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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 = 10ms2 and the universal gas constant =8.3Jmol1K1).


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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=10ms2

Universal gas constant R=8.3Jmol1K1

Step 2: Find the distance of the partition from the top:

Let the area of partition be Athen pressure by the partition will be, mgA.

Let v1 be the volume of cylinder for Pressure P1 and v2 be the volume of cylinder for Pressure P2.

Let his the height of the cylinder at equilibrium and x is the distance of partition from the top as shown in figure.

Since the pressure P2 is the balanced pressure of P1 and pressure mgA by the partition hence,

We can write,

P2=P1+mgA

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,

nRTv2=nRTv1+mgAmgA=nRT1v2-1v1

We know that, Volume of the Cylinder V=Area of the Circle πr2×Height of the Cylinder H

Since the partition is also a circle hence its Area A=πr2

Therefore, we can write v2=Ah-x and v1=Ah+x (where his 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

mgA=0.1×8.3×3001A4-x-1A4+x

8.3×10A=0.1×8.3×300A14-x-14+x

13=4+x-4+x16-x2

6x=16x2

0=x2+6x16

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 6m.


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