a pump contains 0.5L of air at 203kPa, chemistry homework

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Science

Description

expanding the volume until the pressure reads 50.8kPa  what is the new volume of air in the pump?

2.0L

0.2L

1.0L

4.0L

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Explanation & Answer

Thank you for the opportunity to help you with your question!

from Boyle's law, PV=K

this implies that P1*V1 = P2*V2

P1 = 203kPa, V1 = 0.5l

P2 = 50.8, V2= ?

applying the law,

203*0.5 = 50.8V2

thus V2 = (203*0.5)/50.8 =1.998L

thus your answer is 2.00L

Please let me know if you need any clarification. I'm always happy to answer your questions.


Anonymous
Very useful material for studying!

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