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JEE Advanced 2026Paper 1
Exam
JEE Advanced
Year
2026
Paper
Paper 1
Subject
Chemistry
Chapter
States of Matter

JEE Advanced 2026 · Paper 1 · Chemistry · States of Matter

Two cylinders, both fitted with frictionless pistons, are filled...

Two 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 m1m_1 and m2m_2, respectively. In the second cylinder, the masses of He and Ar are m2m_2 and m1m_1, respectively. The molar mass of Ar is 1010 times the molar mass of He. The external pressure applied by the piston on the first cylinder needs to be 55 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 behave like ideal gases, the value of (m1/m2)(m_1/m_2) is ____.

Correct answer

0

Step-by-step solution

Let the molar mass of He be MM. Then the molar mass of Ar is 10M10M.

In the first cylinder, the total number of moles is given by:

n1=m1M+m210M=10m1+m210Mn_1 = \dfrac{m_1}{M} + \dfrac{m_2}{10M} = \dfrac{10m_1 + m_2}{10M}

In the second cylinder, the total number of moles is given by:

n2=m2M+m110M=10m2+m110Mn_2 = \dfrac{m_2}{M} + \dfrac{m_1}{10M} = \dfrac{10m_2 + m_1}{10M}

Using the ideal gas equation PV=nRTPV = nRT, since the volume and temperature are the same for both cylinders, the pressure is directly proportional to the number of moles. Thus, P1P2=n1n2\dfrac{P_1}{P_2} = \dfrac{n_1}{n_2}.

Given that P1=5P2P_1 = 5P_2, we have:

n1n2=5\dfrac{n_1}{n_2} = 5

Substituting the expressions for n1n_1 and n2n_2:

10m1+m210m2+m1=5\dfrac{10m_1 + m_2}{10m_2 + m_1} = 5

10m1+m2=50m2+5m110m_1 + m_2 = 50m_2 + 5m_1

5m1=49m25m_1 = 49m_2

m1m2=495=9.8\dfrac{m_1}{m_2} = \dfrac{49}{5} = 9.8

Answer: 9.89.8
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