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

Chemistry
Tutor: None Selected Time limit: 1 Day

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

Jan 15th, 2016

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.
Jan 15th, 2016

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