Question

A gas with a volume of 4.0 liters at a pressure of 205 kilopascals is allowed to expand to a volume of 12.0 liters. What will the pressure in the container be if the temperature remains constant?

Answer 1

boyle's law p1V1=p2V2

p2=(p1V1)/V2

p2=(205*10^3 Pa * 4*10^-3 m^3 ) / (12*10^-3 m^3)

p2= 68333 Pa