Two monoatomic ideal gases 1 and 2 of molecular masses m1 and m2 respectively are enclosed in separa

Q: Why does the speed of sound depend on the molecular mass of the gas?

Sound is a pressure wave propagated by molecular collisions. Lighter molecules move faster at a given temperature, allowing them to transfer momentum and propagate the wave more quickly.

Q: What is the adiabatic index $\gamma$, and why is it the same for both gases?

The adiabatic index $\gamma$ is the ratio of specific heats ($C_p/C_v$). For all monoatomic ideal gases, it is exactly $5/3$ because they only have translational degrees of freedom.

Q: Does the pressure of the gas affect the speed of sound?

No! Although it seems like higher pressure should increase speed, it also increases density proportionally at a constant temperature. These two effects cancel out completely, leaving the speed of sound independent of pressure.

Q: Why is the speed of sound inversely proportional to the square root of mass, rather than just mass?

This comes from kinetic theory. The average kinetic energy of gas molecules is proportional to temperature ($1/2 m v^2 \propto T$). Thus, molecular speed (and consequently sound speed) scales with $1/\sqrt{m}$.

Q: What would happen if one gas was diatomic and the other was monoatomic?

If the gases had different atomicities, their adiabatic indices $\gamma$ would be different ($\gamma = 7/5$ for diatomic, $5/3$ for monoatomic). You would have to include the ratio of their $\gamma$ values in the final calculation.

Question

Two monoatomic ideal gases 1 and 2 of molecular masses $m_1$ and $m_2$ respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to the gas 2 is given by

Options:

(A) $\sqrt{\frac{m_1}{m_2}}$

(B) $\sqrt{\frac{m_2}{m_1}}$

(C) $\frac{m_1}{m_2}$

(D) $\frac{m_2}{m_1}$

Sigma Percentile · Accepted Answer

### Introduction: The Symphony of Sound in Gases

Imagine standing in a room filled with a mysterious gas, listening to the voice of a friend.

Depending on the gas, their voice might sound incredibly high-pitched or deeply resonant.

This fascinating phenomenon is entirely governed by how fast sound waves can travel through different gaseous mediums.

In this problem, we explore the elegant physics behind the speed of sound in two different monoatomic ideal gases kept at the exact same temperature.

Let's dive into the microscopic world of gas molecules and discover how their mass dictates the speed of sound!

### The Physics of Sound Speed: Laplace's Breakthrough

Sound is a longitudinal pressure wave that propagates through a medium via molecular collisions.

To mathematically describe how fast these pressure disturbances travel, we use the famous **Laplace formula** for the speed of sound in an ideal gas:

$$v = \sqrt{\frac{\gamma RT}{M}}$$

Here, $v$ represents the speed of sound, $\gamma$ is the adiabatic index (the ratio of specific heats $C_p/C_v$), $R$ is the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass (or molecular mass $m$) of the gas.

This formula beautifully bridges macroscopic thermodynamic properties with microscopic molecular characteristics.

### Analyzing the Setup: What Stays Constant?

In our problem, we are comparing two different gases under very specific conditions:

1. **Both gases are monoatomic:** This is a crucial piece of information! For all monoatomic ideal gases, the adiabatic index is a constant determined by their degrees of freedom:

$$\gamma_1 = \gamma_2 = \gamma = \frac{5}{3}$$

2. **Same Temperature:** Both containers are maintained at the exact same absolute temperature $T$:

$$T_1 = T_2 = T$$

3. **Universal Constant:** The gas constant $R$ is, of course, identical for both systems.

### The Power of Proportionality

Since $\gamma$, $R$, and $T$ are identical for both gases, we can establish a direct proportionality relation.

By grouping all the constant terms together, we find that the speed of sound is inversely proportional to the square root of the molecular mass:

$$v \propto \frac{1}{\sqrt{m}}$$

This tells us something deeply intuitive: **lighter molecules move faster** at a given temperature, allowing them to collide more frequently and propagate pressure waves much more rapidly than heavier, sluggish molecules.

### The Final Calculation: A Beautiful Cancellation

Let's set up the ratio of the speed of sound in Gas 1 ($v_1$) to that in Gas 2 ($v_2$):

$$\frac{v_1}{v_2} = \frac{\sqrt{\frac{\gamma RT}{m_1}}}{\sqrt{\frac{\gamma RT}{m_2}}}$$

Because the terms $\gamma$, $R$, and $T$ are identical in both the numerator and the denominator, they cancel out completely!

This leaves us with a remarkably simple and elegant result:

$$\frac{v_1}{v_2} = \sqrt{\frac{m_2}{m_1}}$$

Thus, the ratio of the speed of sound in Gas 1 to Gas 2 is indeed **$\sqrt{\frac{m_2}{m_1}}$**, which perfectly matches **Option (b)**.

Animated Solution for Physics - Waves: Two monoatomic ideal gases 1 and 2 of molecular masses $m_{1}$ and $m_{2}$ respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to the gas 2 is given by

Select Answer:

Visualized Solution

Visualizing the Two Gas Systems

The Speed of Sound in a Gas

Identifying the Constants

Establishing the Proportionality

Setting up the Ratio

Simplifying the Expression

Final Answer

The Sigma Insight: Wave Equation and Wave Speed

Introduction

The Physics of Sound Speed

Analyzing the Setup

The Power of Proportionality

The Final Calculation

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