Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap03

Temperature, pressure, volume, amount, energy, enthalpy, heat capacity, expansion coefficient, isothermal compressibility, and Joule–Thomson coefficient.. E3.2b One can use the general e

Solutions to exercises

Discussion questions

E3.1(b) The following list includes only those state functions that we have encountered in the first three

chapters More will be encountered in later chapters

Temperature, pressure, volume, amount, energy, enthalpy, heat capacity, expansion coefficient, isothermal compressibility, and Joule–Thomson coefficient

E3.2(b) One can use the general expression forπ T given in Justification 3.3 to derive its specific form for

a van der Waals gas as given in Exercise 3.14(a), that is,π T = a/V2

m (The derivation is carried out in Example 5.1.) For an isothermal expansion in a van der Waals gas dUm = (a/Vm)2 Hence

Um = −a(1/Vm,2 − 1/Vm,1 ) See this derivation in the solution to Exercise 3.14(a) This formula

corresponds to what one would expect for a real gas As the molecules get closer and closer the molar volume gets smaller and smaller and the energy of attraction gets larger and larger

E3.3(b) The solution to Problem 3.23 shows that the Joule–Thomson coefficient can be expressed in terms

of the parameters representing the attractive and repulsive interactions in a real gas If the attractive forces predominate then expanding the gas will reduce its energy and hence its temperature This reduction in temperature could continue until the temperature of the gas falls below its condensation point This is the principle underlying the liquefaction of gases with the Linde Refrigerator which utilizes the Joule–Thomson effect See Section 3.4 for a more complete discussion

Numerical exercises

E3.4(b) A function has an exact differential if its mixed partial derivatives are equal That is,f (x, y) has an

exact differential if

∂

∂x

∂f

∂y



= ∂

∂y

∂f

∂x



(a) ∂f

∂x = 3x2y2 and

∂

∂y

∂f

∂x



= 6x2y

∂f

∂y = 2x3y and

∂

∂x

∂f

∂y



= 6x2y Therefore, exact

(b) ∂f

∂s = te s+ 1 and

∂

∂t

∂f

∂s



= es

∂f

∂t = 2t + e s and

∂

∂s

∂f

∂t



= es Therefore, exact

E3.5(b) dz = ∂z

∂xdx +

∂z

∂y dy =

dx (1 + y)2 − 2x dy

(1 + y)3

E3.6(b) (a) dz = ∂x ∂zdx + ∂y ∂zdy = (3x2− 2y2) dx − 4xy dy

(b) ∂2z

∂y∂x =

∂

∂y (3x2− 2y2) = −4y

and ∂2z

∂x∂y =

∂

∂x (−4xy) = −4y

E3.7(b) dz = ∂z

∂xdx +

∂z

∂ydy = (2xy + y2) dx + (x2+ 2xy) dy

∂2z

∂y∂x =

∂

∂y (2xy + y2) = 2x + 2y

and ∂2z

∂x∂y =

∂

∂x (x2+ 2xy) = 2x + 2y

E3.8(b)

∂C p

∂p





∂

∂p

∂H

∂T



p



T

= ∂2H

∂p∂T



∂

∂T

∂H

∂p



T



p

Because

∂H

∂p



T = 0 for a perfect gas, its temperature derivative also equals zero; thus

∂C

p

∂p



T = 0

E3.9(b)

∂H

∂U



p= (∂H/∂V ) (∂U/∂V ) p



∂(U+pV )

∂V



p

(∂U/∂V ) p =

(∂U/∂V ) p + p

(∂U/∂V ) p = 1 +

p (∂U/∂V ) p

E3.10(b) dp =

∂p

∂V



T dV +

∂p

∂T



V dT

d lnp = dp p = p1

∂p

∂V



T dV + p1

∂p

∂T



V dT

We express

∂p

∂V



T in terms of the isothermal compressibilityκ T

κ T = −1

V

∂V

∂p



T = −



V

∂p

∂V



T

−1 so

∂p

∂V



κ T V

We express

∂p

∂T



V in terms ofκ T and the expansion coefficientα = V1

∂V

∂T



p

∂p

∂T



V

∂T

∂V



p

∂V

∂p



T = −1 so

∂p

∂T



V = −(∂V /∂T ) p

(∂V /∂p) T =

α

κ T

so d lnp = − pV κ1

T +pκ α

T = pκ1

T

α dT −dV V



E3.11(b) U =

3 2



nRT so

∂U

∂p



T = 0 by direct differentiation

H = U + pV =3

2



nRT + nRT =5

2



nRT ,

so

∂H

∂p



T = 0 by direct differentiation

E3.12(b) α =

1

V



∂V

∂T



p

∂V

∂T



p = nR

p =

V T

α =

1

V



×

V

T



= 1

T

κ T = −

1

V



∂V

∂p



T

∂V

∂p



T = −nRT

p2

κ T = −

1

V



×

−nRT

p2



= 1

p

E3.13(b) The Joule–Thomson coefficient µ is the ratio of temperature change to pressure change under

conditions of isenthalpic expansion So

µ =

∂T

∂p



p =

−10 K

(1.00 − 22) atm = 0.48 K atm−1

E3.14(b) Um = Um(T , Vm) dUm=

dUm

∂T



Vm

dT +

∂Um

∂Vm



dVm

dT = 0 in an isothermal process, so

dUm=

∂Um

∂Vm



T dVm = a

V2 m

dVm

Um =

Vm2

Vm1

dUm=

Vm2

Vm1

a

V2 m

dVm= a

22.1 L mol−1

1.00 L mol−1

dVm

V2 m

= − a

Vm

22.1 L mol−1

1.00 L mol−1

22.1 L mol−1 + a

1.00 L mol−1 = 21.1a

22.1 L mol−1 = 0.95475a L−1mol

a = 1.337 atm L2mol−2

Um= (0.95475 mol L−1) × (1.337 atm L2mol−2)

= (1.2765 atm L mol−1) × (1.01325 × 105Pa atm−1) ×

1 m3

103L

= 129 Pa m3mol−1 = 129 J mol−1

w = −

pexdVm and p = RT

Vm− b −

a

V2 m

for a van der Waals gas

sow = −

RT

Vm− b



dVm+

V2 m

dVm= −q + Um

Thus

q = +

22.1 L mol−1

1.00 L mol−1

RT

Vm− b



dVm = +RT ln(Vm− b) 22.1 L mol−1

1.00 L mol−1

= +(8.314 J K−1mol−1) × (298 K)

× ln 22.1 − 3.20 × 10−2

1.00 − 3.20 × 10−2



= +7.7465 kJ mol−1

w = −7747 J mol−1+ 129 J mol−1= −761¯8 J mol−1 = −7.62 kJ mol−1

E3.15(b) The expansion coefficient is

α = V1

∂V

∂T



p =V(3.7 × 10−4K−1+ 2 × 1.52 × 10 V −6T K−2)

= V[3.7 × 10−4+ 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+ 2 × 1.52 × 10−6(310)] K−1

0.77 + 3.7 × 10−4(310) + 1.52 × 10−6(310)2 = 1.27 × 10−3K−1

E3.16(b) Isothermal compressibility is

κ T = −V1

∂V

∂p



T ≈ −V p V so p = − V κ V

T

A density increase 0.08 per cent means V /V = −0.0008 So the additional pressure that must be

applied is

p = 0.0008

2.21 × 10−6atm−1 = 3.6 × 102atm

E3.17(b) The isothermal Joule–Thomson coefficient is

∂H

∂p



T = −µC p = −(1.11 K atm−1) × (37.11 J K−1mol−1) = −41.2 J atm−1mol−1

If this coefficient is constant in an isothermal Joule–Thomson experiment, then the heat which must

be supplied to maintain constant temperature is H in the following relationship

H/n

p = −41.2 J atm−1mol−1 so H = −(41.2 J atm−1mol−1)n p

H = −(41.2 J atm−1mol−1) × (12.0 mol) × (−55 atm) = 27.2 × 103J

E3.18(b) The Joule–Thomson coefficient is

µ =

∂T

∂p



p so p =

T

µ =

−4.5 K

13.3 × 10−3K kPa−1 = −3.4 × 102kPa

Solutions to problems

Assume that all gases are perfect and that all data refer to 298 K unless stated otherwise

Solutions to numerical problems

P3.1 κ T = (2.21 × 10−6atm−1) ×

1 atm

1.013 × 105Pa



= 2.18 × 10−11Pa−1

For the change of volume with pressure, we use

dV =

∂V

∂p



T dp[constant temperature] = −κ T V dp



κ T = −1

V

∂V

∂p



T



V = −κ T V p [If change in V is small compared to V ]

p = (1.03 × 103kg m−3) × (9.81 m s−2) × (1000 m) = 1.010 × 107Pa.

Consequently, sinceV = 1000 cm3= 1.0 × 10−3m3,

V ≈ (−2.18 × 10−11Pa−1) × (1.0 × 10−3m3) × (1.010 × 107Pa)

= −2.2 × 10−7m3, or −0.220 cm3 For the change of volume with temperature, we use

dV =

∂V

∂T



p dT [constant pressure] = αV dT



α = 1 V

∂V

∂T



p



V = αV T [if change in V is small compared to V ]

≈ (8.61 × 10−5K−1) × (1.0 × 10−3m3) × (−30 K)

≈ −2.6 × 10−6m3, or − 2.6 cm3

Overall, V ≈ −2.8 cm3 V = 997.2 cm3

Comment A more exact calculation of the change of volume as a result of simultaneous pressure

and temperature changes would be based on the relationship

dV =

∂V

∂p



T dp +

∂V

∂p



p dT = −κ T V dp + αV dT

This would require information not given in the problem statement

P3.5 Use the formula derived in Problem 3.25

C p,m − C V,m = λR 1

λ = 1 −

(3Vr− 1)2

4V3

rTr

which givesγ = C C p,m

V,m =C V,m C + λR

V,m = 1 + C λR

V,m

In conjunction withC V,m= 3

2R for a monatomic, perfect gas, this gives

γ = 1 +2

3λ

For a van der Waals gas Vr = V Vm

3b , Tr = T T

c = 27RbT

8a (Table 1.6) with a =

4.137 L2atm mol−2 and b = 5.16 × 10−2L mol−1 (Table 1.6) Hence, at 100◦C and 1.00 atm,

whereVm ≈RT

p = 30.6 L mol−1

Vr≈ 30.6 L mol−1

(3) × (5.16 × 10−2L mol−1) = 198

Tr≈ (27) × (8.206 × 10−2L atm K−1mol−1) × (5.16 × 10−2L mol−1) × (373 K)

(8) × (4.317 L2atm mol−2) ≈ 1.29

Hence

1

λ = 1 −

[(3) × (198) − (1)]2

(4) × (198)3× (1.29) = 1 − 0.0088 = 0.9912, λ = 1.009

γ ≈ (1) +2

3



× (1.009) = 1.67

Comment At 100◦C and 1.00 atm xenon is expected to be close to perfect, so it is not surprising

thatγ differs only slightly from the perfect gas value of 5

3

P3.7 See the solution to Problem 3.6 It does not matter whether the piston between chambers 2 and 3 is

diathermic or adiabatic as long as the piston between chambers 1 and 2 is adiabatic The answers are the same as for Problem 3.6 However, if both pistons are diathermic, the result is different The solution for both pistons being diathermic follows

See Fig 3.1

Diathermic piston

Diathermic piston

Figure 3.1

Initial equilibrium state.

n = 1.00 mol diatomic gas in each section

pi = 1.00 bar

Ti = 298 K

For each section

Vi =nRTi

pi =(1 mol) × (0.083 145 L bar K−1mol−1) × (298 K)

1.00 bar

= 24.8 L

Vtotal = 3Vi= 74.3 L = constant

Final equilibrium state The diathermic walls allow the passage of heat Consequently, at equlibrium

all chambers will have the same temperatureT1= T2= T3 = 348 K The chambers will also be at mechanical equlibrium so

p1= p2= p3= (n1+ n2+ n3)RT1

Vtotal

= (3 mol) × (0.083 145 L bar K−1mol−1) × (348 K)

74.3 L

= 1.17 bar = p2= p3

The chambers will have equal volume

3 = Vi= 24.8 L = V1= V2= V3

U1= n1C V T1= n1



5

2R T1

= (1 mol) ×5

2



× (8.314 51 J K−1mol−1) × (348 K − 298 K)

U1= 1.04 kJ = U2= U3

Utotal = 3 U1= 3.12 kJ = Utotal

Solutions to theoretical problems

∂w

∂x



y,z dx +

∂w

∂y



x,z dy +

∂w

∂z



x,y dz

dw = (y + z) dx + (x + z) dy + (x + y) dz

This is the total differential of the functionw, and a total differential is necessarily exact, but here

we will demonstrate its exactness showing that its integral is independent of path

Path a

dw = 2x dx + 2y dy + 2z dz = 6x dx

(1,1,1)

(0,0,0) dw =

1

0

6x dx = 3 Path b

dw = 2x2

dx + (y1/2 + y) dy + (z1/2 + z) dz = (2x2+ 2x + 2x1/2 ) dx

(1,1,1)

(0,0,0) dw =

1

0 (2x2+ 2x + 2x1/2 ) dx = 2

3 + 1 +4

3 = 3 Therefore, dw is exact.

P3.12 U = U(T , V )

dU =

∂U

∂T



V dT +

∂U

∂V



T dV = C V dT +

∂U

∂V



T dV

ForU = constant, dU = 0, and

C V dT = −

∂U

∂V



T dV or C V = −

∂U

∂V



T

dV

dT



∂U

∂V



T

∂V

∂T



U This relationship is essentially the permuter [Relation 3, Further information 1.7].

P3.13 H = H (T , p)

dH =

∂H

∂T



p dT +

∂H

∂p



T dp = C pdT +

∂H

∂p



T dp

ForH = constant, dH = 0, and

∂H

∂p



T dp = −C pdT

∂H

∂p



T = −C p

dT

dp



H = −C p

∂T

∂p



H = −C p µ = −µC p

This relationship is essentially the permuter [Relation 3, Further information 1.7].

P3.16 The reasoning here is that an exact differential is always exact If the differential of heat can be shown

to be inexact in one instance, then its differential is in general inexact, and heat is not a state function Consider the cycle shown in Fig 3.2

Isotherm at

Isotherm at

Isotherm at

2 3

B

A

Figure 3.2

The following perfect gas relations apply at points labelled 1, 2, 3 and 4 in Fig 3.2.

(1) p1V1= p2V2= nRT , (2) p2V1= nRT, (3) p1V2= nRT

Define T= T − T, T= T− T

Subtract (2) from (1)

−nRT+ nRT = −p2V1+ p1V1

giving T= V1(p1− p2)

RT

Subtracting (1) from (3) we obtain

T =V2(p1− p2)

RT

SinceV1 = V2, T = T

qA= C p T− C V T= (C p − C V ) T

qB= −C V T+ C p T = (C p − C V ) T

givingqA = qBandq(cycle) = qA− qB = 0

Therefore

dq = 0 and dq is not exact.

P3.18 p = p(T , V ) = V RT

a

V2 m

dp =

∂p

∂T



V dT +

∂p

∂V



T dV

In what follows adopt the notationVm= V

∂p

∂T



V − b;

∂p

∂V



(V − b)2 + 2a

V3

then, dp =

V − b



dT +



2a

V3 − RT

(V − b)2



dV

Because the van der Waals equation is a cubic inV,

∂V

∂T



pis more readily evaluated with the use

of the permuter

∂V

∂T



p = −



∂p

∂T



V



∂p

∂V



T

= − V −b R

−(V −b) RT 2 +2a

V3

 = RT V RV3− 2a(V − b)3(V − b) 2

For path a

T2,V2

T1,V1

dp =

T2

T1

R

V1− bdT +

V2

V1



− RT2

(V − b)2 + 2a

V3



dV

V1− b (T2− T1) + RT2

(V2− b) −

RT2

(V1− b) − a

1

V2 2

− 1

V2 1

= − RT1

V1− b +

RT2

V2− b − a

1

V2 2

− 1

V2 1

For path b

T2,V2

T1,V1

dp =

V2

V1



− RT1

(V − b)2 + 2a

V3



dV +

T2

T1

R

V2− bdT

= RT1

V2− b−

RT1

V1− b − a

1

V2 2

− 1

V2 1

+V R

2− b (T2− T1)

= − RT1

V1− b +

RT2

V2− b − a

1

V2 2

− 1

V2 1

Thus, they are the same and dp satisfies the condition of an exact differential, namely, that its integral

between limits is independent of path

P3.20 p = p(V, T )

Therefore,

dp =

∂p

∂V



T dV +

∂p

∂T



V dT with p = nRT

V − nb−

n2a

V2 [Table 1.6]

∂p

∂V



(V − nb)2 +2n2a

V3 = −p

V − nb+

n2a

V3

×

V − 2nb

V − nb



∂p

∂T



V =V − nb nR = T p + n2a

T V2

Therefore, upon substitution

dp =

−p dV

V − nb

 +

n2a

V3

× (V − 2nb) ×

dV

V − nb

 +

p dT

T

 +

n2a

V2

×

dT T



=

(n2a) × (V − nb)/V3− p

V − nb

dV +

p + n2a/V2

T

dT

=

a(Vm− b)/V3

m− p

Vm− b

dVm+

p + a/V2

m

T

dT

Comment This result may be compared to the expression for dp obtained in Problem 3.18.

P3.21 p = V − nb nRT −n2a

V2 (Table 1.6)

Hence T =  p

nR



× (V − nb) +

na

RV2



× (V − nb)

∂T

∂p



V = V − nb

nR =

Vm− b

1



∂p

∂T



V

For Euler’s chain relation, we need to show that

∂T

∂p



V

∂p

∂V



T

∂V

∂T



p = −1 Hence, in addition to

∂T

∂p



V and

∂p

∂V



T [Problem 3.20] we need

∂V

∂T



p = 1

∂T

∂V



p

which can be found from

∂T

∂V



p = p

nR

 +

na

RV2



−

2na

RV3



× (V − nb)

=

V − nb



−

2na

RV3



× (V − nb)

Therefore,

∂T

∂p



V

∂p

∂V



T

∂V

∂T





∂T

∂p



V



∂p

∂V



T



∂T

∂V



p

=



V −nb nR



×(V −nb) −nRT2 +2n2a

V3





T

V −nb



−2na

RV3



× (V − nb) =



−T

V −nb

 +2na

RV3



× (V − nb)



T

V −nb



−2na

RV3



× (V − nb)

= −1

P3.23 µC p = T

∂V

∂T



p − V =  T

∂T

∂V



p

− V [Relation 2, Further information 1.7]

∂T

∂V



V − nb−

2na

RV3(V − nb) [Problem 3.21]

Introduction of this expression followed by rearrangement leads to

µC p = (2na) × (V − nb)2− nbRT V2

RT V3− 2na(V − nb)2 × V

Then, introducingζ = RT V3

2na(V − nb)2 to simplify the appearance of the expression

µC p =

1−nbζ V

ζ − 1

V =

1−V bζm

ζ − 1

V

For xenon,Vm = 24.6 L mol−1, T = 298 K, a = 4.137 L2atm mol−2, b = 5.16 × 10−2L mol−1,

nb

V =

b

Vm = 5.16 × 10−2L mol−1

24.6 L mol−1 = 2.09 × 10−3

ζ = (8.206 × 10−2L atm K−1mol−1) × (298 K) × (24.6 L mol−1)3

(2) × (4.137 L2atm mol−2) × (24.6 L mol−1− 5.16 × 10−2L mol−1)2 = 73.0

Therefore,µC p= 1− (73.0) × (2.09 × 10−3)

72.0 × (24.6 L mol−1) = 0.290 L mol−1

C p = 20.79 J K−1mol−1[Table 2.6], so

µ = 0.290 L mol−1

20.79 J K−1mol−1 =0.290 × 10−3m3mol−1

20.79 J K−1mol−1

= 1.393 × 10−5K m3J−1 = 1.393 × 10−5K Pa−1

= (1.393 × 10−5) × (1.013 × 105K atm−1) = 1.41 K atm−1

The value ofµ changes at T = T1and when the sign of the numerator 1−nbζ

V changes sign (ζ − 1

is positive) Hence

bζ

Vm = 1 at T = T1 or RT1bV3

2na(V − nb)2Vm

= 1 implying that T1=2a(Vm− b)2

RbV2 m

that is,T1=

2a Rb



×

1− b

Vm

2

= 27

4 Tc

1− b

Vm

2

For xenon, 2a

Rb=

(2) × (4.137 L2atm mol−2) (8.206 × 10−2L atm K−1mol−1) × (5.16 × 10−2L mol−1) = 1954 K

and soT1= (1954 K) ×

1−5.16 × 10−2

24.6

2

= 1946 K

Question An approximate relationship forµ of a van der Waals gas was obtained in Problem 3.17.

Use it to obtain an expression for the inversion temperature, calculate it for xenon, and compare to the result above

P3.25 C p,m − C V,m =α2κ T V

T [3.21] = αT V

∂p

∂T



V [Justification 3 3]

∂p

∂T



V − nb [Problem 3.20]

αV =

∂V

∂T



p=  1

∂T

∂V



p

Substituting,

C p,m − C V,m =T



∂p

∂T



V



∂T

∂V



p

so

∂T

∂V



p= V − nb T − 2na

RV3(V − nb) [Problem 3.21]

Substituting,

C p,m − C V,m =

nRT (V −nb) T

(V −nb)−2na

RV3



× (V − nb) = nλR with λ =

1

1− 2na

RT V3



× (V − nb)2

For molar quantities,

C p,m − C V,m = λR with1λ = 1 −2a(Vm− b)2

RT V3 m

Now introduce the reduced variables and useTc = 8a

27Rb , Vc= 3b.

After rearrangement,

1

λ = 1 −

(3Vr− 1)2

4TrV3 r

For xenon,Vc = 118.1 cm3

mol−1, Tc= 289.8 K The perfect gas value for Vmmay be used as any error introduced by this approximation occurs only in the correction term for 1

λ.

Hence,Vm ≈ 2.45 L mol−1, Vc = 118.8 cm3

mol−1, Tc= 289.8 K, and Vr = 20.6 and Tr = 1.03;

therefore

1

λ = 1 −

(61.8 − 1)2

(4) × (1.03) × (20.6)3 = 0.90, giving λ ≈ 1.1

and

C p,m − C V,m ≈ 1.1R = 9.2 J K−1mol−1

P3.27 (a) µ = − C1

p

∂H

∂p



T =C1

∂V

m

∂T



p − Vm



[Justification 3 1 and Problem 3.24]

Vm =RT

p + aT2

∂Vm

∂T



p =R

p + 2aT

Solutions to problems

Assume that all gases are perfect and that all data refer to 298 K unless stated otherwise

Solutions to numerical problems... pistons are diathermic, the result is different The solution for both pistons being diathermic follows

See Fig 3.1

Diathermic piston

Diathermic piston...

3

P3.7 See the solution to Problem 3.6 It does not matter whether the piston between chambers and is

diathermic or adiabatic as long as the piston between chambers and is