1.

Figure shows a horizontal cylindrical container of length 30 cm, 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 100K and that on right part is 400K. Initially the separator is in equilibrium. As heat is conducted from right to left part, separator displaces to the right. Find the displacement of separator after a long time when gases on the two parts of cylinder are in thermal equilibrium.

Answer»

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Solution :It is given that initially the separator is in equilibrium thus pressure in both sides of GAS are equal say, it is `P_(1)`,. If A be the area of cross-section of cylinder, number of moles of gas in left and right part `n_(1)" and "n_(2)` can be given as
`n_(1)=(P_(i)(10A))/(R(100))" and "n_(2)=(P_(i)(20A))/(R(400))` Finally 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))`
If `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 initially and final state we GET
`n_(1)/n_(2)=((10A//100))/((20A//400))=((10+x)A)/((20-x)A)" or "2(20-x)=10+x" (or) "x=10cm`
Thus in final state when gases in both parts are in thermal equilibrium, the piston is displaced to 10 cm right from its initial position.


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