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The Heat Capacity of a Diatomic Gas

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15.1 Introduction Statistical thermodynamics provides deep insight into the classical description of a MONATOMIC ideal gas. In classical thermodynamics, the principle of equipartition of energy fails to give the observed value of the specific heat capacity for diatomic gases. The explanation of the above discrepancy was considered to be the most important challenge in statistical theory.

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15.1 The quantized linear oscillator A linear oscillator is a particle constrained to move along a straight line & acted on by a restoring force F= -Kx = ma = If displaced from its equilibrium position and released, the particle oscillates with simple harmonic motion of frequency, given by Note that the frequency depends on K and m, and is independent of the amplitude X.

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Consider an assembly of N one-dimensional harmonic oscillators, in which the oscillators are loosely coupled so that the energy exchange among them is small. In classical mechanics, a particle can oscillate with any amplitude and energy. From quantum mechanics, the single particle energy levels are given by E J = (J + ½)h, J = 0, 1, 2, ….. The energies are equally spaced and the ground state has non-zero energy.

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The internal degrees of freedom include vibrations, rotations, and electronic excitations. For internal degrees of freedom, Boltzmann Statistics applies. The distinguishable property arises from the fact that those diatomic molecules have different translational energy. The states are nondegenerate, i.e. g j = 1 The partition function of an oscillator

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Introducing the characteristic temperature θ = h /k The solution for the above eq. is (in class derivation)

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The internal energy is: U = NkT 2 since

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For T 0 For thus

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15.3Vibrational Modes of Diatomic Molecules The most important application of the above result is to the molecules of a diatomic gas From classical thermodynamics for a reversible process!

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Since Or At high temperatures

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At low temperature limit On has So approaching zero faster than the growth of (θ/T) 2 as T 0

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ThereforeC v 0 as T 0 The total energy of a diatomic molecule is made up of four contributions that can be separately treated:

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1.The kinetic energy associated with the translational motion 2.The vibrational motion 3.Rotation motion (To be discussed later) Example: 15.1 a) Calculate the fractional number of oscillators in the three lowest quantum states (j=0, 1, 2,) for Sol:

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J = 0

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15.2) a) For a system of localized distinguishable oscillators, Boltzmann statistics applies. Show that the entropy S is given by Solution: according to Boltzmann statistics So

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15.4 Rotational modes of diatomic molecules The moment of inertia, where μ is the reduced mass r 0 is the equilibrium value of the distance between the nuclei.

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From quantum mechanics, the allowed angular momentum states are where l = 0, 1, 2, 3… … From classical mechanics, the rotational energy equalswith w is angular velocity. The angular momentum therefore, the energy

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Define a characteristic temperature for rotation θ rot can be found from infrared spectroscopy experiments, in which the energies required to excite the molecules to higher rotational states are measured. Different from vibrational motion, the energy levels of the above equation are degenerate. for level

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Now, one can get the partition function For, virtually all the molecules are in the few lowest rotational states. As a result, the series of can be truncated with negligible errors after the first two or three terms!

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For all diatomic gases, except hydrogen, the rotational characteristics temperature is of the order of 10 k (Kelvin degree). At ordinary temperature, Therefore, many closely spaced energy states are excited. The sum ofmay be replaced by an integral. Define:

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Note that the above result is too large for homonuclear molecules such as H 2, O 2 and N 2 by a factor of 2… why? The slight modification has no effect on the thermodynamics properties of the system such as the internal energy and the heat capacity!

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Using (Note: ) again At low temperature Keeping the first two terms

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Using the relationship (for ) And (for )

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Characteristic Temperatures Characteristic temperature of rotation of diatomic molecules Substance θ rot (K) H 2 85.4 O 2 2.1 N 2 2.9 HCl 15.2 CO 2.8 NO 2.4 Cl 2 0.36 Characteristic temperature of vibration of diatomic molecules Substance θ vib (K) H 2 6140 O 2 2239 N 2 3352 HCl 4150 CO 3080 NO 2690 Cl 2 810

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15.5 Electronic Excitation The electronic partition function is where g 0 and g 1, are, respectively, the degeneracies of the ground state and the first excited state. E 1 is the energy separation of the two lowest states. Introducing

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For most gases, the higher electronic states are not excited (θ e ~ 120, 000k for hydrogen). therefore, At practical temperature, electronic excitation makes no contribution to the external energy or heat capacity!

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15.6 The total heat capacity For a diatomic molecule system Since Discussing the relationship of T and C v (p. 288-289)

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Heat capacity for diatomic molecules

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Example I (problem 15.7) Consider a diatomic gas near room temperature. Show that the entropy is Solution: For diatomic molecules

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For translational motion, the molecules are treated as non-distinguishable assemblies

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For rotational motion (they are distinguishable in terms of kinetic energy)

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Example II (problem 15-8) For a kilomole of nitrogen (N 2 ) at standard temperature and pressure, compute (a) the internal energy U; (b) the Helmholtz function F; and (c) the entropy S. Solution: calculate the characteristic temperature first!

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