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 m₁ and m₂, respectively. In the second cylinder, the masses of He and Ar are m₂ and m₁, 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

$\frac{m_1}{m_2}$

is ____.

The correct answer is 49/5.

Let the molar mass of Helium (He) be $M_{\text{He}}$ and the molar mass of Argon (Ar) be $M_{\text{Ar}}$.
Given that the molar mass of Ar is 10 times the molar mass of He:

$M_{\text{Ar}}=10M_{\text{He}}$

Step 1: Calculate the number of moles in Cylinder 1 and Cylinder 2

In Cylinder 1, the mass of He is $m_1$ and the mass of Ar is $m_2$.
The total number of moles in Cylinder 1, $n_1$, is:

$n_1=\frac{m_1}{M_{\text{He}}}+\frac{m_2}{M_{\text{Ar}}}=\frac{m_1}{M_{\text{He}}}+\frac{m_2}{10M_{\text{He}}}=\frac{10m_1+m_2}{10M_{\text{He}}}$

In Cylinder 2, the mass of He is $m_2$ and the mass of Ar is $m_1$.
The total number of moles in Cylinder 2, $n_2$, is:

$n_2=\frac{m_2}{M_{\text{He}}}+\frac{m_1}{M_{\text{Ar}}}=\frac{m_2}{M_{\text{He}}}+\frac{m_1}{10M_{\text{He}}}=\frac{10m_2+m_1}{10M_{\text{He}}}$

Step 2: Apply the Ideal Gas Equation

From the ideal gas equation, $PV=nRT$.
Since both cylinders are at the same temperature ($T$) and have equal volume ($V$), the pressure is directly proportional to the total number of moles ($P\propto n$).

$\frac{P_1}{P_2}=\frac{n_1}{n_2}$

Given that the pressure in the first cylinder is 5 times that in the second cylinder ($P_1=5P_2$):

$5=\frac{n_1}{n_2}$

Substituting the values of $n_1$ and $n_2$:

$5=\frac{\frac{10m_1+m_2}{10M_{\text{He}}}}{\frac{10m_2+m_1}{10M_{\text{He}}}}$

$5=\frac{10m_1+m_2}{10m_2+m_1}$

Step 3: Solve for $\frac{m_1}{m_2}$

Cross-multiplying the equation gives:

$5(10m_2+m_1)=10m_1+m_2$

$50m_2+5m_1=10m_1+m_2$

Rearranging the terms:

$50m_2-m_2=10m_1-5m_1$

$49m_2=5m_1$

$\frac{m_1}{m_2}=\frac{49}{5}$

Thus, the value of $\frac{m_1}{m_2}$ is 49/5.