InterviewSolution
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InterviewSolution
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The rectangular box shown in Fig has partition which can slide without friction along the length of the box. Initially each of the two chambers of the box has one mole of a mono-atomic ideal gas `(lambda=5//3)` at a pressure `P_0`, volume `V_0` and temperature `T_0`. The chamber on the left is slowly heated by an electric heater. The walls of the box and the lead wires of the heater is negligible. The gas in the left chamber expands pushing the partition until the final pressure in both chambers becomes `243P_0//32`. Determine (i) the final temperature of the gas in each chamber and (ii) the work done by the gas in the right chamber. A. `2.25 T_(0)`B. `4.5 T_(0)`C. `8.75 T_(0)`D. `12.93 T_(0)` |
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Answer» Correct Answer - A For the left chamber, the process of heating is slow initially parameters are `P_(0),V_(0),T_(0)` Finally prarameters are `(243)/(32)P_(0),V_(1),T_(1)` `(P_(0)V_(0))/(T_(0))=((243P_(0))/(32))(V_(1))/(T_(1))` `:. T_(1)=(243)/(32)(V_(1))/(V_(0))T_(0)............(i)` Adiabatic compression ocurs in right chamber initial parameters are `P_(0),V_(0),T_(0)` Final parameters are `(243)/(32)P_(0),V_(2),T_(2)` `P_(0)V_(0)^(gamma)=(P_(0)243)/(32)V_(2)^(gamma) (V_(2)/V_(0))^(5//3)=(32)/(243)` `(V_(2))/(V_(0))=(8)/(27)V_(0) , V_(1)+V_(2)=2V_(0)` or ` V_(1)=2V_(0)-2V_(0)-(8)/(27)V_(0)=(46)/(27)V_(0)....(ii)` (i)and (ii)`impliesT_(1)=12.93T_(0)` For right chamber `(T_(0)/(T_(2)))^(gamma) ((P_(0)xx32)/(243 P_(0)))^(gamma-1)(T_(0))/(T_(2))=((2)/(3))^(2)=(4)/(9)T_(2)=2.25T_(0)` Work done by gas in right chamber `W=(P_(2)V_(2)-P_(0)V_(0))/(1-gamma)=-(3)/(2)[(9)/(4)-1]P_(0)V_(0)` `=-(3)/(2)xx(5)/(4)RT_(0)=-(15)/(8)xx8.3 T_(0)J=-15.58T_(0)J` |
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