Engenheiro

Chapter 1 Solutions Engineering and Chemical Thermodynamics

Wyatt Tenhaeff Milo Koretsky Department of Chemical Engineering Oregon State University [e-mail]

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

⎧ m 2⎫ 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 is larger than ma, the molecules of species A move faster on average.

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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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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