Monatomic Gases

The molecule of a monatomic gas has only three translational degrees of freedom. Thus, the average energy of a molecule at temperature \(T\) is \((3 / 2) k_{\mathrm{B}} T\). The total internal energy of a mole of such a gas is
\(
U=\frac{3}{2} k_B T \times N_A=\frac{3}{2} R T \dots(13.27)
\)
The molar specific heat at constant volume, \(C_v,\) is
\(
\left.C_v \text { (monatomic gas }\right)=\frac{\mathrm{d} U}{\mathrm{~d} T}=\frac{3}{2} R T \dots(13.28)
\)
For an ideal gas,
\(
C_p-C_v=R \dots(13.29)
\)
where \(C_p\) is the molar-specific heat at constant pressure. Thus,
\(
C_p=\frac{5}{2} R \dots(13.30)
\)
The ratio of specific heats \(\gamma=\frac{C_{\mathrm{p}}}{C_{\mathrm{v}}}=\frac{5}{3} \dots(13.31)\)

Diatomic Gases

As explained earlier, a diatomic molecule treated as a rigid rotator, like a dumbbell, has 5 degrees of freedom: 3 translational and 2 rotational. Using the law of equipartition of energy, the total internal energy of a mole of such a gas is
\(
U=\frac{5}{2} k_B T \times N_A=\frac{5}{2} R T \dots(13.32)
\)
The molar-specific heats are then given by
\(
\begin{aligned}
& C_v(\text { rigid diatomic })=\frac{5}{2} R, C_p=\frac{7}{2} R \dots(13.33)\\
& \gamma \text { (rigid diatomic })=\frac{7}{5} \dots(13.34)
\end{aligned}
\)
If the diatomic molecule is not rigid but has, in addition, a vibrational mode
\(
\begin{aligned}
& U=\left(\frac{5}{2} k_B T+k_B T\right) N_A=\frac{7}{2} R T \\
& C_v=\frac{7}{2} R, C_p=\frac{9}{2} R, \gamma=\frac{9}{7} R \dots(13.35)
\end{aligned}
\)

Polyatomic Gases

In general, a polyatomic molecule has 3 translational, 3 rotational degrees of freedom, and a certain number \((f)\) of vibrational modes. According to the law of equipartition of energy, it is easily seen that one mole of such a gas has
\(
U=\left(\frac{3}{2} k_B T+\frac{3}{2} k_B T+f k_B T\right) N_A
\)
\(
\begin{aligned}
& \text { 1.e., } C_v=(3+f) R, C_p=(4+f) R, \\
& \gamma=\frac{(4+f)}{(3+f)} \dots(13.36)
\end{aligned}
\)
Note that \(C_p-C_v=R\) is true for any ideal gas, whether mono, di or polyatomic.

Example 1: 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 \(15.0^{\circ} \mathrm{C} ?\left(R=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\).

Solution:

Using the gas law \(P V=\mu R T\), you can easily show that \(1 \mathrm{~mol}\) of any (ideal) gas at a standard temperature \((273 \mathrm{~K})\) and pressure \(\left(1 \mathrm{~atm}=1.01 \times 10^5 \mathrm{~Pa}\right)\) occupies a volume of 22.4 litres. This universal volume is called molar volume. Thus the cylinder in this example contains 2 mol of helium. Further, since helium is monatomic, its predicted (and observed) molar specific heat at constant volume, \(C_v=(3 / 2) R\), and molar specific heat at constant pressure, \(C_p=(3 / 2) R+R=(5 / 2) R\). Since the volume of the cylinder is fixed, the heat required is determined by \(C_{v}\).
Therefore, Heat required \(=\) no. of moles \(\times\) molar specific heat \(\times\) rise in temperature
\(
=2 \times 1.5 R \times 15.0=45 R
\)
\(
=45 \times 8.31=374 \mathrm{~J}
\)

Specific Heat Capacity of Solids

We can use the law of equipartition of energy to determine specific heats of solids. Consider a solid of \(N\) atoms, each vibrating about its mean position. An oscillation in one dimension has average energy of \(2 \times 1 / 2 k_B T=k_B T\). In three dimensions, the average energy is \(3 k_B T\). For a mole of solid, \(N=N_A\), and the total energy is
\(
U=3 k_B T \times N_A=3 R T
\)
Now at constant pressure \(\Delta Q=\Delta U+P \Delta V\) \(=\Delta U\), since for a solid \(\Delta V\) is negligible.
Hence, \(C=\frac{\Delta Q}{\Delta T}=\frac{\Delta U}{\Delta T}=3 R \dots(13.37)\)

Specific Heat Capacity of Water

We treat water like a solid. For each atom average energy is \(3 k_B T\). The water molecule has three atoms, two hydrogen and one oxygen. So it has
\(
\begin{aligned}
& U=3 \times 3 k_B T \times N_A=9 R T \\
& \text { and } C=\Delta Q / \Delta T=\Delta U / \Delta T=9 R .
\end{aligned}
\)

In the calorie, gram, and degree units, water is defined to have unit-specific heat. As 1 calorie \(=4.179\) joules and one mole of water is 18 grams, the heat capacity per mole is \(\sim 75 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \sim\) 9R.