JEE Advanced · 2026 · 17 May 2026 · Paper 1 · States of Matter

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The cylinders, both fitted with frictionless pistons, are filled with mixtures of He and Ar gases. In the first cylinder, the masses of He and Ar are

m_{1}

and

m_{2}

, respectively. In the second cylinder, the masses of He and Ar are

m_{2}

and

m_{1}

, respectively the molar mass of Ar is 10 times the molar mass of He . The external pressure applied by the piston on the first cylinder needs to be 5 times that on the second cylinder so that the volume of the gas mixtures in both the cylinders are equal at the same temperature. Assuming He and Ar behaves like ideal gases, the value of

(m_{1} / m_{2})

is ____.

Answer: 9.8

Solution

Let the molar mass of He be M, then molar mass of Ar = 10M.

Given:

P_{1} = 5 P_{2}

V_{1} = V_{2} = V

T_{1} = T_{2} = T

Applying ideal gas equation:

Vessel (I):

P_{1} V = [\frac{m _{1}}{M} + \frac{m _{2}}{10 M}] R T

Vessel (II):

P_{2} V = [\frac{m _{2}}{M} + \frac{m _{1}}{10 M}] R T

Dividing (II) by (I):

\frac{P _{2}}{P _{1}} = \frac{\frac{m _{2}}{M} + \frac{m _{1}}{10 M}}{\frac{m _{1}}{M} + \frac{m _{2}}{10 M}} = \frac{1}{5}

Simplifying:

5 m_{2} + \frac{m _{1}}{2} = m_{1} + \frac{m _{2}}{2}

5 m_{2} - \frac{m _{2}}{2} = m_{1} - \frac{m _{1}}{2}

\frac{49 m _{2}}{10} = \frac{m _{1}}{2}

\frac{m _{1}}{m _{2}} = 9.8

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Numerical answer

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JEE Advanced · 2026 · 17 May 2026 · Paper 1 · States of Matter

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