For an O2molecule, the constant is approximately 0.00018eV. Estimate the rotational partition function for an O2molecule at room temperature.

Short Answer

Expert verified

The rotational partition function is 72.

Step by step solution

01

Step 1. Given Information

We are given a oxygen molecule at room temperature.

02

Step 2. Rotational partition function 

The rotational partition function is given by,

Zrot=kT2

Here, k is the Boltzmann constant, T is the absolute temperature, and is the rotational constant.

Substitute k=8.617×10-5eV/K,

T=300k and 0.00018eVin the equation, we get

Zrot=(8.617×10-5eV/K)(300K)(2)(0.00018eV)=72

Hence, the rotational partition function of an oxygen molecule at room temperature is 72.

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Most popular questions from this chapter

In this problem you will investigate the behavior of ordinary hydrogen, H2, at low temperatures. The constant εis0.0076eV. As noted in the text, only half of the terms in the rotational partition function, equation6.3, contribute for any given molecule. More precisely, the set of allowed jvalues is determined by the spin configuration of the two atomic nuclei. There are four independent spin configurations, classified as a single "singlet" state and three "triplet" states. The time required for a molecule to convert between the singlet and triplet configurations is ordinarily quite long, so the properties of the two types of molecules can be studied independently. The singlet molecules are known as parahydrogen while the triplet molecules are known as orthohydrogen.

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(b) Evaluate the partition function for a single harmonic oscillator. Use the result of part (a) to simplify your answer as much as possible.

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