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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 unnecessary 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.
 Spring '10
 bill
 Organic chemistry

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