An ideal gas at `NTP` is enclosed in an adiabatic vertical cylinder havin an area of cross section `A=27 cm^(2)` between two light movable pistons as shown in figure. Spring with force constant `K=37-- N//m` is in a relaxed state initially. Now the lower position is moved upwards a distance `h//2`, `h` being the initial length of gas column . It is observed that the upper piston moves up by a distance `h//16`. Final temperature of gas` 4//3 xx 273 K`. Take `gamma` for the gas to be `3//2`.

When the lower pistone is moved upwards a distance `h//2`, the compression.

Answer

`1.6m , 364K`

Explanation

New length of gas column =`(h)/(2) + (h)/(16) = (9)/(16)h`
New volume of gas = `((9)/(16)h)A = (9)/(16)V_(0)`
`P = P_(0) + (kx)/(A)`
As `PV^(gamma)` = constant so `P_(0)V_(0)^(1.5) = P((9)/(16)V_(0))^(1.5)`
`implies P = (64)/(27)P_(0) = P_(0) + (kx)/(A) implies (kx)/(A) = (37)/(27)P_(0)`
But `(kx)/(A) = (3700)((h)/(16)) = ((37)/(27))(10^(5)) h= 1.6m`
Now `T_(1)V_(1)^(gamma-1) = T_(2)V_(2)^(gamma-1)`
`implies (273)(V_(0))^(0.5) = (T_(2))((9)/(16)V_(0))^(0.5)`
`implies T_(2) = (4)/(3) xx 273 = 364K`