A certain of a perfect gas is enclosed in a cylinder of volume `V_(0)` fitted with a smooth heavy piston of mass m and area a. The piston is displaced through a small distance downwards so as to compress the gas isothermally , and find its period . Take the atmospheric pressure as `P_(atom)`.

Explanation

Let A be the area of cross section of the cylinder. When the poston is in euqilibrium, force on it must balance.
`P_(0)A = p_(sin) A + mg`
Let us now displace. As the compression of the gas isothermal , the find pressure P is given by
`P_(0) V_(0) = P (V_(0) - xA)`

`p = (P_(0) V_(0))/(V_(0) - xA) = P_(0) (1 - (xA)/(V_(0))) ^(-1)`
`P_(0) = (1 - (xA)/(V_(0)))` `[(1 + t)^(n) = 1 + nt for t gtgt1]`
Considering the force on the displacement piston, the thrust of the gas presure now exceds its weight , and hence , magnitude of net upward force is `F = P_(0) = (1 - (xA)/(V_(0))) A - mg - P_(sin)A`
`= (P_(0) A^(2))/(V_(0))` [using Eq. (i)]
Hence the magnitude of force is proportional to x and its direction is towards equilibrium position.
Comparing with `F = kx, "we get" k = (P_(0) A^(2))/(V_(0))`
Time period `T = 2 pi sqrt((m)/(k)) = 2 pi sqrt((m V_(0))/(P_(0) A^(2)))`
`T = sqrt((m V_(0))/((p_(sin) + (mg)/(A))^(2))) implies T = 2 pi sqrt((m V_(0))/(P_(sin) A^(2) + mg A))`