The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300 K is
(a) √(2/7)
(b) √(

Q: Why did we use the Laplace correction instead of Newton's formula for the speed of sound?

Newton assumed that sound propagation in a gas is an isothermal process, which yielded a wave speed significantly lower than experimental values. Laplace corrected this by showing that compressions and rarefactions happen so rapidly that heat has no time to flow, making the process adiabatic. Thus, we must use the adiabatic index $\gamma$ in our formula.

Q: How do we know the adiabatic index of Nitrogen is 7/5 and Helium is 5/3?

The adiabatic index is given by $\gamma = 1 + 2/f$, where $f$ is the number of active degrees of freedom. Nitrogen ($N_2$) is diatomic and has 5 degrees of freedom at room temperature, giving $\gamma = 1 + 2/5 = 7/5$. Helium ($He$) is monoatomic and has 3 degrees of freedom, giving $\gamma = 1 + 2/3 = 5/3$.

Q: Do we need to convert the molar masses from grams per mole to kilograms per mole?

No, because we are taking a ratio of the two speeds. The conversion factor of $10^{-3}$ to convert grams to kilograms is present in both the numerator and the denominator, so it cancels out completely.

Q: What would happen to the speed of sound if we doubled the absolute temperature of both gases?

Since the speed of sound is proportional to the square root of the absolute temperature ($v \propto \sqrt{T}$), doubling the temperature would increase the speed of sound in both gases by a factor of $\sqrt{2}$. However, their ratio would remain exactly the same.

Q: Why does sound travel faster in Helium than in Nitrogen even though Helium has a lower adiabatic index?

Even though Helium has a slightly different adiabatic index, its molar mass ($4\text{ g/mol}$) is much smaller than that of Nitrogen ($28\text{ g/mol}$). Because speed is inversely proportional to the square root of molar mass, the lighter mass of Helium dominates, making sound travel much faster in it.

Question

The ratio of the speed of sound in nitrogen gas to that in helium gas, at $300	ext{ K}$ is

Options:

(A) $\sqrt{\frac{2}{7}}$

(B) $\sqrt{\frac{1}{7}}$

(C) $\frac{\sqrt{3}}{5}$

(D) $\frac{\sqrt{6}}{5}$

Sigma Percentile · Accepted Answer

### Introduction: The Symphony of Sound in Gases

Have you ever wondered why your voice sounds incredibly high-pitched and squeaky after inhaling helium from a balloon?

It is not just a funny party trick; it is a direct demonstration of fundamental thermodynamics and wave mechanics in action!

The speed of sound is not a universal constant; it is a dynamic property that depends heavily on the medium through which it propagates.

In this problem, we are going to explore a classic JEE question from 1999 that asks us to find the exact ratio of the speed of sound in Nitrogen gas ($N_2$) to that in Helium gas ($He$) at a constant temperature of $300	ext{ K}$.

Let's dive deep into the physics behind sound propagation and see how molecular structure dictates wave speed.

---

### The Physics of Sound Propagation

Sound is a longitudinal mechanical wave that propagates through a medium via a series of compressions and rarefactions.

When a sound wave travels through a gas, the compressions and rarefactions happen so rapidly that there is no time for heat to exchange with the surroundings.

Therefore, the process is highly **adiabatic**, not isothermal!

This crucial realization was first pointed out by Pierre-Simon Laplace, correcting Isaac Newton's earlier isothermal assumption.

Laplace's correction gives us the master formula for the speed of sound in an ideal gas:

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

Where:
- $\gamma$ is the adiabatic index (the ratio of specific heats, $C_p/C_v$),
- $R$ is the universal gas constant ($8.314	ext{ J/mol}\cdot	ext{K}$),
- $T$ is the absolute temperature in Kelvin,
- $M$ is the molar mass of the gas.

---

### Analyzing the Competitors: Nitrogen vs. Helium

To solve our problem, we need to look at the specific molecular properties of our two competing gases.

#### 1. Nitrogen Gas ($N_2$)
Nitrogen is a **diatomic** gas.

At room temperature ($300	ext{ K}$), a diatomic molecule has 5 degrees of freedom (3 translational and 2 rotational).

This gives it an adiabatic index of:

$$\gamma_{N_2} = 1 + \frac{2}{f} = 1 + \frac{2}{5} = \frac{7}{5}$$

The molar mass of Nitrogen ($N_2$) is:

$$M_{N_2} = 28	ext{ g/mol} = 28 	imes 10^{-3}	ext{ kg/mol}$$

#### 2. Helium Gas ($He$)
Helium is a **monoatomic** noble gas.

It has only 3 translational degrees of freedom.

This gives it an adiabatic index of:

$$\gamma_{He} = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$$

The molar mass of Helium ($He$) is:

$$M_{He} = 4	ext{ g/mol} = 4 	imes 10^{-3}	ext{ kg/mol}$$

---

### Setting Up the Mathematical Duel

Since both gases are at the exact same temperature ($T = 300	ext{ K}$), the terms $R$ and $T$ are constant and will cancel out when we take the ratio.

This means the speed of sound in this scenario is proportional to:

$$v \propto \sqrt{\frac{\gamma}{M}}$$

Let's set up our ratio of the speed of sound in Nitrogen to that in Helium:

$$\frac{v_{N_2}}{v_{He}} = \sqrt{\frac{\gamma_{N_2}}{\gamma_{He}} 	imes \frac{M_{He}}{M_{N_2}}}$$

Notice how the molar masses are inverted because $M$ is in the denominator of our master formula!

---

### Step-by-Step Calculation

Let's substitute our values into the ratio equation:

$$\frac{v_{N_2}}{v_{He}} = \sqrt{\frac{7/5}{5/3} 	imes \frac{4}{28}}$$

Now, let's simplify this step-by-step to avoid any silly algebraic mistakes.

First, let's simplify the ratio of the adiabatic indices:

$$\frac{7/5}{5/3} = \frac{7}{5} 	imes \frac{3}{5} = \frac{21}{25}$$

Next, let's simplify the ratio of the molar masses:

$$\frac{4}{28} = \frac{1}{7}$$

Now, let's multiply these two simplified fractions together inside our square root:

$$\frac{v_{N_2}}{v_{He}} = \sqrt{\frac{21}{25} 	imes \frac{1}{7}}$$

Notice that $21$ divided by $7$ is exactly $3$:

$$\frac{v_{N_2}}{v_{He}} = \sqrt{\frac{3}{25}}$$

Finally, we take the square root of the numerator and the denominator separately:

$$\frac{v_{N_2}}{v_{He}} = \frac{\sqrt{3}}{\sqrt{25}} = \frac{\sqrt{3}}{5}$$

This elegant result perfectly matches **Option (c)**!

Animated Solution for Physics - Waves: The ratio of the speed of sound in nitrogen gas to that in helium gas, at $300 K$ is

Select Answer:

Visualized Solution

Visualizing the Two Gases

The Speed of Sound Formula

Properties of Nitrogen Gas ( $N_{2}$ )

Properties of Helium Gas ( $H e$ )

Setting up the Ratio

Substituting the Values

Simplifying the Gamma Ratio

Simplifying the Molar Mass Ratio

Multiplying the Simplified Ratios

Extracting the Final Ratio

The Way Forward

The Sigma Insight: Wave Equation and Wave Speed

Introduction

The Physics of Sound Propagation

Analyzing the Competitors

# 1

# 2

Setting Up the Mathematical Duel

Step-by-Step Calculation

Similar Questions

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

A point source emits sound equally in all directions in a non-absorbing medium. Two points $P$ and $Q$ are at a distance $9 m$ and $25 m$ respectively from the source. The ratio of amplitudes of the waves at $P$ and $Q$ is ……

A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle $6 0^{\circ}$ with ground level. But he finds the aeroplane right vertically above his position. If $v$ is the speed of sound, then speed of the plane is

When temperature increases, the frequency of a tuning fork

The displacement $y$ of a particle in a medium can be expressed as $y = 1 0^{- 6} sin (100 t + 20 x + \frac{π}{4}) m$ where, $t$ is in second and $x$ in metre. The speed of the wave is

A transverse sinusoidal wave of amplitude $a$ , wavelength $λ$ and frequency $f$ is travelling on a stretched string. The maximum speed of any point on the string is $v /10$ , where $v$ is the speed of propagation of the wave. If $a = 1 0^{- 3} m$ and $v = 10 m/s$ , then $λ$ and $f$ are given by

The displacement $y$ of a wave travelling in the $x$ -direction is given by $y = 1 0^{- 4} sin (600 t - 2 x + \frac{π}{3})$ metre where, $x$ is expressed in metres and $t$ in seconds. The speed of the wave-motion, in $ms^{- 1}$ is

Speed of a transverse wave on a straight wire (mass $6.0 g$ , length $60 cm$ and area of cross-section $1.0 mm^{2}$ ) is $90 ms^{- 1}$ . If the Young's modulus of wire is $16 \times 1 0^{11} Nm^{- 2}$ , the extension of wire over its natural length is

Animated Solution for Physics - Waves: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300 K is