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 and (ii) the work done by the gas in the right chamber.

Options

A. `4 //9 T _(0)`
B. `4//7 T_(0)`
C. `1// 5 T_(0)`
D. `9 // 4 T _(0)`

Answer

D

Explanation

For the right part,
`P_(0)V_(0)^(gamma)=PV_(2)^(gamma)`
`V_(2)=((P_(0))/(P))^(1//gamma)V_(0)=((32)/( 243))^(3//5)V_(0)`
`V_(2)=(8)/(27)V_(0)`
Monatomic gas
`gamma = 5//3`
`V_(1)=2V_(0)-(8)/(27)V_(0)=(46)/(27)V_(0)`
For left part,
`=(P_(0)V_(0))/(T_(0))=(PV_(1))/(T_(1))`
`T_(1)=(PV_(1))/(P_(0)V_(0))T_(0)=((243P_(0))/(32)xx(46)/( 27))/(P_(0)V_(0))T_(0)`
`T_(1)=(207)/(16)T_(0)`
Now applying ideal gas equation in right part,
`T_(2)=(PV_(2))/(P_(0)V_(0))T_(0)=(9)/(4)T_(0)`
For right part,
`Delta Q=0=dU+dW`
`dW=-Du=-1C_(v)RT`
`=-(3)/(2) R((9)/(4)T_(0)-T_(0))=-(15)/(8)RT_(0)`