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

The physical system consists of one mole of gas confined in the smooth vertical tube. Let `m_1` and `m_2` be the masses of upper and lower pistons and `S_1` and `S_2` are their respective areas.
For the lower piston
`p S_2 + m_2 g = p_0 S_2 + T`,
or `T = (p - p_0) S_2 + m_2 g` ...(1)
Similarly for the uper piston
`p_0 S_1 + T + m_1 g = p S_1`,
or, `T = (p - p_0) S_1 - m_1 g` ....(2)
From (1) and (2)
`(p - p_0)(S_1 - S_2) = (m_1 + m_2) g`
or, `(p - p_0) Delta S = mg`
so, `p = (mg)/(Delta S) + p_0 = constant`
From the gas law, `p V = vRT`
`p Delta V = v R Delta T` (because `p` is constant)
So, `(p_0 + (mg)/(Delta S)) Delta S l = R Delta T`
Hence, `Delta T = (1)/(R) (p_0 Delta S + ,g) l = 0.9 K`.
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