A Textbook of Physical Chemistry by Arthur Adamson (Auth.)

By Arthur Adamson (Auth.)

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0. We are dealing with very large numbers and it is permissible to replace factorials by Stirling's approximation, In x\ = χ In χ — χ. (2-15) Equation (2-14) becomes 0 = χ Ô(7V, In Ni - Ni) i = Σ[^~- + N<) Mi - Σ (In Nt) 8N{ = 0. Mt], (2-16) i Equations (2-12)—(2-14) impose three different conditions on the ôTV's and are to be obeyed even though the system makes small, arbitrary fluctuations. A way of handling such a situation is Lagrange's method of undetermined multipliers [see, for example, Blinder (1969)].

1-52) We turn now to the van der Waals equation; referring to Fig. 1-12, it seems clear that the section labeled Vv must represent the molar volume of the gaseous state of the substance, while that labeled Vt should correspond to the liquid state. The van der Waals equation connects these two branches with the section showing a maximum and a minimum, but a real substance takes the short cut of direct condensation when Ρ reaches P° as illustrated in Fig. 1-13. We would like to know how to locate the horizontal line of this short cut, and hence the liquid vapor pressure P°.

Ans. 5 atm pressure. Ans. 158 x l O ^ g c m " . In the Dumas method one determines the molecular weight of a gas by a direct measurement of its density. 01613 g when filled with a hydrocarbon gas at 25°C and 100 Torr pressure. Calculate the molecular weight of the gas, assuming ideal behavior. - 1 Ans. 9 g m o l e . Calculate V and p for dry air at STP. Repeat the calculation for air saturated with water vapor at 25°C and at 1 atm total pressure. Assume ideal behavior. 3 3 Ans. 836 x 1(T g cm" .

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