A rectangualr box (shown in figure) has a movable and smooth portition which can slide along the length of the box. Both chambers contains 1 mole of monoatomic gas (`gamma=(5)/(3)`) at a pressure walls of box and partition are thermally insulated. Due to heating, gas in left chamber expands until pressure in both chambers become `32P_(0)` determine
(a) The final temperature of gas in each chamber
(b) The work done by the gas in the right chamber.

Answer

A, B

Explanation

For the left chamber
`(P_0V_0)/(T_0)=(P_0xx243)/(32xxT_1)xxV_1`
`rArrT_1=234/32xx(V_1T_0)/(V_0)…..(i)`
For the right chamber for adiabatic compression
We get, `P_0V_0^(gamma)=P_0xx243/32xxV_2^gamma`
`rArr(V_2)/(V_0)=(32/243)^(3//5) rArr V_2=8/27 V_0`
But `V_1+V_2=2V_0`
`:. V_1=2V_0-V_2=2V_0-8/27V_0=46/27V_0....(ii)`
From (i) and (ii) `T_1=243/32xx(46xxV_0)/(V_0xx27)xxT_0`
or, `T_1=207/16T_0=12.9T_0 (approx.)
To find the temperature in the second chamber (right), we
apply
`((T_1)/(T_2))^(gamma)=((P_2)/(P_1))^(1-gamma)`
`rArr((T_0)/(T_2))^(5//3)=((243P_0)/(32P_0))^(1-5//3) rArr T_2=2.25 T_0`
Work done in right chamber (adiabatic process)
`W=1/(1-gamma)(P_2V_2-P_0V_0)`
`=-3/2[243/32P_0xx8/27V_0-P_0V_0]`
`=-3/2(9/4-1)P_0V_0=-15/8xxRT_0=-15.8T_0`