A ring shaped tube contains two ideal gases with equal masses and relative molar masses M1 = 32 and

Q: Why is the pressure equal on both sides?

The stopper is free to move without friction. If the pressure on one side was higher, it would exert a net force on the stopper, causing it to move. It only stops when the forces (and thus pressures) are perfectly balanced.

Q: Why do we use the ratio of moles instead of mass?

The Ideal Gas Law ($pV = nRT$) depends on the number of particles (moles), not their total mass. Lighter molecules will have more moles for the exact same mass, thereby occupying more volume.

Q: How do we know the temperature is the same for both gases?

The problem states they are in the same tube and doesn't mention any thermal insulation between them. In such thermodynamic equilibrium problems, thermal equilibrium (equal temperature) is assumed unless stated otherwise.

Q: Why is the volume proportional to the angle?

The tube has a uniform cross-sectional area. The volume of a cylindrical segment is Area $\times$ Length. Since the arc Length is equal to Radius $\times$ Angle, the volume is directly proportional to the angle.

Q: Why does gas 2 have a larger angle even though its molar mass is smaller?

Because its molar mass is smaller, a given mass of gas 2 contains more individual molecules (moles) than the exact same mass of gas 1. More molecules at the same pressure and temperature require more volume.

Question

A ring shaped tube contains two ideal gases with equal masses and relative molar masses $M_1 = 32$ and $M_2 = 28$. The gases are separated by one fixed partition and another movable stopper $S$ which can move freely without friction inside the ring. The angle $\alpha$ as shown in the figure is …… degrees.

Sigma Percentile · Accepted Answer

### The Setup: A Delicate Balance

Imagine a perfectly circular, ring-shaped tube. Inside this tube, two different ideal gases are trapped. They are separated by a rigid, fixed partition at the top and a movable stopper at the bottom. This stopper is completely frictionless, meaning it can slide freely along the ring.

We are given that both gases have the exact same mass, $m$. However, they are different gases: Gas 1 has a molar mass of $M_1 = 32$, and Gas 2 has a molar mass of $M_2 = 28$. Our goal is to find the angle $\alpha$ that Gas 2 subtends at the center of the ring when the system reaches equilibrium.

### The Master Equation: Ideal Gas Law

Because the stopper is free to move, it will slide until the forces on both sides are perfectly balanced. Since the cross-sectional area of the tube is uniform, balanced forces mean balanced pressures. Therefore, at equilibrium, the pressure of Gas 1 must equal the pressure of Gas 2:

$$p_1 = p_2$$

To relate this to the volume each gas occupies, we bring in the Ideal Gas Law, $pV = nRT$. Rearranging this for volume, we get:

$$V = \frac{nRT}{p}$$

Since both gases are in the same environment, they share the same temperature $T$. We've also established they share the same pressure $p$. The universal gas constant $R$ is, of course, constant. This means the volume of each gas is directly proportional to its number of moles, $n$.

### The Molar Advantage

We don't know the number of moles directly, but we do know the masses are equal. The number of moles is simply the total mass divided by the molar mass ($n = \frac{m}{M}$). Let's set up a ratio of their volumes:

$$\frac{V_1}{V_2} = \frac{n_1}{n_2} = \frac{m / M_1}{m / M_2}$$

Notice how beautifully the mass $m$ cancels out! The ratio of their volumes is simply the inverse ratio of their molar masses:

$$\frac{V_1}{V_2} = \frac{M_2}{M_1}$$

Plugging in the given values ($M_1 = 32$ and $M_2 = 28$):

$$\frac{V_1}{V_2} = \frac{28}{32} = \frac{7}{8}$$

This tells us a crucial physical fact: even though they weigh the same, Gas 2 has lighter molecules, meaning there are *more* of them. More molecules require more space to maintain the same pressure, so Gas 2 occupies a larger volume.

### Geometry Meets Thermodynamics

Now, let's translate this volume ratio into the geometry of the ring. In a uniform circular tube, the volume a gas occupies is directly proportional to the angle $	heta$ it sweeps out from the center. Therefore, the ratio of their angles is identical to the ratio of their volumes:

$$\frac{	heta_1}{	heta_2} = \frac{7}{8}$$

We also know that the two gases completely fill the ring. This means their individual angles must add up to a full circle:

$$	heta_1 + 	heta_2 = 360^\circ$$

### Final Calculation

Looking at the diagram, the angle $\alpha$ corresponds to the region occupied by Gas 2 ($M_2$). So, we need to find $	heta_2$. Using the ratio we established, Gas 2 takes up 8 parts out of a total of $7 + 8 = 15$ parts of the circle.

$$\alpha = 	heta_2 = \left(\frac{8}{15}ight) 	imes 360^\circ$$

Let's do the final arithmetic:

$$\alpha = 8 	imes 24^\circ$$

**$$\alpha = 192^\circ$$**

The frictionless stopper will settle at a position where Gas 2 sweeps out exactly $192^\circ$ of the ring.

Enter Numerical Value:

Visualized Solution

$M_{1} = 32, M_{2} = 28$

$p_{1} = p_{2}$

$p V = n R T$

$\frac{V _{1}}{V _{2}} = \frac{n _{1}}{n _{2}}$

$n = \frac{m}{M}$

$\frac{V _{1}}{V _{2}} = \frac{7}{8}$

$V \propto θ$

$θ_{1} + θ_{2} = 36 0^{\circ}$

$α = θ_{2}$

$α = 19 2^{\circ}$

$What if T_{1} = 2 T_{2} ?$

The Sigma Insight: Kinetic Theory of Gases

The Setup

The Master Equation

The Molar Advantage

Geometry Meets Thermodynamics

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M1​=32,M2​=28

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p1​=p2​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V2​V1​​=n2​n1​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=Mm​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V2​V1​​=87​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ1​+θ2​=360∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=θ2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=192∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if T1​=2T2​?