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 1, and that below the piston is 2, such that 1 > 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) nRT/g[(₁ - ₂)/(₁ ₂)] (b) RT/g[(2₁ + ₂)/(₁ ₂)] (c) nRT/g[(1)/(₂) + (1)/(₁)] (d) RT/ng[(₁ - 3₂)/(₁ ₂)]

Answer: (a). With cross-section A: pressure above P₁ = (nRT)/(A₁), pressure below P₂ = (nRT)/(A₂) Piston in equilibrium: P₂A = P₁A + mg mg = nRT((1)/(₂) - (1)/(₁)) = nRT (₁ - ₂)/(₁₂) So the answer is (a) m = nRT/g[(₁ - ₂)/(₁₂)].

A vertical closed cylinder is separated into two parts by a… | JEE Main 2019 (12 Jan, Shift 2)

Q18JEE Main 2019 · 12 JAN · SHIFT 2Difficult

Heat, Work & First Law of Thermodynamics

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 $\ell$ 1, and that below the piston is $\ell$ 2, such that $\ell$ 1 > $\ell$ 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)

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Answer: (a)

With cross-section $A$ : pressure above $P_1 = \frac{nRT}{A\ell_1}$ , pressure below $P_2 = \frac{nRT}{A\ell_2}$
Piston in equilibrium: $P_2A = P_1A + mg$
$mg = nRT\left(\frac{1}{\ell_2} - \frac{1}{\ell_1}\right) = nRT\,\frac{\ell_1 - \ell_2}{\ell_1\ell_2}$
So the answer is (a) $m = \frac{nRT}{g}\left[\frac{\ell_1 - \ell_2}{\ell_1\ell_2}\right]$ .

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Heat, Work & First Law of Thermodynamics question from JEE Main 2019 (12 Jan, Shift 2)

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