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 $\mathrm{l}_{1}$, and that below the piston is $\mathrm{l}_{2}$, such that $\mathrm{I}_{1}>\mathrm{I}_{2}$. 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)

(D) $\frac{\mathrm{nRT}}{\mathrm{g}}\left[\frac{\mathrm{I}_{1}-\mathrm{I}_{2}}{\mathrm{I}_{1} \mathrm{I}_{2}}\right]$

(A) $\frac{\mathrm{RT}}{\mathrm{ng}}\left[\frac{\mathrm{I}_{1}-3 \mathrm{I}_{2}}{\mathrm{I}_{1} \mathrm{I}_{2}}\right]$

(B) $\frac{\mathrm{RT}}{\mathrm{g}}\left[\frac{2 \mathrm{I}_{1}+\mathrm{I}_{2}}{\mathrm{I}_{1} \mathrm{I}_{2}}\right]$

(C) $\frac{\mathrm{nRT}}{\mathrm{ng}}\left[\frac{1}{\mathrm{I}_{2}}+\frac{1}{\mathrm{l}_{1}}\right]$

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Heat & Thermodynamics

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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

l_{1}

, and that below the piston is

l_{2}

, such that

I_{1} > I_{2}

. 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)

(A) $\frac{RT}{ng} [\frac{I _{1} - 3 I _{2}}{I _{1} I _{2}}]$
(B) $\frac{RT}{g} [\frac{2 I _{1} + I _{2}}{I _{1} I _{2}}]$
(C) $\frac{nRT}{ng} [\frac{1}{I _{2}} + \frac{1}{l _{1}}]$
(D) $\frac{nRT}{g} [\frac{I _{1} - I _{2}}{I _{1} I _{2}}]$ ✓ Correct

Answer: (D)

Solution

Balance the piston between the gas below pushing up, the gas above pushing down and gravity:

P_{2} A = P_{1} A + m g

, where

P_{2}

is the lower pressure and

P_{1}

the upper. Each column holds

n

moles at the same temperature

T

, so with

P V = n R T

and

V = A ℓ

P_{1} = \frac{n R T}{A ℓ _{1}}

and

P_{2} = \frac{n R T}{A ℓ _{2}}

. Substituting and cancelling the area,

m g = n R T (\frac{1}{ℓ _{2}} - \frac{1}{ℓ _{1}}) = n R T \frac{ℓ _{1} - ℓ _{2}}{ℓ _{1} ℓ _{2}}

, hence

m = \frac{n R T}{g} (\frac{ℓ _{1} - ℓ _{2}}{ℓ _{1} ℓ _{2}})

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Heat & Thermodynamics

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