# 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?

Sep 13, 2017

Well, we use old Boyle's law.....$P V = k$, given constant temperature, and CONSTANT amount of gas.....and gets ${P}_{1} = 60 \cdot k P a$.
And since $P V = k$, and thus for a PISTON at constant temperature, we write.....
${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$.............
And we solve for P_1=(120*kPaxx0.25*m^3)/(0.5*m^3)=??*kPa.