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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?

User Dombesz
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1 Answer

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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
User Bob Stout
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