Calculate the work done by the system in an irreversible (sing step) adiabatic expansion of 2 mole of a polyatomic gas `(gamma= 4//3)` from 300K and pressure 10atm to 1 atm: (in KJ) (Give your answer after multiplying with 2.08).

Answer

3

Explanation

In irreversible adiabatic process work done `w= -p_("ext") (v_(2) - v_(1))` (or) `w= (nR)/(gamma -1) [T_(2) - T_(1)]`
For irreversible process `pv^(gamma)` is not valid and final temperature is to be obtained by equating the above two expression
So, `-p_(2) (v_(2) - v_(1)) = (nR)/(gamma-1) [T_(2) - T_(1)]`
`p_(2) ((n T_(1)R)/(P_(1)) - (nT_(2) T)/(P_(2))) = (nR)/(gamma-1) [T_(2) - T_(1)]`
`(P_(2)T_(1))/(P_(1)) - T_(2) = (1)/(gamma-1) [T_(2) - T_(1)]`
So, `T_(1) [(P_(2))/(P_(1)) + (1)/(gamma-1)], T_(2) [1 + (1)/(gamma-1)]`
`T_(2) = T_(1) ((gamma-1)/(gamma)) [(P_(2))/(P_(1)) + (1)/(gamma-1)]`
`= 300 [(1//3)/(4//3) ] [(1)/(10) + (1)/(1//3)]`
`= 300 xx (1)/(4) xx [(31)/(10)] = 232.5`
`w= (2 xx 8.314 xx 10^(-3))/(1//3) [232.5- 300]` So, Ans 3