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InterviewSolution
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When an ideal gas is taken from state `a` to `b`, along a path `acb, 84 kJ` of heat flows into the gas and the gas does `32 kJ` of work. The following conclusions are drawn. Mark the one which is not correct. A. If the work done along the path `adb` is `10.5 kJ`, the heat will flow into the gas is `62.5 kJ`B. When the gas is returned from `b` to `a` along the curved path, the work done on the gas is `21 kJ`, and the system absorbs `73 kJ` of heatC. If `U_(a) = 0, U_(d) = 42 kJ`, and the work done along the path `adb` is `10.5 kJ` then the heat absorbed in the process `ad` is `52.4 kJ`.D. If `U _(a) = 0, U_(d) = 42 kJ`, heat absorbed in the precess `db` is `10 kJ`. |
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Answer» Correct Answer - B b. For path `acb`: `Delta Q = Delta U + Delta W` `implies 84 = Delta U + 32 implies Delta U = 52 kJ` Hence `Delta U_(acb) = Delta U_(ab) = Delta U_(adb) = 52 J` For path `adb`: `Delta Q = Delta U + Delta W` `= 52 + 10.5 = 52.5 kJ` So option (a) is correct. For process `ba`, system will release the heat. So option (b) is wrong. For path `ad`: `Delta W_(adb) = Delta W_(ad) + Delta W_(db)` `implies 10.5 = Delta W_(ad) + 0` `implies Delta W_(ad) = 10.5 kJ` `Delta Q_(ad) = Delta U_(ad) + Delta W_(ad)` `= (42 - 0) + 10.5` `= 52.5 kJ` So option (c ) is correct. `Delta Q_(adb) = 52 + 10.5 = 62.5 kJ` `Delta Q_(db) = Delta Q_(adb) - Delta Q_(ad)` `= 62.5 52.5` `= 10 kJ` So option (d) is correct. Hence answer of this question is (b) |
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