Fig. shows a horzontal cylindrical container of length `30 cm`, which is partitioned by a tight-fiting separator. The separator is diathermic but conductws heat very slowly. Initially the separator is the state shown in the figure. The temperature of left part of cylinder is `100 K` and that on right part is `400 K`. Initially the separator is in equilibrium. As heat is conducted from right to left part, of separator after a long when gases on the two displacement of separator after a long when gases on the two parts of cylinder are in thermal equilibrium.

Explanation

It is given that initially the separator is in equilibrium thus pressure on both sides of gas 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 part `n_(1)` and `n_(2)` can be given as
`n_(1)=(P_(1)(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 moles `(n_(1))/(n_(2))` in initial 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=10" cm"`
Thus in final state when gases in both parts are in thermal equilibrium, the piston is displaced to 10 cm right from its intial position.