# Two gases separated by a movable piston in a cylindrical container

We have a cylinder compartment separated by a fixed piston in the middle. In the left volume we have a gas with pressure $P_L = P_0$ and on the right volume the same gas but with pressure $P_R = \frac{P_0}{2}$. The cylinder is in thermal equilibrium with a outside heat bath at temperature T.

Now the piston gets unfixed so it is able to move in the left or right direction.

I want to know, how can I calculate the position of the pistol after new equilibrium has been reached?

I suppose this is an adiabatic process since the total volume doesn't change and the heat bath keeping the Volume at constant temperature.

The system reachs equilibrium when pressure on both side of cylinder are equal, that's the condition you need to apply in order to solve the problem (and apply volume conservation as well)

Assume the initial volume on either side of the piston to be $V$.

After the piston is released and we allow everything to come to thermal equilibrium (the temperature is then $T$ on both sides), the pressure $P$ is now equal on both sides but the volumes are not.

Call the new volumes $V_L$ and $V_R$ respectively.

Because the transition is isothermal, with the Ideal Gas Law we can write for the left hand side:

$$P_0V=PV_L$$

And for the right hand side:

$$\frac{P_0}{2}V=PV_R$$

Multiply the last equation by $2$:

$$P_0V=2PV_R$$

Comparing the top and bottom equations we get:

$$PV_L=2PV_R$$

So we have, with the factor $P$ dropping out:

$$V_L=2V_R$$

This means that the position of the (middle of the) piston will now be two thirds to the right of the whole length of the cylinder (measured from the left hand side edge).

I suppose this is an adiabatic process since the total volume doesn't change and the heat bath keeping the Volume at constant temperature.

No. Allowing for thermal equilibrium to be established as stated in the problem, this is a purely isothermal transition.