Chapter 6 - Section B - Non-Numerical Solutions

6.1 By Eq. (6.8),

(∂ H

∂S

)P

= T and isobars have positive slope

Differentiate the preceding equation:

(∂2 H

∂S2

)P

=(

∂T

∂S

)P

Combine with Eq. (6.17):

(∂2 H

∂S2

)P

= T

CPand isobars have positive curvature.

6.2 (a) Application of Eq. (6.12) to Eq. (6.20) yields:(∂CP

∂ P

)T

=[∂{V − T (∂V/∂T )P}

∂T

]P

or(

∂CP

∂ P

)T

=(

∂V

∂T

)P

− T

(∂2V

∂T 2

)P

−(

∂V

∂T

)P

Whence,(

∂CP

∂ P

)T

= −T

(∂2V

∂T 2

)P

For an ideal gas:(

∂V

∂T

)P

= R

Pand

(∂2V

∂T 2

)P

= 0

(b) Equations (6.21) and (6.33) are both general expressions for d S, and for a given change of stateboth must give the same value of d S. They may therefore be equated to yield:

(CP − CV )dT

T=

(∂ P

∂T

)V

dV +(

∂V

∂T

)P

d P

Restrict to constant P: CP = CV + T

(∂ P

∂T

)V

(∂V

∂T

)P

By Eqs. (3.2) and (6.34):

(∂V

∂T

)P

= βV and

(∂ P

∂T

)V

= β

κ

Combine with the boxed equation: CP − CV = βT V

(β

κ

)

6.3 By the definition of H , U = H − PV . Differentiate:

(∂U

∂T

)P

=(

∂ H

∂T

)P

− P

(∂V

∂T

)P

or

(∂U

∂T

)P

= CP − P

(∂V

∂T

)P

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Substitute for the final derivative by Eq. (3.2), the definition of β:

(∂U

∂T

)P

= CP − β PV

Divide Eq. (6.32) by dT and restrict to constant P . The immediate result is:

(∂U

∂T

)P

= CV +[

T

(∂ P

∂T

)V

− P

] (∂V

∂T

)P

Solve for the two derivatives by Eqs. (6.34) and (3.2); substitution gives:

(∂U

∂T

)P

= CV + β

κ(βT − κ P)V

6.4 (a) In general, dU = CV dT +[

T

(∂ P

∂T

)V

− P

]dV (6.32)

By the equation of state, P = RT

V − bwhence

(∂ P

∂T

)V

= R

V − b= P

T

Substituting this derivative into Eq. (6.32) yields dU = CV dT , indicating that U = f (T ) only.

(b) From the definition of H , d H = dU + d(PV )

From the equation of state, d(PV ) = R dT + b d P

Combining these two equations and the definition of part (a) gives:

d H = CV dT + R dT + b d P = (CV + R)dT + b d P

Then,

(∂ H

∂T

)P

= CV + R

By definition, this derivative is CP . Therefore CP = CV + R. Given that CV is constant, thenso is CP and so is γ ≡ CP/CV .

(c) For a mechanically reversible adiabatic process, dU = dW . Whence, by the equation of state,

CV dT = −P dV = − RT

V − bdV = −RT

d(V − b)

V − b

ordT

T= − R

CVd ln(V − b)

But from part (b), R/CV = (CP − CV )/CV = γ − 1. Then

d ln T = −(γ − 1)d ln(V − b) or d ln T + d ln(V − b)γ−1 = 0

From which: T (V − b)γ−1 = const.

Substitution for T by the equation of state gives

P(V − b)(V − b)γ−1

R= const. or P(V − b)γ = const.

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6.5 It follows immediately from Eq. (6.10) that:

V =(

∂G

∂ P

)T

and S = −(

∂G

∂T

)P

Differentation of the given equation of state yields:

V = RT

Pand S = − d(T )

dT− R ln P

Once V and S (as well as G) are known, we can apply the equations:

H = G + T S and U = H − PV = H − RT

These become:

H = (T ) − Td(T )

dTand U = (T ) − T

d(T )

dT− RT

By Eqs. (2.16) and (2.20),

CP =(

∂ H

∂T

)P

and CV =(

∂U

∂T

)V

Because is a function of temperature only, these become:

CP = −Td2

dT 2and CV = −T

d2

dT 2− R = CP − R

The equation for V gives the ideal-gas value. The equations for H and U show these properties to befunctions of T only, which conforms to ideal-gas behavior. The equation for S shows its relation toP to be that of an ideal gas. The equations for CP and CV show these properties to be functions of Tonly, which conforms to ideal-gas behavior, as does the result, CP = CV + R. We conclude that thegiven equation of state is consistent with the model of ideal-gas behavior.

6.6 It follows immediately from Eq. (6.10) that:

V =(

∂G

∂ P

)T

and S = −(

∂G

∂T

)P

Differentation of the given equation of state yields:

V = K and S = − d F(T )

dT

Once V and S (as well as G) are known, we can apply the equations:

H = G + T S and U = H − PV = H − P K

These become:

H = F(T ) + K P − Td F(T )

dTand U = F(T ) − T

d F(T )

dT

By Eqs. (2.16) and (2.20),

CP =(

∂ H

∂T

)P

and CV =(

∂U

∂T

)V

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Because F is a function of temperature only, these become:

CP = −Td2 F

dT 2and CV = −T

d2 F

dT 2= CP

The equation for V shows it to be constant, independent of both T and P . This is the definition ofan incompressible fluid. H is seen to be a function of both T and P , whereas U , S, CP , and CV arefunctions of T only. We also have the result that CP = CV . All of this is consistent with the model ofan incompressible fluid, as discussed in Ex. 6.2.

6.11 Results for this problem are given in the text on page 213 by Eqs. (6.59) and (6.60) for G R and H R .Eq. (6.45), page 206, then yields SR .

6.12 Parameter values for the van der Waals equation are given by the first line of Table 3.1, page 99. Atthe bottom of page 214, it is shown that I = β/Z . Equation (6.63b) therefore becomes:

G R

RT= Z − 1 − ln(Z − β) − qβ

Z

For given T and P , Z is found by solution of Eq. (3.49) for a vapor phase or Eq. (3.53) for a liquidphase with σ = ε = 0. Equations (3.50) and (3.51) for the van der Waals equation are:

β = Pr

8Trand q = 27

8Tr

With appropriate substitutions, Eqs. (6.64) and (6.65) become:

H R

RT= Z − 1 − qβ

Zand

SR

R= ln(Z − β)

6.13 This equation does not fall within the compass of the generic cubic, Eq. (3.41); so we start anew. First,multiply the given equqation of state by V/RT :

PV

RT= V

V − bexp

( −a

V RT

)

Substitute: Z ≡ PV

RTV = 1

ρ

a

bRT≡ q

Then, Z = 1

1 − bρexp(−qbρ)

With the definition, ξ ≡ bρ, this becomes:

Z = 1

1 − ξexp(−qξ) (A)

Because ρ = P/Z RT , ξ = bP

Z RT

Given T and P , these two equations may be solved iteratively for Z and ξ .

Because b is a constant, Eqs. (6.57) and (6.58) may be rewritten as:

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G R

RT=

∫ ξ

0(Z − 1)

dξ

ξ+ Z − 1 − ln Z (B)

H R

RT=

∫ ξ

0

(∂ Z

∂T

)ξ

dξ

ξ+ Z − 1 (C)

In these equations, Z is given by Eq. (A), from which is also obtained:

ln Z = − ln(1 − ξ) − qξ and

(∂ Z

∂T

)ξ

= qξ

T (1 − ξ)exp(−qξ)

The integrals in Eqs. (B) and (C) must be evaluated through the exponential integral, E(x), a specialfunction whose values are tabulated in handbooks and are also found from such software packages asMAPLE R©. The necessary equations, as found from MAPLE R©, are:

∫ ξ

0(Z − 1)

dξ

ξ= exp(−q){E[−q(1 − ξ)] − E(−q)} − E(qξ) − ln(qξ) − γ

where γ is Euler’s constant, equal to 0.57721566. . . .

and −T∫ ξ

0

(∂ Z

∂T

)ξ

sξ

ξ= q exp(−q){E[−q(1 − ξ)] − E(−q)}

Once values for G R/RT and H R/RT are known, values for SR/R come from Eq. (6.45). The diffi-culties of integration here are one reason that cubic equations have found greater favor.

6.18 Assume the validity for purposes of interpolation of Eq. (6.70), and write it for T2, T , and T1:

ln P sat2 = A − B

T2(A)

ln P sat = A − B

T(B)

ln P sat1 = A − B

T1(C)

Subtract (C) from (A): lnP sat

2

P sat1

= B

(1

T1− 1

T2

)= B

(T2 − T1)

T1T2

Subtract (C) from (B): lnP sat

P sat1

= B

(1

T1− 1

T

)= B

(T − T1)

T1T

The ratio of these two equations, upon rearrangement, yields the required result.

6.19 Write Eq. (6.70) in log10 form: log P sat = A − B

T(A)

Apply at the critical point: log Pc = A − B

Tc(B)

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By difference, log P satr = B

(1

Tc− 1

T

)= B

(Tr − 1

T

)(C)

If P sat is in (atm), then application of (A) at the normal boiling point yields:

log 1 = A − B

Tnor A = B

Tn

With θ ≡ Tn/Tc, Eq. (B) can now be written:

log Pc = B

(1

Tn− 1

Tc

)= B

(Tc − Tn

TnTc

)= B

(1 − θ

Tn

)

Whence, B =(

Tn

1 − θ

)log Pc

Equation (C) then becomes:

log P satr =

(Tn

1 − θ

) (Tr − 1

T

)log Pc =

(θ

1 − θ

) (Tr − 1

Tr

)log Pc

Apply at Tr = 0.7: log(P satr )Tr =0.7 = − 3

7

(θ

1 − θ

)log Pc

By Eq. (3.45), ω = −1.0 − log(P satr )Tr =0.7

Whence, ω = 3

7

(θ

1 − θ

)log Pc − 1

6.83 The slopes of isobars and isochores on a T S diagram are given by Eqs. (6.17) and (6.30):(∂T

∂S

)P

= T

CPand

(∂T

∂S

)V

= T

CV

Both slopes are necessarily positive. With CP > CV , isochores are steeper.

An expression for the curvature of isobars results from differentiation of the first equation above:(

∂2T

∂S2

)P

= 1

CP

(∂T

∂S

)P

− T

C2P

(∂CP

∂S

)P

= T

C2P

− T

C2P

(∂CP

∂T

)P

(∂T

∂S

)P

= T

C2P

[1 − T

CP

(∂CP

∂T

)P

]

With CP = a + bT ,

(∂CP

∂T

)P

= b and 1 − T

CP

(∂CP

∂T

)P

= 1 − bT

a + bT= a

a + bT

Because this quantity is positive, so then is the curvature of an isobar.

6.84 Division of Eq. (6.8) by d S and restriction to constant T yields:(∂ H

∂S

)T

= T + V

(∂ P

∂S

)T

By Eq. (6.25),

(∂ P

∂S

)T

= −1

βV

Therefore,

(∂ H

∂S

)T

= T − 1

β= 1

β(βT − 1)

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Also,

(∂2 H

∂S2

)T

= 1

β2

(∂β

∂S

)T

= 1

β2

(∂β

∂ P

)T

(∂ P

∂S

)T

= 1

β2

(∂β

∂ P

)T

( −1

βV

)

Whence,

(∂2 H

∂S2

)T

= − 1

β3V

(∂β

∂ P

)T

By Eqs. (3.2) and (3.37): β = 1

V

(∂V

∂T

)P

and V = RT

P+ B

Whence,

(∂V

∂T

)P

= R

P+ d B

dTand β = 1

V

(R

P+ d B

dT

)

Differentiation of the second preceding equation yields:(

∂β

∂ P

)T

= − R

V P2−

(R

P+ d B

dT

)1

V 2

(∂V

∂ P

)T

= − R

V P2− (βV )

1

V 2

(∂V

∂ P

)T

From the equation of state,(

∂V

∂ P

)T

= − RT

P2

Whence,

(∂β

∂ P

)T

= − R

V P2+ β

V

RT

P2= R

V P2(βT − 1)

Clearly, the signs of quantity (βT − 1) and the derivative on the left are the same. The sign is deter-mined from the relation of β and V to B and d B/dT :

βT − 1 = T

V

(R

P+ d B

dT

)− 1 =

RT

P+ T

d B

dTRT

P+ B

− 1 =T

d B

dT− B

RT

P+ B

In this equation d B/dT is positive and B is negative. Because RT/P is greater than |B|, the quantityβT − 1 is positive. This makes the derivative in the first boxed equation positive, and the secondderivative in the second boxed equation negative.

6.85 Since a reduced temperature of Tr = 2.7 is well above ”normal” temperatures for most gases, weexpect on the basis of Fig. 3.11 that B is (−) and that d B/dT is (+). Moreover, d2 B/dT 2 is (−).

By Eqs. (6.53) and (6.55), G R = B P and SR = −P(d B/dT )

Whence, both G R and SR are (−). From the definition of G R , H R = G R + T SR , and H R is (−).

By Eqs. (3.37) and (6.40), V R = B, and V R is (−).

Combine the equations above for G R , SR , and H R:

H R = P

(B − T

d B

dT

)Whence,

(∂ H R

∂T

)P

= P

(d B

dT− T

d2 B

dT 2− d B

dT

)= −PT

d2 B

dT 2

Therefore, C RP ≡

(∂ H R

∂T

)P

is (+). (See Fig. 6.5.)

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