An insulting container of volume `2V_(0)` is divided in two equal parts by a diathermic (conducting) fixed piston. The left part contains one mole of monoatomic ideal gas whereas right part contains two moles of the same gas. Initial pressue on each side is `P_(0)`. The system is left for sufficient time so that a steady state is reached. Find (a) the work done by the gas in the left part during the process, (b) initial temperature in the two sides, (c) the final common temperature reached by the gases, (d) the heat given to the gas in the right part and (e) the increase in the internal energy of the gas in the left part.

Answer

(a) zero
(b) `(P_(0)V_(0))/(R ) (left), (P_(0)V_(0))/(2R) (right)`
(c ) `(2P_(0)V_(0))/(3R)`
(d) `(P_(0)V_(0))/(2)`
(e) `(-P_(0)V_(0))/(2)`

Explanation

(a) Change in volume `DeltaV = 0`
Work doen by the gas in the left part during the process `= int PdV =0`
(b) Temperature in the two part in the benginning
`T_(1) = (P_(0)V_(0))/(1xxR) = (P_(0)V_(0))/(R), T_(2) = (P_(0)V_(0))/(2R) = (P_(0)V_(0))/(2R)`

outer wall is adiabatic therefore no heat transfer form wall.
`U_(1) +U_(2) = U'_(1) +U'_(2)`
`(f)/(2) n_(1)RT_(1) +(f)/(2) n_(2)RT_(2) = (f)/(2)n_(1)RT +(f)/(2)n_(2)RT`
`T = (n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2)) = (1((P_(0)V_(0))/(R))+2((P_(0)V_(0))/(2R)))/(1+2)`
`rArr T = (2P_(0)V_(0))/(3R)`
(d) Heat given to the gas in right part is
`Q = DeltaU = (3)/(2)nR (T - T_(2))`
`Q = (3)/(2) xx 2R. ((2P_(0)V_(0))/(3R) -(P_(0)V_(0))/(2R)) = (P_(0)V_(0))/(2)`
(e) Increases in the internal energy of the gas in the left part is `DeltaU = (3)/(2) nR (T - T_(1))`
`DeltaU = (3)/(2) xx1 xx R ((2P_(0)V_(0))/(3R)-(P_(0)V_(0))/(R)) =- (P_(0)V_(0))/(2)`