Question 16 of 16advanced🔧 ApplyNumerical4 marks

The oxygen molecule has a mass of 5.30×105.30 \times 10⁻²⁶ kg and a moment of inertia of 1.94×101.94 \times 10⁻⁴⁶ kg m² about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

Correct Answer

The average angular velocity of the oxygen molecule is 6.75×106.75 \times 10¹² rad/s.

Exercise: EXERCISES | Q: 6.17 | (Page 35)
For More Understanding

Explanation

This problem connects translational and rotational kinetic energies. From the context, we use the formula for translational kinetic energy K = (1/2)mv² and rotational kinetic energy K = (1/2)Iω². The problem states that rotational kinetic energy is 2/3 of translational kinetic energy, which allows us to find ω.

Solution Steps

  1. Step 1: Calculate Kinetic Energy of Translation K_trans = (1/2)mv² K_trans = (1/2) × (5.30×105.30 \times 10⁻²⁶ kg) × (500 m/s)² K_trans = (1/2) × (5.30×105.30 \times 10⁻²⁶) × (2.5×102.5 \times 10⁵) K_trans = 6.625×106.625 \times 10⁻²¹ J

  2. Step 2: Calculate Kinetic Energy of Rotation Given: K_rot = (2/3) × K_trans K_rot = (2/3) × 6.625×106.625 \times 10⁻²¹ J K_rot = 4.417×104.417 \times 10⁻²¹ J

  3. Step 3: Apply Rotational Kinetic Energy Formula K_rot = (1/2)Iω² 4.417×104.417 \times 10⁻²¹ = (1/2) × (1.94×101.94 \times 10⁻⁴⁶) × ω² 4.417×104.417 \times 10⁻²¹ = 0.97×100.97 \times 10⁻⁴⁶ × ω²

  4. Step 4: Solve for Angular Velocity (ω) ω² = (4.417×104.417 \times 10⁻²¹) / (0.97×100.97 \times 10⁻⁴⁶) ω² = 4.55×104.55 \times 10²⁵ ω = \sqrt{$$4.55 \times 10$$²⁵} ω = 6.75×106.75 \times 10¹² rad/s

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