A vertical cylinder closed from both ends is equipped with an easily moving piston dividing the volume into two parts, each containing one mole of air. In equilibrium at `T_0 = 300 K` the volume of the upper part is `eta = 4.0` times greater than that of the lower part. At what temperature will the ratio of these volumes be equal to `eta' = 3.0` ?

Explanation

Let `p_1` and `p_2` be the pressure in the upper and lower part of the cylinder respectively at temperature `T_0`. At the equilibrium position for the piston :
`p_1 S + mg = p_2 S` or, `p_1 + (mg)/(S) = p_2` (`m` is the mass of the piston)
But `p_1 = (R T_0)/(eta V_0)` (where `V_0` is the initial volume of the lower part)
So, `( RT_o)/(eta V_0)+(mg)/(S) = (RT_0)/(V_0)` or, `(mg)/(S) = (RT_0)/(V_0) (1 - (1)/(eta))` ....(1)
Let `T` be the sought temperature and at this temperature the volume of the lower part becomes `V'`, then according to the problem the volume of the upper part becomes `eta' V'`
Hence, `(mg)/(S) = (RT)/(V')(1 - (1)/(eta'))` ...(2)
From (1) and (2).
`(RT_0)/(V_0)(1 - (1)/(eta)) = (RT')/(V') (1 -(1)/(eta'))` or, `T' = (T_0(1 - (1)/(eta))V')/(V_0(1 - (1)/(eta')))`
As, the total volume must be constant,
`V_0 =(1 + eta) = V' (1 + eta')` or `V' (V_0(1 + eta))/((1 + eta'))`
Putting the value of `V'` in Eq. (3), we get
`T' = (T_0(1 -(1)/(eta))V_0 ((1+ eta))/((1 + eta')))/(V_0(1 - (1)/(eta')))`
=`(T_0(eta^2 - 1)eta')/((eta'^2 - 1)eta) = 0.42 k K`.