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Unformatted text preview: unit of time, thus the pressure increases. 6. P 1 =103 kPa, V 1 =5.2 L, P 2 =400 kPa, V 2 =? Using Boyle’s law: P 1 V 1 = P 2 V 2 , 103 kPa x 5.2 L = 400 kPa x V 2 V = (103 kPa x 5.2 L) / 400 kPa = 1.34 L 7. P 1 = 700 mm Hg x 1 atm / 760 mm Hg = 0.921 atm , V 1 = 200 ml, P 2 = ?, V 2 = 950 ml Using Boyle’s law: P 1 V 1 = P 2 V 2 0.921 atm x 200 ml = P 2 x 950 ml P 2 = (0.921 atm x 200 ml)/950 ml = 0.194 atm Conclusion • By comparing values in the PV column and via the graph we see that PV = constant • This is called Boyle’s law (after the scientist that first discovered the relationship) • We will be looking more closely at Boyle’s law • For example, using the form P 1 V 1 = P 2 V 2 • Read “Results of Boyle’s Law Experiment” on handout. Do questions 1 – 6. For more lessons, visit www.chalkbored.com...
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 Fall '10
 Sullivan
 Boyle’s law, PV Column

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