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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

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.)

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u(x)=u0x0x12-2x0x6

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