Question

Diagram shows a horizontal cylindrical 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 the diagram. 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 when gases on the two parts of cylinder are in thermal equilibrium.

Answer 1

Answer: 10

It is given that initially the separator is in equilibrium; thus pressure on both sides of the gas is 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, $<br/>n_1=\frac{P_i(10A)}{R(100)}<br/>$
and $n_2=\frac{P_i(20A)}{R(400)}$
Finally, if separator is displaced to right by a distance $x$, we have
$<br/>n_1=\frac{P_f(10+x)A}{R{T}_f}<br/>$
and $n_2=\frac{P_f(20-x)A}{R{T}_f}$
Here $P_f$ and $T_f$ are the final pressure and temperature on both sides after a long time.
Now if we equate the ratio of moles in initial and final states, we get
$<br/>\frac{n_1}{n_2}=\frac{(10A/100)}{(20A/400)}=\frac{(10+x)A}{(20-x)A}<br/>$
$<br/>2(20-x)=10+x<br/>$
$<br/>x=10cm<br/>$