Solutions to numerical problems

Draw up the following table and then plot p/p versus p to find the zero pressure limit of p/p where all gases behave ideally. [Pg.13]

Rearrangement yields the desired relation, that is [Pg.13]

This relation is valid in the limit of zero pressure (for a gas behaving perfectly). [Pg.14]

In a proper series of experiments one should reduce the pressure (e.g. by adjusting the balanced [Pg.14]

We assume that no Hi remains after the reaction has gone to completion. The balanced equation is [Pg.14]

Since the Neptunians know about perfect gas behavior, we may assume that they will write pV = nRT at both temperatures. We may also assume that they will establish the size of their absolute unit to be the same as the °N, just as we write IK = 1°C. Thus [Pg.13]

As in the relationship between our Kelvin scale and (Celsius scale T 6— absolute zero(°N) so absolute zero (°N) = -233°N. [Pg.13]

COMMENT. To facilitate communication with Earth students we have converted the Neptunians units of the pV product to units familiar to humans, which are dm atm. However, we see from the soiution that oniy the ratio of pV products is required, and that will be the same in any civilization. [Pg.13]

Question. If the Neptunians unit of volume is the lagoon (L), their unit of pressure is the poseidon (P), their unit of amount is the nereid (n). and their unit of absolute temperature is the titan (T). what is the value of the Neptunians gas constant (R) in units of L, P. n, and T [Pg.13]

The value of absolute zero can be expressed in terms of a by using the requirement that the volume of a perfect gas becomes zero at the absolute zero of temperature. Hence [Pg.13]

COMMENT. This method of the determination of the molar masses of gaseous compounds is due to Can-nizarro who presented it at the Karlsruhe conference of 1860 which had been called to resolve the problem ot the determination of the molar masses of atoms and molecules and the molecular formulas ot compounds. [Pg.14]

The mass of displaced gas is pV, where V is the volume of the buib and p is the density of the gas. The balance condition for the two gases is //t(bulb) = pl/fbulb), wCbulb) = p V(bulb) [Pg.14]

As specified by the relationship = fl/4 )v, the density of cross-links would be half (2/0) of this value, [Pg.63]

The standard relationship (Eq. (1.19)) for swelling with A = would give [Pg.63]

It is a pleasure to acknowledge the financial support provided by the National Science Foundation through grants DMR-0075198 and DMR-0314760 (Polymers Program, Division of Materials Research), and by the Dow Corning Corporation. [Pg.63]

Vincent and J. D. Currey (Cambridge University Press, Cambridge, 1980), p. 331. [Pg.68]

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