The diagram 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 the diagram. The temperature of the left part of the cylinder is 100 K and that on the right part is 400 K. Initially the separator is in equilibrium. As heat is conducted from the right to left part, separator displaces to the right. Find the displacement of separator (in cm) after a long when gases on the two parts of the cylinder are in thermal equilibrium

Explanation

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.