A cylindrical container having non-conducting walls is partitioned in two equal parts such that the volume of the each parts is eqaul to `V_(0)` A movable non-conducting piston is kept between the two parts. Gas on left is slowly heated so that the gas on right is compressed upto volume `V_(0)//8`. Find pressure and temperature on both sides if initial pressure and temperature, were `P_(0)` and `T_(0)` respectively. Also find heat given by the heater to the gas (number of moles in each parts is n)

Explanation

Since the process on right is adiabatic therefore
`PV^(gamma) = constant`
`rArr P_(0)V_(0)^(gamma) = P_(final) (V_(0)//8)^(gamma) rArr P_("final") = 32 P_(0)`
`T_(0)V_(0^(gamma-1) = T_("final") (V_(0)//8)^(gamma-1) rArr T_("final") = 4T_(0)`
Let volume of the left part is `V_(1)`
`rArr 2V_(0) = V_(1) +(V_(0))/(8) rArr V_(1) = (15V_(0))/(8)`.
Since number of moles on left part remains constant therefore for the left part `(PV)/(T) =` constant.
Final pressure on both sides will be same
`rArr (P_(0)V_(0))/(T_(0)) = (P_("final")V_(1))/(T_("final")) rArr T_("final") = 60 T_(0)`
`DeltaQ = DeltaU +W`
`DeltaQ = n (5R)/(2) (60T_(0) -T_(0)) +n(3R)/(2) (4T_(0)-T_(0))`
`DeltaQ = (5nR)/(2) xx 59T_(0) +(nRT)/(2) xx 3T_(0) = 152 nRT_(0)`