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

label Chemistry
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

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

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Jan 15th, 2016
...
Jan 15th, 2016
Sep 26th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer