The insulated box shown in figure has an insulated partition which can slide without friction along the length of the box. Initially each of the two chambers of the box has `1 mil` of a monatomic ideal gas `( gamma= 5//3)` at a pressure `P_(0)`, volume `V_(0)` and temperature `T_(0)`. The chamber of the left is slowly heated by an electric heater so that its gas, pushing the partition, expands until the final pressure in both the chambers becomes `243 P_(0)//32`. Determine

Final temperature fo the gas in each chamber.

Options

A. `(15)/(8) RT_(0)`
B. `-(15)/(8) RT_(0)`
C. `(7)/(4) RT_(0)`
D. `-(7)/(4) RT_(0)`

Answer

B

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)`