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:
$\small{\gamma\,=\,\frac{C_p}{C_v}\,=\,\frac{5}{3}}$

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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:
$\small{\gamma\,=\,\frac{C_p}{C_v}\,=\,\frac{7}{5}}$

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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:
$\small{\gamma\,=\,\frac{C_p}{C_v}\,=\,\frac{9}{5}}$

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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:
$\small{\gamma\,=\,\frac{C_p}{C_v}\,=\,\frac{4+f}{3+f}}$

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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
= $\small{\frac{3}{2} R~\times~2~\times~15}$

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

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

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

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