Correct Answer - (a) `sqrt(2) xx 10^(5)Pa` (b) `200sqrt(2)K` (c ) `402 (2-sqrt(2))J`
(a) For adiabatic `P_(1)V_(1)^(gamma) = P_(2)V_(2)^(gamma)`
`P_(2) = 4 xx 10^(5) ((V)/(2V)) = 4 xx 10^(5) xx2^(-3//2)`
`P_(2) = sqrt(2) xx 10^(5) Pa`
(b) `T_(1) V_(1)^(gamma-1) = T_(2) V_(2)^(gamma-1)`
`T_(2) = 400 ((V)/(2V))^((3)/(2)-1) = 400 xx2^(-1//2)`
`T_(2) = 200 sqrt(2)`
(c ) `W = (-nRDeltaT)/((gamma-1))`
As `n = (PV)/(RT) = (4 xx 10^(5) xx 100 xx 10^(-6))/((25)/(3)xx400) = (3)/(25)` moles
`:. W =- (3 xx 25 xx (200sqrt(2)-400))/(250xx3((3)/(2)-1))`
`=- (1 xx 100(2sqrt(2)-4)xx2)/(10)`
`W = 40 (2-sqrt(2))J`