Given below are observations on molar specific heats at room temperature of some common gases. | Gas | Molar specific heat $(C_v)$ $(\text{cal mol}^{-1}\ \text{K}^{-1})$ | | --- | --- | | Hydrogen | $4.87$ | | Nitrogen | $4.97$ | | Oxygen | $5.02$ | | Nitric oxide | $4.99$ | | Carbon monoxide | $5.01$ | | Chlorine | $6.17$ | The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is $2.92\ \text{cal/mol K}$. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

Given below are observations on molar specific hea – Class 11 Physics

Exercises · 10.15

Given below are observations on molar specific heats at room temperature of some common gases.

Gas	Molar specific heat $(C_v)$ $(\text{cal mol}^{-1}\ \text{K}^{-1})$
Hydrogen	$4.87$
Nitrogen	$4.97$
Oxygen	$5.02$
Nitric oxide	$4.99$
Carbon monoxide	$5.01$
Chlorine	$6.17$

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is $2.92\ \text{cal/mol K}$ . Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

CBSENCERTSubjective· 3mImportance★★★★★est

42% · 23/55 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Equipartition Of Energy

Equipartition of Energy: From Intuition to the Precise Statement

Imagine you have a box of gas — a bunch of tiny molecules zipping around, spinning, and vibrating. Each molecule has some energy. But here's the question: How is that total energy shared among all the different ways a molecule can move?

That's exactly what the equipartition of energy answers.

The Core Intuition

Think of a room full of people dancing. Some are jumping, some are spinning, some are waving their arms. If the music is steady and the room is crowded, eventually everyone will be moving with roughly the same average energy per "type" of motion. No one type of motion hogs all the energy.

In a gas at thermal equilibrium, nature does the same thing. Every independent way a molecule can store energy — every "degree of freedom" — gets the same average amount of energy.

Note

A degree of freedom is an independent way a molecule can move or store energy. For a single atom, that's just three directions of translation (x, y, z). For a diatomic molecule, you also get rotations and vibrations.

The Precise Statement

\text{Average energy per degree of freedom} = \frac{1}{2} k_B T

Where:

$k_B$ is Boltzmann's constant ( $1.38 \times 10^{-23} \ \text{J/K}$ )
$T$ is the absolute temperature in Kelvin

So if a molecule has $f$ degrees of freedom, its total average energy is:

\langle E \rangle = \frac{f}{2} k_B T

Why "Half $k_B T$ "?

This comes from a deeper statistical mechanics result. For any quadratic term in the energy expression (like $\frac{1}{2}mv_x^2$ for translation, or $\frac{1}{2}I\omega^2$ for rotation), the average energy contributed by that term is exactly $\frac{1}{2}k_B T$ .

Tip

| Molecule type | Degrees of freedom ( $f$ ) | Average energy |

|:---|:---:|:---:|

| Monatomic (He, Ar) | 3 (all translational) | $\frac{3}{2}k_B T$ |

| Diatomic (N₂, O₂) at room temp | 5 (3 trans + 2 rot) | $\frac{5}{2}k_B T$ |

| Diatomic at high temp | 7 (3 trans + 2 rot + 2 vib) | $\frac{7}{2}k_B T$ |

A Concrete Example

Take a helium atom in a box at 300 K. It can only move in three directions. Each direction contributes $\frac{1}{2}k_B T$ of energy.

So the average kinetic energy of one helium atom is:

\langle E \rangle = 3 \times \frac{1}{2} k_B T = \frac{3}{2} k_B T

For a mole of helium (Avogadro's number $N_A$ of atoms), the total internal energy becomes:

U = N_A \cdot \frac{3}{2} k_B T = \frac{3}{2} RT

where $R = N_A k_B$ is the universal gas constant. …

Start your 14-day free trial to unlock the full solution →

The equipartition of energy explains why diatomic gases have higher molar specific heats than monatomic gases because they have additional rotational degrees of freedom. Chlorine’s larger value suggests its vibrational modes are also partially excited at room temperature.

The equipartition of energy is the key idea here. It says that each quadratic term in a molecule’s energy contributes $\frac{1}{2}k_BT$ per molecule (or $\frac{1}{2}RT$ per mole) to the internal energy. A monatomic gas has only three translational degrees of freedom — motion along x, y, and z — so its molar internal energy is $U = \frac{3}{2}RT$ , giving $C_v = \frac{3}{2}R \approx 2.98\ \text{cal mol}^{-1}\ \text{K}^{-1}$ . The observed value of $2.92$ is close, confirming this.

Now look at the gases in the table: hydrogen, nitrogen, oxygen, nitric oxide, carbon monoxide, and chlorine. All are diatomic molecules (two atoms). A diatomic molecule can do more than just translate. It can also rotate about two perpendicular axes (like a dumbbell spinning), adding two rotational degrees of freedom. Each contributes $\frac{1}{2}RT$ to the molar internal energy. So for a diatomic gas with translation and rotation active:

U = \frac{3}{2}RT + \frac{2}{2}RT = \frac{5}{2}RT

Then:

C_v = \frac{dU}{dT} = \frac{5}{2}R

With $R \approx 1.987\ \text{cal mol}^{-1}\ \text{K}^{-1}$ , this gives:

C_v = \frac{5}{2} \times 1.987 \approx 4.97\ \text{cal mol}^{-1}\ \text{K}^{-1}

That matches the values for nitrogen, oxygen, nitric oxide, and carbon monoxide almost exactly. Hydrogen’s $4.87$ is slightly lower — a subtle quantum effect: at room temperature, hydrogen’s rotational levels are not fully populated because its moment of inertia is very small, so the equipartition prediction isn’t fully realised.

Why the difference from monatomic gases?

Monatomic gases have only 3 translational degrees of freedom. Diatomic gases have 3 translational + 2 rotational = 5 active degrees of freedom at room temperature. Each degree contributes $\frac{1}{2}R$ to $C_v$ , so diatomic $C_v$ is $\frac{5}{2}R \approx 4.97$ , while monatomic $C_v$ is $\frac{3}{2}R \approx 2.98$ . The table confirms this: most diatomic gases cluster around $4.97$ , far above $2.92$ .
What about chlorine’s larger value ( $6.17$ )? …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 73 on this concept.

TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
$3 : 5$
›Reveal solutionSolution
The ratio of internal energy changes for a monoatomic vs. diatomic gas at constant pressure with equal heat input is found by relating heat to temperature change via molar specific heats, then converting temperature change to internal energy change. The result is 21 : 25, so option (C) is correct.

Concept & Intuition

When equal amounts of heat are supplied at constant pressure, the temperature rise depends on the molar specific heat at constant pressure, $C_p$ . For an ideal gas, the change in internal energy $\Delta U$ depends only on the temperature change and the molar specific heat at constant volume, $C_v$ . The ratio of $\Delta U$ for two gases thus involves both $C_p$ (to relate heat to $\Delta T$ ) and $C_v$ (to relate $\Delta T$ to $\Delta U$ ). The key is that monoatomic and diatomic gases have different degrees of freedom, giving different $C_v$ and $C_p$ values.

Step-by-step solution

Recall the relevant specific heats For an ideal monoatomic gas:

$C_v = \frac{3}{2}R, \quad C_p = \frac{5}{2}R$
For a rigid diatomic gas (no vibration, only translation and rotation):
$C_v = \frac{5}{2}R, \quad C_p = \frac{7}{2}R$

Relate heat supplied to temperature change at constant pressure At constant pressure, the heat $Q$ supplied to $n$ moles is:

$Q = n C_p \Delta T$
Since equal amounts of heat are supplied to equal numbers of moles (implied by "equal amounts of heat" and same amount of gas), we have:
$n C_{p,\text{mono}} \Delta T_{\text{mono}} = n C_{p,\text{dia}} \Delta T_{\text{dia}}$
Cancelling $n$ :
$\frac{5}{2}R \cdot \Delta T_{\text{mono}} = \frac{7}{2}R \cdot \Delta T_{\text{dia}}$
So:
$\Delta T_{\text{mono}} = \frac{7}{5} \Delta T_{\text{dia}}$

Relate temperature change to internal energy change For an ideal gas, the change in internal energy is:

$\Delta U = n C_v \Delta T$
Thus:
$\Delta U_{\text{mono}} = n \cdot \frac{3}{2}R \cdot \Delta T_{\text{mono}}$ $\Delta U_{\text{dia}} = n \cdot \frac{5}{2}R \cdot \Delta T_{\text{dia}} …$
Start your 14-day free trial to see the full solution →
TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
$^\circ$
›Reveal solutionSolution
The root mean square (RMS) speed of gas molecules depends on the square root of the absolute temperature and inversely on the square root of the molar mass. By converting temperatures to Kelvin and applying the formula, the ratio of RMS speeds for helium and oxygen is found to be $\boxed{2:1}$ .

The root mean square (RMS) speed of gas molecules is a measure of the average speed of the molecules in a gas. It's not a simple average, but rather a statistical average that gives more weight to faster molecules, which is relevant for phenomena like diffusion and pressure. The kinetic theory of gases provides a direct relationship between the RMS speed, the absolute temperature of the gas, and its molar mass.

The intuition behind this relationship is that at higher temperatures, molecules have more kinetic energy, and thus move faster. Conversely, for a given kinetic energy, lighter molecules will move faster than heavier ones.

The formula for the RMS speed ( $v_{rms}$ ) of gas molecules is:
$v_{rms} = \sqrt{\frac{3RT}{M}}$
To find the ratio of RMS speeds, we will apply this formula to both helium and oxygen, ensuring all units are consistent, especially temperature in Kelvin.

Here's how to solve the problem step-by-step:

Convert temperatures to absolute scale (Kelvin).

The formula for RMS speed requires temperature in Kelvin. We convert the given Celsius temperatures by adding $273.15$ (or simply $273$ for exam purposes).

For helium: $T_{He} = 127^\circ C + 273 = 400 \, K$

For oxygen: $T_{O_2} = 527^\circ C + 273 = 800 \, K$

Watch out

A common mistake is to use Celsius temperatures directly in the RMS speed formula. Always convert to Kelvin!

Write down the RMS speed expressions for both gases.

Using the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$ :

For helium: $v_{rms, He} = \sqrt{\frac{3RT_{He}}{M_{He}}}$

For oxygen: $v_{rms, O_2} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$

We are given the molar masses: $M_{He} = 4$ and $M_{O_2} = 32$ . While molar mass is typically in $g/mol$ , for a ratio, as long as both are in the same units (e.g., $g/mol$ or $kg/mol$ ), the units will cancel out.

Formulate the ratio of the RMS speeds.

We need to find the ratio of RMS speeds of helium to oxygen: …
Start your 14-day free trial to see the full solution →
TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
$24RT$
›Reveal solutionSolution
The internal energy of an ideal gas depends only on its degrees of freedom and temperature. For a mixture, we sum the contributions of each gas. The total internal energy is $30RT$ , so option (C) is correct.

The key idea here is that internal energy of an ideal gas is purely kinetic — it comes from the random motion of molecules. For a diatomic gas like nitrogen ( $N_2$ ), oxygen ( $O_2$ ), and hydrogen ( $H_2$ ), each molecule has 5 degrees of freedom at ordinary temperatures (3 translational + 2 rotational). The internal energy per mole is $\frac{f}{2}RT$ , where $f$ is the number of degrees of freedom.

Since all three gases are diatomic and at the same temperature $T$ , each mole contributes $\frac{5}{2}RT$ to the total internal energy. The mixture is just the sum of the contributions from each gas.

Identify the degrees of freedom.

Nitrogen, oxygen, and hydrogen are all diatomic molecules. At temperatures where vibrational modes are not excited (which is the case in most standard problems unless stated otherwise), each has $f = 5$ degrees of freedom.

Watch out

A common mistake is to treat hydrogen as monatomic or to add vibrational degrees of freedom. Unless the problem explicitly mentions high temperature, stick with $f = 5$ for diatomic gases.

Write the internal energy per mole.

For one mole of a diatomic ideal gas:

$U_{\text{per mole}} = \frac{f}{2}RT = \frac{5}{2}RT$

Calculate the contribution of each gas. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$Q. If the ratio of specific heats of a gas is 1.4, then the gas is (A) Monoatomic (B) Diatomic (C) Triatomic (D) Polyatomic$
›Reveal solutionSolution
The ratio of specific heats $\gamma$ is fixed entirely by the number of degrees of freedom of a gas molecule; $\gamma=1.4$ is the textbook signature of diatomic gases like $N_2$ , $O_2$ , and air. Answer: (B) Diatomic.

Concept and Intuition

From the equipartition theorem, each degree of freedom of a gas molecule contributes $\tfrac12RT$ to its internal energy per mole, giving $C_v=\tfrac{f}{2}R$ and $C_p=C_v+R=\tfrac{f+2}{2}R$ , so
$\gamma=\frac{C_p}{C_v}=\frac{f+2}{f}=1+\frac{2}{f}.$
Monoatomic gases (only translational, $f=3$ ) give $\gamma=5/3\approx1.67$ . Diatomic gases (3 translational + 2 rotational at room temperature, $f=5$ ) give $\gamma=7/5=1.4$ . Polyatomic gases (more rotational/vibrational modes, $f=6$ or more) give $\gamma$ closer to $1.33$ or lower.

Step-by-Step Solution

Formula: $\gamma=1+\dfrac{2}{f}$ .

Given $\gamma=1.4$ : $1.4-1=0.4=\dfrac{2}{f} \Rightarrow f=\dfrac{2}{0.4}=5$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$Q. If the temperature of 3 moles of helium gas is increased by 2K, then the change in the internal energy of helium gas is (A) 70.0 J (B) 68.2 J (C) 74.8 J (D) 78.2 J$
›Reveal solutionSolution
Internal energy of an ideal gas is a state function depending only on temperature; for a monatomic gas, $\Delta U = \tfrac{3}{2}nR\Delta T$ , giving 74.8 J for 3 mol of He heated by 2 K.

Concept and Intuition

For an ideal gas, internal energy is purely a function of temperature (a consequence of there being no intermolecular potential energy in the ideal-gas model). So no matter how the temperature changed (at constant volume, pressure, or otherwise), the change in internal energy for a given $\Delta T$ is fixed by $\Delta U = nC_v\Delta T$ . Helium is monatomic, with only translational kinetic energy degrees of freedom, so $C_v = \tfrac{3}{2}R$ per mole.

Step-by-Step Solution

Helium is a monatomic ideal gas: $C_v = \dfrac{3}{2}R$ .

Given: $n = 3\ \text{mol}$ , $\Delta T = 2\ \text{K}$ , $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ .

$\Delta U = nC_v\Delta T = 3 \times \dfrac{3}{2}(8.314) \times 2$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$P$
›Reveal solutionSolution
This tests the fundamental property of an ideal gas that internal energy is a function of temperature alone, since there are no intermolecular potential energy contributions.

Concept and Intuition

In the kinetic theory of an ideal gas, molecules are treated as point particles (or rigid bodies with no long-range interactions) that don't exert any potential-energy-storing forces on each other except during instantaneous collisions. So the internal energy of an ideal gas is entirely the kinetic energy of translational (and, for polyatomic gases, rotational/vibrational) motion of its molecules, which is directly proportional to the absolute temperature: $U = \dfrac{f}{2}nRT$ , where $f$ is the number of degrees of freedom. Since $P$ , $V$ , and $T$ are related for a fixed amount of ideal gas by $PV=nRT$ , one might think $U$ could depend on $P$ or $V$ too — but for an ideal gas specifically, $U$ depends on $T$ alone, not on $P$ or $V$ individually (e.g., you can change $P$ and $V$ in opposite ways at constant $T$ and $U$ stays the same).

Step-by-Step Solution

For an ideal gas, intermolecular potential energy is zero (no interaction forces assumed).

Internal energy is therefore purely the sum of kinetic energies of all molecules: $U = \dfrac{f}{2}nRT$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$Q. The ratio of the degrees of freedom of a monoatomic gas and a nonlinear polyatomic gas having 2 vibrational modes is (A) 3 : 10 (B) 3 : 7 (C) 3 : 5 (D) 3 : 11$
›Reveal solutionSolution
A monoatomic gas has 3 degrees of freedom; a nonlinear polyatomic gas with 2 vibrational modes has $3+3+2(2)=10$ ; the ratio is 3:10. Answer: (A).

Concept and Intuition

Degrees of freedom count the independent ways a molecule can store kinetic (and, for vibration, potential) energy. A monoatomic gas molecule (like Ar or He) can only translate in 3 independent directions — it has no internal structure to rotate or vibrate. A nonlinear polyatomic molecule, having a genuine 3-D shape, adds 3 rotational degrees of freedom about three independent axes. Each vibrational mode contributes 2 degrees of freedom (one for kinetic energy of vibration, one for the potential energy stored in the "spring" of the bond), not just 1 — this is the classic trap.

Step-by-Step Solution

Monoatomic gas: $f_{mono} = 3$ (translational only; point-like, no rotation or vibration).

Nonlinear polyatomic gas: translational $=3$ , rotational $=3$ (three independent rotation axes since it's nonlinear/3-D shaped). …
Start your 14-day free trial to see the full solution →
COMEDK 2026Set 2026-A1 markMCQ
$1.882 \times 10^{23}$
›Reveal solutionSolution

Each $\text{O}_2$ molecule has 5 degrees of freedom, so $N=\tfrac{1.882\times10^{23}}{5}$ molecules $=0.0625$ mol $=2$ g — option (D).

A diatomic oxygen molecule has $f=3$ translational $+\,2$ rotational $=5$ degrees of freedom per molecule.

Total degrees of freedom $=N\times f$ , so the number of molecules is
$N=\frac{1.882\times10^{23}}{5}=3.764\times10^{22}.$
Number of moles: …
Start your 14-day free trial to see the full solution →
MHT-CET 2026Set pcm-2026-04-11-E1 markMCQ
$Q. For a gas, 𝑅𝐶𝑣=0.4 where R is the universal gas constant and 𝐶𝑣 is the molar specific heat at constant volume. The gas is made up of molecules which are (A) monoatomic (B) polyatomic (C) non rigid diatomic (D) rigid diatomic$
›Reveal solutionSolution

Solve for Cv from the given ratio and match it to the standard value for each gas type.

R/Cv = 0.4 → Cv = R/0.4 = 2.5R = 5R/2. …
Start your 14-day free trial to see the full solution →
MHT-CET 2026Set pcm-2026-04-13-E1 markMCQ
$Q. According to law of equipartition of energy the molar specific heat of a non rigid diatomic gas at constant volume is (A) (9/2)𝑅 (B) (5/2)𝑅 (C) (3/2)𝑅 (D) (7/2)𝑅$
›Reveal solutionSolution

Non-rigid diatomic: 7 total degrees of freedom, so Cv = (7/2)R.

A diatomic molecule considered non-rigid (i.e., vibration is allowed, not just a rigid rotator) has: 3 translational degrees of freedom, 2 rotational degrees of freedom, and 2 vibrational degrees of freedom (1 kinetic + 1 potential), totaling $f=7$ . …
Start your 14-day free trial to see the full solution →
MHT-CET 2026Set pcm-2026-04-13-E1 markMCQ
$Q. Let 𝛾1 be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a mono-atomic gas and 𝛾2 be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio, 𝛾1/𝛾2 is (A) 27/35 (B) 35/27 (C) 25/21 (D) 21/25$
›Reveal solutionSolution

γ1/γ2 = (5/3)/(7/5) = 25/21.

For a monoatomic gas ( $f=3$ ): $\gamma_1 = 1+\dfrac{2}{f} = 1+\dfrac23 = \dfrac53$

For a rigid-rotator diatomic gas ( $f=5$ ): $\gamma_2 = 1+\dfrac{2}{f} = 1+\dfrac25 = \dfrac75$ …
Start your 14-day free trial to see the full solution →
MHT-CET 2026Set pcm-2026-04-16-E1 markMCQ
$Q. A rigid diatomic gas having molar mass m is contained in an insulated container. The container is moving with velocity V. If it is stopped suddenly, the change in temperature is (R - gas constant) (A) 𝑚𝑉2𝑅 (B) 𝑚𝑉23𝑅 (C) 𝑚𝑉25𝑅 (D) 𝑚𝑉27𝑅$
›Reveal solutionSolution

$\Delta T=\dfrac{mV^2}{5R}$ .

For one mole (molar mass m, treated as the mass of the moving gas), the bulk kinetic energy is $\tfrac12mV^2$ . On sudden stopping, this converts fully into internal energy: $\tfrac12mV^2=\Delta U=\dfrac52R\Delta T$ (rig …
Start your 14-day free trial to see the full solution →

←Previous question10.14 — In an experiment on the specific heat of a metal, a 0.20kg block of th…→Next question10.16 — A child running a temperature of 101circF is given an antipyrin (i.e. …

🎓Unlock everything free for 14 days

Unlock everything free for 14 days