Show a vertical cylindrical vesse seperated in two parts by a frictionless piston free to move along the length of vessel. The length of the cylilender is 90 cm and the piston divides the cylinder in the ratio of 5:4. Each of the two parts of the vessel contains 0.1 mole of an ideal gas. The temoerature of the gas is 300K in each part. Calculate the mass of the piston.(figure)

Explanation

Let `l_(1)` and `l_(2)` be the lengths of the upper part and the lower part of the cylinder respectively. Clearly, `l_(1)` = 50cm and `l_(2)` = 40 cm. Let the perssures in the upper and lower parts be `p_(1)` and `p_(2)` respectively Let the area of coss- section of the cylinder be a. The temperature in both parts is T = 300 K.
Consider the equilibrium of the piston. The forces acting on the pistion are
(a) Its weight mg
(b) `p_(1)` a downward, by the upper part of the gas and (c) `p_(2)` a upward, by the lower part of the gas.
Thus, `p_(2A) = p_1 A + mg` ...(i)`
Using `p_(V) = nRT` for the upper and the lower parts
`p_(1)l_(1) A = nRT` ...(ii)
and `p_(2)l_(2)A = nRT`. ...(iii)
Putting `p_(1) A` and `p_(2)` A from (ii) and (iii) in to (i),
`(nRT)/(l_2)= (nRT)/(l_1) + mg`
Thus, `m = (nRT)/(g)[(1)/(l_1)-(1)/(l_2)]`
`((0.1 mol)(8.3J//mol-K)(300K)/(9.8 m//s^2)[(1)/(0.4m)-(1)/(0.5m)]`
=12.7 kg.