Question

How do you calculate the initial pressure this situation: a sealed container of gas is squashed, reducing the volume from 0.5 `m^3` to 0.25 `m^3`, with a final pressure of 120 kPa?

Answer 1

Well, we use old Boyle's law.....`PV=k`, given constant temperature, and CONSTANT amount of gas.....and gets `P_1=60*kPa`.

Explanation:

And since `PV=k`, and thus for a PISTON at constant temperature, we write.....

`P_1V_1=P_2V_2`.............

And we solve for `P_1=(120*kPaxx0.25*m^3)/(0.5*m^3)=??*kPa`.