A cylindrical tube with adiabatic walls having volume `2V_(0)`contains an ideal monoatomic gas as shown in figure. The tube is divided into two equal parts by a fixed super conducting wall. Initially, the pressure and the temperature are `P_(0), T_(0)` on the left and `2P_(0), 2T_(0)` on the rigid. When system is left for sufficient amount of time the temperature on both sides becomes equal

Options

A. Find work done by the gas on the right part?
B. Find the final pressure on the two sides
C. Find the final equilibrium temperature
D. How much heat has flows from the gas on the right to the gas on the left?

Answer

A, B, C, D

Explanation

(a) As change in volume `DeltaV = 0`
`:.` Work done by gas on right part `= int p dV = 0`
(b) and (c ) Due to adiabatic walls, no heat transfer form cylindrical tube.
`U_(1)+U_(2) = U_(1) +U_(2)`
`(3)/(2)n_(1)RT_(1) +(3)/(2) n_(2)RT_(2) =(3)/(2)n_(1)RT+(3)/(2)n_(2)RT`

`T = (n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2))`
`= (((P_(0)V_(0))T_(0))/(RT_(0))+(2P_(0)(V_(0))2T_(0))/(R(2T_(0))))/((P_(0)V_(0))/(RT_(0))+(2P_(0)V_(0))/(R(2T_(0))))=(3T_(0))/(2)`
From mole conservation
`(P_(0)(V_(0)))/(RT_(0)) = (P_(1)(V_(0)))/(RT) rArr p' = (3P_(0))/(2)`
`rArr` similarly `p'_(2) = (3P_(0))/(2)`
(d) Heat flow form the gas right to left.
`Q = DeltaU +W ( :' W = 0)`
`Q = DeltaU = (3)/(2)n_(1)R (T-T_(1))`
`= (3)/(2) (P_(0)V_(0)R)/(RT_(0)) ((3T_(0))/(2)-T_(0)) = (3P_(0)V_(0))/(4)`