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Chapter 6: Q.6.25 (page 237)

The analysis of this section applies also to liner polyatomic molecules, for which no rotation about the axis of symmetry is possible. An example is CO2, with ∈=0.000049eV. Estimate the rotational partition function for a CO2molecule at room temperature. (Note that the arrangement of the atoms isOCO, and the two oxygen atoms are identical.)

Short Answer

Expert verified

The rotational partition function for a CO2molecule at room temperature is localid="1649329748565" 264.

Step by step solution

01

Step 1. Given Information 

We are given aCO2molecule at room temperature.

02

Step 2. Rotational partition function  

The arrangement of the atoms in CO2molecule is same as that of the oxygen atoms,

Zrot=kT2ε

Substituting the values, we get

Zrot=kT2ε=(8.617×10-5eV)(300K)2(0.000049eV)=263.78

Rounding off to three significant figures, the rotational partition function at room temperature is 264.

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Most popular questions from this chapter

Prove that, for any system in equilibrium with a reservoir at temperature T, the average value of the energy is

E¯=-1Z∂Z∂β=-∂∂βlnZ

where β=1/kT. These formulas can be extremely useful when you have an explicit formula for the partition function.

Some advances textbooks define entropy by the formula

S=-k∑sPslnPs

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.

Use a computer to sum the exact rotational partition function numerically, and plot the result as a function ofkT∈. Keep enough terms in the sum to be confident that the series has converged. Show that the approximation in equation 6.31 is a bit low, and estimate by how much. Explain the discrepancy

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.

Estimate the probability that a hydrogen atom at room temperature is in one of its first excited states (relative to the probability of being in the ground state). Don't forget to take degeneracy into account. Then repeat the calculation for a hydrogen atom in the atmosphere of the starγ UMa, whose surface temperature is approximately 9500 K.
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