`14 g` oxygen at `0^(@)C` and `10 atm` is subjected to reversible adiabatic expasnion to a pressure of `1atm`. Calculate the work done in
a. Litre atomsphere.
b. Caloride (given, `C_(P)//C_(V) = 1.4)`.

Explanation

`P_(1) = 10atm` at `T = 273 K` for `(14)/(32) mol O_(2)`.
`P_(2) atm` at `T = T_(2)` for `(14)/(32) mol O^(2)`.
For adiabatic expansion, we have `T^(gamma).P^(1-gamma) =` constant
`:. ((T_(1))/(T_(2)))^(gamma) = ((P_(2))/(P_(1)))^(1-gamma)` or `gamma "log" (T_(1))/(T_(2)) - (1-gamma) "log"(P_(2))/(P_(1))`
or `1.4 "log"(273)/(T_(2)) = (1-1.4)"log" (1)/(10) rArr T_(2) = 141.14K`
Work done in adiabatic expansion `= (nR)/((gamma -1)) (T_(2) - T_(1))`
In `L-atm = (14)/(32) xx (0.0821(141.4-273))/((1.4-1))`
`W_(rev) = 11.82L-atm`
In `cal = (14)/(32) xx (2xx(141.4-273))/((1.4-1))`
`W_(rev) =- 287.88cal`