JEE Advanced 2026 — Paper 1 — CHEM-III solution

JEE Advanced2026Paper 1CHEM-III

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 _____.

Your answer

Solution

Set up the ideal-gas relation for each cylinder. Let $M_{He} = M$ , so $M_{Ar} = 10 M$ .

Cylinder (I) at pressure $P_{1}$ , volume $V$ , temperature $T$ : $P_{1} V = (\frac{m _{1}}{M} + \frac{m _{2}}{10 M}) R T$ .

Cylinder (II) at pressure $P_{2}$ , volume $V$ , temperature $T$ : $P_{2} V = (\frac{m _{2}}{M} + \frac{m _{1}}{10 M}) R T$ .

Take the ratio $\frac{P _{2}}{P _{1}} = \frac{1}{5}$ :

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

Solve. $10 m_{1} + m_{2} = 5 (10 m_{2} + m_{1}) = 50 m_{2} + 5 m_{1}$ . So $5 m_{1} = 49 m_{2}$ , giving $\frac{m _{1}}{m _{2}} = \frac{49}{5} = 9.80$ .

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