# 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?

(1) Answers

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

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