By Arthur Adamson (Auth.)
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31 km. 1-8 A good vacuum for many purposes has a pressure of 1 0 atm. Treating air as a single gas of molecular weight 29, at what elevation will this 2 pressure be found? Assume — 70°C. - 01 Ans. 368 x 10 km (assuming g to remain constant). Derive the van der Waals equation for η moles of gas. Ans. 22 [Ρ + (an /v )](v - nb) = nRT. 1-10 Calculate the second and third virial coefficients for C 0 assuming it to be a van der Waals 2 3 2 gas. Ans. 82 χ 10" liter . 1-11 What is the Boyle temperature of C 0 assuming it to be a van der Waals gas?
A very qualitative rationale of the treatment is as follows. It was noted in Section 1-5 that the barometric equation could be regarded as a particular appli kT that the probab cation of the Boltzmann principle Eq. (1-35). The principle stated ility of a molecule having an energy e is proportional to e~^ . SPECIAL TOPICS 29 If now molecules experience mutual attractive and repulsive forces,
It is ordinarily of more interest to deal with the net velocity c than with the separate components. That is, although the u and ν velocity components represent 2 independent ways in which the molecule can have kinetic energy, it is the net velocity of the molecule and its total kinetic energy mc /2 that are needed for most applications. N o w the sum of all the possible ways in which the net velocity can 46 CHAPTER 2: KINETIC MOLECULAR THEORY OF GASES increase from c to c + dc, that is, by adding increments du and dv to various combinations of u and is given by the area of the annulus shown in Fig.
A Textbook of Physical Chemistry by Arthur Adamson (Auth.)