Question

What is the new volume of a gas in a piston that was originally under `760*mm*Hg`, and expressed a volume of `818*mL`, when the pressure is reduced to `589*mm*Hg`?

Answer 1

`V_2~=1*L`

Explanation:

`P_1V_1=P_2V_2` under conditions of constant temperature.

Standard pressure specifies `1*atm`.

We also note that `1*atm-=760*mm*Hg`, i.e. an atmosphere will support a column of mercury that is `760*mm` high...

But here we do not even have to use this conversion...and solve for `V_2` directly.

`V_2=(P_1V_1)/P_2=(760*mm*Hgxx818*mL)/(589*mm*Hg)=1055.5*mL`.

That the volume HAS INCREASED is consistent with REDUCED pressure.