Match the Cp/CV ratio for ideal gases with different type of molecules: (A) Monatomic molecules I. 7

Q: Why do we ignore vibrational degrees of freedom at room temperature?

At room temperature, the thermal energy ($kT$) is usually not sufficient to excite the higher energy quantum states of vibration. Therefore, vibrational modes remain "frozen" and do not contribute to the specific heat until the gas reaches significantly higher temperatures.

Q: How is the formula $\gamma = 1 + 2/f$ derived?

From the law of equipartition of energy, the molar heat capacity at constant volume is $C_V = \frac{f}{2}R$. Using Mayer's relation $C_p = C_V + R$, we get $C_p = \frac{f}{2}R + R = R(\frac{f}{2} + 1)$. The ratio $\gamma = \frac{C_p}{C_V}$ then simplifies exactly to $1 + \frac{2}{f}$.

Q: Why does a rigid diatomic molecule have 5 degrees of freedom instead of 6?

A diatomic molecule has 3 translational degrees of freedom. While it can rotate in 3D space, the moment of inertia along the axis connecting the two atoms is negligibly small. Thus, rotational kinetic energy is only significant about the two axes perpendicular to the bond, giving $3 + 2 = 5$ degrees of freedom.

Q: What does "non-rigid" mean in the context of a diatomic gas?

"Non-rigid" implies that the bond between the atoms is not fixed; it can stretch and compress like a spring. This introduces a vibrational mode. According to classical mechanics, a 1D harmonic oscillator has two degrees of freedom: one for kinetic energy and one for potential energy, bringing the total to $5 + 2 = 7$.

Q: Are all triatomic gases rigid with 6 degrees of freedom?

Not necessarily! A non-linear triatomic gas (like $H_2O$) has 3 translational and 3 rotational degrees of freedom ($f=6$). However, a linear triatomic gas (like $CO_2$) behaves rotationally like a diatomic molecule, having only 2 rotational degrees of freedom ($f=5$), assuming it is rigid.

Question

Match the $C_p/C_V$ ratio for ideal gases with different type of molecules:
\begin{array}{ll}
	ext{Molecule type} & C_p/C_V \
	ext{(A) Monatomic molecules} & 	ext{I. } 7/5 \
	ext{(B) Diatomic rigid molecules} & 	ext{II. } 9/7 \
	ext{(C) Diatomic non-rigid molecules} & 	ext{III. } 4/3 \
	ext{(D) Triatomic rigid molecules} & 	ext{IV. } 5/3
\end{array}

Options:

(A) A $ightarrow$ IV, B $ightarrow$ I, C $ightarrow$ II, D $ightarrow$ III

(B) A $ightarrow$ III, B $ightarrow$ IV, C $ightarrow$ II, D $ightarrow$ I

(C) A $ightarrow$ II, B $ightarrow$ III, C $ightarrow$ I, D $ightarrow$ IV

(D) A $ightarrow$ IV, B $ightarrow$ II, C $ightarrow$ I, D $ightarrow$ III

Sigma Percentile · Accepted Answer

Have you ever wondered why different gases heat up differently? Why does it take a different amount of energy to raise the temperature of a balloon filled with helium compared to one filled with oxygen? The secret lies hidden in the microscopic dance of their molecules, governed by a beautiful principle known as the **Law of Equipartition of Energy**.

In thermodynamics, one of the most critical parameters of a gas is the ratio of its specific heats, denoted by $\gamma$ (gamma). This ratio, $\gamma = \frac{C_p}{C_V}$, dictates how a gas behaves during adiabatic processes, like the rapid compression of air in a diesel engine or the propagation of sound waves.

### The Magic of Gamma and Degrees of Freedom

To find $\gamma$, we don't need to memorize endless tables of data. We just need to understand the geometry of the gas molecules! The master formula connecting thermodynamics to molecular geometry is:

$$\gamma = 1 + \frac{2}{f}$$

Here, $f$ represents the **degrees of freedom** of the gas molecule. Think of a degree of freedom as an independent way a molecule can possess energy. It can translate (move in straight lines), rotate (spin), or vibrate (jiggle like a spring). Let's embark on a journey through different types of gases and decode their degrees of freedom.

### The Lone Wanderer: Monatomic Gas

Imagine a monatomic gas like Helium or Neon. It consists of single, isolated atoms flying around in space. Because it's just a tiny point mass, spinning it doesn't store any meaningful rotational kinetic energy.

Therefore, it can only move along the x, y, and z axes. This gives it exactly **3 translational degrees of freedom** ($f = 3$).

Plugging this into our master formula:

$$\gamma = 1 + \frac{2}{3} = \frac{5}{3}$$

This perfectly matches our first molecule type (A) with the ratio $\frac{5}{3}$ (IV).

### The Tumbling Dumbbell: Rigid Diatomic Gas

Now, let's upgrade to a rigid diatomic gas, like Oxygen ($O_2$) or Nitrogen ($N_2$) at room temperature. Picture it as a rigid dumbbell.

Like the monatomic gas, the center of mass can translate in 3 directions. But now, the dumbbell can also tumble! It can rotate end-over-end about two independent axes perpendicular to the bond. (Rotation along the bond axis itself is ignored because the moment of inertia is infinitesimally small).

This gives us $3 	ext{ (translational)} + 2 	ext{ (rotational)} = \mathbf{5 	ext{ degrees of freedom}}$ ($f = 5$).

Substituting this into our formula:

$$\gamma = 1 + \frac{2}{5} = \frac{7}{5}$$

Thus, the rigid diatomic molecule (B) matches with the ratio $\frac{7}{5}$ (I).

### The Jiggling Spring: Non-Rigid Diatomic Gas

What happens if we crank up the heat? At high temperatures, the rigid bond between the two atoms begins to act like a flexible spring. The molecule is no longer rigid; it's a **non-rigid diatomic gas**.

This "spring" introduces a vibrational mode. According to classical mechanics, a 1D harmonic oscillator possesses two degrees of freedom: one for its kinetic energy (as the atoms move) and one for its potential energy (stored in the stretched/compressed bond).

Adding these 2 vibrational degrees to our previous 5 gives us a total of **7 degrees of freedom** ($f = 7$).

Let's calculate gamma:

$$\gamma = 1 + \frac{2}{7} = \frac{9}{7}$$

This aligns the non-rigid diatomic molecule (C) with the ratio $\frac{9}{7}$ (II).

### The Spinning Triangle: Rigid Triatomic Gas

Finally, let's visualize a rigid triatomic gas. Unless specified as linear (like $CO_2$), we generally assume a non-linear geometry, like a triangle (e.g., $H_2O$ vapor or $NO_2$).

This triangular molecule can translate in 3 directions. Because it has a substantial spread of mass in 3D space, it can now meaningfully rotate about all 3 independent spatial axes (x, y, and z).

This yields $3 	ext{ (translational)} + 3 	ext{ (rotational)} = \mathbf{6 	ext{ degrees of freedom}}$ ($f = 6$).

Plugging this final value into our equation:

$$\gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3}$$

So, the rigid triatomic molecule (D) matches with the ratio $\frac{4}{3}$ (III).

### Bringing It All Together

By simply visualizing the geometry and motion of these microscopic particles, we have successfully mapped their macroscopic thermodynamic properties.

Our final matching sequence is:
- **(A) Monatomic** $ightarrow$ **IV** ($\frac{5}{3}$)
- **(B) Diatomic rigid** $ightarrow$ **I** ($\frac{7}{5}$)
- **(C) Diatomic non-rigid** $ightarrow$ **II** ($\frac{9}{7}$)
- **(D) Triatomic rigid** $ightarrow$ **III** ($\frac{4}{3}$)

This logical deduction leads us straight to the correct option. Physics is truly beautiful when you can see the invisible mechanics driving the visible world!

Select Answer:

Visualized Solution

$γ and Degree of Freedom$

$Monatomic Gas$

$Diatomic Rigid Gas$

$Diatomic Non-Rigid Gas$

$Triatomic Rigid Gas$

$Final Match$

The Sigma Insight: Kinetic Theory of Gases

The Magic of Gamma and Degrees of Freedom

The Lone Wanderer

The Tumbling Dumbbell

The Jiggling Spring

The Spinning Triangle

Bringing It All Together

Similar Questions

$C_{V}$ and $C_{p}$ denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then,

The internal energy ( $U$ ), pressure ( $p$ ) and volume ( $V$ ) of an ideal gas are related as $U = 3 p V + 4$ . The gas is

Two moles of an ideal gas with $\frac{C _{p}}{C _{V}} = \frac{5}{3}$ are mixed with 3 mol of another ideal gas with $\frac{C _{p}}{C _{V}} = \frac{4}{3}$ . The value of $\frac{C _{p}}{C _{V}}$ for the mixture is

Two non-reactive monoatomic ideal gases have their atomic masses in the ratio $2 : 3$ . The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is $4 : 3$ . The ratio of their densities is

Consider a gas of triatomic molecules. The molecules are assumed to be triangular and made of massless rigid rods whose vertices are occupied by atoms. The internal energy of a mole of the gas at temperature $T$ is

For a gas $C_{p} - C_{V} = R$ in a state $P$ and $C_{p} - C_{V} = 1.10 R$ in a state $Q$ . $T_{P}$ and $T_{Q}$ are the temperatures in two different states $P$ and $Q$ , respectively. Then,

Let $\overset{v}{ˉ}$ , $v_{rms}$ and $v_{p}$ respectively denote the mean speed, root mean square speed and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature $T$ . The mass of a molecule is $m$ . Then,

One mole of ideal monoatomic gas $(γ = \frac{5}{3})$ is mixed with one mole of diatomic gas $(γ = \frac{7}{5})$ . What is $γ$ for the mixture? $γ$ denotes the ratio of specific heat at constant pressure, to that at constant volume.

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γ and Degree of Freedom

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Diatomic Non-Rigid Gas

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

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