Chapter 10 - Section B - Non-Numerical Solutions
10.5 For a binary system, the next equation following Eq. (10.2) shows that P is linear in x1 . Thus no maximum or minimum can exist in this relation. Since such an extremum is required for the existence of
Chapter 11 - Section A - Mathcad Solutions
11.1 For an ideal gas mole fraction = volume fraction CO2 (1): N2 (2): S := R 11.2 x1 := 0.7 x2 := 0.3 i := 1 . 2 J mol K
i
xi ln ( xi)
S = 5.079
Ans.
For a closed, adiabatic, fixed-volume system, U =0. Also, for
Chapter 11 - Section B - Non-Numerical Solutions
11.6 Apply Eq. (11.7): (nT ) Ti ni =T n ni =T (n P ) Pi ni =P n ni =P
P ,T ,n j
T , P ,n j
P ,T ,n j
T , P ,n j
11.7 (a) Let m be the mass of the solution, and dene the partial molar mass by: mi Let Mk be t
Chapter 16 - Section B - Non-Numerical Solutions
16.1 The potential is displayed as follows. Note that K is used in place of k as a parameter to avoid confusion with Boltzmanns constant.
Combination of the potential with Eq. (16.10) yields on piecewise in
Chapter 14 - Section B - Non-Numerical Solutions
14.2 Start with the equation immediately following Eq. (14.49), which can be modied slightly to read: ln i = (nG R/ RT ) (n Z ) ln Z +n +1 ni ni ni
where the partial derivatives written here and in the foll
Chapter 13 - Section A - Mathcad Solutions
Note: For the following problems the variable kelvin is used for the SI unit of absolute temperature so as not to conflict with the variable K used for the equilibrium constant 13.4 H2(g) + CO2(g) = H2O(g) + CO(g
SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 7
SONNTAG BORGNAKKE VAN WYLEN
FUNDAMENTALS
of
Thermodynamics
Sixth Edition
Sonntag, Borgnakke and van Wylen
CONTENT CHAPTER 7
SUBSECTION Correspondence table Concept-Study guide problems Heat engines and refrigerat
Chapter 5 - Section A - Mathcad Solutions
5.2 Let the symbols Q and Work represent rates in kJ/s. Then by Eq. (5.8) = TC Work = 1 QH TH TH := 798.15 K TC QH := 250 kJ s kJ s
TC := 323.15 K Work := Q H 1
TH
or
Work = 148.78
Work = 148.78 kW which is t
Chapter 2 Homework
2.2 The law of definite proportions are different samples of a given compound always contain
the same elements in the same mass ratio. For example, a sample of carbon dioxide gas obtained
from different sources, such as the exhaust from
Nicole Mobley
Chapter 5 Homework
5.1 Kinetic energy is the energy associated with motion. Potential energy is energy possessed by
virtue of position. Thermal energy is a form of kinetic energy. Chemical energy is a form of
potential energy. The law of con
Chapter 3 Homework
3.1 Molecular mass is calculated by summing the masses of all atoms in a molecule. Molecular
weight is another term for molecular mass. The molecular mass that we calculate is generally an
average molecular mass because the atomic masse
Chapter 4 Homework
4.1 A solution is a homogeneous mixture consisting of a solvent and one or more dissolved
species called solutes. If you dissolve a teaspoon of sugar in a glass of water, water is the solvent
and sugar is the solute.
4.2 An electrolyte
Chapter 9 - Section B - Non-Numerical Solutions
9.1 Since the object of doing work |W | on a heat pump is to transfer heat | Q H | to a heat sink, then: What you get = | Q H | What you pay for = |W | Whence For a Carnot heat pump, = |Q H | |W |
|Q H | TH
SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 2
SONNTAG BORGNAKKE VAN WYLEN
FUNDAMENTALS
of
Thermodynamics
Sixth Edition
CONTENT
SUBSECTION Correspondence table Concept-Study Guide Problems Properties and Units Force and Energy Specific Volume Pressure Manomet
Chapter 1 - Section B - Non-Numerical Solutions
1.1 This system of units is the English-system equivalent of SI. Thus, gc = 1(lbm )(ft)(poundal)1 (s)2 1.2 (a) Power is power, electrical included. Thus, Nm kgm2 energy [=] [=] time s s3 (b) Electric current
Chapter 2 - Section A - Mathcad Solutions
2.1 (a) Mwt := 35 kg Work := Mwt g z (b) Utotal := Work g := 9.8 m s
2
z := 5 m Work = 1.715 kJ Ans. Utotal = 1.715 kJ dU + d ( PV) = CP dT Ans.
(c) By Eqs. (2.14) and (2.21): Since P is constant, this can be writ
Chapter 2 - Section B - Non-Numerical Solutions
2.3 Equation (2.2) is here written: Ut + EP + EK = Q + W
(a) In this equation W does not include work done by the force of gravity on the system. This is accounted for by the E K term. Thus, W = 0. (b) Since
Chapter 3 - Section A - Mathcad Solutions
3.1 = 1 d dT = P 1 d dP
T
At constant T, the 2nd equation can be written: d = dP ln ( 1.01) ln
2 = P 1
:= 44.18 10
6
bar
1
2 = 1.01 1 Ans.
P :=
P = 225.2bar
3
P2 = 226.2 bar
3.4 b := 2700 bar
c := 0.125
V
c
Chapter 3 - Section B - Non-Numerical Solutions
3.2 Differentiate Eq. (3.2) with respect to P and Eq. (3.3) with respect to T : P T =
T
1 V2
V P V T
T
V T V P
+
P
1 V
2V PT 2V T P
= +
2V PT 2V PT
=
P
1 V2
T
P
1 V
=
Addition of these two equations leads i
Chapter 4 - Section B - Non-Numerical Solutions
4.5 For consistency with the problem statement, we rewrite Eq. (4.8) as: CP = A + B C T1 ( + 1) + T12 ( 2 + + 1) 2 3
where T2 / T1 . Dene C Pam as the value of C P evaluated at the arithmetic mean temperatur
Chapter 5 - Section B - Non-Numerical Solutions
5.1 Shown to the right is a P V diagram with two adiabatic lines 1 2 and 2 3, assumed to intersect at point 2. A cycle is formed by an isothermal line from 3 1. An engine traversing this cycle would produce
Chapter 6 - Section A - Mathcad Solutions
6.7 At constant temperature Eqs. (6.25) and (6.26) can be written: dS = V dP and dH = ( 1 T) V dP
For an estimate, assume properties independent of pressure. T := 270 K V := 1.551 10 P1 := 381 kPa
3 3 m
P2 := 1200
Chapter 6 - Section B - Non-Numerical Solutions
H S =T
P
6.1 By Eq. (6.8),
and isobars have positive slope T S
Differentiate the preceding equation: 2 H S2
2 H S2 =
P
=
P
P
Combine with Eq. (6.17):
T CP
and isobars have positive curvature.
6.2 (a) Applica