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# Solve this problem A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder . The length of the cylinder above the piston is l1, and

A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder . The length of the cylinder above the piston is l1 , and that below the piston is l2 , such that l1>l2 . Each part of the cylinder contains n moles of an ideal gas at equal temperature T . If the piston is stationary , its mass , m , will be given by :

( R is universal gas constant and g is the acceleration due to gravity )

• Option 1)

$\frac{nRT}{g}\left [ \frac{1}{l_{2}} + \frac{1}{l_{1}} \right ]$

• Option 2)

$\frac{RT}{ng}\left [ \frac{l_{1} - 3l_{2}}{l_{1}l_{2}} \right ]$

• Option 3)

$\frac{nRT}{g}\left [ \frac{l_{1} - l_{2}}{l_{1}l_{2}} \right ]$

• Option 4)

$\frac{RT}{g}\left [ \frac{2l_{1} +l_{2}}{l_{1}l_{2}} \right ]$

Views

Ideal gas equation -

$PV = nRT$

- wherein

T= Temprature

P= pressure of ideal gas

V= volume

n= numbers of mole

R = universal gas constant

Draw the FBD on piston

so

P1A + mg =  P2A ........(1)

and $PV = nRT$ .......(2) → put in equatopn (1)

volume = l X A

$\frac{nRT}{v_{1}}\times A+mg =\frac{nRT}{V_{2}}A$

$\frac{nRT}{l_{1}A}\times A+mg = \frac{nRT}{l_{2}A}\times A$

$mg = \frac{nRT}{g}\left ( \frac{l_{1}-l_{2}}{l_{1}l_{2}} \right )$

Option 1)

$\frac{nRT}{g}\left [ \frac{1}{l_{2}} + \frac{1}{l_{1}} \right ]$

Option 2)

$\frac{RT}{ng}\left [ \frac{l_{1} - 3l_{2}}{l_{1}l_{2}} \right ]$

Option 3)

$\frac{nRT}{g}\left [ \frac{l_{1} - l_{2}}{l_{1}l_{2}} \right ]$

Option 4)

$\frac{RT}{g}\left [ \frac{2l_{1} +l_{2}}{l_{1}l_{2}} \right ]$

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