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)
$W=\vec{F}\bullet\vec{\Delta l}$
$W=F\Delta lcos(0)$
And the cosine of 0 is 1
$W=F\Delta l$ and $F=PS$
Where S is the cross sectional area
$W=PS\Delta l$
The initial volume of the gas is:
$V=Sl$
And the final volume:
$2V=Sl'$
$2V-V=S(l'-l)$
Then $V=S\Delta l$
Substituting this in the work we'll have:
$W=PV$
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 $P=mV + n$, 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 $W=\frac{3PV}{2}$