Scan_Doc0100 - T2 TI T2 T 2 V I (359 K)(3.8 L) = 4.9L...

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192 Chapter Five Gases EXAMPLE 5.8 A sample of methane gas that has a volume of 3.8 L at 5°C is heated to 86°C at constant pressure. Calculate its new volume. Solution Where are we going? To use the ideal gas equation to determine the final volume What do we know? ./ TI = 5°C + 273 = 278 K ./ VI = 3.8 L ./ T2 = 86°C + 273 = 359 K ./ V 2 = ? What information do we need? ./ Ideal gas law PV = nRT ./ R = 0.08206 L . atm/K . mol How do we get there? What are the variables that change? V,T What are the variables that remain constant? n,R,P 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. V nR -=- T P which leads to Combining these gives VI nR V 2 nR -=- and -=- TI P T2 P VI nR V 2 VI V 2 -=-=- or -=- T) P
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Unformatted text preview: T2 TI T2 T 2 V I (359 K)(3.8 L) = 4.9L V?=-=-Tl 278K Solving for V 2 : Reality Check: Is the answer sensible? In this case the temperature increased (at constant pressure), so the volume should increase. Thus the answer makes sense. SEE EXERCISES 5.55 AND 5.57 The problem in Example 5.8 could be described as a Charles's law problem, whereas the problem in Example 5.7 might be called a Boyle's law problem. In both cases, how-ever, we started with the ideal gas law. The real advantage of using the ideal gas law is that it applies to virtually any problem dealing with gases and is easy to remember....
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