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Question

Physics Question on kinetic theory

A vertical closed cylinder is separated into two parts by a frictionless piston of mass mm 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\ell_1, and that below the piston is 2\ell_2 , such that 1\ell_1>2\ell_2. Each part of the cylinder contains nn moles of an ideal gas at equal temperature TT. If the piston is stationary, its mass, mm, will be given by : (R is universal gas constant and gg is the acceleration due to gravity)

A

nRTg[12+11]\frac{nRT}{g} \bigg[\frac{1}{\ell_2} + \frac{1}{\ell_1}\bigg]

B

nRTg[1212]\frac{nRT}{g} \bigg[\frac{\ell_1 - \ell_2}{\ell_1\ell_2} \bigg]

C

RTg[21+212]\frac{RT}{g} \bigg[\frac{2\ell_1 + \ell_2}{\ell_1 \ell_2} \bigg]

D

RTg[13212]\frac{RT}{g} \bigg[\frac{\ell_1 - 3\ell_2}{\ell_1\ell_2} \bigg]

Answer

nRTg[1212]\frac{nRT}{g} \bigg[\frac{\ell_1 - \ell_2}{\ell_1\ell_2} \bigg]

Explanation

Solution

P2A=P1A+mgP_2 A \, = \, P_1A \, + \, mg
nRT.AA2=nRT.AA2+mg\frac{nRT.A}{A\ell_2} \, = \frac{nRT.A}{A\ell_2} +mg
nRT(1211)=mgnRT \bigg(\frac{1}{\ell_2} \, - \, \frac{1}{\ell_1}\bigg) = mg
m=nRTg(121.2)m \, = \, \frac{nRT}{g} \, \bigg(\frac{\ell_1\ell_2}{\ell_1 .\ell_2}\bigg)