Chapter6 B

7/30/2019 Chapter6 B

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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:

2H

S2

P

=

T

S

P

Combine with Eq. (6.17):

2H

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 Por

CP

P

T

=

V

T

P

T

2V

T2

P

V

T

P

Whence,

CP

P

T

= T

2V

T2

P

For an ideal gas:

V

T

P

=R

Pand

2V

T2

P

= 0

(b) Equations (6.21) and (6.33) are both general expressions for d S, and for a given change of state

both must give the same value ofd S. They may therefore be equated to yield:

(CP CV)d T

T=

P

T

V

d V+

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

= V and

P

T

V

=

Combine with the boxed equation: CP CV = T V

6.3 By the definition of H, U = H P V. 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 :

U

T

P

= CP P V

Divide Eq. (6.32) by d T 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 +

(T P)V

6.4 (a) In general, dU = CV

d T+ T P

TV P d V (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 d T, indicating that U = f(T)

only.

(b) From the definition ofH, d H = dU + d(P V)

From the equation of state, d(P V) = R d T+ b d P

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

d H = CVd T+ R d T+ b d P = (CV + R)d T+ b d P

Then,

H

T

P

= CV + R

By definition, this derivative is CP . Therefore CP = CV + R. Given that CV is constant, then

so is CP and so is CP /CV.

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

CVd T = P d V = RT

V bd V = RT

d(V b)

V b

ord T

T=

R

CV

dln(V b)

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

dln T = ( 1)dln(V b) or dln T+ dln(V b)1 = 0

From which: T(V b)1 = const.

Substitution for T by the equation of state gives

P(V b)(V b)1

R= const. or P(V b) = 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)

d T 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 P V = H RT

These become:

H = (T) Td(T)

d Tand U = (T) T

d(T)

d T 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

d T2and CV = T

d2

d T2 R = CP R

The equation for V gives the ideal-gas value. The equations for H and U show these properties to

be functions ofT only, which conforms to ideal-gas behavior. The equation for S shows its relation

to P to be that of an ideal gas. The equations for CP and CV show these properties to be functions

of T only, which conforms to ideal-gas behavior, as does the result, CP = CV + R. We conclude

that the given 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)

d T

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

H = G + T S and U = H P V = H P K

These become:

H = F(T) + K P Td F(T)

d Tand U = F(T) T

d F(T)

d T

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

d T2and CV = T

d2 F

d T2= CP

The equation for V shows it to be constant, independent of both T and P . This is the definition of

an incompressible fluid. H is seen to be a function of both T and P , whereas U, S, CP , and CV arefunctions ofT only. We also have the result that CP = CV. All of this is consistent with the model

of an incompressible fluid, as discussed in Ex. 6.2.

6.11 Results for this problem are given in the text on page 217 by Eqs. (6.61), (6.62) and (6.63) for GR ,

HR , and SR respectively.

6.12 Parameter values for the van der Waals equation are given by the first line of Table 3.1, page 98. At

the bottom of page 215, it is shown that I = /Z. Equation (6.66b) therefore becomes:

GR

RT= Z 1 ln(Z )

q

Z

For given T and P , Z is found by solution of Eq. (3.52) for a vapor phase or Eq. (3.56) for a liquid

phase with = = 0. Equations (3.53) and (3.54) for the van der Waals equation are:

=Pr

8Trand q =

27

8Tr

With appropriate substitutions, Eqs. (6.67) and (6.68) become:

HR

RT= Z 1

q

Zand

SR

R= ln(Z )

6.13 This equation does not fall within the compass of the generic cubic, Eq. (3.42); so we start anew.

First, multiply the given equation of state by V/RT:

P V

RT=

V

V bexp

a

V R T

Substitute: Z P V

RTV =

1

a

b RT q

Then, Z=1

1 bexp(qb)

With the definition, b, this becomes:

Z=1

1 exp(q ) (A)

Because = P/Z RT, =b P

Z RT

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

Because b is a constant, Eqs. (6.58) and (6.59) may be rewritten as:

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GR

RT=

0

(Z 1)d

+ Z 1 ln Z (B)

HR

RT=

0

Z

T

d

+ Z 1 (C)

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

ln Z = ln(1 ) q and

Z

T

=q

T(1 )exp(q )

The integrals in Eqs. (B) and (C) must be evaluated through the exponential integral, E(x ), a special

function whose values are tabulated in handbooks and are also found from such software packages

as MAPLE R. The necessary equations, as found from MAPLE R, are:0

(Z 1)d

= exp(q){E[q(1 )] E(q)} E(q ) ln(q )

where is Eulers constant, equal to 0.57721566. . . .

and T

0

Z

T

s

= q exp(q){E[q(1 )] E(q)}

Once values for GR/RT and HR /RT are known, values for SR /R come from Eq. (6.47). The

difficulties of integration here are one reason that cubic equations have found greater favor.

6.18 Assume the validity for purposes of interpolation of Eq. (6.75), 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 sat2P sat1

= B

1

T1

1

T2

= B

(T2 T1)

T1T2

Subtract (C) from (B): lnP sat

Psat

1

= 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. (6.75) in log10 form: log Psat = 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:

log1 = A B

Tnor A =

B

Tn

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

log Pc = B

1

Tn

1

Tc

= B

Tc Tn

Tn Tc

= B

1

Tn

Whence, B =

Tn

1

log Pc

Equation (C) then becomes:

log P satr =

Tn

1

Tr 1

T

log Pc =

1

Tr 1

Tr

log Pc

Apply at Tr = 0.7: log(Psat

r )Tr=0.7 = 3

7

1

log Pc

By Eq. (3.48), = 1.0 log(P satr )Tr=0.7

Whence, =3

7

1

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

= TCP

and T S

V

= TCV

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

V

Therefore,

H

S

T

= T1

=

1

( T 1)

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Also,

2H

S2

T

=1

2

S

T

=1

2

P

T

P

S

T

=1

2

P

T

1

V

Whence,

2H

S2

T

= 1

3V

P

T

By Eqs. (3.2) and (3.38): =1

V

V

T

P

and V =RT

P+ B

Whence,

V

T

P

=R

P+

d B

d Tand =

1

V

R

P+

d B

d T

Differentiation of the second preceding equation yields:

P

T

= R

V P2

R

P+

d B

d T

1

V2

V

P

T

= R

V P 2 ( V)

1

V2

V

P

T

From the equation of state, V

P

T

= RT

P 2

Whence,

P

T

= R

V P 2+

V

RT

P2=

R

V P2(T 1)

Clearly, the signs of quantity ( T 1) and the derivative on the left are the same. The sign is

determined from the relation of and V to B and d B/d T:

T 1 =

T

V R

P +

d B

d T 1 =RT

P+ T

d B

d T

RT

P+ B 1 =

Td B

d T B

RT

P+ B

In this equation d B/d T is positive and B is negative. Because RT/P is greater than |B|, the quantity

T 1 is positive. This makes the derivative in the first boxed equation positive, and the second

derivative in the second boxed equation negative.

6.85 Since a reduced temperature of Tr = 2.7 is well above normal temperatures for most gases, we

expect on the basis of Fig. 3.10 that B is () and that d B/d T is (+). Moreover, d2B/d T2 is ().

By Eqs. (6.54) and (6.56), GR = B P and SR = P(d B/d T)

Whence, both GR and SR are (). From the definition ofGR , HR = GR + T SR , and HR is ().

By Eqs. (3.38) and (6.40), VR = B, and VR is ().

Combine the equations above for GR , SR , and HR:

HR = P

B T

d B

d T

Whence,

HR

T

P

= P

d B

d T T

d2B

d T2

d B

d T

= P T

d2B

d T2

Therefore, CRP

HR

T

P

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

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6.89 By Eq. (3.5) at constant T: P =1

ln

V

V1 P1 (A)

(a) Work d W = P d V =

1

ln

V

V1 P1

d V =

1

ln V d V

P1 +

1

ln V1

d V

W = 1V2

V1

ln V d V P1 + 1

ln V1 (V2 V1)W =

1

[(V2 ln V2 V2) (V1 ln V1 V1)] P1(V2 V1)

1

(V2 ln V1 V1 ln V1)

=1

V2 ln

V2

V1+ V1 V2

P1(V2 V1)

By Eq. (3.5), lnV2

V1= (P2 P1) whence W = P1V1 P2V2

V2 V1

(b) Entropy By Eq. (6.29), d S= V d P

By Eq. (A), P =ln V

ln V1

P1 and d P =

1

dln V

d S= V

dln V =

d V and S=

(V2 V1)

(c) Enthalpy By Eq. (6.28), d H = (1 T)V d P

Substitute for d P: d H = (1 T)V 1

dln V =

1 T

d V

H =1 T

(V1 V2)

These equations are so simple that little is gained through use of an average V. For the conditions

given in Pb. 6.9, calculations give:

W = 4.855 kJ kg1 S= 0.036348 kJ kg1 K1 H = 134.55 kJ kg1

6.90 The given equation will be true if and only if

M

P Td P = 0

The two circumstances for which this condition holds are when (M/ P)T = 0 or when d P = 0.

The former is a property feature and the latter is a process feature.

6.91

Hig

P

V

=

Hig

P

T

+

Hig

T

P

T

P

V

= Cig

P

T

P

V

Neither Cig

P nor ( T/ P)V is in general zero for an ideal gas.Hig

P

S

=

Hig

P

T

+

Hig

T

P

T

P

S

= Cig

P

T

P

S

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T

P

S

=

T

Sig

P

Sig

P

T

=T

Cig

P

Sig

P

T

Hig

P

S

= T

Sig

P

T

Neither T nor ( Sig

/ P)T is in general zero for an ideal gas. The difficulty here is that theexpression independent of pressure is imprecise.

6.92 For S= S(P, V): d S=

S

P

V

d P +

S

V

P

d V

By the chain rule for partial derivatives,

d S=

S

T

V

T

P

V

d P +

S

T

P

T

V

P

d V

With Eqs. (6.30) and (6.17), this becomes:

d S= CVT T

P

V

d P + CPT T

V

P

d V

6.93 By Eq. (6.31), P = T

P

T

V

U

V

T

(a) For an ideal gas, P =RT

Vand

P

T

V

=R

V

ThereforeRT

V=

RT

V

U

V

T

and

U

V

T

= 0

(b) For a van der Waals gas, P =RT

V b

a

V2and

P

T

V

=R

V b

ThereforeRT

V b

a

V2=

RT

V b

U

V

T

and

U

V

T

=a

V2

(c) Similarly, for a Redlich/Kwong fluid find:

U

V

T

=(3/2)A

T1/2V(V+ b)

where A = a(Tc) T12

c

6.94 (a) The derivatives ofG with respect to T and P follow from Eq, (6.10):

S=

G

T

P

and V =

G

P

T

Combining the definition of Z with the second of these gives:

Z P V

RT=

P

RT

G

P

T

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Combining Eqs. (2.11) and (3.63) and solving for U gives U = G + T S P V.

Replacing S and V by their derivatives gives: U = G T

G

T

P

P

G

P

T

Developing an equation for CV

is much less direct. First differentiate the above equation for U

with respect to T and then with respect to P: The two resulting equations are:U

T

P

= T

2G

T2

P

P

2G

T P

U

P

T

= T

2G

T P

P

2G

P 2

T

From the definition ofCV and an equation relating partial derivatives:

CV

U

T

V

=

U

T

P

+

U

P

T

P

T

V

Combining the three equations yields:

CV = T

2G

T2

P

P

2G

T P

T

2G

T P

+ P

2G

P 2

T

P

T

V

Evaluate ( P/ T)V through use of the chain rule: P

T

V

=

P

V

T

V

T

P

=( V/ T)P

( V/ P)T

The two derivatives of the final term come from differentiation of V = ( G/ P)T:

V TP = 2G

P T and V PT = 2G

P 2TThen

P

T

V

=(2G/ T)P

(2G/ P 2)T

and CV = T

2G

T2

P

P

2G

T P

+

T

2G

T P

+ P

2G

P2

T

(2G/ P T)

(2G/ P2)T

Some algebra transforms this equation into a more compact form:

CV = T

2G

T2

P

+ T(2G/ T P)2

( 2G/ P 2)T

(b) The solution here is analogous to that of part (a), but starting with the derivatives inherent in

Eq. (6.9).

6.97 Equation (6.74) is exact:dln P sat

d(1/T)=

Hlv

RZlv

The right side is approximately constant owing to the qualitatively similar behaviior of Hlv and

Zlv. Both decrease monotonically as T increases, becoming zero at the critical point.

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6.98 By the Clapeyron equation:d P sat

d T=

Ssl

Vsl=

Hsl

TVsl

If the ratio Ssl to Vsl is assumed approximately constant, then

P sat = A + BT

If the ratio Hsl to Vsl is assumed approximately constant, then

P sat = A + B ln T

6.99 By Eq, (6.73) and its analog for sv equilibrium:d P satsv

d T

t

=PtH

svt

RT2t Zsvt

PtH

svt

RT2td P satlv

d T

t

=PtH

lvt

RT2t Zlvt

PtH

lvt

RT2td P satsvd T

t

d P satlv

d T

t

Pt

RT2t

Hsvt H

lvt

Because

Hsvt H

lvt

= Hslt is positive, then so is the left side of the preceding equation.

6.100 By Eq. (6.72):d P sat

d T=

Hlv

TVlv

But Vlv =RT

P satZlv whence

dln P sat

d T=

Hlv

RT2Zlv(6.73)

dln P satr

d Tr=

TcHlv

RT2Zlv=

Hlv

RTc

1

T2r Zlv=

HlvT2r Z

lv

6.102 Convert c to reduced conditions:

c

dln P sat

dln T

T=Tc

=

dln P satr

dln Tr

Tr=1

= Tr

dln P satr

d Tr

Tr=1

=

dln Psatr

d Tr

Tr=1

From the Lee/Kesler equation, find that

dln Psat

r

d Tr Tr=1 = 5.8239 + 4.8300 Thus, c(L/K) = 5.82 for = 0, and increases with increasing molecular complexity as quantified

by .

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Chapter6 B