿255 K0 .014 m3 =3 . 8×105 Pa

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 ΔV =−( 110×103 Pa )×(15 cm )×( 22 cm2)(100 cm m−1 )3

=−36 J

ΔU=q+w=−36 J w=CV ΔT =n (C p,m-R ) ΔT so

ΔT =wn (C p,m-R )

¿−36 J

(3 .0 mol )×(29 . 355−8. 3145 ) J K -1 mol-1=−0 .57 K

ΔH=ΔU+ Δ ( pV )=ΔU +nRΔT¿−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 H5 OH+7O2→6CO2+3H2 O

Δc HΘ=6 Δf HΘ (CO2 )+3 Δf HΘ (H2 O )−Δf HΘ (C6 H 5 OH)−7 Δf H Θ (O2)¿ [6 (−393 . 51 )+3 (−285 . 83 )−(−165 .0 )−7 (0 ) ]K J mol−1

¿−3053 .6 kJ mol−1

27

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

(1 ) B2 H 6 ( g )+3O2 ( g )→B2O3 (s )+3H2O (g ) Δr Hθ=−1941 kJ mol−1

(2 ) 2B ( s)+32

O2 ( g )→B2O3 ( s ) Δr Hθ=−2368 kJ mol−1

(3 ) H2 ( g )+12

O2 ( g )→H2 O (g ) Δr Hθ=−241 .8 kJ mol−1

Solution: We need Δf HΘ for the reaction

(4 ) 2B (s )+3H2 ( g )→B2 H6 ( g )reaction (4 )=reaction(2)+3×reaction(3 ) -reaction(1)Thus, Δf HΘ=Δr HΘ { reaction(2)}+3×Δr HΘ {reaction (3 )}−Δr HΘ {reaction(1) }¿ {−2368+3×(−241 .8 )−(−1941 ) }K J mol−1=−1152 K 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 )+332

O2 ( g )→14CO2 ( g )+5H2O ( l )

Δc UΘ=Δc HΘ−Δng RT [ 26 ] Δng=−52

mol

Δc HΘ=−7163 kJ mol−1( Hand book of chemistry and physics)

Δc UΘ=−7163 kJ mol−1−(−52×8 .3×10−3kJ K -1 mol−1×298 K ) ( assume T=298 K )

=-7157 kJ mol−1

|q|=|qV|=|nΔc UΘ|=(2 .25×10−3 g172 .23 g mol−1 )×(7157 kJ mol−1)

=0 .0935kJ

C=|q|ΔT

=0 .0935 kJ1.35 K

=0 .0693 kJ K−1=69 .3 J K−1

When phenol is used the reaction is C6 H5OH (s )+152

O2 (g )→6CO2 ( g )+3H2O (l )

Δc HΘ=−3054 kJ mol−1 (Table2 .5 )

Δc UΘ=Δc HΘ−Δng RT , Δng=−32

mol

¿ (−3054 kJ mol−1)+(32 )× (8 .314×10−3kJ K -1 mol−1 )×(298 K )

¿−3050 kJ mol−1

|q|=(135×10−3 g

94 .12 g mol−1 )( 3050 kJ mol−1 )=4 .37 5 kJ

ΔT =|q|C

=4 .37 5 kJ

0.0693 kJ K−1=+ 63.1 K

Comment . In this case Δc U Θand Δc HΘ differed by≈0.1 per cent . Thus, to within 3 significant

figures, it would not have mattered if we had used Δc HΘ instead of 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 graphitediamond transition.

Solution: The difference of the equations is C (gr )→C (d )

Δtrans HΘ=[−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 )+H2O ( l )

Solution:

(a ) Δr HΘ=Δf HΘ ( propene , g )−Δf HΘ (cycloropane , g )=[ (20 .42 )−(53 .30 ) ] kJ mol−1

¿−32 . 88 kJ mol−1

(b ) The met ionic reaction is obtained fromH+ (aq )+Cl− ( aq )+Na+ (aq )+OH− ( aq )→Na+ (aq )+Cl− ( aq )+H2O ( l )and is H+ (aq )+OH− (aq )→H2 O (l )

Δr HΘ=Δf HΘ (H2O,l )−Δf HΘ ( H+ ,aq )−Δf HΘ (OH− ,aq )¿ [ (−285. 83 )−(0 )− (−229. 99 ) ] kJ mol−1

¿−55 . 84 kJ mol−1

31

2.40(b) Given the reactions(1) and (2) below, determine (a) Δr HΘ and Δr U Θ

for reaction (3), (b)Δf HΘ

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

(1 ) H2 ( g )+ I 2 (g )→2HI ( g ) Δr Hθ =+52. 96 kJ mol−1

(2 ) 2H2 (g )+O2 ( g )→2H2 O (g ) Δr Hθ=−483 .64 kJ mol−1

(3 ) 4HI ( g )+O2 ( g )→ I 2 (s )+2H2 O (g )

Solution:

reaction (3 )=reaction (2 )−2 (reaction (1 ) )(a ) Δr HΘ (3 )=Δr HΘ (2 )−2 Δr HΘ (1 )

=−483 . 64 kJ mol−1−2 (52. 96 kJ mol−1 ) =-589 . 56kJ mol−1

Δr UΘ=Δr HΘ−Δng RT

=-589 . 56kJ mol−1−(−3 )×(8 .314 J K -1 mol−1 )×(298 K ) =-589 . 56kJ mol−1+7 . 43 kJ mol−1=−582 .13 kJ mol−1

(b ) Δf HΘ ( HI )=12

( 52.96 kJ mol−1 )=26 .48 kJ mol−1

Δf HΘ (H2 O )=−12

( 483.64 kJ mol−1 )=−241 . 82 kJ mol−1

32

2.44(b) Calculate Δr HΘand Δr U Θ

at 298 K and Δr HΘat 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 ) C2 H2 ( g )+H 2 (g )→C2 H 4 ( g ) Δr HΘ (Tc )=?

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

Are

(2 ) H2 ( g )+12

O2 ( g )→H2O (l ) ΔC HΘ (2 )=−285 .83 kJ mol−1

(3 ) C2H 4 ( g )+3O2 ( g )→2H2O (l ) +2CO2 (g ) ΔC HΘ (3 )=−1411 kJ mol−1

(4 ) C2 H2 ( g )+52

O2 (g )→H 2O (l )+2CO2 ( g ) ΔC HΘ (4 )=−1300 kJ mol−1

reaction (1 )=reaction(2)-reaction(3 )+reaction( 4 )Hence,(a ) Δr HΘ (T )=Δr HΘ (2 )−Δc HΘ (3 )+ Δc HΘ (4 )¿ {(−285 .83 )− (−1411)+(−1300 ) }kJ mol−1

−175 kJ mol−1

Δr UΘ (T )=Δr HΘ (T )−Δng RT [26 ] Δng=−1

¿ (-175kJ mol−1+2. 48 kJ mol−1 )=−173 kJ mol−1

(b ) Δr HΘ (348 K )=Δr HΘ (298 K )−Δr C p (348 K−298 K ) [ Example2. 7 ]Δr C p=∑

J

vJ C p , m ( J ) [ 47 ]=C p, m (C2 H4 , g )−C p , m (C2 H2 , g )−C p, m (H2 , g )

¿ (43 .56−43 .93−28 .82 )×10−3kJ K -1 mol−1=−29 . 19×10−3 kJ K -1 mol−1

Δr HΘ (348 K )=(−175 kJ mol−1)−( 29.19×10−3kJ K -1 mol−1)×(50 K )¿−176 kJ mol−1

33

Problems2.2 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 (H2O )=75 .3 J K−1 mol−1 Δvap HΘ (H2O )=44 .0 kJ mol−1

n (H2O )=65 kg0 . 018 kg mol−1

=3 . 6×103mol

From ΔH=nC p,m ΔT , we obtain

ΔT =ΔHnC p,m

=1 . 0×104kJ

(3 . 6×103mol )×( 0. 0753 kJ K -1 mol−1)=+37 K

From ΔH=nΔvap HΘ=mM

Δvap HΘ

m=M×ΔHΔvap HΘ

=(0 . 018 kg mol−1)×( 1. 0×104 kJ )44 .0kJ mol−1

=4 . 09 kg

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 μ is the ratio of temperature change to pressure change under conditions of isenthalpic expansion. So

μ=(∂T∂ p )

H≈ ΔT

Δp= −10 K

(1. 00−22 ) atm=0. 48 K atm−1

35

3.11(b) For a van der Waals gas,

πT=a

Vm2 . Calculate

ΔU 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:

Um=Um (T,V m ) dU m=(dUm

∂ T )V m

dT+(∂U m

∂V m) dV m

dT=0 in an isothermal process, so

dU m=(∂Um

∂V m)T

dV m=aV

m2

dV m

ΔU m=∫V m1m

V m2dUm =∫V m1

m

V m2 aV

m2

dV m =a∫1 .00 L mol-1

22. 1 L mol-1 dV m

Vm2

=−aV

m2

|1. 00 L mol-122. 1L mol-1

=−a22. 1L mol-1 +a

1 . 00 L mol-1 =21 . 1 a22 . 1L mol-1 =0. 954 75 L mol -1

a=1 .345 atm L2 mol−2

ΔU m =(0 . 95475 mol L−1 )¿ (1. 345 atm L2 mol−2)=( 0 .28 41 atm L2 mol−2)¿ (1 . 01325×105 Pa atm−1 )

¿ (1 .01325×105 Pa atm−1 )¿(1m3

103 L )=130P am3 mol-1=130 J mol−1

w=∫ pex d V m and p=RTV m−b

−aV m2

for a van der Waals gas

so w=−∫(RTV m−b )dV m +∫ a

V m2

dV m =−q+ΔU m

Thus

q =+∫1 .00 L mol-1

22. 1 L mol-1 (RTV m−b )dV m =+ RT ln (V m−b )|

1. 00 L mol−122. 1 L mol−1

=+ (8 . 314 J K-1 mol-1 )¿ (298 K )×ln {22 . 1−3 . 22×10−2

1 . 00−3 . 22×10−2 } =+ 7 . 74 69 kJ mol -1

w=−774 7 J mol -1+130 J mol-1 =−761 7 J mol-1 =−7 . 62 kJ mol -1

36

3.12(b) The volume of a certain liquid varies with temperature as

V=V ' {0 .77+3.7×10−4 (T/K )+1 .52×10−6 (T/K )2}

Where V ' is its volume at 298 K. Calculate its expansion coefficient, α , at 310 K

Solution:

The expansion coefficient is

α=1V (∂V

∂T )p=

V' (3 . 7×10−4 K−1+2×1 . 52×10−6 T K−2)V

¿V' [3 .7×10−4 K−1+2×1 .52×10−6 ( T/K ) ] K−1

V ' [0 .77+3 .7×10−4 (T/K )+1.52×10−6 (T/K )2]

¿[3 .7×10−4 K−1+2×1 .52×10−6 (310 ) ] K−1

0 .77+3 .7×10−4 (310 )+1 .52×10−6 (310 )2=1 .27×10−3 K−1

37