InterviewSolution
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InterviewSolution
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A cylindrical chamber `A` of uniform cross section is divided into two parts `X` and `Y` by a movalbe piston `P` which can slide without friction inside the chamber. Initially part `X` contains `1 "mol"` of a monochromatic gas and `Y` contains `2 mol` of a diatomic gas, and the volumes of `X` and `Y` are in the ratio `1:2` with both parts `X` and `Y` being at the same temperature `T`. Assuming the gases to be ideal, the work `W` that will be done in moving the piston slowly to the position where the ratio of the volumes of `X` and `Y` is `2:1` will be A. `-5.8T`B. `8.3 T`C. `12.3 T`D. zero |
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Answer» Correct Answer - A a. Since the piston is moved slowly we assume isothermal condition for both the gases as thermal equilibrium is maintained throughout. Let the total of chamber be. `V`. Then volume of gas in `X` increase (expand) from `V//3` to `2V//3`. The work done is positve and given (for 1 mol of monatomic gas by `W_(x) = + RT log_(e) .(2 V//3)/(V//3) = RT log_(e) 2` The volume of gas in `Y` decrease (compressed) from `2V//3` to `V//3`. The work done (isothermally) is negative and given (for 2 mol of diatomic gas) by `W_(x) = 2 RT log_(e) (V//3)/(2V//3) = - RT log_(e) 2` Hence the total work done on the system is `W = W_(x) + W_(Y) = - RT log_(e) 2` Substituting `R = 8.3, log_(2) 2 = 0.6996` and simplifiying `W = - 5.8 T`. |
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