In given figure, an adiabatic cylindrical tube of volume `2V_(0)` is divided in two equal parts by a frictionless adiabatic separator. An ideal gas in left side of a tube having pressure `P_(1)` and temperature `T_(1)` where as in the right side having pressure `P_(2)` and temperature `T_(2).C_(p)//C_(v) =gamma` is the same for both the gases. The separator is slid slowly and is released at a position where it can stay in equilibrium. Find (a) the final volumes of the two parts (b) the heat given to the gas in the left part and (c) the final common pressure of the gases,

Answer

(a) `(2p_(1)^(1//gamma)V_(0))/(A).(2p_(2)^(1//gamma)V_(0))/(A)`,
(b) zero,
(c ) `(A//2)^(gamma)` where `A = p_(1)^(1//gamma) +p_(2)^(1//gamma)`

Explanation

for left part
`P_(1) (V_(0))^(gamma) = PV^(gamma) ….(1)`
for right part
`P_(2) (V_(0))^(gamma) =P(2V_(0)-V)^(gamma) ….(2)`
from eq. (1) & (2), we get
`V_(1) = V = (2V_(0)P_(1)^(1//gamma))/(P_(1)^(1//gamma)+P_(2)^(1//gamma)) , V_(2) =2V_(0)-V = (2V_(0)P_(2)^(1//gamma))/(P_(1)^(1//gamma)+P_(2)^(1//gamma))`
(b) heat given to the gas in left par is zero
[As process is adiabatic]
(c ) fro eq (1)
`P_(1)(V_(0))^(gamma) = ((2V_(0)P_(1)^(1//gamma))/(P_(1)^(1//gamma)+P_(2)^(1//gamma)))^(gamma)P`
`rArr P = ((P_(1)^(1//gamma)+P_(2)^(1//gamma))/(2))^(gamma)`