Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 17, Problem 20P
To determine
The average translational kinetic energy of the oxygen and average speed of the nitrogen molecule.
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Let us consider a bacterial cell that depends on
chemical reactions using oxygen to produce energy.
The cell needs to obtain molecular oxygen from the
extra-cellular medium by diffusion through the cell
membrane.
We model the cell as a water-filled sphere of radius 1 µm. This means that we neglect the
presence of the internal structures of the cell: nucleus, nutrients, etc. The oxygen diffusion
coefficient (O2) in water at 25°C is 1.0x10-5 cm².s'.
Figure 1: Transmission electron microscope
Reminder: The distance travelled by diffusion can be picture of a bacterium. [from Wikipedia:
written as L(t) = V6Dt, where D is the diffusion
coefficient and t the diffusion time.
"Transmission Electron Microscopy"|
2. Diffusion as a function of time
People have measured the distance travelled by the oxygen molecule in water as a function of
time. The results are given in the table below.
0.1
t(s)
L (x10-“m)
log1o(t)
log10(L)
0.001
0.005
0.01
0.05
2.45
5.4
7.7
17.3
25.5
Draw the plot of L as a…
The average translational kinetic energy for a molecule (Erans) is given
by the following equation:
Etrans =mv?
Where m is the mass of the molecule and v? is the average of the square
of the velocity. Given v2 = 3kT/m where k is the Boltzmann's constant,
calculate the ratio of the kinetic energies at 200 °C and 100 °C
The principal components of the atmosphere of the Earth are diatomic molecules, which can rotate as well as translate. Given that the translational kinetic energy density of the atmosphere is 0.15 J cm−3, what is the total kinetic energy density, including rotation?
Chapter 17 Solutions
Physics for Scientists and Engineers
Ch. 17 - Prob. 1PCh. 17 - Prob. 2PCh. 17 - Prob. 3PCh. 17 - Prob. 4PCh. 17 - Prob. 5PCh. 17 - Prob. 6PCh. 17 - Prob. 7PCh. 17 - Prob. 8PCh. 17 - Prob. 9PCh. 17 - Prob. 10P
Ch. 17 - Prob. 11PCh. 17 - Prob. 12PCh. 17 - Prob. 13PCh. 17 - Prob. 14PCh. 17 - Prob. 15PCh. 17 - Prob. 16PCh. 17 - Prob. 17PCh. 17 - Prob. 18PCh. 17 - Prob. 19PCh. 17 - Prob. 20PCh. 17 - Prob. 21PCh. 17 - Prob. 22PCh. 17 - Prob. 23PCh. 17 - Prob. 24PCh. 17 - Prob. 25PCh. 17 - Prob. 26PCh. 17 - Prob. 27PCh. 17 - Prob. 28PCh. 17 - Prob. 29PCh. 17 - Prob. 30PCh. 17 - Prob. 31PCh. 17 - Prob. 32PCh. 17 - Prob. 33PCh. 17 - Prob. 34PCh. 17 - Prob. 35PCh. 17 - Prob. 36PCh. 17 - Prob. 37PCh. 17 - Prob. 38PCh. 17 - Prob. 39PCh. 17 - Prob. 40PCh. 17 - Prob. 41PCh. 17 - Prob. 42PCh. 17 - Prob. 43PCh. 17 - Prob. 44PCh. 17 - Prob. 45PCh. 17 - Prob. 46PCh. 17 - Prob. 47PCh. 17 - Prob. 48PCh. 17 - Prob. 49PCh. 17 - Prob. 50PCh. 17 - Prob. 51PCh. 17 - Prob. 52PCh. 17 - Prob. 53PCh. 17 - Prob. 54PCh. 17 - Prob. 55PCh. 17 - Prob. 56PCh. 17 - Prob. 57PCh. 17 - Prob. 58PCh. 17 - Prob. 59PCh. 17 - Prob. 60PCh. 17 - Prob. 61PCh. 17 - Prob. 62PCh. 17 - Prob. 63PCh. 17 - Prob. 64PCh. 17 - Prob. 65PCh. 17 - Prob. 66PCh. 17 - Prob. 67PCh. 17 - Prob. 68PCh. 17 - Prob. 69PCh. 17 - Prob. 70PCh. 17 - Prob. 71PCh. 17 - Prob. 72PCh. 17 - Prob. 73PCh. 17 - Prob. 74PCh. 17 - Prob. 75PCh. 17 - Prob. 76PCh. 17 - Prob. 77PCh. 17 - Prob. 78PCh. 17 - Prob. 79PCh. 17 - Prob. 80P
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