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`1 mol` of `NH_(3)` gas at `27^(@)C` is expanded under adiabatic condition to make volume `8` times `(gamma=1.33)`. Final temperature and work done, respectively, areA. `150K, 900 cal`B. `150K, 400cal`C. `250K, 100cal`D. `200K, 800cal` |
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Answer» `T_(2)= T_(1) ((V_(1))/(V_(2)))^(gamma-1) = 150K` `w =- C_(V) DeltaT =- C_(V) (T_(2) -T_(1))` `=- 3 xx 2xx (150 -300)` `= 900 kcal` |
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