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For a COmolecule, the constant is approximately 0.00024eV.(This number is measured using microwave spectroscopy, that is, by measuring the microwave frequencies needed to excite the molecules into higher rotational states.) Calculate the rotational partition function for a COmolecule at room temperature (300K), first using the exact formula 6.30 and then using the approximate formula 6.31

Short Answer

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The rotational partition function of a heterogeneous diatomic molecule

Step by step solution

01

Rotational partition function:

The equation is

Zrot=j=0(2j+1)exp-j(j+1)kT

Here, is the rotational constant, kis the Boltzmann constant, and Tis the absolute temperature.

At higher temperatures, for kT>>, the rotational partition function becomes as follows:

Zrot=kT

Substitute 8.617×10-5eV/Kfork,300Kfor T, and 0.00024eVin the equationZrot=kT

Zrot=(8.617×10-5eV/K)(300K)0.00024eV=107.7

Therefore, the rotational partition function of a COmolecule is107.7

02

The rotational partition function of a heterogeneous diatomic molecule:

The equations are

Zrot=j=0(2j+1)exp-j(j+1)kT

Expand the above summation from j=0to j=50:

Zrot=1+3exp-2kT+5exp-6kT+7exp-12kT+...101exp-2550kT

Substitute 107.7for kTin the above equation.

Zrot=1+3exp-2107.7+5exp-6107.7+7exp-12107.7+...101exp-2550107.7

=108.03

Therefore, the exact value of rotational partition function of a COmolecule is108.03

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Most popular questions from this chapter

Some advances textbooks define entropy by the formula

S=-ksPslnPs

where the sum runs over all microstates accessible to the system and Psis the probability of the system being in microstate s.

(a) For an isolated system, role="math" localid="1647056883940" Ps=1Ωfor all accessible states s. Show that in this case the preceding formula reduces to our familiar definition of entropy.

(b) For a system in thermal equilibrium with a reservoir at temperatureT,role="math" localid="1647057328146" Ps=e-EskTZ. Show that in this case as well, the preceding formula agrees with what we already know about entropy.

Consider a classical "degree of freedom" that is linear rather than quadratic E=cqfor some constant c. (As example would be the kinetic energy of a highly relativistic particle in one dimension, written in terms of its momentum.) Repeat derivation of the equipartition theorem for this system, and show that the average energy isrole="math" localid="1646903677918" E-=kT.

Consider a hypothetical atom that has just two states: a ground state with energy zero and an excited state with energy 2 eV. Draw a graph of the partition function for this system as a function of temperature, and evaluate the partition function numerically at T = 300 K, 3000 K, 30,000 K, and 300,000 K.

Although an ordinary H2 molecule consists of two identical atoms, this is not the case for the molecule HD, with one atom of deuterium (i.e., heavy hydrogen, 2H). Because of its small moment of inertia, the HD molecule has a relatively large value of ϵ:0.0057eV At approximately what temperature would you expect the rotational heat capacity of a gas of HD molecules to "freeze out," that is, to fall significantly below the constant value predicted by the equipartition theorem?

Imagine a particle that can be in only three states, with energies -0.05 eV, 0, and 0.05 eV. This particle is in equilibrium with a reservoir at 300 K.

(a) Calculate the partition function for this particle.

(b) Calculate the probability for this particle to be in each of the three states.

(c) Because the zero point for measuring energies is arbitrary, we could just as well say that the energies of the three states are 0, +0.05 eV, and +0.10 eV, respectively. Repeat parts (a) and (b) using these numbers. Explain what changes and what doesn't.

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