Question

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

Answer 1

`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.