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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

Fill in the steps between equations 6.51 and 6.52, to determine the average speed of the molecules in an ideal gas.

In this section we computed the single-particle translational partition function,Ztr, by summing over all definite-energy wave functions. An alternative approach, however, is to sum over all position and momentum vectors, as we did in Section 2.5. Because position and momentum are continuous variables, the sums are really integrals, and we need to slip a factor of 1h3to get a unitless number that actually counts the independent wavefunctions. Thus we might guess the formula

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(a) For each atom in the five-atom toy model of Figure 6.5, compute the deviation of the energy from the average energy, that is, Ei-E¯,fori=1to5. Call these deviations ΔEi.

(b) Compute the average of the squares of the five deviations, that is, ΔEi2¯. Then compute the square root of this quantity, which is the root-mean- square (rms) deviation, or standard deviation. Call this number σE. Does σEgive a reasonable measure of how far the individual values tend to stray from the average?

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