Scan_Doc0101 - 2 : T 2 P I VI (309 R)(345 tetf)(3.48 L) V 2...

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5.3 The Ideal Gas Law 193 EXAMPLE 5.9 A sample of diborane gas (B 2 H 6 ), a substance that bursts into flame when exposed to air, has a pressure of 345 torr at a temperature of -15°C and a volume of 3.48 L. If condi- tions are changed so that the temperature is 36°C and the pressure is 468 torr, what will be the volume of the sample? SoLution Where are we going? To use the ideal gas equation to determine the final volume What do we know? ./ t; = 15°C + 273 = 258 K ./ V j = 3.48 L ./ P, = 345 torr ./ T2 = 36°C + 273 = 309 K ./ V 2 = ? ./ P 2 = 468 torr What information do we need? ./ Ideal gas law PV = nRT ./ R = 0.08206 L . atmJK . mol How do we get there? What are the variables that change? P, V,T What are the variables that remain constant? n,R Write the ideal gas law, collecting the change variables on one side of the equal sign and the variables that do not change on the other. PV = nR T which leads to P1V 1 P 2 V 2 -=nR=- or t, T2 PjVj Tl P 2 V 2 T2 Solving for V
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Unformatted text preview: 2 : T 2 P I VI (309 R)(345 tetf)(3.48 L) V 2 = T I P 2 = (258 R)(468 torr) = 3.07 L SEE EXERCISES 5.59 AND 5.60 Always convert the temperature to the Kelvin scale when applying the ideal gas law. Since the equation used in Example 5.9 involves a ratio of pressures, it was unnec-essary to convert pressures to units of atmospheres. The units of torrs cancel. (You will . .. 345 468. . obtam the same answer by msertmg PI = 760 and P 2 = 760 mto the equation.) However, temperature must always be converted to the Kelvin scale; since this conversion involves addition of 273, the conversion factor does not cancel. Be careful. One of the many other types of problems dealing with gases that can be solved using the ideal gas law is illustrated in Example 5.10....
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This note was uploaded on 12/13/2010 for the course CHEM 2301 taught by Professor Bill during the Spring '10 term at South Texas College.

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