Q9 / 48 · JEE Advanced 2026 (Paper 1) · Chemistry
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 $m_1$ 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 behave like ideal gases, the value of $(m_1/m_2)$ is ____.
Correct answer: 9.8
Solution
Let the molar mass of He be $M$. Then the molar mass of Ar is $10M$.
In the first cylinder, the total number of moles is given by:
$n_1 = \dfrac{m_1}{M} + \dfrac{m_2}{10M} = \dfrac{10m_1 + m_2}{10M}$
In the second cylinder, the total number of moles is given by:
$n_2 = \dfrac{m_2}{M} + \dfrac{m_1}{10M} = \dfrac{10m_2 + m_1}{10M}$
Using the ideal gas equation $PV = nRT$, since the volume and temperature are the same for both cylinders, the pressure is directly proportional to the number of moles. Thus, $\dfrac{P_1}{P_2} = \dfrac{n_1}{n_2}$.
Given that $P_1 = 5P_2$, we have:
$\dfrac{n_1}{n_2} = 5$
Substituting the expressions for $n_1$ and $n_2$:
$\dfrac{10m_1 + m_2}{10m_2 + m_1} = 5$
$10m_1 + m_2 = 50m_2 + 5m_1$
$5m_1 = 49m_2$
$\dfrac{m_1}{m_2} = \dfrac{49}{5} = 9.8$
Answer: $9.8$
In the first cylinder, the total number of moles is given by:
$n_1 = \dfrac{m_1}{M} + \dfrac{m_2}{10M} = \dfrac{10m_1 + m_2}{10M}$
In the second cylinder, the total number of moles is given by:
$n_2 = \dfrac{m_2}{M} + \dfrac{m_1}{10M} = \dfrac{10m_2 + m_1}{10M}$
Using the ideal gas equation $PV = nRT$, since the volume and temperature are the same for both cylinders, the pressure is directly proportional to the number of moles. Thus, $\dfrac{P_1}{P_2} = \dfrac{n_1}{n_2}$.
Given that $P_1 = 5P_2$, we have:
$\dfrac{n_1}{n_2} = 5$
Substituting the expressions for $n_1$ and $n_2$:
$\dfrac{10m_1 + m_2}{10m_2 + m_1} = 5$
$10m_1 + m_2 = 50m_2 + 5m_1$
$5m_1 = 49m_2$
$\dfrac{m_1}{m_2} = \dfrac{49}{5} = 9.8$
Answer: $9.8$
Hence, the correct answer is 9.8.