The cylinders, both fitted with frictionless pistons, are filled with mixtures of He and Ar gases. I

Q: Why do we add the moles of the two gases?

According to Dalton's Law of Partial Pressures, in a mixture of non-reacting ideal gases, the total pressure is determined by the total number of moles of all gases present.

Q: Can we just use the masses directly in the ideal gas equation?

No, the ideal gas equation $PV=nRT$ strictly requires the amount of substance to be in moles ($n$), not mass. You must divide mass by molar mass.

Q: Why did we divide the two equations instead of solving them individually?

Dividing the equations is a powerful mathematical trick to eliminate unknown or constant variables (like $V$, $R$, and $T$) that are identical in both states, leaving a simpler relationship between the variables we care about.

Q: What if the volumes were not equal?

If the volumes were different, say $V_1$ and $V_2$, they wouldn't cancel out when dividing the equations. You would need the ratio of the volumes to proceed with the calculation.

Q: Does the fact that they are He and Ar matter?

The specific gases only matter to establish the ratio of their molar masses ($1:10$). As long as they behave ideally and don't react, any two gases with a $1:10$ molar mass ratio would yield the same result.

Question

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 _____.

Sigma Percentile · Accepted Answer

### Visualizing the Gas Cylinders

Imagine you are standing in a laboratory looking at two identical, robust cylinders. Both are fitted with frictionless pistons, ensuring that any external pressure applied is perfectly balanced by the internal gas pressure.

We are told that both cylinders have the exact same volume $V$ and are kept at the exact same temperature $T$. This physical constraint is our biggest clue! It means that any difference in pressure between the two cylinders must come entirely from the amount of gas inside them.

In the first cylinder, we have a mixture of Helium with mass $m_1$ and Argon with mass $m_2$. In the second cylinder, the masses are playfully swapped: Helium has mass $m_2$ and Argon has mass $m_1$.

We are also given a crucial piece of data: the molar mass of Argon is 10 times that of Helium. If we let the molar mass of Helium be $M$, then the molar mass of Argon is $10M$. Finally, the external pressure on the first cylinder $P_1$ is 5 times the pressure on the second cylinder $P_2$.

### The Master Equation: Ideal Gas Law for Mixtures

To connect all these macroscopic properties—pressure, volume, temperature, and mass—we need our trusty **Ideal Gas Equation**.

For a single gas, it is $PV = nRT$. But what happens when we have a mixture of non-reacting gases? According to **Dalton's Law of Partial Pressures**, the total pressure exerted by a mixture of ideal gases is determined by the total number of moles of all the gases combined.

So, our master equation becomes:
$$PV = n_{	ext{total}}RT$$

And how do we find the number of moles? We simply divide the given mass of the gas by its molar mass:
$$n = \frac{	ext{mass}}{	ext{Molar Mass}}$$

### Setting Up the Mathematical Framework

Let's apply this logic to both of our cylinders.

For the first cylinder, the total number of moles is the sum of the moles of Helium and Argon. 
$$n_{	ext{total, 1}} = \frac{m_1}{M} + \frac{m_2}{10M}$$

Plugging this into the ideal gas equation gives us our first major relationship:
$$P_1 V = \left( \frac{m_1}{M} + \frac{m_2}{10M} ight) RT$$

Now, we do the exact same thing for the second cylinder. The only difference is that the masses are swapped.
$$P_2 V = \left( \frac{m_2}{M} + \frac{m_1}{10M} ight) RT$$

### The Power of Ratios

We now have a system of two equations. We could try to solve for individual variables, but that would be a nightmare since we don't know the actual values of $V$, $R$, $T$, or $M$.

Whenever you see identical constants in multiple equations, your instinct should be to divide them! Dividing the second equation by the first is a mathematically elegant move.

$$\frac{P_2 V}{P_1 V} = \frac{\left( \frac{m_2}{M} + \frac{m_1}{10M} ight) RT}{\left( \frac{m_1}{M} + \frac{m_2}{10M} ight) RT}$$

Watch how beautifully the volume $V$, the gas constant $R$, and the temperature $T$ cancel out completely. We are left with a pure ratio of pressures equal to the ratio of their total moles.

$$\frac{P_2}{P_1} = \frac{\frac{m_2}{M} + \frac{m_1}{10M}}{\frac{m_1}{M} + \frac{m_2}{10M}}$$

### Algebraic Simplification

We were given that $P_1 = 5P_2$. This means the ratio $\frac{P_2}{P_1}$ is simply $\frac{1}{5}$.

Let's substitute this into our equation. Also, notice how the molar mass $M$ is present in the denominator of every single term on the right side? We can factor it out and cancel it entirely!

$$\frac{1}{5} = \frac{m_2 + \frac{m_1}{10}}{m_1 + \frac{m_2}{10}}$$

Now, we cross-multiply to get rid of the fractions and flatten the equation.

$$m_1 + \frac{m_2}{10} = 5 \left( m_2 + \frac{m_1}{10} ight)$$

Expanding the right side gives:

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

Which simplifies to:

$$m_1 + \frac{m_2}{10} = 5m_2 + \frac{m_1}{2}$$

### The Final Calculation

We are almost there! The final step is to group the like terms. Let's bring all the $m_1$ terms to the left side and the $m_2$ terms to the right side.

$$m_1 - \frac{m_1}{2} = 5m_2 - \frac{m_2}{10}$$

Subtracting $\frac{m_1}{2}$ from $m_1$ leaves us with exactly half of $m_1$. On the right side, finding a common denominator gives us:

$$\frac{m_1}{2} = \frac{50m_2 - m_2}{10}$$

$$\frac{m_1}{2} = \frac{49m_2}{10}$$

Finally, we isolate the ratio we are looking for, $\frac{m_1}{m_2}$.

$$\frac{m_1}{m_2} = \frac{49 	imes 2}{10}$$

$$\frac{m_1}{m_2} = \frac{98}{10} = 9.80$$

And there we have it! The ratio of the masses is exactly **9.80**.

Enter Numerical Value:

Visualized Solution

$System Setup$

$Ideal Gas Equation for Mixture$

$Applying Gas Law to Cylinders$

$Dividing the Equations$

$Substituting Pressure Ratio$

$Cross Multiplication$

$Grouping Variables$

$Calculating the Ratio$

$Conceptual Reflection$

The Sigma Insight: Gaseous State

Visualizing the Gas Cylinders

The Master Equation

Setting Up the Mathematical Framework

The Power of Ratios

Algebraic Simplification

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ideal Gas Equation for Mixture

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Gas Law to Cylinders

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dividing the Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Pressure Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cross Multiplication

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Grouping Variables

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conceptual Reflection