Thermo 7e SM chap12 1 sloution manual

Limited distribution permitted only to teachers and educators for course Downloaded by Pham Quang Huy ebook4you.online@gmail.com... Limited distribution permitted only to teachers and e

Trang 1

Solutions Manual for

Thermodynamics: An Engineering Approach

Seventh Edition Yunus A Cengel, Michael A Boles

McGraw-Hill, 2011

Chapter 12 THERMODYNAMIC PROPERTY

RELATIONS

PROPRIETARY AND CONFIDENTIAL

This Manual is the proprietary property of The McGraw-Hill Companies, Inc (“McGraw-Hill”) and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual

should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to

authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill

PROPRIETARY MATERIAL

preparation If you are a student using this Manual, you are using it without permission.

Downloaded by Pham Quang Huy (ebook4you.online@gmail.com)

Trang 2

Partial Derivatives and Associated Relations

12-1C For functions that depend on one variable, they are identical For functions that depend on two or more variable, the

partial differential represents the change in the function with one of the variables as the other variables are held constant The ordinary differential for such functions represents the total change as a result of differential changes in all variables

12-2C (a) ( ∂x)y = dx ; (b) (∂z) y ≤ dz; and (c) dz = (∂z)x + (∂z) y

12-3C Yes

12-4C Yes

12-5 Air at a specified temperature and specific volume is considered The changes in pressure corresponding to a certain

increase of different properties are to be determined

Assumptions Air is an ideal gas

Properties The gas constant of air is R = 0.287 kPa·m3/kg·K (Table A-1)

Analysis An ideal gas equation can be expressed as P = RT/v Noting that R is a constant and P = P(T, v),

2v

vv

d T R dT R dv T

P dT T

K)K)(3.0/kgmkPa(0.287

3 3v

v

dT R dP

(b) The change in v can be expressed as dv ≅ ∆v = 1.2 × 0.01 = 0.012 m3

/kg At T = constant,

2 3

3 3

/kg)mK)(0.012K)(300

/kgmkPa(0.287

v

d T R

=+

=(dP) (dP)T 0.7175 ( 0.7175)

dP v

Thus the changes in T and v balance each other

Trang 3

12-6 Helium at a specified temperature and specific volume is considered The changes in pressure corresponding to a

certain increase of different properties are to be determined

Assumptions Helium is an ideal gas

Properties The gas constant of helium is R = 2.0769 kPa·m3/kg·K (Table A-1)

Analysis An ideal gas equation can be expressed as P = RT/v Noting that R is a constant and P = P(T,v),

2v

vv

vvv

d T R dT R d T

P dT T

K)K)(3.0/kgmkPa(2.0769

3 3v

v

dT R dP

(b) The change in v can be expressed as dv ≅ ∆v = 1.2 × 0.01 = 0.012 m3

/kg At T = constant,

2 3

3 3

)mK)(0.012K)(300

/kgmkPa(2.0769

v

d T R

=+

=(dP) (dP)T 5.192 ( 5.192)

dP v

Thus the changes in T and v balance each other

12-7 Nitrogen gas at a specified state is considered The c p and cv of the nitrogen are to be determined using Table A-18,

and to be compared to the values listed in Table A-2b

Analysis The c p and cv of ideal gases depends on temperature only, and are expressed as c p (T) = dh(T)/dT and cv(T) =

du(T)/dT Approximating the differentials as differences about 400 K, the c p and cv values are determined to be

K kJ/kg

kJ/kg.011,347)/28(11,932

K390410

K390K

410

)()

()K400

(

K 400 K

400

h h

T

T h dT

T dh c

T T

kJ/kg08,104)/28

(8,523

K390410

K390K

410

)()

()K400

(

K 400 K

400

u u

T

T u dT

T du c

T T

v

(Compare: Table A-2b at 400 K → cv = 0.747 kJ/kg·K)

Trang 4

12-8E Nitrogen gas at a specified state is considered The c p and cv of the nitrogen are to be determined using Table A-18E,

and to be compared to the values listed in Table A-2Eb

Analysis The c p and cv of ideal gases depends on temperature only, and are expressed as c p (T) = dh(T)/dT and cv(T) =

du(T)/dT Approximating the differentials as differences about 600 R, the c p and cv values are determined to be

R Btu/lbm

Btu/lbm)/28.013

2.5424(5704.7

R807820

R780R

820

)()

()R800

(

R 800 R

800

h h

T

T h dT

T dh c

T T

Btu/lbm)/28.013

2.3875(4076.3

R807820

R780R

820

)()

()R800

(

R 800 R

800

u u

T

T u dT

T du c

T T

v

(Compare: Table A-2Eb at 800 R = 340°F → cv = 0.179 Btu/lbm·R)

Trang 5

12-9 The state of an ideal gas is altered slightly The change in the specific volume of the gas is to be determined using

differential relations and the ideal-gas relation at each state

Assumptions The gas is air and air is an ideal gas

Properties The gas constant of air is R = 0.287 kPa·m3/kg·K (Table A-1)

Analysis ( a) The changes in T and P can be expressed as

kPa4kPa100)96(

K4K400)(404

=+

m(0.0117

kPa)(98

kPa)4K)(

(402kPa98

K4K)/kgmkPa(0.287

3 3

2 3

2

P

dP T R P

dT R dP P

dT T d

T P

vv

v

(b) Using the ideal gas relation at each state,

/kgm1.2078kPa

96

K)K)(404/kg

mkPa(0.287

/kgm1.1480kPa

100

K)K)(400/kg

mkPa(0.287

3 3

2

2 2

3 3

1

1 1

Trang 6

P T

where

R

a P

T R

a P

T

P

R T

a P

RT

a

P a

RT P

a

RT

P

P T

vv

v

)(

R a

P P

T T

P

P T

vv

v

which is the desired result

(b) The reciprocity rule for this gas at v = constant can be expressed as

a

R T

P a

RT

P

R

a P

T R

a P

T

P T T

vv

v v

)(

)/(1

We observe that the first differential is the inverse of the second one Thus the proof is complete

Trang 7

12-11 It is to be proven for an ideal gas that the P = constant lines on a T- diagram are straight lines and that the high

pressure lines are steeper than the low-pressure lines

which remains constant at P = constant Thus the derivative

(∂T/∂v)P , which represents the slope of the P = const lines on a T-v

diagram, remains constant That is, the P = const lines are straight

lines on a T-v diagram

(b) The slope of the P = const lines on a T-v diagram is equal to P/R,

which is proportional to P Therefore, the high pressure lines are

steeper than low pressure lines on the T-v diagram

12-12 A relation is to be derived for the slope of the v = constant lines on a T-P diagram for a gas that obeys the van der

Waals equation of state

Analysis The van der Waals equation of state can be expressed as

( b)

a P R

011

which is the slope of the v = constant lines on a T-P diagram

Trang 8

The Maxwell Relations

12-13 The validity of the last Maxwell relation for refrigerant-134a at a specified state is to be verified

Analysis We do not have exact analytical property relations for refrigerant-134a, and thus we need to replace the

differential quantities in the last Maxwell relation with the corresponding finite quantities Using property values from the tables about the specified state,

K/kgm101.602K

/kgm101.621

C0)3(70

/kgm0.029966)(0.036373

kPa00)5(900

KkJ/kg1.0309)(0.9660

C3070kPa

500900

3 4 3

4

3

?

kPa 700

C 30 C 70

?

C 50

kPa 500 kPa 900

kPa 700

?

C 50

P T

s s

T P

s

T P

s

vvv

Trang 9

LeftSide =DELTAs/DELTAP*Convert(kJ,m^3-kPa) "[m^3/kg-K]" "at T = Const."

RightSide=-DELTAv/DELTAT "[m^3/kg-K]" "at P = Const."

Trang 10

12-15E The validity of the last Maxwell relation for steam at a specified state is to be verified

Analysis We do not have exact analytical property relations for steam, and thus we need to replace the differential

quantities in the last Maxwell relation with the corresponding finite quantities Using property values from the tables about the specified state,

R/lbmft101.635R

/lbmft101.639

F00)7(900

/lbmft.6507)1(1.9777psia

0)35(450

RBtu/lbm1.7009)(1.6706

F007900psia

035450

3 3 3

3

?

psia 400

F 700 F 900

?

F 800

psia 350 psia

450

psia 400

?

F 800

P T

s s

T P

s

T P

s

vvvv

since 1 Btu ≡ 5.4039 psia·ft3, and R ≡ °F for temperature differences Thus the fourth Maxwell relation is satisfied

12-16 Using the Maxwell relations, a relation for (∂s/∂P)T for a gas whose equation of state is P(v-b) = RT is to be

a b

From the third Maxwell relation,

T

Trang 11

R T

T

Trang 12

k T

P

Analysis Using the definition of cv ,

v v v

s T T

s T

c

Substituting the first Maxwell relation

s

T P

c

s

Using the definition of c p,

P P P

p

T

s T T

s T

T T

P T

P T T c

2

Also,

v

c c

c k

P

P s

P

T T

P P

T

T T P

v

v2

T T

P T

P

T P s s

But, according to the cyclic relation, the last three terms are equal to −1 Then,

s

P T

Trang 13

12-20 It is to be shown how T, v, u, a, and g could be evaluated from the thermodynamic function h = h(s, P)

Analysis Forming the differential of the given expression for h produces

dP P

h ds s

h dh

s P

both of which can be evaluated for a given P and s

From the definition of the enthalpy,

s

P

h h-P P h

h P h Ts u

Trang 14

The Clapeyron Equation

12-21C It enables us to determine the enthalpy of vaporization from h fg at a given temperature from the P, v, T data alone

12-22C It is assumed that vfg ≅ vg ≅ RT/P, and hfg ≅ constant for small temperature intervals

12-23 Using the Clapeyron equation, the enthalpy of vaporization of steam at a specified pressure is to be estimated and to

be compared to the tabulated data

Analysis From the Clapeyron equation,

kJ/kg 2159.9

kPa50/kg)

m0.001073K)(0.60582

273.15(133.52

kPa75)2(325)

(

)(

3

kPa

@275 sat kPa 25 3

@ sat kPa 00 3

@ kPa

300

@ sat

kPa 300 sat, kPa

300

@ sat

T T

T

P T

dT

dP T

h

f g

fg fg

vv

vvv

The tabulated value of h fg at 300 kPa is 2163.5 kJ/kg

12-24 The h fg and s fg of steam at a specified temperature are to be calculated using the Clapeyron equation and to be compared to the tabulated data

kg kJ 8

2206 /

K10

kPa.18)169(232.23/kg)

m0.001060K)(0.89133

273.15(120

C115C125)

(

)(

3

C 115

@ sat C 125

@ sat C 120

@

C 120 sat, C

120

@ sat

T

P T

dT

dP T

h

f g

fg fg

vv

vvv

kJ/kg2206.8

Trang 15

R10

psia.793)23(29.759/lbm)

ft01201.0R)(1.7345459.67

(10

F5F15)

(

)(

3

F 5

@ sat F 15

@ sat F 0

@

F 10 sat, F

0

@ sat

Btu/lbm 89.31

T

P T

dT

dP T

h

f g

fg fg

vv

vvv

since 1 Btu = 5.4039 psia·ft3

(b) From the Clapeyron-Clausius equation,

error)(7.6%

R 67.4595

1R

67.45915

1R

Btu/lbm0.01946

psia29.759

psia793

23

ln

11ln

sat 2 1 sat

1 2

Btu/lbm 96.04

h

T T R

h P

P

The tabulated value of h fg at 10°F is 89.23 Btu/lbm

Trang 16

∆

= vClapeyron

,

At 100ºC, for an increment of 5ºC, we obtain

/kgm6710.1001043.06720.1

/kgm6720.1

/kgm001043.0

kPa29.3661.8490.120

∆

C1095105

∆

kPa90.120

kPa61.84

C1055100

C955100

3

3 C

100

@

3 C

100

@

1 2

C 105

@ sat 2

C 95

@ sat 1

increment 2

increment 1

=+

f

P P P

T T T

P P

T T T

vvv

v

Substituting,

kJ/kg 2262.8

=+

kPa36.29/kg)mK)(1.671015

.273100

h fg vfg

The enthalpy of vaporization from steam table is

/kg m 2256.4 3

=

°C 100

4.22568.2262orPercentErr

We repeat the analysis over the temperature range 10 to 200ºC using EES Below, the copy of EES solution is provided:

Trang 17

PercentError [%]

T [C]

Trang 18

12-27E A substance is cooled in a piston-cylinder device until it turns from saturated vapor to saturated liquid at a constant

pressure and temperature The boiling temperature of this substance at a different pressure is to be estimated

psia/R896.1lbm)

)/(0.5ftR)(1.5(475

lbm) /(0.5Btu

1

ftpsia5.404Btu)(250

3 3

h dT

dP

v

Weight

50 psia 15°F 0.5 lbm Sat vapor

Q

Using the finite difference approximation,

sat 1 2

1 2 sat ≈⎜⎜⎛ −− ⎟⎟⎞

P P dT

dP

Solving for T2,

R 480.3

=

−+

=

−+

=

psia/R1.896

psia)5060(R475/

1 2 1

2

dT dP

P P T

T

pressure and temperature The saturation pressure of this substance at a different temperature is to be estimated

psia/R896.1lbm)

)/(0.5ftR)(1.5(475

lbm) /(0.5Btu

1

ftpsia5.404Btu)(250

3 3

h dT

dP

v

Weight

Q

Using the finite difference approximation,

sat 1 2

1 2

P P dT

dP

Solving for P2,

psia 40.52

=

−+

=

−+

= 1 ( 2 1) 50psia (1.896psia/R)(470 475)R

dT

dP P

P

Trang 19

pressure and temperature The s fg of this substance at the given temperature is to be estimated

Q

fg fg fg

fg s T

h dT

lbm)Btu)/(0.5(250

.1ftpsia5.404

Btu1lbm0.5

ft1.5)psia/R896.1(

3 3

12-30E Saturation properties for R-134a at a specified temperature are given The saturation pressure is to be estimated at

two different temperatures

psia/R4979.0Btu

1

ftpsia5.404/lbm)ftR)(2.1446(460

Btu/lbm90.886 3

3 sat

h dT

dP

vUsing the finite difference approximation,

sat 1 2

1 2

P P dT

dP

Solving for P2 at −15°F

psia 13.72

=

−+

=

−+

= 1 ( 2 1) 21.185psia (0.4979psia/R)(445 460)R

dT

dP P

P

Solving for P2 at −30°F

psia 6.25

=

−+

=

−+

= 1 ( 2 1) 21.185psia (0.4979psia/R)(430 460)R

dT

dP P

P

Trang 20

h dT

T

P P dT

1 2 sat

Solving this for the second pressure gives for T2 = 110°F

psia 134.0

=

−+

=

R)100110(Btu1

ftpsia404.5/lbm)ftR)(0.86332(560

Btu/lbm154.85

psia7.116

)(

3 3

1 2 1

T

h P

=

−+

=

R)10090(Btu1

ftpsia404.5/lbm)ftR)(0.86332(560

Btu/lbm154.85

psia7.116

)(

3 3

1 2 1

T

h P

Trang 21

12-32 It is to be shown that

sat ,

,

)/(

T h T c

P

fg f

p

Analysis The definition of specific heat and Clapeyron equation are

fg fg P p

T

h dT

dP

T

h c

,

2 2

11

//

/

T

h h T

c T c

T

h T

h T T

h T

T

T h T

T h

f g f p g p

f P

f g

P g

P f P g P fg

,

)/(

T h T c

P

fg f

p

Trang 22

General Relations for du, dh, ds, c v , and c p

12-33C Yes, through the relation

P T

p

T

T P

12-34E The specific heat difference c p − cv for liquid water at 1000 psia and 300°F is to be estimated

Analysis The specific heat difference c p − cv is given as

T P p

P T T c

2

Approximating differentials by differences about the specified state,

)ftpsia5.4039Btu

(1

R/lbmftpsia986.0

/lbmft0.017417)(0.017345

psia1000R

50

/lbmft0.017151)(0.017633

R)609.67(

psia500)(1500F

)275325()R67.459300(

3 3

3

2 3

F 300 psia 500 psia 1500 2

psia 1000

F 275 F 325

F 300 2

psia 1000

T P

p

P T

T c

c

vv

v

vv

Properties are obtained from Table A-7E

Trang 23

Approximating differentials by differences about the specified state,

1

K 00381

/kgm0.11418)(0.12322

/kgm0.11874

1

C)2040(

11

3 3

kPa 200

C 20 C 40 kPa

P

T

vvv

vv

β

and

1

kPa 00482

/kgm0.13248)(0.09812

/kgm0.11874

1

kPa180)(240

11

3 3

C 30

kPa 180 kPa 240 C

T P

vv

v

vv

α

Trang 24

12-36 The internal energy change of air between two specified states is to be compared for two equations of states

Assumptions Constant specific heats for air can be used

Properties For air at the average temperature (20+300)/2=160°C=433 K, cv = 0.731 kJ/kg⋅K (Table A-2b)

Analysis Solving the equation of state for P gives

Using equation 12-29,

vv

T

P T dT c

=

Substituting,

dT c

d a

RT a

RT dT c

du

v

vv

=

Integrating this result between the two states with constant specific heats gives

kJ/kg 205

du= v

which gives the same answer

Trang 25

12-37 The enthalpy change of air between two specified states is to be compared for two equations of states

Properties For air at the average temperature (20+300)/2=160°C=433 K, cp = 1.018 kJ/kg⋅K (Table A-2b)

Analysis Solving the equation of state for v gives

a P

= v v

Substituting,

adP dT c

dP P

RT a P

RT dT c

=

Integrating this result between the two states with constant specific heats gives

kJ/kg 290.0

=

−+

−

⋅

=

−+

−

=

−

100)kPa/kg)(600

m(0.01K)20K)(300kJ/kg

018.1(

)()(

3 1

2 1 2 1

2 h c T T a P P

For an ideal gas,

dT c

dh= p

which when integrated gives

kJ/kg 285.0

Trang 26

12-38 The entropy change of air between two specified states is to be compared for two equations of states

Properties For air at the average temperature (20+300)/2=160°C=433 K, cp = 1.018 kJ/kg⋅K (Table A-2b) and R = 0.287 kJ/kg⋅K (Table A-1)

a P

dT c

dP T T

dT c

ds

p

P p

kPa600K)lnkJ/kg287.0(K293

K573K)lnkJ/kg018.1(

lnln

1 2 1

2 1

2

P

P R T

T c s

Trang 27

12-39 The internal energy change of helium between two specified states is to be compared for two equations of states

Properties For helium, cv = 3.1156 kJ/kg⋅K (Table A-2a)

Analysis Solving the equation of state for P gives

Using equation 12-29,

vv

T

P T dT c

=

Substituting,

dT c

d a

RT a

RT dT c

du

v

vv

=

Integrating this result between the two states gives

kJ/kg 872.4

du= v

which gives the same answer

Trang 28

12-40 The enthalpy change of helium between two specified states is to be compared for two equations of states

Properties For helium, c p = 5.1926 kJ/kg⋅K (Table A-2a)

a P

= v v

Substituting,

adP dT c

dP P

RT a P

RT dT c

=

Integrating this result between the two states gives

kJ/kg 1459

=

−+

−

⋅

=

−+

−

=

−

100)kPa/kg)(600

m(0.01K)20K)(300kJ/kg

1926.5(

)()(

3 1

2 1 2 1

2 h c T T a P P

For an ideal gas,

dT c

dh= p

which when integrated gives

kJ/kg 1454

Trang 29

12-41 The entropy change of helium between two specified states is to be compared for two equations of states

Properties For helium, c p = 5.1926 kJ/kg⋅K and R = 2.0769 kJ/kg⋅K (Table A-2a)

a P

dT c

dP T T

dT c

ds

p

P p

kPa600K)lnkJ/kg0769.2(K293

K573K)lnkJ/kg1926.5(

lnln

1 2 1

2 1

2

P

P R T

T c s

Trang 30

1 2

c u u

RT P T

P

T

a

R T

P a

RT P

v

vv

1

1 2

c h h

h

∂

∂vThe equation of state for the specified gas can be expressed as

Thus,

a a P

R T T

v T

P

R T

a P RT

=

)(vvv

v

vv

Substituting,

( 2 1)

2 1 2

1 2

1

P P a dT c dP a dT c

T p P

P T

1

1 2

P

T T

p

dP T

dT T

c s

2 1 2

1 2

P R dT T

c dP P

R dT

T

c

T p P

2 , and ln

,

0

P

P R s P

P a h

u= ∆ = − ∆ =−

∆

Trang 31

12-43 General expressions for (∂u/∂P) T and (∂h/∂v)T in terms of P, v, and T only are to be derived

Analysis The general relation for du is

vv

T

P T dT c

du= +⎜⎜⎛ ⎜⎝⎛∂∂ ⎟⎠⎞ − ⎟⎟⎞

Differentiating each term in this equation with respect to P at T = constant yields

T T

T

P T P

P T

P T P

0

Using the properties P, T, v, the cyclic relation can be expressed as

P T

T

P P

T T

T P

T

P T T P P

T T h

vv

T

P P

T

Substituting, we get

vv

T T

Trang 32

12-44 It is to be shown that

P p

T T

P T c

d s dT T

s ds

v

d T

P dT T

=

Taking the entropy to be a function of pressure and temperature,

dP P

s dT T

s ds

T P

dT T

v

T

P dP T T dT c

d P dT T

P dP

vv

v v

T

P P T T dT T

P T T dT c

c

T P P

T T

P T c

Trang 33

12-45 It is to be proven that the definition for temperature T =(∂u/∂s)v reduces the net entropy change of two volume systems filled with simple compressible substances to zero as the two systems approach thermal equilibrium

constant-Analysis The two constant-volume systems form an isolated system shown here

For the isolated system

0

B A tot =dS +dS ≥

Assume S =S(u,v)

Then,

vvv

d s du u

s ds

A A

T

du m T

du m

dStot = +

The first law applied to the isolated system yields

A A B B B

B A

m dU

dU E

out in

Now, the entropy change may be expressed as

A B A A B A A A

T T

T T du m T T du m

dStot 1 1

As the two systems approach thermal equilibrium,

0lim tot

Trang 34

12-46 An expression for the volume expansivity of a substance whose equation of state is P(v− )a =RT is to be derived

a P

P(v− )=

Analysis The specific heat difference is expressed by

T P p

P T T c

P P

R T RT

P P

R T c

2 v

Trang 35

RT aP

RT

P P

=

α

12-49 An expression for the isothermal compressibility of a substance whose equation of state is

2 / 1)( b T

a b

RT P

2)

b T

a b

RT P

++

vv

v

Substituting,

2 2

/ 1 2 2

2 2 / 1 2

)(

2)

(

1)

(

2)

(

11

1

b

b T

a b

RT b

b T

a b

RT

P T

+

++

vv

v

vv

v

vv

α

Trang 36

12-50 An expression for the volume expansivity of a substance whose equation of state is

2 / 1)( b T

a b

RT P

v

P

T P

a b

R T

v

and

2 2 2 / 1 2

2 2

2 / 1 2

)(

2)

(

)(

11)

(

b

b T

a b RT

b bT

a b

RT P

T

+

++

vv

vvv

v

Substituting these results into the definition of the volume expansivity produces

2 2 2 / 1 2

2 / 3

)(

2)

(

)(21

b

b T

a b RT

T b

a b

R

+

++

vv

vvv

v

Trang 37

12-51 An expression for the volume expansivity of a substance whose equation of state is

T

a b

RT P

2v

v

P

T P

T

a b

R T

P

vvv

RT P

2)

RT T

a b R

3 2

2 22)(

1

vv

Trang 38

P

T P

which on rearrangement becomes

v

vv

Trang 39

12-53 It is to be demonstrated that c p ( P)s

c k

vv

Analysis The relations for entropy differential are

dP T T

dT c

ds

d T

P T

dT c

=

v

vv v

For fixed s, these basic equations reduce to

dP T T

dT

c

d T

P T

vv v

Also, when s is fixed,

v

P

T P

Solving this for the numerator of the specific heat ratio expression and substituting the result into this numerator produces

s s

T

P P

v

α

Trang 40

ln

T

T T

T cT RT

a

v

It is to be shown how

to obtain P, h, s, cv, and c p from this expression

Analysis Taking the Helmholtz function to be a function of temperature and specific volume yields

vvv

d a dT T

a da

a P

T

Taking the indicated partial derivatives of the Helmholtz function given in the problem statement reduces these expressions

to

0 0

lnln

T

T c R

The definition of the enthalpy (h = u + Pv) and Helmholtz function (a = u−Ts) may be combined to give

RT cT cT

RT T

T cT RT

T

T T

T cT RT

a T

a T a

P Ts a

P u

h

T

++

=

+

−+

0 0 0 0 0

lnln

ln1

ln

v

vv

v

vvvv

v

According to

T

c T

c T T

s T

The preceding expression for the temperature indicates that the equation of state for the substance is the same as that of an ideal gas Then,

c R c R

c p = + v = +

Tiêu đề	Thermodynamic Property Relations
Tác giả	Yunus A. Cengel, Michael A. Boles
Trường học	McGraw-Hill
Chuyên ngành	Thermodynamics
Thể loại	solutions manual
Năm xuất bản	2011
Thành phố	New York

Định dạng
Số trang	86
Dung lượng	862,53 KB