A cylindrical container of volume `V_(0)` is divided into two parts by a thin conducting separator of negligible mass. The walls of the container are adiabatic. Ideal gases are filled in the two parts such that the pressures are `P_(0) " and " 2 P_(0)` when the separator is held in the middle of the container (see figure). Now the separator is slowly slid and released in a position where it stays in equilibrium. Find the volume of the two parts.

Explanation

As the separtor is conducting, the temperatures in the two parts will be the same. Suppose the common temperature 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)/(2) = n_(2)RT`
Thus, `(n_1)/(n_2) = (p_1)/(p_2)`....(i)
The separator will Stay in equilibrium at a position Where 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
and `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))`