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# Scan_Doc0111 - lr L u 5.5 Dalton's Law of Partial Pressures...

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lr---- L , u EXAMPLE 5.17 Vapor pressure will be discussed in detail in Chapter 10.Atable of water vapor pressure values is given in Section 10.8. 5.5 Dalton's Law of Partial Pressures 203 What do we know? ./ Po, = 156 torr ./ P TOTAL = 743 torr How do we get there? The mole fraction of O 2 can be calculated from the equation Po, 156 torr Xo, = -- = = 0.210 - P TOTAL 743 terr Note that the mole fraction has no units. SEE EXERCISE 5.85 The expression for the mole fraction, PI Xl = P TOTAL can be rearranged to give PI = Xl X P TOTAL That is, the partial pressure of a particular component of a gaseous mixture is the mole fraction of that component times the total pressure. " III The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N2 in air when the atmospheric pressure is 760. torr. Solution The partial pressure of N2 can be calculated as follows: P N , = X N , X P TOTAL = 0.7808 X 760. torr = 593 torr SEE EXERCISE 5.86 Collecting a Gas over Water
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