InterviewSolution
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InterviewSolution
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`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)`. |
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Answer» `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` |
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