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The most common error is not being careful with the units Grams must be converted to moles, and milliliters must be converted to liters Another way to prepare a molar solution is by dilution of a more concentrated solution to a more dilute one by adding solvent The following equation can be used: (Mbefore)(Vbefore) = (Mafter)(Vafter) In the preceding equation, before refers to before dilution and after refers to after dilution Let s see how to apply this relationship Determine the final concentration when 5000 mL of water is added to 4000 mL of a 01111 M solution of HC1 Assume the volumes are additive Mbefore = 01111 M Vbefore = 4000 mL Mafter = Vbefore = (4000 + 5000) mL

Play on tradition Tradition motivates Tradition helps players subordinate their individual interests to the needs of the team If you don t have traditions, make some up Understand the dangers of winning If winning comes too hard, your team may get demoralized But if winning comes too easily, it will be harder to be motivated, even as the world is expecting more and more from you What s your personal and organizational plan for the day after you succeed

The first limit of the last member is Henry's constant k i ; the second limit is evaluated from Eq (14104), written n / n i = z i R T / A ; thus,

In combination with Eq (14118) these equations give:

Similarly, equating Eqs (14115) and (14117) yields:

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Perhaps my father s most famous quote which I ll return to in a later chapter was Winning isn t everything; it s the only thing This quote conjures up a bad picture in some people s minds They envision an out-of-control father at a Little League game, yelling at his son for not being a winner, or an overheated coach berating the kids on a Pop Warner football team Believe me, that wasn t what my father was about at all The following excerpt makes his point far more effectively, I think:

= Ti(T)

Mafter = (Mbefore)(Vbefore)/(Vafter) = (01111 M) (4000 mL)/(9000 mL) = 004938 M The most common error is forgetting to add the two volumes

Then by Eq (14119), These equations show that equality of fugacities is not a proper criterion for gasladsorbate equilibrium This is also evident from the fact that the units of gas-phase fugacities are those of pressure, while the units of adsorbate fugacities are those of spreading pressure In most applications the fugacities appear as ratios, and the factor ki R T I A cancels Nevertheless it is instructive to note that equality of chemical potentials, not fugacities, is the fundamental criterion of phase equilibrium An activity coefficient for the constituent species of a mixed-gas adsorbate is defined by the equation:

where f i and f," are evaluated at the same T and spreading pressure n The degree sign ( O ) denotes values for the equilibrium adsorption of pure i at the spreading pressure of the mixture Substitution for the fugacities by Eqs (14120) and (14121) gives:

W H AT I T TA K E S T O B E # 1

f;(p) xifig(PiO) The fugacities are evaluated at the pressures indicated in parentheses, where P is the equilibrium mixed-gas pressure and Pi is the equilibrium pure-gas pressure that produces the same spreading pressure If the gas-phase fugacities are eliminated in favor of fugacity coefficients then: [Eqs (1132) and (1148)],

Sometimes the varying volumes of a solution s liquid component(s) due to changes in temperature present a problem Many times volumes are not additive, but mass is additive The chemist then resorts to defining concentration in terms of the molality Molality (m) is defined as the moles of solute per kilogram of solvent m = moles solute/kilograms solvent Notice that this equation uses kilograms of solvent, not solution The other concentration units use mass or volume of the entire solution Molal solutions use only the mass of the solvent For dilute aqueous solutions, the molarity and the molality will be close to the same numerical value For example, ethylene glycol (C2H6O2) is used in antifreeze Determine the molality of ethylene glycol in a solution prepared by adding 621 g of ethylene glycol to 1000 g of water molality = 621 g C 2H6O2 1000 g 1 mol C 2H6O2 = 10 0 m C 2H6O2 1000 g H 2O 1 kg 621 g C 2H6O2

The usual assumption is that the gas phase is ideal, in which case the fugacity coefficients are unity:

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