Question

A thermally isolated cylindrical closed vessel of height $$8 m$$ is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass $$8.3 kg$$. Thus the partition is held initially at a distance of $$4 m$$ from the top, as shown in the schematic figure below. Each of the two parts of the vessel contains $$0.1$$ mole of an ideal gas at temperature $$300 K$$. The partition is now released and moves without any gas leaking from one part of the vessel to the other. When equilibrium is reached, the distance of the partition from the top (in m) will be _______. (take the acceleration due to gravity $$= 10 ms^{-2}$$ and the universal gas constant $$= 8.3 J \ mol ^{-1} K^{-1}$$)

Answer 1

Initial Condition

$$PA\times 4=0.1R\times 300$$..........(i)

For both partition $$'A'$$ and $$'B'$$

After release

For $$A$$

$$P_AA\times x=0.1\times R\times 300$$............(ii)

For $$B$$

$$P_BA(8-x)=0.1\times R\times 30$$............(iii)

At equilibrium

$$P_AA+mg=P_BA$$

$$P_AA+8.3=P_BA$$...........(iv)

Put the value of equation (ii) and equation (iii) in equation (iv)

$$\dfrac{0.1\times 8.3\times 300}{y}+8.3g=\dfrac{0.1\times 8.3\times 300}{8-y}$$

$$y^2-2y-24=0$$

$$y=6m$$