Solution

We have a cylindrical vessel, with a movable piston with a gas on its left side and vacuum on the right side. The gas is at pressure $P$ and volume $V$. We need the work done by the gas on the piston, if the volume of the gas when moving the piston will double, and its pressure will be:
a)remain constant
b)increase with increasing volume linearly yo a pressure of $2P$
Assuming that we're working with ideal gases, and gas pushes the piston with a constant force, the work will be:
Let's go with a)
$W=\vec{F}\bullet\Delta l\vect{l}$
$W=F*Delta l*cos(0)$
And the cosine of 0 is 1
$W=F*Delta l$ and $F=P*S$
$W=P*S*Delta l$
The initial volume of the gas is:
$V=S*l$
And the final volume:
$2V=S*l'$
$2V-V=S*(l'-l)$
Then $V=S*Delta l$
Substituting this in the work we'll have:
$W=P*V$
For b)
We know that for a linear function
$m=\frac{y_2 -y_1}{x_2 -x_1} In our case, we have a$P=f(V)$function, the final pressure is$2P$and the final volume$2V$, so$m=\frac{2P-P}{2V-V}$$m=\frac{P}{V}$And substituting this in$P=m*V + n$, we get$P=P+n$, no$n=0$If we graph this function, the work will be the area of the trapezoid under this two points, wich is (bass minor + base major) times height over 2$W=\frac{(2P+P)*(2V-V)}{2}$So$W=\frac{3PV}{2}$\