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)` are the cross-sectional areas of the larger and smaller piston then for equilibrium of the two pistons we have
For larger piston
`P_(0)S_(1)+m_(1)g+T=PS_(1)`
[If T is the tension in string] ...(2.16)
For smaller piston
`PS_(2)+m_(2)g=T+P_(0)S_(2)" "...(2.17)`
Adding equation-(2.16) and-(2.17), we get
`P_(0)(S_(1)-S_(2))+m_(1)g+m_(2)g=P(S_(1)-S_(2))`
or `" "P_(0)+((m_(1)+m_(2))/(DeltaS))g=P" "...(2.18)`
If gas temperature is increased from `T_(1)` to `T_(2)` the volume of gas increases from V to `V+l DeltaS` as l is the displacement of pistons, then from gas law we must have
`P.V=RT_(1)" "["For initial state"]" "...(2.19)`
`P(V+l DeltaS)=RT_(2)" "["For final state"]" "...(2.20)`
According to equation-(2.18)
Pressure of gas does not change as it does not depend on temperature
From equation-(2.19) and (2.20), if we subtract these equation, we get
`P.l DeltaS=R(T_(2)-T_(1))`
or `" "T_(2)-T_(1)=(Pl DeltaS)/(R)`
`=(P_(0)+((m_(1)+m_(2)))/(DeltaS)g)(l DeltaS)/(R)`
`=[P_(0)DeltaS+(m_(1)+m_(2))g](l)/(R)`