The rectangular box shown in Fig has partition which can slide without friction along the length of the box. Initially each of the two chambers of the box has one mole of a mono-atomic ideal gas `(lambda=5//3)` at a pressure `P_0`, volume `V_0` and temperature `T_0`. The chamber on the left is slowly heated by an electric heater. The walls of the box and the lead wires of the heater is negligible. The gas in the left chamber expands pushing the partition until the final pressure in both chambers becomes `243P_0//32`. Determine (i) the final temperature of the gas in each chamber.

Answer

(i) `T =12.94 T_(0) T_(2) = 2.25 T_(0)`
(ii)` -1.875RT_(0)`

Explanation

(i) For the right chamber `P_(0)^(1-gamma)T_(0)^(gamma) = ((243P_(0))/(32))^(1-gamma) . T_(2)^(gamma)`
`implies T_(2) = (9)/(4)T_(0)` and `(P_(0)V_(0))/(T_(0)) = (((243P_(0))/(32)). V_(2))/(((9)/(4)T_(0)))implies V_(2) = (9)/(4) xx (32)/(243)V_(0) = (8)/(27)V_(0)`
For the left chamber `(P_(0)V_(0))/(T_(0)) = (P_(1).V_(1))/(T_(1)) "where" P_(1) = (243)/(32)P_(0)`
`V_(1) = 2V_(0) - (8)/(27)V_(0) = (46)/(27)V_(0)`
`implies T_(1) = (T_(0))/(P_(0)V_(0)) xx ((243)/(32)P_(0))((46)/(27)V_(0)) = (9xx 23)/(16)T_(0) = 12.90 T_(0)`
(ii) Work done by the gas in the right chamber : `(P_(1)V_(1) - P_(2)V_(2))/(gamma-1) = (nR(T_(1) - T_(2)))/(gamma-1) = (1xxR(T_(0) - (9)/(4)T_(0)))/(((5)/(3)- 1)) =-(3)/(2)xx R xx (5)/(4)T_(0) = -1.875RT_(0)`