Question

Part A

Calculate the total rotational kinetic energy of the molecules in 1.00 mol of a diatomic gas at 300 K.

Krot = ? J

Part B

Calculate the moment of inertia of an oxygen molecule (O2) for rotation about either the y- or z-axis shown in Figure 18.18 in the textbook. Treat the molecule as two massive points (representing the oxygen atoms) separated by a distance of 1.21×10?10m. The molar mass of oxygen atoms is 16.0 g/mol.

I = ? kg.m^2

Part C

Find the rms angular velocity of rotation of an oxygen molecule about either the y- or z-axis shown in Figure 18.15 in the textbook.

? = ? rad/s

18.15 In a time dt a molecule with radius will collide with any other molecule within a cylindrical volume of radius 2r and length v dt 18.15 In a time dt a molecule with radius r 2r U dt

Answer 1

(a) A diatomic molecule has two degrees of freedom due to rotation. The total rotational kinetic energy is \(K_{\text {rot }}=n R T=(1.00 \mathrm{~mol})(8.314 \mathrm{~J} / \mathrm{mol} \mathrm{K})(300 \mathrm{~K})=2.49 \times 10^{3} \mathrm{~J}\)

(b) The moment of inertia of the oxygen molecule is, \(I=2 m L^{2}=2\left(\frac{16 \times 10^{-3} \mathrm{~kg} / \mathrm{mol}}{6.023 \times 10^{23} \mathrm{molecules} / \mathrm{mol}}\right)\left(\frac{1.21 \times 10^{-10} \mathrm{~m}}{2}\right)^{2}\)

\(I=2\left(2.65 \times 10^{-26} \mathrm{~kg}\right)\left(6.05 \times 10^{-11} \mathrm{~m}\right)^{2}=1.95 \times 10^{-46} \mathrm{~kg} \cdot \mathrm{m}^{2}\)

(c) The ms angular velocity of rotational of oxygen molecule is, \(\omega=\sqrt{\frac{2 K}{N_{A} I}}=\sqrt{\frac{2\left(2.494 \times 10^{3} \mathrm{~J}\right)}{\left(1.95 \times 10^{-46} \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\left(6.023 \times 10^{23}\right)}}=6.52 \times 10^{12} \mathrm{rad} / \mathrm{s}\)