At 300 K, the position of the piston in the cylinder o volume, say V, is represented as :
Let `p_(1)` and `p_(2)` be the pressure at the upper and lower part of the cylinder respectively. Let the pressure at the lower part due to the weight of the piston of the cylinder be `p_(0)`.
`therefore p_(2) = p_(1) + p_(0)`. ....(1)
In the two parts of the cylinder, the no. of moles of air are same. As the temperature is also same (300 K),
We have `p_(1) xx (4V)/(5) = p_(2) xx (V)/(5)`,
or `4p_(1) = p_(2) = p_(1) + p_(0), p_(1) = (p_(0))/(3)` ......(2)
Now let the temperature be T at which the volume of the upper part will be three times than that of the lower part.
Let the pressures at the uppr and lower parts be `p._(1)` and `p._(2)` respectively. Thus,
`p._(2) = p._(1) + p_(0)` ........(3)
Again in both the parts, temperature and no. of moles are the same, we have,
`p._(1) xx (3V)/(4) = p._(2) xx (V)/(4)`.
or `3p._(1) = p._(2) = p._(1) + p_(0)`
or `p._(1) = (p_(0))/(2)`. .........(4)
From (2) and (4), we have , `p._(1) = (3)/(2)p_(1)` ......(5)
Now, for the upper part of the cylinder at temperature T,
We have, `p._(1) xx (3V)/(4) = RT " "(n = 1)`
Substituting `p._(1)` from (5), we have,
`(3)/(2)p_(1) xx (3V)/(4) = RT`
`(9)/(8) p_(1) V = RT` .......(6)
Further, for the upper part of the cylinder at 300 K,
We have, `p_(1) xx (4V)/(5) = R(300)` ......(7)
From (6) and (7), we get , T = 421.9 K.
Second method At 300 K,
`P_(1) xx (4V)/(5) = R xx 300` and `P_(2) xx (V)/(5) = R xx 300`
At a temperature T (say),
`P._(1) xx (3V)/(4) = RT` and `P._(2) xx (V)/(4) = RT`
Substituting `P_(1), P_(2), P._(1)` and `P._(2)` from the above equations in
`P_(2) - P_(1) = P._(2) - P._(1)`
we get, T = 421.9 K.
