Show a cylindrical tube of volume V_(0) divided in two parts by a frictionless separator. The walls of the tube are adiabatic but the separator is conducting, Ideal gases ate filled in the two parts. When the swparator is kept in the middle, the pressure are p_(1) and p-(2) in the left part and the right part respectively. The separator is slowly slid and is released at a position where it can stay in equilibrium. Find the volumes if the two parts. (figure)

Explanation

As the separator is conducting, the tenperature is T when the separator is in the middle. Let n_(1) and n_(2) be the number of moles of the gas in the left part and the right part respectively. Using ideal gas equation,
p_(1)(v_(0))/(2)=n_(1)RT
and p_(2)(v_(0)=n_(2)RT.
Thus, (n_(1))/(n_(2))=p_(1)/p_(2) ...(i)
The separator will stay in equilibrium at a position wher the pressures on the two sides are equal. Suppose the volume of the left part is V_(1) and of the right part is V_(2) in this situation. Let the common pressure be p'. Also, let the common temperature in this situation be T'. Using ideal gas equation,
`p'V_(1)=n_(1)RT'`
and `p'V_(2)=n_(2)RT'`
or, `(V_(1))/(V_(2))=(n_(1))/(n_(2))=(p_(1))/(p_(2))` [using(i)]
Also, `V_(1)+V_(2)=V_(0)`
Thus, `V_(1)=(p_(1)V_(0))/(p_(1)+p_(2))` and `V_(2)=(p_(2)V_(0))/(p_(1)+p_(2))`.