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The pressure of air in a 2.25-L container is 1.20 atm. What is the new pressure if the sample is transferred to a 6.50-L container, assuming temperature is constant?

1 Answer

Feb 24, 2016

Using Boyle's law, #P_1V_1=P_2V_2#, #P_2~=0.40*atm#.

Explanation:

From Boyle's law, #P_1V_1=P_2V_2#.

#P_2 = (P_1V_1)/V_2# #=# #(1.20*atmxx2.25*cancelL)/(6.50*cancelL)# #=# #??atm#

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