Asmooth vertical tibe having two different sections, is open at both ends and equipped with two pistons of different areas. Each piston slides within a respective tube section. The pistons are tied with an inextensible weightless string. One mole of ideal gas is enclosed between them. The cross-section area of upper piston is `Delta A = 10 cm^(2)`greater than that of lower one. The combined mass of two piston is equal to `M = 5.0 kg`. The external pressure is `P_(0) = 1.0 atm`. By how much must the gas between the pistons the heated to shift pistons through `l = 5.0 cm`. `(1 atm = 1.0 xx 10^(5) Pa)`

Explanation

Let `m_(1)` and `m_(2)` be the masses of upper and lower pistons. `A_(1)` and `A_(2)` their respective areas and `T` tension in the string.
Equilibrium:
Upper piston: `P_(0)A_(1) + m_(1) g + T = PA_(1)`
`T = (P - P_(0)) A_(1) - m_(1)g` (i)
Lower piston : `PA_(2) + m_(2)g = P_(0)A_(2) + T`
`T = (P - P_(0))A_(2) + m_(2)g` (ii)
`(i) = (ii)`
`(P - P_(0))A_(1) - m_(1)g = (P - P_(0))A_(2) + m_(2)g`
`(P - P_(0)) (A_(1) - A_(2)) = (m_(1) + m_(2))g`
`A_(1) - A_(2) = Delta A = 10 cm^(2), (m_(1) + m_(2)) = M = 5 kg`
`(P - P_(0))Delta A = Mg rArr P = P_(0) + (M g)/(Delta A) = 10^(5) + (5 xx 10)/(10 xx 10^(-4))`
`P = 1.5 xx 10^(5) N//m^(2)`
From ideal gas equation
`PV = nRT`
`PDelta V = n R Delta T`
`P(Delta A l) = nR Delta T`
`Delta T = (P(Delta A l))/(nR) = (1.5 xx 10^(5) xx 10 xx 10^(-4) xx 5 xx 10^(-2))/(1 xx 8.3)`
`= 0.9 K`