Physics for Scientists and Engineers
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?

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