A container of volume `4V_(0)` made of a perfectly non`-` conducting material is divided into two equal parts by a fixed rigid wall whose lower half is non`-` conducting and upper half is purely conducting. The right side of the wall is divided into equal parts `(` initially `0` by means of a massless non`-` conducting piston free to move as shown . Section`A` constains `2 mol` of a gas while the section `B` and `C` contain ` 1 mol` each of the same gas `( gamma = 15. )` at pressure `P_(0)` . The heater in left part is switched on till the final pressure in section `C` becomes `125//27 P_(0)`. Calculate

Final temperature in part `C`.

Options

A. `(P_(0)V_(0))/(R)`
B. `(5P_(0)V_(0))/(3R)`
C. `(P_(0)V_(0))/(3R)`
D. `(5P_(0)V_(0))/(R)`

Answer

B

Explanation

For `'C',P_(0)V_(0)^(gamma)=(125)/(27)P_(0)V_(C)^(gamma)`
`V_(C)=V_(0)((9)/(25))`
`V_(B)=2V_(0)-(9V_(0))/(25)=((50-9))/(25)V_(0)`
` V_(B)=(41)/( 25)V_(0)`
`T_(0A)=T_(0B)=(P_(0)V_(0))/(R)`
`T_(C)=((125)/(27)P_(0)xx(9)/(25)V_(0))/( R)=(5P_(0)V_(0))/(3R)`
`T_(B)=((125)/(27)P_(0)xx(41)/(25)V_(0))/( R)=(205)/( 27)(P_(0)V_(0))/(3R)=T_(A)`
`Q=n_(A)C_(V)(T_(A)-T_(0))+n_(B)C_(V)DeltaT_(B)+n_(C)C_(V)DeltaT_(C)`
`Q=2.2R((205)/(27)-1)(P_(0)V_(0))/(R)+1.2R((205)/(27)-1)(P_(0)V_(0))/(R)+1.2R[(5)/(3)(P_(0)V_(0))/(R)-(P_(0)V_(0))/(R)]`
`=2P_(0)V_(0)[(178xx2)/( 27)+(178)/(27) +(2)/(3)]=(368)/( 9)P_(0)V_(0)`