Solution Found!
(a) Calculate the total rotational kinetic energy of the
Chapter 18, Problem 82P(choose chapter or problem)
(a) Calculate the total rotational kinetic energy of the molecules in 1.00 mol of a diatomic gas at 300 K.
(b) Calculate the moment of inertia of an oxygen molecule \(\left(\mathrm{O}_{2}\right)\) for rotation about either the y- or z-axis shown in Fig. 18.18b. Treat the molecule as two massive points (representing the oxygen atoms) separated by a distance of \(1.21 \times 10^{-10} \mathrm{~m}\). The molar mass of oxygen atoms is 16.0 g/mol
(c) Find the rms angular velocity of rotation of an oxygen molecule about either the y- or z-axis shown in Fig. 18.18b. How does your answer compare to the angular velocity of a typical piece of rapidly rotating machinery (10,000 rev / min)?
Questions & Answers
QUESTION:
(a) Calculate the total rotational kinetic energy of the molecules in 1.00 mol of a diatomic gas at 300 K.
(b) Calculate the moment of inertia of an oxygen molecule \(\left(\mathrm{O}_{2}\right)\) for rotation about either the y- or z-axis shown in Fig. 18.18b. Treat the molecule as two massive points (representing the oxygen atoms) separated by a distance of \(1.21 \times 10^{-10} \mathrm{~m}\). The molar mass of oxygen atoms is 16.0 g/mol
(c) Find the rms angular velocity of rotation of an oxygen molecule about either the y- or z-axis shown in Fig. 18.18b. How does your answer compare to the angular velocity of a typical piece of rapidly rotating machinery (10,000 rev / min)?
ANSWER:Step 1 of 4
Given data:
The number of moles n is 1.00.
The temperature T of the diatomic gas is 300 K.
The separation distance L is given as \(1.21 \times {10^{ - 10}}\;{\rm{m}}\).
The revolution rate is given as 10,000 rev/min).
From the expression of rotational kinetic energy,
\({K_{rot}} = 2m{\left( {\frac{L}{2}} \right)^2}\omega \)
Here , m is inertia, and \(\omega\) is angular velocity.
Step 2 of 4
(a)
For a diatomic molecule, the two degrees of freedom associated with the rotation of a diatomic molecule account for two-fifths of the total kinetic energy.