A piston of #11*L# volume under #42*kPa# pressure is compressed to a volume of #8*L#; what is the new pressure?

1 Answer
anor277
Aug 3, 2016

#P_1V_1=P_2V_2#. #P_2~=58*kPa#

Explanation:

#P_2=(P_1V_1)/V_2# #=# #(42*kPaxx11*L)/(8*L)# #=# #??#

This is not a very realistically described experiment. How does the volume decrease? Was the gas originally in a piston whose volume reduced? Well, the reaction does not say so, but the volume reduced somehow.

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