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.

Explanation

Initially pressure, volume and temperature of chambers are same `(P_(0),V_(0) and T_(0))`. Due to heating partition starts to slide and compresses the right portion. We can represent initial and final states as ,brgt
(a) For right chamber, compression is adiabatic
Hence we can apply
`PV^(gamma)`=constant
`impliesP_(0)(V_(0))^(5//3)=32P_(0)(V_(0)V)^(5//3)`
`((V_(0))/(V_(0)-V))=(32)^(3//5)=2^(3)=8`
`impliesV_(0)=8V_(0)-8V`
`implies8V=7V_(0)`
`impliesV=(7V_(0))/(8)`
For right chamber we can also apply
`TV^(f)-1`=constant
`impliesT_(0)V_(0)^(2//3)=T_(2)(V_(0)-(7V_(0))/(8))^(2//3)`
`impliesT_(0)V_(0)^(2//3)=T_(0)((V_(0))/(8))^(2//3)`
`impliesT_(2)=T_(0)xx8^(2//3)=4T_(0)` . . . (ii)
For left chamber, we can apply conservation of moles
`impliesn=(P_(0)V_(0))/(RT_(0))=(32_(0)(V_(0)+V))/(RT_(1))`
`impliesT_(1)=(T_(0))/(V_(0))(V_(0)+V)=(T_(0))/(V_(0))xx(V_(0)+(7V_(0))/(8))=(T_(0))/(V_(0))xx(15V_(0))/(8)`
`impliesT_(1)=(15T_(0))/(8)`
(b) For gas in right chamber.
`Q=DeltaU+W`
`implies0=DeltaU+W`
`implies0=W=-DeltaU`
`=-nC_(V)(T_(2)-T_(0))`
`=-1xx(3R)/(2)(4T_(0)-T_(0))`
`impliesW=(-9RT_(0))/(2)`