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Chapter 1 Solutions Engineering and Chemical Thermodynamics

Wyatt Tenhaeff Milo Koretsky Department of Chemical Engineering Oregon State University koretsm@engr.orst.edu

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

1 r2 molecular mV = ek 2 3 molecular kT = ek 2
r 3kT ∴V ≈ m

Assume the temperature is 22 ºC. The mass of a single oxygen molecule is m =5.14 × 10 −26 kg . Substitute and solve:
r 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:
⎛ m ⎞ f (v)dv = 4π ⎜ ⎟ ⎝ 2πkT ⎠
3/ 2

⎧ m2⎫ 2 v ⎬v dv exp⎨− ⎩ 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


r V =

∫ f (v)vdv
0 ∞

=

∫ f (v)dv
0

8kT = 449 [m/s] πm

2

1.3 Derive the following expressions by combining Equations 1.4 and 1.5:

r 3kT Va2 = ma Therefore, r Va2 mb r = Vb2 ma

r 3kT Vb2 = mb

Since mb islarger than ma, the molecules of species A move faster on average.

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

4

1.5 (a) After a short time, the temperature gradientin 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 intemperature. The system is in thermal equilibrium.

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 = (1 atm )exp ⎢ ⎢ ⎥ ⎛ ⎡ J ⎤⎞ ⎟(273.15 K ) ⎜ 8.314 ⎢ ⎢ ⎥ ⎟ ⎜ ⎣ mol ⋅ K ⎥ ⎠ ⎦ ⎝ ⎣ ⎦
P = 0.330 atm = 33.4 kPa

Interpolate steamtable data:

T sat = 71.4 º C

for P sat = 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.

6

1.7 To solve these problems, the steam tables wereused. 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)

(b)

(c)

(d)

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. Saturatedvapor-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 vaporliquid mixture. Superheated vapor Explanation: the specific volume of the saturated vapor at P = 60 [bar] = 6 [MPa], is 0.03244 [m3/kg] and thesaturated 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. 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...
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