`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)`.

Answer

In litre -atm `=(14)/(32) xx (0.0821 (141.4 - 273))/((1.4 - 1))`
`W_("rev") = - 11.82` litre atm (or) `-288 cal`

Explanation

`p_(1) = 10 atm` at `T = 273 k` for `(14)/(32)` mole `O_(2)`
`p_(2) = 1 atm` at `T = T_(2) k` for `(14)/(32)` mole `O_(2)`
For adiabatic expansion we have `T^(gamma) . P^(1 - gamma) =` cons tan t
`((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) :. T_(2) 141.4 K`
Work done in adiabatic expansion `= (nR)/((gamma - 1)) (T_(2) - T_(1))`