Question

# A thermally isolated cylindrical closed vessel of height $8m$ is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass $8.3kg$. Thus, the partition is held initially at a distance of $4m$ from the top, as shown in the schematic figure below. Each of the two parts of the vessel contains $0.1mole$ of an ideal gas at temperature $300K$. 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 = $10m{s}^{-2}$ and the universal gas constant =$8.3Jmo{l}^{-1}{K}^{-1}$).

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Solution

## Step 1. Given Data:Height of vessel $H=8m$Mass of partition $m=8.3kg$The temperature of ideal gas $T=300K$Acceleration due to gravity $g=10m{s}^{-2}$ Universal gas constant $R=8.3Jmo{l}^{-1}{K}^{-1}$Step 2: Find the distance of the partition from the top:Let the area of partition be $A$then pressure by the partition will be, $\frac{mg}{A}$.Let ${v}_{1}$ be the volume of cylinder for Pressure ${P}_{1}$ and ${v}_{2}$ be the volume of cylinder for Pressure ${P}_{2}$.Let $h$is the height of the cylinder at equilibrium and $x$ is the distance of partition from the top as shown in figure.Since the pressure ${P}_{2}$ is the balanced pressure of ${P}_{1}$ and pressure $\frac{mg}{A}$ by the partition hence,We can write,${P}_{2}={P}_{1}+\frac{mg}{A}$Using Ideal Gas Equation, $PV=nRT$ (where, $P$ is Pressure, $V$ is Volume, $n$ is number of moles, $R$ is Universal gas constant and $T$ is temperature.) we can write the above equation as,$⇒\frac{nRT}{{v}_{2}}=\frac{nRT}{{v}_{1}}+\frac{mg}{A}\phantom{\rule{0ex}{0ex}}⇒\frac{mg}{A}=nRT\left[\frac{1}{{v}_{2}}-\frac{1}{{v}_{1}}\right]$We know that, Volume of the Cylinder $V=$Area of the Circle $\left({\mathrm{\pi r}}^{2}\right)×$Height of the Cylinder $H$Since the partition is also a circle hence its Area $A=\pi {r}^{2}$Therefore, we can write ${v}_{2}=A\left(h-x\right)$ and ${v}_{1}=A\left(h+x\right)$ (where $h$is the height of the cylinder at equilibrium and $x$ is the distance of partition from the top.) Then from figure, we can write, $h=4$$⇒$ $\frac{mg}{A}=0.1×8.3×300\left[\frac{1}{A\left[4-x\right]}-\frac{1}{A\left[4+x\right]}\right]$$⇒$$\frac{8.3×10}{A}=\frac{0.1×8.3×300}{A}\left[\frac{1}{4-x}-\frac{1}{4+x}\right]$$⇒$ $\frac{1}{3}=\frac{4+x-4+x}{16-{x}^{2}}$$⇒$ $6x=16–{x}^{2}$$⇒$ $0={x}^{2}+6x–16$$⇒$ $x=–8,2$ Since $x$ is a distance hence Ignoring the negative value,The distance of the partition from the top at the equilibrium position will be $h+x=4+2=6m$.Hence, the distance of the partition from the top is $\mathbf{6}\mathbf{}\mathbit{m}$.

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