Two containers `A` and `B` of equal volume `V_(0)//2` each are connected by a narrow tube which can be closed by a value. The containers are fitted with pistons which can be moved to change the volumes. Initially the value is open and the containers contain an ideal gas `(C_(p)//C_(v)=gamma)` at atmospheric pressure `P_(0)` and atmospheric temperature `2T_(0)`. The walls of the containers `A` are highly conducting and of `B` are non-conducting. The value is now closed and the pistons are slowly pulled out to increase the volumes of the containers to double the original value. (a) Calculate the temperature and pressure in the two containers. (b) The value is now opened for sufficient time so that the gases acquire a common temperature and pressure. Find the new values of the temperature and the pressure.

Answer

(a) `2T_(0), (p_(0))/(2)` in the vessel `A` and `(T_(0))/(2^(gamma-2)),p_(0)//2^(gamma)` in vessel `B`,
(b) `2T_(0), P_(0)//2]`

Explanation

For vessel `A,` Temperature is constant due to diathermic wall. For vessel `B`, no heat exchange due to adiabatically.
For vessel `A P_(1)V_(1) =P_(2)V_(2) ( :' T = "const")`
`P_(2) = (P_(0)(V_(0)//2))/(V_(0)) = (P_(0))/(2)`
Final temperature and pressue of vessel `A` is `2T` and `(P_(0))/(2)`
For vessel `B` adiabatic process.
`P_(2) = (P_(1)V_(1)^(gamma))/(V_(2)^(gamma)) = (P_(0)(V_(0)//2)^(gamma))/((V_(0)^(gamma)) = (P_(0))/(2^(gamma))`
`T_(2) = (T_(1)V_(1)^(gamma-1))/(V_(2)^(gamma-1)) = (2T_(0)(V_(0)//2)^(gamma-1))/((V_(0)^(gamma-1)) = (T_(0))/(2^(gamma-2))`
Final temperature and pressure of vessel `B` is `(T_(0))/(2^(gamma-1))` and `(P_(0))/(2^(gamma))`
(b) After value is open heat is flow to atmosphere from vessel `A`. So common temperature of both vessel is `2T_(0)`.
From mole conservation.
`n_(1) +n_(2) = n_(1)+n_(2) = (P_(0)(V//2))/(R2T_(0)) +(P_(0)(V_(0)//2))/(R2T_(0))`
`=(PV_(0))/(R2T_(0)) +(PV_(0))/(R2T_(0)) rArr P = (P_(0))/(2)`