13.3 - Specific Heats of Gases

In the previous section, we saw the law of equipartition of energy. In this section, we will apply the law to predict the specific heats of gases.

First we will see monatomic gases. The details can be written in 5 steps:

1. For a monatomic gas molecule, there will not be any rotation or vibration. So we need to consider the translational kinetic energy only.
• Applying law of equipartition, We get:
For each molecule at temperature T, this energy will be $\small{\frac{3}{2}K_B\,T}$

2. In a mole of the gas, there will be $\small{N_A}$ molecules. So the total internal energy (U) of one mole of a monatomic gas can be obtained as:
$\small{U\,=\,\frac{3}{2}K_B\,T\,N_A\,=\,\frac{3}{2}R\,T}$  

3. Specific heat is the change in internal energy per unit change in temperature. So we can obtain the specific heat for one mole at constant volume as:
$\small{C_v\,\text{monatomic gas}\,=\,\frac{3}{2}R}$

4. For an ideal gas, $\small{C_p\,-\,C_v\,=\,R}$
Where $\small{C_p}$ is the specific heat for one mole at constant pressure. We can write:
$\small{C_p\,=\,R\,+\,C_v\,=\,\frac{5}{2}R}$

5. We can write the ratio of specific heats also:
$\small{\gamma\,=\,\frac{C_p}{C_v}\,=\,\frac{5}{3}}$

---

Next we will see diatomic gases whose molecules are rigid. The details can be written in 5 steps:

1. For a diatomic gas molecule, assuming that it is rigid, there will not be any vibration. So we need to consider the translational and rotational kinetic energies only.
• Applying law of equipartition, We get:
For each molecule at temperature T, this energy will be $\small{5\times \frac{1}{2}K_B\,T~=~\frac{5}{2}K_B\,T}$

2. In a mole of the gas, there will be $\small{N_A}$ molecules. So the total internal energy (U) of one mole of a diatomic gas can be obtained as:
$\small{U\,=\,\frac{5}{2}K_B\,T\,N_A\,=\,\frac{5}{2}R\,T}$  

3. Specific heat is the change in internal energy per unit change in temperature. So we can obtain the specific heat for one mole at constant volume as:
$\small{C_v\,\text{rigid diatomic gas}\,=\,\frac{5}{2}R}$

4. For an ideal gas, $\small{C_p\,-\,C_v\,=\,R}$
Where $\small{C_p}$ is the specific heat for one mole at constant pressure. We can write:
$\small{C_p\,=\,R\,+\,C_v\,=\,\frac{7}{2}R}$

5. We can write the ratio of specific heats also:
$\small{\gamma\,=\,\frac{C_p}{C_v}\,=\,\frac{7}{5}}$

---

Next we will see diatomic gases whose molecules are not rigid. The details can be written in 5 steps:

1. For a diatomic gas molecule, if it is not rigid, there will be vibration.
• Applying law of equipartition, We get:
For each molecule at temperature T, this energy will be $\small{5\times \frac{1}{2}K_B\,T\,+\,K_B\,T~=~\frac{7}{2}K_B\,T}$

2. In a mole of the gas, there will be $\small{N_A}$ molecules. So the total internal energy (U) of one mole of a diatomic gas can be obtained as:
$\small{U\,=\,\frac{7}{2}K_B\,T\,N_A\,=\,\frac{7}{2}R\,T}$  

3. Specific heat is the change in internal energy per unit change in temperature. So we can obtain the specific heat for one mole at constant volume as:
$\small{C_v\,\text{non-rigid diatomic gas}\,=\,\frac{7}{2}R}$

4. For an ideal gas, $\small{C_p\,-\,C_v\,=\,R}$
• We can write:
$\small{C_p\,=\,R\,+\,C_v\,=\,\frac{9}{2}R}$

5. We can write the ratio of specific heats also:
$\small{\gamma\,=\,\frac{C_p}{C_v}\,=\,\frac{9}{5}}$

---

Next we will see polyatomic gases. The details can be written in 5 steps:

1. In general, a polyatomic molecule has:
    ♦ 3 translational degrees of freedom
    ♦ 3 rotational degrees of freedom
    ♦ $\small{f}$ number of vibrational modes

• Applying law of equipartition, We get:
For each molecule at temperature T, this energy will be $\small{3\times \frac{1}{2}K_B\,T\,+\,3\times \frac{1}{2}K_B\,T\,+f\,\,K_B\,T~=~(3+f)K_B\,T}$

2. In a mole of the gas, there will be $\small{N_A}$ molecules. So the total internal energy (U) of one mole of a polyatomic gas can be obtained as:
$\small{U\,=\,(3+f)K_B\,T\,N_A\,=\,(3+f)R\,T}$  

3. Specific heat is the change in internal energy per unit change in temperature. So we can obtain the specific heat for one mole at constant volume as:
$\small{C_v\,\text{polyatomic gas}\,=\,(3+f)R}$

4. For an ideal gas, $\small{C_p\,-\,C_v\,=\,R}$
• We can write:
$\small{C_p\,=\,R\,+\,C_v\,=\,(4+f)R}$

5. We can write the ratio of specific heats also:
$\small{\gamma\,=\,\frac{C_p}{C_v}\,=\,\frac{4+f}{3+f}}$

---

Using the above equations, we can predict the specific heats of different types of gases. Two examples are given below. In those examples, vibrational modes are ignored:

Example 1:
• Predicted value of $\small{C_v}$ and $\small{C_p}$ for monatomic gas are 12.5 J mol⁻¹ K⁻¹ and 20.8 J mol⁻¹ K⁻¹ respectively.
• Experimental value of $\small{C_v}$ and $\small{C_p}$ for the monatomic gas helium are 12.5 J mol⁻¹ K⁻¹ and 20.8 J mol⁻¹ K⁻¹ respectively.

Example 2:
• Predicted value of $\small{C_v}$ and $\small{C_p}$ for diatomic gas are 20.8 J mol⁻¹ K⁻¹ and 29.1 J mol⁻¹ K⁻¹ respectively.
• Experimental value of $\small{C_v}$ and $\small{C_p}$ for the diatomic gas oxygen are 21.0 J mol⁻¹ K⁻¹ and 29.3 J mol⁻¹ K⁻¹ respectively.

If vibrational modes are included in the calculations, the predicted values become closer to the experimental values.

Now we will see some solved examples.

Solved example 13.14
A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by 15.0 °C ? (R = 8.31 J mol⁻¹ K⁻¹)
Solution:
1. Helium is a monatomic gas.
• So $\small{C_v\,=\,\frac{3}{2} R}$ and $\small{C_p\,=\,\frac{5}{2} R}$
• Given that, cylinder is of fixed capacity. That means, volume is constant. So we have to use $\small{C_v}$.

2. Heat required
= $\small{C_v~\times~\text{no. of moles}~\times~\text{rise in temperature}}$

3. So our next task is to find the number of moles of helium present in the cylinder.
• Given that, the gas is at STP. We know that, at STP, one mole of an ideal gas will have a volume of 22.4 litres.
• In our present case, the volume is 44.8 litres. So there are two moles.

4. Thus from (2), we get:
Heat required
= $\small{\frac{3}{2} R~\times~2~\times~15}$

= $\small{\frac{3}{2} \left(8.31 \text{J}\, \text{mol}^{-1} \,\text{K}^{-1} \right)~\times~2(\text{mol})~\times~15\left(\text{K} \right)}$

= $\small{373.95 \,\text{J}}$

---

In the next section, we will see specific heat capacity of solids.

Previous

Contents

Next

Copyright©2021 Higher secondary physics.blogspot.com

13.2 - Law of Equipartition of Energy

In the previous section, we explained the gas laws using kinetic theory. In this section, we will see the law of equipartition of energy.

The details can be written in 17 steps:
1. Consider a gas in thermal equilibrium at temperature T.
If 'm' is the mass of each molecule in that gas, then the average value of energy ($\small{\epsilon_t}$) of each molecule can be obtained as:
$\small{\epsilon_t~=~\frac{1}{2}m\,v_x^2~+~\frac{1}{2}m\,v_y^2~+~\frac{1}{2}m\,v_z^2~=~\frac{3}{2}K_B\,T}$

2. $\small{\frac{3}{2}K_B\,T}$ is three times $\small{\frac{1}{2}K_B\,T}$. We know that, there is no preferred direction. So we can write:
    ♦ $\small{\frac{1}{2}m\,v_x^2~=~\frac{1}{2}K_B\,T}$
    ♦ $\small{\frac{1}{2}m\,v_y^2~=~\frac{1}{2}K_B\,T}$
    ♦ $\small{\frac{1}{2}m\,v_z^2~=~\frac{1}{2}K_B\,T}$

3. Now we will write about the number of coordinates required for a molecule.
• If the molecule is free to move in space, we need three coordinates to locate it.
• If the molecule is constrained to move in a plane, we need two coordinates to locate it.
• If the molecule is constrained to move along a line, we need only one coordinate to locate it.

4. The above information in (3) can be written in another form also:
• If the molecule is free to move in space, it has three degrees of freedom.
• If the molecule is constrained to move in a plane, it has two degrees of freedom.
• If the molecule is constrained to move along a line, it has one degree of freedom.

5. The above information in (3) and (4) can be written in yet another form also. Motion of a body as a whole from one point to another is called translation. So we can write:
• If the molecule is free to move in space, it has three translational degrees of freedom.
• If the molecule is constrained to move in a plane, it has two translational degrees of freedom.
• If the molecule is constrained to move along a line, it has one translational degree of freedom.

6. We just saw that, if a molecule is free to move in space, it has three translational degrees of freedom. Each of those three degrees, will contribute a term towards the total kinetic energy of the molecule.
• The degree of freedom in the x-direction will contribute $\small{\frac{1}{2}m\,v_x^2}$
• The degree of freedom in the y-direction will contribute $\small{\frac{1}{2}m\,v_y^2}$
• The degree of freedom in the z-direction will contribute $\small{\frac{1}{2}m\,v_z^2}$

7. Consider a diatomic gas like oxygen ($\small{O_2}$) or nitrogen ($\small{N_2}$). Each molecule of such a gas will have two atoms.
• In fig.13.2 below, the brown spheres represent atoms.
• Two brown spheres are joined together by a yellow line. • This line represents the bond between the two atoms.
    ♦ The green line is perpendicular to the yellow line.
    ♦ The red line is also perpendicular to the yellow line.
    ♦ The green and red lines are perpendicular to each other.

Fig.13.2

8. In the fig. on the left side, the molecule as a whole, is rotating about the green line. So green line is the axis of rotation. It is marked as (1).
• Due to this rotation, the molecule will have a rotational energy equal to $\small{\frac{1}{2}\,I_1\,\omega_1^2}$
    ♦ $\small{I_1}$ is the moment of inertia of the molecule about axis (1)
    ♦ $\small{\omega_1}$ is the angular speed of the molecule about axis (1)

9. In the fig. on the right side, the molecule as a whole, is rotating about the red line. So red line is the axis of rotation. It is marked as (2).
• Due to this rotation, the molecule will have a rotational energy equal to $\small{\frac{1}{2}\,I_2\,\omega_2^2}$
    ♦ $\small{I_2}$ is the moment of inertia of the molecule about axis (2)
    ♦ $\small{\omega_2}$ is the angular speed of the molecule about axis (2)

10. The molecule can rotate about the yellow line also. But the moment of inertia of the molecule about the yellow line is very small. So this rotation will not contribute much to the total rotational energy. We can safely ignore this rotation.

11. Based on (8) and (9), we can write:
Rotational energy of the molecule is given by:
$\small{\epsilon_r~=~\frac{1}{2}\,I_1\,\omega_1^2~+~\frac{1}{2}\,I_2\,\omega_2^2}$

• It is clear that, a diatomic molecule has two rotational degrees of freedom.
    ♦ One along axis (1)
    ♦ The other along axis (2)

12. So a diatomic molecule has a total energy given by:
$\small{\epsilon_t\,+\,\epsilon_r\,=\,\frac{1}{2}m\,v_x^2\,+\,\frac{1}{2}m\,v_y^2\,+\,\frac{1}{2}m\,v_z^2\,+\,\frac{1}{2}\,I_1\,\omega_1^2\,+\,\frac{1}{2}\,I_2\,\omega_2^2}$

• Note that, for a monatomic molecule, the last two terms will be absent.

13. We have seen the energy contributions from translation and rotation. If a molecule experience vibration, that molecule will posses vibrational energy also.
• Consider the two spheres in fig.13.2 above. If the spheres oscillate along the yellow line, then that molecule as a whole, will posses vibrational energy. It is like a spring with the two spheres at it's ends. We will see more details in later chapters.
• The $\small{O_2}$ molecule is rigid. So this energy does not come into effect at moderate temperatures. But a $\small{CO}$ molecule will posses this energy even at moderate temperatures.

14. For a molecule like $\small{CO}$, the vibrational energy is given by:
$\small{\epsilon_v\,=\,\frac{1}{2} m \left(\frac{dy}{dt} \right)^2\,+\,\frac{1}{2} k y^2}$
    ♦ $\small{k}$ is the force constant of the oscillator
    ♦ $\small{y}$ is the vibrational co-ordinate

15. Now we can write the total energy:
$\small{\epsilon\,=\,\epsilon_t\,+\,\epsilon_r\,+\,\epsilon_v}$

• Where,

$\small{\epsilon_t\,=\,\frac{1}{2}m\,v_x^2\,+\,\frac{1}{2}m\,v_y^2\,+\,\frac{1}{2}m\,v_z^2}$

$\small{\epsilon_r\,=\,\frac{1}{2}\,I_1\,\omega_1^2\,+\,\frac{1}{2}\,I_2\,\omega_2^2}$

$\small{\epsilon_v\,=\,\frac{1}{2} m \left(\frac{dy}{dt} \right)^2\,+\,\frac{1}{2} k y^2}$

16. Now we can analyze the contribution from each type of energy.
(i) Each of the translational degree of freedom, contributes one square term.
(ii) Each of the rotational degree of freedom, contributes one square term.
(iii) Each vibrational frequency, contributes two square terms.

17. Now we can write about the general form of contribution:
• We have seen that, each square term in the translational energy is equal to $\small{\frac{1}{2}K_B\,T}$
• The Scottish physicist James Clerk Maxwell proved that, each square term has an average energy equal to $\small{\frac{1}{2}K_B\,T}$. This is known as the law of equipartition of energy.
• So we can write:
    ♦ Each translational degree of freedom contributes $\small{\frac{1}{2}K_B\,T}$ to the total energy.
    ♦ Each rotational degree of freedom contributes $\small{\frac{1}{2}K_B\,T}$ to the total energy.
    ♦ Each vibrational frequency contributes two times $\small{\frac{1}{2}K_B\,T}$ = $\small{K_B\,T}$ to the total energy.

• We will see the detailed proof of the law in higher classes.

---

In the next section, we will apply the law to predict specific heats of gases theoretically.

Previous

Contents

Next

Copyright©2026 Higher secondary physics.blogspot.com