A smooth vertical tube having two different cross sections is open from both the ends but closed by two sliding pistions as shown in Fig. and tied with an inextensible string. One mole of an ideal gas is enclosed between the piston The difference in cross-sectional areas of the two pistons is given `Delta S`. The masses of piston are `m_(1)` and `m_(2)` for larger and smaller one, respectively. Find the temperature by which tube is raised so that the pistons will be displaced by a distance l. Take atmospheric pressure equal to `P_(0)`

Explanation

If initial pressure of gas is P and let `S_(1)` and `S_(2)` be the cross-sectional area of the larger and the smaller piston, respectivley, then for equilibrium of the two piston, we have
For larger piston
`P_(0) S_(1) m_(1) g + T = PS_(1)` (if T is the tension in string) (i)
For smaller piston
`PS_(2) + m_(2) g - (T) + P_(0) S_(2)` (ii)
Adding Eqs. (i) adn (ii), we get
`P_(0) (S_(1) - S_(2)) + m_(1) g + m_(2) g = P (S_(1) - S_(2)) + m_(1) g+ m_(2) g = P (S_(1) - S_(2))`
or `P_(0) ((m_(1) + m_(2))/(Delta S)) g = P` (iii)
In gas temperature is increased fromm `T_(1)` to `T_(2)`, the volume of gas increase from V to `V + l Delta S` as l is the displacement of pistons, them gas law we must have
`PV = RT_(1)` (for initial state) (iv)
`P (V + l Delta S) = RT_(2)` (for final state) (v)
According to Eq. (iii), we find the pressure of the gas does not change, as does not depend on temperature.
From Eqs. (iv) and (v), if we subtract the these equation, we get
`P l Delta S = R (T_(2) - T_(1))`
or `T_(2) - T_(1) = (P l Delta S)/(R )`
`= (P_(0) + ((m_(1) + m_(2)))/(Delta S) g) (l Delta S)/(R)`
`[P_(0) Delta S + (m_(1) + m_(2))g] (1)/(R)`