Solutions Manual for Engineering and Chemical Thermodynamics (2nd Edition) by Milton M. Koretsky

All Chapters Covered




SOLUTION MANUAL

, 1.2
An approximate solution can be found if we combine Equations 1.4 and 1.5:

1 V2 emolecular
2m k
3
kT e molecular
2 k


V 3kT
m
Assume the temperature is 22 ºC. The mass of a single oxygen molecule is m 5.14 10 26 kg . Substitute
and solve:

V 487.6 m/s

The molecules are traveling really, fast (around the length of five football fields every second).

Comment:
We can get a better solution by using the Maxwell-Boltzmann distribution of speeds that is sketched in
Figure 1.4. Looking up the quantitative expression for this expression, we have:
3/ 2 m
f (v)dv 4 m exp v 2 
v 2 dv
2 kT 2kT
where f(v) is the fraction of molecules within dv of the speed v. We can find the average speed by
integrating the expression above


 0 f (v)vdv
V   8kT 449 m/s
m
f (v)dv
0
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2

, 1.3
Derive the following expressions by combining Equations 1.4 and 1.5:
3kT 3kT
V2 V b2
a
a mb
m
Therefore,
Va2 mb
V
2 ma
b


Since mb is larger than m a , the molecules of species A move faster on average.
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3

, 1.4
We have the following two points that relate the Reamur temperature scale to the Celsius scale:
0 º C, 0 º Reamur and 100 º C, 80 º Reamur

Create an equation using the two points:
T º Reamur 0.8 T º Celsius

At 22 ºC,

T 17.6 º Reamur
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4

, 1.5
(a)
After a short time, the temperature gradient in the copper block is changing (unsteady state), so the
system is not in equilibrium.

(b)
After a long time, the temperature gradient in the copper block will become constant (steady state), but
because the temperature is not uniform everywhere, the system is not in equilibrium.

(c)
After a very long time, the temperature of the reservoirs will equilibrate; The system is then
homogenous in temperature. The system is in thermal equilibrium.
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5

, 1.6
We assume the temperature is constant at 0 ºC. The molecular weight of air is

MW 29 g/mol 0.029 kg/mol

Find the pressure at the top of Mount Everest:



0.029 kg/mol 9.81 m/s 8848 m
P 1atm exp J
8.314 273.15 K
mol K
P 0.330 atm 33.4 kPa
Interpolate steam table data:

T sat 71.4 º C for Psat 33.4 kPa

Therefore, the liquid boils at 71.4 ºC. Note: the barometric relationship given assumes that the
temperature remains constant. In reality the temperature decreases with height as we go up the
mountain. However, a solution in which T and P vary with height is not as straight-forward.
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6

, 1.7
To solve these problems, the steam tables were used. The values given for each part constrain the
water to a certain state. In most cases we can look at the saturated table, to determine the state.

(a) Subcooled liquid
Explanation: the saturation pressure at T = 170 [oC] is 0.79 [MPa] (see page 508); Since
the pressure of this state, 10 [bar], is greater than the saturation pressure,
water is a liquid.
(b) Saturated vapor-liquid mixture
Explanation: the specific volume of the saturated vapor at T = 70 [oC] is 5.04 [m3/kg] and
the saturated liquid is 0.001 [m3/kg] (see page 508); Since the volume of this state, 3
[m3/kg], is in between these values we have a saturated vapor- liquid mixture.
(c) Superheated vapor
Explanation: the specific volume of the saturated vapor at P = 60 [bar] = 6 [MPa], is
0.03244 [m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 511); Since the volume
of this state, 0.05 [m3/kg], is greater than this value, it is a vapor.
(d) Superheated vapor
Explanation: the specific entropy of the saturated vapor at P = 5 [bar] = 0.5 [MPa], is
6.8212 [kJ/(kg K)] (see page 510); Since the entropy of this state, 7.0592 [kJ/(kg K)], is
greater than this value, it is a vapor. In fact, if we go to the superheated water vapor
tables for P = 500 [kPa], we see the state is constrained to T = 200 [oC].
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7

, 1.8
From the steam tables in Appendix B.1:

critical m3
v̂ 0.003155 T 374.15 º C, P 22.089 MPa
kg
At 10 bar, we find in the steam tables

sat m3
v̂ 0.001127
l kg
sat
m3
v̂v 0.19444
kg
Because the total mass and volume of the closed, rigid system remain constant as the water
condenses, we can develop the following expression:

v̂critical 1 x v̂ sat xv̂ sat
l v

where x is the quality of the water. Substituting values and solving for the quality, we obtain

x 0.0105 or 1.05 %
A very small percentage of mass in the final state is vapor.
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8

, 1.9
The calculation methods will be shown for part (a), but not parts (b) and (c)

(a)
Use the following equation to estimate the specific volume:
v̂1.9 MPa, 250 º C v̂1.8 MPa, 250 º C 0.5 v̂ 2.0 MPa, 250 º C v̂1.8 MPa, 250 º C
Substituting data from the steam tables,

m3
v̂1.9 MPa, 250 º C 0.11821
kg
From the NIST website:
m3
v̂NIST 1.9 MPa, 250 º C 0.11791
kg
Therefore, assuming the result from NIST is more accurate
ˆ ˆNIST
v v
% Difference v̂NIST 100 % 0.254 %

(b)
Linear interpolation: m3
v̂1.9 MPa, 300 º C 0.13284
kg
NIST website:
m3
v̂NIST 1.9 MPa, 300 º C 0.13249
kg
Therefore,

% Difference 0.264 %

(c)
Linear interpolation: m3
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v̂1.9 MPa, 270 º C 0.12406
kg
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, 1.1
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