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.
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
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 solut
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 o
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
whe
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 e
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 C
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
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 ob
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. Chemic
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 gener
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 t
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 |
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
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
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. (
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 i
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
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
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 v
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 f
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.
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