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  • In a circle model of a rotating diatomic molecule of oxygen <img alt="\left(\mathrm{O}_{2}\right)" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%28%5Cmathrm%7BO%7D_%7B2%7D%5Crig

In a circle model of a rotating diatomic molecule of oxygen \left(\mathrm{O}_{2}\right), the two oxygen atoms are 1.5 \times 10^{-10} \mathrm{~m} apart and rotate about their center of mass with an angular speed \omega=3.0 \times 10^{12} \mathrm{rad} / \mathrm{s}. What is the rotational kinetic energy of one molecule of \mathrm{O}_{2}, which has a molar mass of 32.0 \mathrm{~g} / \mathrm{mol} ?

Option: 1

K_{\text {rot }}=2.43 \times 10^{-29} \text { Joules }


Option: 2

K_{\text {rot }}=2.43 \times 10^{-25} \text { Joules }


Option: 3

K_{\text {rot }}=2.43 \times 10^{-23} \text { Joules }


Option: 4

K_{\text {rot }}=2.43 \times 10^{-22} \text { Joules }


Answers (1)

best_answer

1. Given values: - Distance between oxygen atoms (r)=1.5 \times 10^{-10} \mathrm{~m} 

 Angular speed (\omega)=3.0 \times 10^{12} \mathrm{rad} / \mathrm{s}

 Molar mass of \mathrm{O}_{2}(M)=32.0 \mathrm{~g} / \mathrm{mol} 

 Avogadro's number \left(N_{A}\right)=6.022 \times 10^{23}$ molecules $/ \mathrm{mol}

2. Calculate the reduced mass (\mu) : Since \mathrm{O}_{2} is a diatomic molecule with two identical oxygen atoms, we have:

m_{1}=m_{2}=\frac{M}{2}

\therefore \mu=\frac{m_{1} m_{2}}{m_{1}+m_{2}}=\frac{\frac{M}{2} \cdot \frac{M}{2}}{\frac{M}{2}+\frac{M}{2}}=\frac{M^{2}}{4 M}=\frac{M}{4}

3. Calculate the moment of inertia (I) :

I=\mu r^{2}=\frac{M}{4} \cdot\left(1.5 \times 10^{-10} \mathrm{~m}\right)^{2}

4. Calculate the rotational kinetic energy \left(K_{\mathrm{rot}}\right) :

K_{\text {rot }}=\frac{1}{2} I \omega^{2}=\frac{1}{2} \cdot \frac{M}{4} \cdot\left(1.5 \times 10^{-10} \mathrm{~m}\right)^{2} \cdot\left(3.0 \times 10^{12} \mathrm{rad} / \mathrm{s}\right)^{2}

Now, let's perform the calculations:

1. Calculate \mu :

\mu=\frac{M}{4}=\frac{32.0 \mathrm{~g} / \mathrm{mol}}{4}=8.0 \mathrm{~g} / \mathrm{mol}

2. Calculate I :

I=\mu r^{2}=(8.0 \mathrm{~g} / \mathrm{mol}) \cdot\left(1.5 \times 10^{-10} \mathrm{~m}\right)^{2}=1.8 \times 10^{-44} \mathrm{~kg} \mathrm{~m}{ }^{2}

3. Calculate K_{\text {rot }} :

K_{\text {rot }}=\frac{1}{2} I \omega^{2}=\frac{1}{2} \cdot\left(1.8 \times 10^{-44} \mathrm{~kg} \mathrm{~m}^{2}\right) \cdot\left(3.0 \times 10^{12} \mathrm{rad} / \mathrm{s}\right)^{2}4. Evaluate K_{\text {rot }} :

K_{\text {rot }}=2.43 \times 10^{-29} \text { Joules }

The rotational kinetic energy of one molecule of \mathrm{O}_{2} is equal to 2.43 \times 10^{-29} Joules.

Posted by

Rishi

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