The molecules of a certain gas consist of two different...

The molecules of a certain gas consist of two different atoms, each with zero nuclear spin, bound together. Measurements of the specific heat of this material, over a wide range of temperatures, give the graph shown below. FIGURE CAN'T COPY. (The values marked on the vertical scale correspond to the height of the curve in each of the "plateau" regions.) (a) Account for each of the different results found in the temperature regions: above $T_3$; between $T_2$ and $T_3$; between $T_1$ and $T_2$; below $T_1$, (b) Given that the first excited state of the rotational spectrum of this molecule is at an energy $k T_e$ above the ground rotational state, and $T_e=64 \mathrm{~K}$, calculate from basic theory the rotational contribution to the specific heat capacity of this gas at 20 K at 100 K , at 300 K .

The molecules of a certain gas consist of two different atoms, each with zero nuclear spin, bound together. Measurements of the specific heat of this material, over a wide range of temperatures, give the graph shown below.
FIGURE CAN'T COPY.
(The values marked on the vertical scale correspond to the height of the curve in each of the "plateau" regions.)
(a) Account for each of the different results found in the temperature regions: above $T_3$; between $T_2$ and $T_3$; between $T_1$ and $T_2$; below $T_1$,
(b) Given that the first excited state of the rotational spectrum of this molecule is at an energy $k T_e$ above the ground rotational state, and $T_e=64 \mathrm{~K}$, calculate from basic theory the rotational contribution to the specific heat capacity of this gas at 20 K at 100 K , at 300 K .

Key Concepts

Vibrational Motion in Diatomic Molecules

Vibrational degrees of freedom in diatomic molecules involve oscillatory motion within the chemical bond and are quantized with relatively large energy separations compared to rotational states. Consequently, at low temperatures these modes are typically not excited and do not contribute to the specific heat until T becomes comparable to the vibrational energy spacing. When vibrational modes are activated, they contribute significantly because each vibrational mode effectively provides two degrees of freedom (one for kinetic and one for potential energy).

Rotational Motion in Diatomic Molecules

For a diatomic molecule, rotational motion is characterized by quantized energy states separated by an energy gap. At low temperatures, if kT is less than the gap to the first excited rotational state, rotations do not contribute significantly to the specific heat. As temperature increases and approaches or exceeds the characteristic rotational energy spacing, the rotational degrees of freedom become thermally activated, contributing additional energy and thus increasing the specific heat.

Molecular Degrees of Freedom

In statistical mechanics, a molecule’s degrees of freedom are the independent modes by which it can store energy. For a diatomic molecule, this includes three translational modes, two rotational modes, and one vibrational mode (which itself has both kinetic and potential components, effectively counting as two degrees of freedom when fully excited). The activation of these modes depends on the temperature, with some degrees of freedom ‘freezing out’ at lower temperatures due to the quantized nature of their energy levels.

Equipartition Theorem

The equipartition theorem states that energy is equally distributed among all active quadratic degrees of freedom, contributing (1/2)kT per degree of freedom per molecule (or (1/2)R per mole). This theorem provides a classical prediction for the specific heat contributions of various modes. However, in systems with quantized energy levels, deviations occur at lower temperatures when the thermal energy is insufficient to excite higher-energy states.

Quantization of Energy Levels

Quantum mechanics tells us that energy in atoms and molecules is not continuous but quantized. For molecular rotations and vibrations the allowed energy levels are discrete, and only those levels with energies comparable to or lower than kT are appreciably populated. This quantization explains why specific heat measurements show different 'plateau' regions as temperature increases, corresponding to the sequential activation of translational, rotational, and vibrational degrees of freedom.

Transcript

00:01 For this problem on the topic of condensed matter, we are to consider a gas of diatomic molecules with the moment of inertia i at a temperature t.

00:09 Eg is the ground state energy and ex is the energy of an excited state, and we want to explain why the ratio of the number of molecules in the l rotational energy, energy level to the number of molecules in the ground state is given by nl over n0, which is 2l plus 1, e to the minus l into l plus 1h bar squared over 2i kt.

00:30 We then want to determine the ratio of nl over n0 for a gas of carbon monoxide molecules at 300 kelvin for values of l, 1 to 10, 20, and then 50.

00:41 We then want to explain why the results in b show that as l is increased, the ratio nl over n0 first increases and then decreases.

00:52 Now the excited state energy is given as l squared over two times the same moment of inertia i, which is h -bar -squared l into l plus 1 divided by 2i, and the ground state eg is equal to 0, since al is equal to 0.

01:20 And there is an additional multiplicative factor of 2l plus 1, because for each l -state, they are really 2l plus 1 into ml states with the same energy.

01:30 So for part a, we have the ratio that we require nl over n0 to be 2l plus 1, e to the minus h bar squared times l into l plus 1 divided by 2i kt.

01:59 Now for part b1, for e, for e, which, when l is equal to 1, we have this to be h bar squared into 1, into 1 plus 1, divided by 2 times the moment of inertia given to be 1 .449 times 10 to the minus 46 kg meter squared for carbon monoxide.

02:28 We get this energy to be 7 .67 times 10 to the minus 20.

02:36 And so this energy divided by k t is equal to 7 .67 times 10 to the minus 23 joules divided by balsman's constant 1 .38 times 10 to the minus 23 joules per calvin times the temperature of 300 calvin so this ratio is 0 .0185 and so we get 2 l plus 1 equal to 3 which means that the ratio n and l is equal to 1 over the ground state n is equal to 3 times e to the minus 0 .0 .0 which is equal to 2 .95.

03:45 And now we want to do the same for the different values of l...

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