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Figure shows a horizontal cylindical container of length 30cm, which is partitioned by a tight fitting separator. The separator is diathermic but conducts heat very slowly. Initially the separator is in the state shown in figure. The temperature of left part of cylinder is 100 K and that on right part is 400K Initially the separator is in equilibrium. As heat is conducted from right to left part, seprator displaces to the right. Find the displacment of separaJor after a long time, when gases on the parts of cylinder are in thermal equilibrium. |
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Answer» Solution : It is given that initially the SEPARATOR is in equli- brium thus pressures of the GAS on both sides of the separator are equal say, it is `P_(i)`.,, If A be the area of cross-section of CYLINDER, number of moles of gas in left and right PARTS, `n_(1) and n_(2)` can be given as `n_(1) = (P_(i) (10A))/(R (100)) and n_(2) = (P_(i) (20A))/(R (400))` Finially, if separator is displaced to right by a distance x, we have `n_(1) = (P_(f) (10 + x) A )/(RT_(f)) and n_(2) = (P_(f) (20 - x)A)/(RT_(f))` Where `P_(f) and T_(f)` be the final pressure and temperature on both sides after a long time. Now if we equate the RATIO of moles `(n_(1))/(n_(2))` in initial and final states, we get `(n_(1))/(n_(2)) = ((10 A //100))/((20 A // 400 )) = ((10 + x)A)/((20 - x ) A )`or 2(20 - x) = 10 + x (or ) x = 10 cm Thus in final state, when gases in both parts are in thermal equilibrium, the pistion is displaced to 10 cm right from its initial postion. |
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